## Gravitation Class 11 Important Extra Questions Physics Chapter 8

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 8 Gravitation. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

## Class 11 Physics Chapter 8 Important Extra Questions Gravitation

### Gravitation Important Extra Questions Very Short Answer Type

Question 1.
What velocity will you give to a donkey and what velocity to a monkey so that both escape the gravitational field of Earth?
We will give them the same velocity as escape velocity is independent of the mass of the body.

Question 2.
How does Earth retain most of the atmosphere?
Due to force of gravity.

Question 3.
Earth is continuously pulling the moon towards its center. Why does not then, the moon falls on the Earth?
The gravitational force between the Earth and the moon provides the necessary centripetal force to the moon to move around the Earth. This centripetal force avoids the moon to fall onto the Earth.

Question 4.
Which is greater out of the following:
(a) The attraction of Earth for 5 kg of copper.
(b) The attraction of 5 kg copper for Earth?
Same.

Question 5.
Where does a body weigh more – at the surface of Earth or in a mine?
At the surface of Earth, a body weighs more.

Question 6.
How is it that we learn more about the shape of Earth by studying the motion of an artificial satellite than by studying the motion of the moon?
This is because an artificial satellite is closer to the Earth than Moon.

Question 7.
If the Earth is regarded as a hollow sphere, then what is the weight of an object below the surface of Earth?
Zero.

Question 8.
What is the formula for escape velocity in terms of g and R?
Ve = $$\sqrt{2gR}$$ .

Question 9.
What is the orbital period of revolution of an artificial satellite revolving in a geostationary orbit?
It is 24 hours.

Question 10.
Can we determine the mass of a satellite by measuring its time period?
Yes.

Question 11.
Is it possible to put a satellite into an orbit by firing it from a huge canon?
Tins can be possible only if we can ignore air friction and technical difficulties.

Question 12.
What is the amount of work done in bringing a mass from the surface of Earth on one side to a point diametrically opposite on the other side? Why?
The work done is zero. Because the gravitational potential difference is zero.

Question 13.
Name one factor on which the period of revolution of a planet around the Sun depends.
Mean distance of the planet from the Sun.

Question 14.
The gravitational potential energy of a body of mass m is -107 J. What is the energy required to project the body out of the gravitational field of Earth?
107 J.

Question 15.
Does the force of friction and other contact forces arise due to gravitational attraction? If not, what is the origin of these forces?
No. The contact forces have an electrical origin.

Question 16.
Two satellites are at different heights. Which would have greater orbital velocity? Why?
The satellite at the smaller height would have greater velocity.
This is because v0 ∝ $$\frac{1}{\sqrt{\mathrm{r}}}$$ .

Question 17.
How much energy is required by a satellite to keep it orbiting? Neglect air resistance? Why?
No energy is required by a satellite to keep it orbiting. This is because the work done by the centripetal force is zero.

Question 18.
At noon the attractions of the Earth and Sun on a body on the surface of Earth are in opposite directions. But at midnight, they are in the same direction. Does a body weigh more at mid-night?
No. The weight of the body is due to the Earth’s gravity only. What is the full form of the geostationary satellite APPLE? The full form of APPLE is the Ariane Passenger PayLoad Experiment.

Question 19.
What is geodesic?
It is the shortest distance between any two points.

Question 20.
Why is G called a universal constant?
G is called a universal constant because its value is the same everywhere.

Question 22.
Where does a body weigh more at the pole or at the equator?
It weighs more at the pole.

Question 23.
What is the relation between orbital and escape velocity?
ve = $$\sqrt{2}$$v0.

Question 24.
The weight of a body is 20N, What is the gravitational pull of the body on the Earth?
20N.

Question 25.
Why Newton’s law of gravitation is said to be universal- law?
It is called so because this law holds good irrespective of the nature of the interacting bodies at all places and at all times.

Question 26.
A particle is to be placed, in turn, outside four objects, each of mass m;
(a) a large uniform solid sphere,
(b) a large uniform spherical shell,
(c) a small uniform solid sphere,
(d) a small uniform shell.
In each situation, the distance between the particle and the center of the object is d. Rank the objects according to the magnitude of the gravitational force they apply on the particle, greatest first.
They all apply the same gravitational force and hence all will tie.

Question 27.
Can we determine the mass of a satellite by measuring its time-period?
No.

Question 28.
Does the gravitational force between two particles depends upon the medium between them?
No, it does not depend upon the medium between the two particles.

Question 29.
Two artificial satellites of different masses are moving in the same orbit around the Earth. Can they have the same speed?
Yes, they can have the same speed as orbital speed is independent of the mass of the satellite.

Question 30.
In an imaginary system, the central star has the same mass as our Sun but is much brighter so that only a planet twice the distance between Earth and the Sun can support life. Assuming biological evolution (including aging processes etc.) on that planet similar to ours, what would be the average life span of a ‘human’ on that planet in terms of its natural year? The average life span of a human on the Earth may be taken to be 70 years.
25 planet years.

Question 31.
If the force of gravity acts on all bodies in proportion to their masses, then why does not a heavy body fall faster than a light body?
Acceleration due to gravity is independent of the mass of the body.

Question 32.
What is the weight of a body in a geostationary satellite?
The weight of a body is zero in a geostationary satellite.

Question 33.
A satellite does not need any fuel to circle around the Earth. Why?
The gravitational force between satellite and Earth provides the centripetal force required by the satellite to move in a circular orbit.

Question 34.
What will be the effect on the weight of the bodies if Earth stops rotating about its axis?
The weight of the bodies will increase.

Question 35.
Name the natural satellite of Earth.
Moon.

Question 36.
Why a multi-stage rocket is required to launch a satellite?
This is done to save fuel.

Question 37.
Mention one difference between g and G.
The value of ‘G’ remains the same throughout the universe while the value of ‘g’ varies from place to place.

Question 38.
When an apple falls towards the Earth, the Earth moves up to meet the apple. Is the statement true? If yes, why is the Earth’s motion not noticeable?
Yes, the statement is true. The acceleration of Earth is very small as compared to that of apple as the mass of Earth is very large.

Question 39.
Which of the following observations point to the equivalence of inertial and gravitational mass? Why?
(a) Two spheres of different masses dropped from the top of a long evacuated tube reach the bottom of the tube at the same time.
(b) The time period of a simple pendulum is independent of its mass.
(c) The gravitational force on a particle inside a hollow isolated sphere is zero.
(d) For a man in a closed cabin that is falling freely under gravity, gravity disappears.
(e) An astronaut inside a spaceship orbiting around the Earth feels weightless.
(f) Planets orbiting around the Sun obey Kepler’s Third Law.
(g) Gravitational force on a body due to the Earth is equal and opposite to the gravitational force on the Earth due to the body.
As bodies are in motion in observations (a), (b), (d), (e), (f), thus, these point to the equivalence of inertial and gravitational masses.

Question 40.
How will the value of acceleration due to gravity be affected if the Earth begins to rotate at a speed greater than its present speed?
Acceleration due to gravity will decrease if the angular speed of rotation of Earth increases.

Question 41.
How do we choose zero levels of gravitational potential energy?
It corresponds to the infinite separation between two interacting masses.

Question 42.
Is it possible to put an artificial satellite in an orbit in such a way that it always remains visible directly over New Delhi? Why?
It is not possible. This is because New Delhi is not in the equatorial plane.

Question 43.
A body is suspended with a spring balance attached to the ceiling of an elevator. The balance shows a reading of 5 divisions when the elevator is stationary. During the downward acceleration of the elevator, the balance shows zero reading. Do you think that the inertial mass and gravitational mass of the body are equal? Justify your answer.
Yes, they will be equal. The two masses differ only when the velocity of the lift approaches the velocity of light.

Question 44.
Does a comet move faster at aphelion or perihelion?
At the perihelion where it is close to Sun, the comet moves faster.

Question 45.
Due to some unforeseen event, the planet of the I0 satellites of Jupiter does not pass through the center of Jupiter. Will the orbit of I0 be stable? Why?
No. For a stable orbit, the plane of the orbit must pass through the center (equatorial plane) of the planet.

Question 46.
Is it appropriate to describe the condition of weightlessness as the condition of masslessness? Why?
No. Weight and mass are different physical quantities. A body may have zero weight but never zero mass. So it is not appropriate to describe the condition of weightlessness as a condition of masslessness.

Question 47.
Define central force.
It is defined as the force which is always directed along the position vector of the point of application of the force w.r.t. the fixed point i.e. away or towards a fixed point.

Question 48.
The angular momentum is always conserved in the motion under a central force. Name the two results that follow from this.

1. The motion of a particle under the central force is always confined to a plane.
2. The position vector of the particle w.r.t. the center of force has a constant areal velocity i.e. the position vector sweeps out equal areas in equal times as the particle moves under the influence of the central force.

Question 49.
Why an astronaut has a sense of weightlessness in a satellite revolving around the Earth?
For revolving around the Earth, the astronaut and the satellite require the centripetal force, and their weight is used up in providing the necessary centripetal force. So the astronaut feels weightlessness in space.

Question 50.
Why an astronaut in an orbiting spacecraft is not in zero-gravity although he is in a state of weightlessness?
We know that acceleration due to gravity depends on the height, so there is some value of acceleration due to gravity at every point on the orbit of the spacecraft and his weight is used up in providing the necessary centripetal force for the orbital motion.

### Gravitation Important Extra Questions Short Answer Type

Question 1.
Explain how the weight of the body varies en route from the Earth to the moon. Would its mass change?
When a body is taken from Earth to the moon, then its weight slowly decreases to zero and then increases till it becomes $$\frac{1}{6}$$th of the weight of the body on the surface of the moon.

We know that mgh = mg (1 – $$\frac{2 h}{R}$$)

As h increases, gh, and hence mgh, decreases. When R = $$\frac{R}{2}$$ the force of attraction of Earth is equal to the force of attraction of the moon.

Then gh = 0, so mg becomes zero, and the value of g on the moon’s surface is $$\frac{1}{6}$$th of its value on the surface of Earth. Hence on increasing h beyond $$\frac{R}{2}$$, mg starts increasing due to the gravity of the moon. f ts mass remains constant.

Question 2.
Among the known type of forces in nature, the gravitational force ¡s the weakest. Why then does ¡t play a dominant role in the motion of bodies on the terrestrial, astronomical, and cosmological scale?
Electrical forces are stronger than gravitational forces for a given distance, but they can be attractive as well as repulsive, unlike gravitational force which is always attractive. As a consequence, the forces between massive neutral bodies are predominantly gravitational and hence play a dominant role at long distances. The strong nuclear forces dominate only over a range of distances of the order of 10-14 m to 10-15m.

Question 3.
Show that the average life span of humans on a planet in terms of its natural years is 25 planet years if the average span of life on Earth is taken to be 70 years.
Take the distance between Earth and Sun twice the distance between Earth and planet. According to Kepler’s third law of planetary motion,

where Te, TR is the average life span on Earth and planet respectively.
Rg = distance between Earth and Sun.
Rp = distance between Earth and planet.
Here, Re = 2Rp

Question 4.
Hydrogen escapes faster from the Earth than oxygen. Why?
The thermal speed of hydrogen is much larger than oxygen. Therefore a large number of hydrogen molecules are able to acquire escape velocity than that of oxygen molecules. Hence hydrogen escapes faster from the Earth than oxygen.

Question 5.
In a spaceship moving in a gravity-free region, the astronaut will not be able to distinguish between up and down. Explain why?
The upward and downward sense is due to the gravitational force of attraction between the body and the earth. In a spaceship, the gravitational force is counterbalanced by the centripetal force needed by the satellite to move around the Earth in a circular orbit. Hence in the absence of zero force, the astronaut will not be able to distinguish between up and down.

Question 6.
Why the space rockets are generally launched from west to east?
Since the Earth revolves from west to east around the Sun, so when the rocket is launched from west to east, the relative velocity of the rocket = launching velocity of rocket + linear velocity of Earth. Thus the velocity of the rocket increases which helps it to rise without much consumption of the fuel. Also, the linear velocity of Earth is maximum in the equatorial plane.

Question 7.
Explain why the weight of a body becomes zero at the center of Earth.
We know that the weight of a body at a place below Earth’s surface is given by
W = mgd …. (i)
Where gd = acceleration due to gravity at a place at a depth ‘d’ below Earth’s surface and is given

From Eqn. (i) W = 0 at the center of Earth.
i.e., g decreased with depth and hence becomes zero at the center of Earth, so W = 0 at Earth’s center.

Question 8.
We cannot move even our little fingers without disturbing the whole universe. Explain why.
According to Newton’s law of gravitation, every particle of this universe attracts every other particle with a force that is inversely proportional to the square of the distance between them. When we move our fingers, the distance between the particles changes, and hence the force of attraction changes which in turn disturbs the whole universe.

Question 9.
Explain why tennis ball bounces higher on hills than on plains.
The value of acceleration due to gravity is lesser on hills than on the plains, so the weight of the tennis ball at hills is lesser than that on the plains and hence it re-bounces more. In other words, the force with which the Earth attracts the ball on hills will be lesser than that on the plains.

Question 10.
Why is the atmosphere comparatively rarer on some of the planets as compared to that on earth?
The escape velocity of some of the planets is very small as compared to that on the surface of Earth. Most of these gases have their root mean square (r.m.s.) velocities more than the escape velocity on these planets and hence they have escaped from the surface of these planets and hence the atmosphere is rarer on some of the planets than on Earth.

Question 11.
Why moon has no atmosphere? Explain.
An atmosphere means the presence of a mixture of a number of gases. The molecules of these gases are in the state of continuous random motion moving with different velocities. As the value of escape velocity On the surface of the moon is small (only 2.5 km s-1), the molecules of gases with velocities greater than the escape velocity moved out of the atmosphere. As time passed, nearly all the molecules escaped from the moon’s atmosphere.

Question 12.
Under what conditions a satellite will be called geo-stationary?
The following are the conditions to be fulfilled by a satellite to be geostationary:

1. The time period of the satellite around the Earth must be equal to the rotational period of the satellite i.e. 24 hours.
2. The direction of motion of the satellite must be the same as that of the Earth.
3. The height of the geostationary satellite must be about 36000 km.

Question 13.
What are the uses of Artificial Satellites?
Following are some of the important uses of Artificial Satellites:

1. They are used as communication satellites to send messages to distant places.
2. They are used as weather satellites to forecast weather.
3. They are used to know the exact shape of Earth.
4. They are used to telecast T.V. programs to distant places.
5. They are used to explore the upper region of the atmosphere.

Question 14.
Under what condition, a rocket fired from Earth will launch an artificial satellite around Earth?
Following are the basic conditions:

1. The rocket must take the satellite to a suitable height above the surface of Earth.
2. From the desired height, the satellite must be projected with a suitable velocity called orbital velocity.
3. In the orbital path of the satellite, the air resistance should be negligible so that its velocity does not decrease and it does not bum due to the heat produced.

Question 15.
The acceleration due to gravity on a planet is 1.96 ms-2. If it is safe to jump from a height of 2m on the Earth, what will be the corresponding safe height on the planet?
The safety of a person depends upon the momentum with which the person hits the planet. Since the mass of the person is constant therefore the maximum velocity (v) is the limiting factor.
In case of earth, v2 = 2gehe …. (i)
In case of planet, v2 = 2gphp …. (ii)
where he, hp are safe heights of jumping from Earth and planet respectively,
Here, he = 2m; gp = 1.96 ms-2, ge = 9.8 ms-2

∴ From (i) and (ii), we get
2ge he = 2gp hp

Question 16.
Is it correct to state that we are living at the bottom of a gravitational well? Why?
Yes. The gravitational force varies with distance from the center of Earth as shown in the fig. below. The graph clearly shows a minimum at a point on the surface of Earth. Thus, it is correct to state that we are living at the bottom of a gravitational well.

Question 17.
Prove that if two spheres of the same material, mass, and radius, are put in contact, the gravitational attraction between them is directly proportional to the fourth power of their radius.
Let m = mass of each sphere.
R = radius of each sphere,
p = density of the material of the spheres.
∴ If F be the gravitational attraction between them, then according to Newton’s law of gravitation,

∴ From (1) and (2), we get

Question 18.
Prove that if the gravitational attraction of the Sun on the planets varies as nth power of the distance (of the planet from the Sun), then a year of the planet will be proportional to R(n + 1)/2.
Let m = mass of the planet
R = radius of its orbit around Sun
M = mass of the sun
V = orbital speed of the planet
Then,

Question 19.
The figure below shows four arrangements of three particles of equal masses:

(a) Rank the arrangements according to the magnitude of the net gravitational force on the particle labeled m, greatest first, and explain.
(b) In arrangement 2, in the direction of the net force closer to the line of length d or to the line of length D?
(a) 1,2 and 4 tie, 3 (b) line d.
Explanation: (a) An arrangement (1), the forces exerted by both the particles are in the same direction. So these forces get add up to give maximum force. In arrangements (2) and (4), the angle between individual forces as well as the distances are the same. So the magnitudes of the net forces in (2) and (4) are the same and greater than in arrangement (3) and less than in (1).

In arrangement (3), the angle between the separate forces is maximum i.e. 180. So the net force is minimum.

(b) Due to the smaller distance, force is greater the net force always makes a smaller angle with the larger force as compared to the angle with a smaller force.

Question 20.
Why the atmosphere of Jupiter contains light gases (generally hydrogen) whereas the Earth’s atmosphere has little hydrogen gas?
The escape velocity of Jupiter is much larger than the escape velocity of Earth. So to escape from the surface of Jupiter, a very large velocity is required. Since the thermal velocity of hydrogen gas molecules is lesser than the escape velocity of Jupiter, therefore hydrogen can’t escape from the surface of Jupiter.

Question 21.
Why does not an astronaut use a simple pendulum clock in a satellite revolving around the Earth?
Acceleration due to gravity is zero inside a satellite therefore the time period of vibration of the simple pendulum
T = 2π $$\sqrt{\frac{l}{g}}$$ will be infinity

So the pendulum will do not vibrate inside the satellite and hence the clock will not work. So astronaut does not use a simple pendulum clock in a satellite.

Question 22.
Show that Kepler’s Second law is the law of conservation of angular momentum.
The second law states that the areal velocity is constant i.e. area covered by the radius vector is the same in equal intervals of time. If the velocity and radius at the time, t1 is v1 and r1 while at another place these are v2 and r2 in the same time, then the area covered by the planet in these intervals are

This shows that the law leads to the conservation of angular momentum law.

Question 23.
Astronomical observations show that Mercury moves faster and Pluto slower, why is it so?
Planets obey Kepler’s Third law of motion.

TM = time of revolution of Mercury around Sun at distance RM.
TP = time of revolution of Pluto around Sun at distance RP.

Pluto moves slower as compared to Mercury due to the reason that RP > RM.

Question 24.
Calculate the density of Earth.
Let us assume the Earth to be a homogeneous sphere of radius R, density ρ, and mass M.

Question 25.
Find an expression for the torque on a body due to gravity. Show that this torque about the center of mass of a body is zero.
Suppose a body is made up of a large number of particles of masses m1, m2, m3, ….. etc. situated at r1, r2, r3,… etc. positions from the center of mass of the body. Each particle is pulled by gravity with a force F = mg. Let this force on ith particle be f = mg. Since the position vector of this particle is r1, hence the torque on this particle due to a national poll of the Earth is
τi = ri × mig
Then total torque on the body

Then torque about the center of mass is eventually 0 as rcm = 0 from the center of mass,
∴ τ = 0.

Question 26.
How do Kepler’s laws lead to Newton’s universal law of gravitation?
Kepler’s laws are applicable to the motion of planets around the Sun. Almost all planets revolve around the Sun in nearly circular orbits.

Let m = mass of a planet.
M = mass of Sun.
r = radius of the circular orbit of the planet, around the Sun.
v = linear velocity of the planet in its orbit.
T = Time period of the planet.

The centripetal force required by the planet is.
F = $$\frac{\mathrm{mv}^{2}}{\mathrm{r}}$$
But v = circumference of the orbit / period of revolution = $$\frac{2πr}{T}$$

This centripetal force is provided by the force of attraction between the Sun and the planet. According to Newton, the force of attraction between the planet and the Sun is mutual.

which is Newton’s law of gravitation.

Question 27.
Explain how the mass of the Sun can be determined by studying the motion of Earth around it.
The Sun’s gravitational force provides the necessary centripetal force to the Earth for its revolution around the Sun in an orbit of radius r. Let the mass of Sun be Ms and that of Earth Mc. Thus centripetal force is then

where R = radius of orbit of Earth.
Let T = Time period of revolution of Earth around Sun, then

we know R, G, and T so the mass of the Sun can be found out by putting their values.

Question 28.
Derive an expression for the gravitational potential energy above the surface of Earth.
Let the body of mass m be taken at a height h above the surface of Earth. At any instant of time t, it reaches at a distance x from the center of Earth. The work done in raising the body through dx is

Hence, the work done in taking the body from the surface of the Earth (x = R) to a height h above the earth’s surface (x = R + h) is given by

This work done is stored in the body in the form of gravitational
P.E. i.e. P.E. = W = mgh
P.E. above the surface of Earth = mgh.

Question 29.
What is the binding energy of the satellite?
The minimum energy required to free a satellite from the gravitational attraction is called its binding energy. The binding energy is the negative value of the total energy of the satellite. Let a satellite of mass m be revolving around the earth of mass M and radius R.

∴ The total energy of satellite = P.E. + K.E.

Bindind energy of satellite = – (total energy of satellite)
$$\frac{GMm}{2R}$$

Question 30.
In a cavendish experiment, two spheres of mass 10 g each are suspended with a torsion rod of length 2m when two lead spheres of mass 10 mg each are brought near the suspended sphere as shown in the figure, the spheres are displaced through 2mm each. The torsional rod is deflected through an angle of 0.02 radian. Calculate the value of G if the torsional couple per unit twist of the suspension wire is 1.66 × 10-4 Nm.

Force exerted by a bigger sphere on the small sphere is
F= $$\frac{GMm}{2R}$$
where r = separation between bigger and small spheres.
M = mass of the bigger sphere.
m = mass of the smaller sphere.

The force on the two sides produces torque and the suspension wire produces opposite torque = cθ = restoring couple produced.
where c = Torsional couple per unit twist.
θ = Twist

If l be the length of torsion rod, the deflecting couple
= Fl = $$\frac{\mathrm{GMm}}{\mathrm{R}^{2}}$$ l

Since the rod is in equilibrium after the deflection
∴ deflecting couple = restoring couple

### Gravitation Important Extra Questions Long Answer Type

Question 1.
(a) Derive the expression for the orbital velocity of an artificial Earth’s satellite. Also, derive its value for an orbit near Earth’s surface.
Let m = mass of the satellite.
M, R = mass and radius of Earth.
h = height of the satellite above the surface of Earth.
r = radius of the robot of the satellite
= R + h.
v0 = orbital velocity of the satellite.

The centripetal force $$\frac{\mathrm{mv}_{0}^{2}}{\mathrm{r}}$$ required by the satellite to move in a circular orbit is proved by the gravitational force between satellite and the Earth.

If the satellite is close to the earth’s surface, then h ≈ 0

(b) Derive the expression for escape velocity of a body from the surface of Earth and show that it $$\sqrt{2}$$ times the orbital velocity close to the surface of the Earth. Derive its value for Earth.
Escape velocity is the minimum velocity with which a body is projected from Earth’s surface so as to just escape its gravitational pull or of any other planet. It is denoted by ve.

Expression: Consider the earth to be a homogenous sphere of radius R, mass M, center O, and density p.
Let m = mass of the body projected from point A on the surface of Earth with vel. ve.

∴ K.E. of the body at point A = $$\frac{1}{2}$$ mve2 …(i)
Let it reaches a point P at a distance x from O. If F be the gravitational force of attraction on the body at P, then

F = $$\frac{\mathrm{GMm}}{\mathrm{x}^{2}}$$ …(ii)

Let it further moves to Q by a distance dx.
If dW be the work done in moving from P to Q, then

If w be the total work done in moving the body from A to ∞,
Then

∴ According to the law of conservation of energy
K.E. = RE

Relation between ve and vo: Also we know that the orbital velocity around Earth close to its surface is given
by v0 = $$\sqrt{gR}$$
and ve = $$\sqrt{2gR}$$ = $$\sqrt{2}$$ $$\sqrt{gR}$$
= $$\sqrt{2}$$v0
Hence proved.

Question 2.
(a) Explain Newton’s law of gravitation.
We know that Newton’s law of gravitation is expressed mathematically as:

where r̂ = unit vector along F
It was found that law is equally applicable anywhere in the universe between small and big objects like stars and galaxies. The value of G remains the same everywhere. (Some scientists have claimed that as the size of the object under consideration becomes big like a galaxy, the value of G also changes). Hence this law of Newton is also called Newton’s universal law of gravitation.

The force of attraction is called the force of gravitation or gravitational force. This force is only attractive and is never repulsive. The force is both ways i.e., particle 1 attracts particle 2 and so does particle 2 attracts particle 1.
Hence F12 = – F21.

The law is a direct outcome of the study of acceleration of bodies. Newton wondered how Moon revolves around the Earth or other planets revolve around the Sun. His calculations showed that the Moon is accelerated by the same amount as does any other object towards the Earth.

His famous narration of the apple falling from the tree and noticing every other object fall towards Earth led to the announcement of his famous law of gravitation about 50 years later in his book ‘Principia’.

Out of the known forces in nature, the Gravitational force is the weakest, yet it is the most apparent one as it acts for long distances and between objects which are visible to us. The law of gravitation has been used to determine the mass of heavenly bodies. It has been used to study the atmosphere of planets. Man-made satellites remain in the orbits due to gravitation.

(b) Define gravitational field intensity. Derive its expression at a point at a distance x from the center of Earth. How is it related to acceleration due to gravity?
The gravitational field intensity at a point is defined as the force acting on a unit mass placed at that point in the field.

Thus, the gravitational field intensity is given by:
E = $$\frac{F}{m}$$
Now at distance x from the center of Earth, the gravitational force is

So, the intensity of the gravitational field at the surface of Earth is equal to the acceleration du&to gravity.

Question 3.
Discuss the variation of acceleration due to gravity with:
(a) Altitude or height
(b) Depth
(c) Latitude i.e. due to rotation of Earth.
Let M, R be the mass and radius of the earth with center O.
g = acceleration due to gravity at a point
An on Earth’s surface.

(a) Variation of g with height: Let g0h be the acceleration due to gravity at a point B at a height h above the earth’s surface

If h << R, then using Binomial Expansion, we get

Thus, from Eqn. (3), we conclude that acceleration due to gravity decreases with height.

(b) With depth: Let the Earth be a uniform sphere.

Let gd = acceleration due to gravity at a depth d below earth’s surface i.e. at point B.

Let ρ = density of Earth of mass M.

Also, let M’ = mass of Earth at a depth d, then

From equation (iv), we see that acceleration due to gravity decreases with depth.
Special case: At the center of Earth, d = R
∴ gd = 0

Hence an object at the center of Earth is in a state of weightlessness.

(c) Variation of g with latitude:
Let m = mass of a particle at a place P of latitude X.
ω = angular speed of Earth about axis NS.

As the earth rotates about the NS axis, the particle at P also rotates and describes a horizontal circle of radius r,
where r = PC = OP cos λ, = R cos λ

Let g’ be the acceleration due to gravity at P when the rotation of Earth is taken into account. Now due to the rotation of the earth, two forces that act on the particle at P are:

1. Its weight mg, acting along with PO.
2. Centrifugal force mroo2 along PO’.

∴ The angle between them = 180 – λ
∴ According to the parallelogram law of vector addition

[As $$\frac{\mathrm{R} \omega^{2}}{\mathrm{~g}}$$ is very small (= $$\frac{1}{289}$$)so its square and higher powers are neglected.]

Using binomial expansion, we get

⇒ g decreases with the rotation of the earth.

1. At poles, λ = 90°, ∴ g’ = gp = g
2. At equator, λ = 0, g’ = ge = g- Rω2.

Clearly gp > ge.

Numerical Problems:

Question 1.
At what height from the surface of Earth will the value of g be reduced by 36% from the value at the surface? The radius of Earth, R = 6400 km.
Let h be the height above the surface at which g reduces by 36% i.e. becomes 64% of that at the surface i.e.

Question 2.
A mass of 5 kg is weighed on a balance at the top of a tower 20 m high. The mass is then suspended from the pan of the balance by a fine wire 20 m long and is reweighed. Find the change in the weight, R = 6400 km.
h = 20 m
m = mass = 5 kg
R = 6400 km = 64 × 105m
g = 9.8 ms-2

As h < R, so from relation,

Now weight at the foot of tower, W = mg
Now weight at the top of tower, Wh = mgh

Question 3.
How much faster than its present rate should the Earth rotate about its axis in order that the weight of the body on the equator may be zero? What will be the duration of the day? The radius of Earth, R = 6400 km.
We know that the value of ‘g’ at latitude λ is
gλ = g – R ω2 cos2λ
at equator λ = 0,
∴ cos 0 = 1
∴ gλ = g – Rω2

Let ω1 = new angular speed of rotation so that weight at equator becomes zero

in equation (1), we get

i.e. the Earth must rotate 17 times faster thair its present speed. Now the duration of the day 24 hours i.e. earth makes one rotation in 24 hours. When it rotates 17 times faster, it will make 17 rotations in 24 hours

∴ Time for 1 rotation (Duration of Day) = $$\frac{24}{17}$$ = 1.412 hours.

Question 4.
Show the moon would escape if its speed were increased by 42%.
Let M = mass of Earth.
m = mass of the moon,
r = radius of the circular orbit of the moon around the Earth.
vo = orbital velocity of the moon.

The centripetal force needed by moon = Gravitational force between the Earth and moon

∴ % increase in the velocity of the moon

Question 5.
If the Earth has a mass 9 times and a radius twice of the planet Mars, calculate the minimum velocity required by a rocket to pull out of the gravitational force of Mars. Escape velocity on the surface of Earth is 11.2 km s-1.
Let Mm and be the mass and radius of Mars,
and M, R = mass and radius of Earth.

Also, let v and Vm be the escape velocities from Earth and Mars respectively.

Question 6.
What would be the duration of the year ¡f the distance between the Earth and the Sun gets doubled?
Let T1 = time period of Earth about the Sun when the distance between Sun and Earth is R and T2 = time period of Earth about the Sun when the distance between Earth and Sun is 2R.

Now using the relation,

Question 7.
The period of the moon around Earth is 27.3 days and the radius of its orbit is 3.9 × 105 km. Find the mass of Earth, G = 6.67 × 10-11 Nm2 Kg-2.
T = time period = 27.3 days
= 27.3 × 24 × 60 × 60s.

= 3.9 × 105 km = 3.9 × 108m.

G = 6.67 × 10-11 N m2 kg-2.
M = mass of Earth = ?

Let m, v be the mass and linear velocity of the moon. Then the centripetal force to the moon is provided by the force of attraction between Earth and the moon.

Question 8.
Find the percentage decrease in weight of a body when taken 16 km below the surface of Earth. Take the radius of Earth = 6400 km.
Here, R = 6400 km, d = 16km

If m = mass of the body, then

Question 9.
A mass M is broken into two parts m and M – m. How are m and M related so that the force of gravitational attraction between the two parts is maximum?
Let r = distance between two parts i.e. m and M – m
If F be the gravitational attraction between m and M – m, then from Newton’s law of gravitation,

eqn. (i) may be rewritten as:

∴ each piece has equal mass.

Aliter:
For F to be maximum,

Question 10.
A satellite orbits the Earth at a height of 500 km from its surface.
Calculate (i) K.E.
(ii) P.E.
(iii) Total energy.
Mass of satellite is 300 kg, mass of Earth is 6 × 1024 kg, radius of Earth = 6.4 × 106 m, G = 6.67 × 10-11 Nm2 kg-2. Will your answer alter if the Earth were to shrink suddenly to half its size?

If the earth shrinks suddenly to half its radius (R becomes R/2) but r = (R + h) remains unchanged, then the answer will not alter.

Question 11.
If a body is projected with a velocity v greater than ve (escape velocity from Earth’s surface), find its velocity in interstellar space.
Let v’ be the velocity of the body of mass m in interstellar space where the gravitational field of Earth is zero; so the P.E. of the body is also zero.

∴ According to the law of conservation of energy,
we have, K.E. + RE. at the surface of Earth

Question 12.
What will be the acceleration due to gravity on the surface of the moon if its radius is 1/4th of the radius of Earth and its mass l/80th of the mass of Earth?
Answer:The value of acceleration due to gravity is g =$$\frac{\mathrm{GM}}{\mathrm{R}^{2}}$$

Question 13.
Consider an Earth satellite so positioned that it appears stationary to an observer on Earth and serves the purpose of a fixed relay station for intercontinental transmission of television and other communications. What should be the height at which the satellite should be positioned and what would be the direction of its motion? (Radius of Earth R = 6400 km.)
Suppose the satellite is stationed at height h from the surface of the Earth, then

Question 14.
If the moon followed a circular orbit of radius r around the Earth with a uniform angular velocity co, so that ω2r2 = gR2, where R is the radius of the Earth and g the acceleration due to gravity. If r = 60R and the period of revolution of the moon around the Earth is 27.3 days. Find R.

Question 15.
The mass of asteroid Ceres is approximately 7 × 1020 kg and its diameter is 1100 km. What is the value of acceleration due to gravity at its surface? What would be the weight of an 80 kg astronaut on this asteroid?
The radius of the asteroid Ceres,

∴ Weight of astronaut, W = mg = 80 × 0.15 ms-2 = 12 N.

Question 16.
The mass of the Sun is 2 × 1030 kg and its radius 7 × 108 m. Calculate the rate of emission of radiation from it if its radius shrinks at the rate of 1 × 104 ms-1.
The self-energy of the Sun, considering it spherical in shape, is given by

Question 17.
Two stars of mass 3 × 1031 kg each, in a double star, rotate about their common center of mass 1011 m part
(i) Calculate their common angular speed.
Let one-star move in a circular orbit of radius R, then gravitational pull provides the necessary centripetal force. Also, the distance between two stars = x, and they have equal mass then x = 2R.
Therefore,

Substituting the given values, we have

(ii) If a meteorite passes through its center of mass moving at right angles to the line joining the stars. In order to escape from the gravitational field of the double star what should be the speed of the meteorite?
The meteorite can escape from the gravitational field if the sum of its K.E. and P.E. due to the two stars is greater than 0 i.e.

Question 18.
Neglecting the presence of other planets and satellites, calculate the binding energy of the Sun-Earth system. Mass of Earth ME = 6 × 1024 kg, mass of Sun = 1.98 × 1030 kg, Orbital radius of Earth R= 1.5 × 1011 m.
The binding energy is equal to the amount of energy spent in bringing the Earth-Sun system from infinity to the distance R. In other words, this is equal to the energy required to separate them to infinity hence it is equal to (P.E.)g

Question 19.
If the radius of the Earth decreases by 10%, the mass remaining unchanged, what will happen to the acceleration due to gravity?
Here, let M, R be the mass and radius of the earth. If g be the acce. to due to gravity on earth’s surface, then

If g’ be the new value of acceleration.

Question 20.
The ratio of the radii of the planets P1 and P2 is k and the ratio of the acceleration due to gravity on them is r. Calculate the ratio of the escape velocities from them.
Let r1 and r2 be the radii of the planet P1 and P2 respectively having respective accelerations due to gravity g1, and g2.

If v1 and v2 be the escape velocities from P1 and P2 respectively, then using the relation,

Question 21.
Calculate the temperature at which oxygen molecules can escape from the surface of Earth. Assume the radius of Earth as 6.4 × 106 m and g = 9.8 ms-2, the universal gas constant, R = 8.4 J mol-1 k-1.
Here, Re = radius of Earth = 6.4 × 106 m
If a ve be the escape velocity of oxygen molecules from Earth’s surface, then

According to the kinetic theory of gases, the most probable velocity of the molecules of a gas is given by,

where M = molecular weight of the gas = 32 × 10-3 kg for O2.
R = 8.4J mol-1 k-1
For the escape of the gas molecules,

Question 22.
Find the gravitational force of attraction on a uniform rod of length L and mass m due to uniform earth as shown in Fig. How the rod will behave if L << r.

Let M, R be the mass and radius of earth respectively. As the earth is uniform, so its whole mass may be supposed to be concentrated at its center O.

Let dF be the force due to earth on the mass element dx at a distance x from O.

If F be the total force on the rod, then

Now if the rod is very small as compared to its distance from the earth i.e. if r >> L, then r + L ≈ r.

∴ From (2), we get F = $$\frac{\mathrm{GMm}}{\mathrm{r}^{2}}$$
i.e. the rod behaves as a point mass lying at a distance r from the center of the earth.

Question 23.
Calculate the scale reading of a spring balance having a body suspended to it and is in a ship sailing along the equator with a speed v.
Let m = mass of the body suspended
W0 = scale reading of the spring balance when the ship is at rest

Let ω = angular speed of rotation of the Earth of radius R
∴ The value of g at equator is given by
gθ = (g – Rω2)
∴ W0 = mgθ = m(g – rω2) ….(1)

when the ship sails from west to east with the speed v on Earth’s surface, then its angular speed $$\frac{v}{R}$$ gets added to ω i.e. ω’ = ω + $$\frac{v}{R}$$

similarly if the ship sails from east to west, then the scale reading is given by

Question 24.
Calculate the minimum work done in bringing a spaceship of mass 1000 kg from the surface of Earth to the moon.
Let M1, M = masses of earth and moon respectively.
R1, R2 = radius of earth and moon respectively,
m = mass of spaceship.

If U1 and U2 be the gravitational P.E. of the spaceship on Earth and the moon respectively, then

The total work (W) will have to be done on the spaceship first against the gravity of Earth and-then against the gravity of the moon through the zero-gravity region while going from the Earth to theme on.

Question 25.
Taking moon’s period of revolution about the Earth as 30 days, calculate its distance from the Earth;
G = 6.67 × 10-11 Nm2 Kg-2, M = mass of Earth = 6 × 1024 kg.
Here, T = 30 days = 30 × 24 × 3600s
= 30 × 86400 s
G = 6.67 × 1011 Nm2Kg-2
M = 6 × 1024 kg

Let m = mass of moon.
Let x = distance of moon from Earth = ?

Here, the force of attraction between moon and Earth provides the centripetal force to the moon i.e.

Value-Based Type:

Question 1.
Suresh was struggling to understand Kepler’s second law of planetary motion. Then his friend Raman who came to him explained how the planet moves around the sun obeying Kepler’s law of planetary motion.
(a) Comment upon the values of Raman.
Raman shares his friends and wants to improve his knowledge in the subjects, has a concern towards his friends.

(b) State Kepler’s laws of planetary motion.
Kepler’s 1st law of planetary motion: Every planet revolves around Sun in an elliptical orbit with Sun at one of its foci. BacondlawsThe radius vector joining the center of Sun and planet sweeps out equal areas in equal intervals of time i.e. areal velocity of the planet around the Sun always remains constant. Third law: The square of the time period T of revolution of a planet around the Sun is proportional to the cube of the semi-major axis R of its elliptical orbit i.e.
T2 ∝ R3

Question 2.
Suresh went to a picnic on a hill station. The teacher told all the students to weigh by weighing machine before going to the hill station. His weight was 56 Kg. When we reached the hill station along with his friends, our teacher again told us to find their weight using a weighing machine installed in front of a shopping mall. He found that his weight was lesser than 56 Kg i.e 52 Kg, He was surprised and asked the teacher the reason behind it. The teacher explained that the acceleration due to gravity decreases as we go up from the ground. Hence, your weight will be lesser than on the ground.
(i) What values of Suresh are displayed here?
Awareness, intelligence, and logic.

(ii) Why does the teacher tell everyone to find their weight?
The teacher wanted to explain the acceleration due to gravity through the demonstration method.

(iii) If a body is taken to a height equal to the radius of the earth from its surface, how will the weight of a body change?
We know that:

Hence, the weight of the body will reduce to one-fourth of its original weight on the surface of the earth.

Question 3.
Two friends Jagat and Ram are discussing escape velocity, Jagat told that if a body is projected vertically upwards with escape velocity, it will cross the earth’s gravitational field and will never come back to the earth’s surface again. Ram was not convinced with Jagat. So, they went to ask their physics teacher about escape velocity. The teacher explained to him about escape velocity:
(i) What values are displayed by Jagat and Ram? Is Jagat’s definition was correct?
Yes, Jagat’s definition was correct.
Values are: Curious to know, intelligent, co-operative, helping and sharing their ideas.

(ii) What is the value of escape velocity?
We know that:
Ve= $$\sqrt{2 \mathrm{~g} \mathrm{R}_{\mathrm{E}}}$$
[ On substituting the values g and RE]
= 112 km / s
This is the value of escape speed, sometimes loosely called the escape velocity.

Question 4.
Abhinav and Ankit are in the same section of class 11. Ankit fell sick for two days. So, he could not attend the classes. In two days his physics teacher explained about “Earth Satellites” and “Weightlessness”. Abhinav explained these two topics to Ankit later on Sunday.
(i) Which values are displayed by Abhinav?
The values displayed by Abhinav are:
Helping nature, Ability, and Willingness to explain, patience

(ii) Find an expression for the weight of a body at the center of the Earth.
We know that the value of acceleration due to gravity at a depth “d” below the surface of the earth is given by at the center of the earth

∴ Weight of a body at the center of the earth mg’ = m × o = o
So, the body will be weightless.

## Work, Energy and Power Class 11 Important Extra Questions Physics Chapter 6

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 6 Work, Energy and Power. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

## Class 11 Physics Chapter 6 Important Extra Questions Work, Energy and Power

### Work, Energy and Power Important Extra Questions Very Short Answer Type

Question 1.
What is the source of the kinetic energy of the falling raindrops?
It is the gravitational potential energy that is converted into kinetic energy.

Question 2.
A spring is stretched. Is the work done by the stretching force positive or negative?
Positive because the force and the displacement are in the same direction.

Question 3.
What is the type of collision when?
(a) Does a negatively charged body collide with a positively charged body?
Perfectlyjnelastic collision.

(b) Do macroscopic bodies collide?
Inelastic collision.

(c) Do two quartz balls collide?
Perfectly elastic collision.

Question 4.
(a) Give two examples of potential energy other than gravitational potential energy.
Electrostatic P.E. and elastic P.E.

(b) Give an example of a device that converts chemical energy into electrical energy.
Daniell cell.

(c) Heat energy is converted into which type of energy in a steam engine?
Mechanical energy.

(d) Where is the speed of the swinging pendulum maximum?
At the bottom of the swing.

(e) A heavy stone is lowered to the ground. Is the work done by the applied force positive or negative?
Negative work.

Question 5.
What is the work done by the centripetal force? Why?
Zero. This is because the centripetal is always perpendicular to the displacement.

Question 6.
(a) What is the work done by the tension in the string of simple pendulum?
zero

(b) What is the work done by a porter against the force of gravity when he is carrying a load on his hand and walking on a horizontal platform?
zero

(c) Name the force against which the porter in part (A) is doing some work.
Frictional force.

Question 7.
When an arrow is shot, wherefrom the arrow will acquire its K.E.?
It is the potential energy of the bent bow which is converted into K.E.

Question 8.
When is the exchange of energy maximum during an elastic collision?
When two colliding bodies are of the same mass, there will be a maximum exchange of energy.

Question 9.
Does the work done in raising a load onto a platform depend upon how fast it is raised?
The work done is independent of time.

Question 10.
Name the parameter which is a measure of the degree of elasticity of a body.
Coefficient of restitution.

Question 11.
When a ball is thrown up, the magnitude of its momentum first decreases and then increases. Does this violate the conservation of momentum principle?
No. This is because the momentum of the system (ball and air molecules) remains constant. When the momentum of the ball decreases, the momentum of air molecules in contact increases and vice-versa.

Question 12.
In a tug of war, one team is slowly giving way to the other. What work is being done and by whom?
+ve. The winning team is performing work over the losing team.

Question 13.
A light and a heavy body have equal momentum. Which one of them has more K.E.?
The lighter body has more K.E.

Question 14.
By using simple mechanical devices such as a lever, wedge, inclined plane, pulley, wheel, etc. we can do work even by applying small force. What makes it possible to do so?
These mechanical devices multiply forces.

Question 15.
Two protons are brought closer. What is the effect on the potential energy of the system?
Work is done to overcome the force of repulsion, so the potential energy of the system increases.

Question 16.
Where is the energy of a vertically projected body maximum?
It is the same at all points. This is in accordance with the law of conservation of energy.

Question 17.
Out of a pair of identical springs of force constants, 240Nm-1 one is compressed by 10 cm and the other is stretched by 10 cm. What is the difference in the potential energies stored in the two springs?
The difference in potential energies is zero. This is because
Ep = $$\frac{1}{2}$$ ky2. In both cases, k and y2 are the same.

Question 18.
What should be the angle between the force and the displacement for maximum and minimum work?
For maximum work, θ = 0° and for minimum work θ = 90°.

Question 19.
Does the P.E. of a spring decreases/increase when it is compressed or stretched? Why?
P.E. of a spring increases in both cases. This is because work is done by us in compression as well as in stretching the spring.

Question 20.
Can a body have momentum without energy?
No, because for momentum, it must have some velocity and hence energy.

Question 21.
What type of energy is stored in the spring of the watch?
Potential energy.

Question 22.
Can a body have energy without momentum?
Yes, everybody has some internal energy due to the thermal agitation of the particles of the body. But the vector sum of linear momenta of the moving particles may be zero. It may have energy due to its position (P.E.) and thus momentum is zero.

Question 23.
In which motion, momentum changes but not the K.E.?
In uniform circular motion.

Question 24.
Is the whole of K.E. lost in any perfectly inelastic collision?
No, only as much K.E. is lost as is necessary for the conservation of linear momentum.

Question 25.
Can the P.E. of an object be negative?
Yes, it can be negative when the forces involved are attractive.

Question 26.
The momentum of a body is increased by 50%. What is the percentage change in its K.E.?
When the momentum is increased by 50%, velocity increases by
i.e. velocity becomes $$\frac{3}{2}$$ times, so K.E. becomes $$\frac{9}{4}$$ times
i.e. $$\frac{9}{4}$$ × 100 = 225%. Hence increase in K.E. = 225 – 100 = 125%.

Question 27.
What is the work done by Earth’s gravitational force in keeping the moon in its orbit in a complete revolution?
Zero, because gravitational force is a conservative force.

Question 28.
A spring is cut into two equal halves. How is the spring constant of each half affected?
The Spring constant of each half becomes twice the spring constant of the original spring.

Question 29.
What happens to the energy of our watch which we wind once a day?
The work done in winding the watch is stored as P.E. in the spring which is converted into K.E. of the moving parts of the watch.

Question 30.
Is collision possible even without actual contact of the colliding particles? Give example.
Yes, such collision is called collision at a distance. The collision between subatomic particles (like proton and neutrons) are examples of such collisions.

Question 31.
A cake of mud is thrown on a wall where it sticks. What happens to its initial K.E.?
A part of the K.E. is used in deforming the cake and the remaining part is converted into heat and sound energy.

Question 32.
Nuclear fission and .fusion reactions are the examples of conversion of mass into energy. Can we say that strictly speaking, mass is converted into energy even in an exothermic chemical reaction?
Yes, mass is converted into energy in an exothermic chemical reaction also. But the mass change in a chemical reaction is about a million times less than in a nuclear reaction.

Question 33.
What type of energy is lost in doing work against friction?
The kinetic energy is lost in doing work against the friction.

Question 34.
Define 1 Joule.
A joule of work is said to be done when a force of 1 N displaces a body through 1 m.

Question 35.
Define an erg.
An erg of work is equal to work done by a force of 1 dyne of displacing a body through one centimeter.

Question 36.
Derive the relation between 1 J and 1 erg.
We know that 1 J = 1 N × 1 m = 1 Nm

Question 37.
What is the nature of the work done by the resistive force of air on a vibrating pendulum in bringing it to rest and why?
Negative, because the air always opposes the displacement of the pendulum bob.

Question 38.
What is the work done on a body in moving it in a circular path with a constant speed? Why?
Zero, because it is acted upon by centripetal force in a direction perpendicular to its motion.
i.e. W = FS cos 90° = 0.

Question 39.
A truck and car are moving with the same K.E. and are brought to rest by the application of brakes that provide equal retarding force. Which one will come to rest in a shorter distance? Why?
Both of them come to rest at the same distance as
K.E. = F.S1 = F.S2
or
FS1 cos 180 = F.S2 cos 180
or
S1 = S2.

Question 40.
A conservative force does positive work on a body. What happens to the potential energy of the body? Give an example.
It (P.E.) decreases.
e.g. a body falling freely under gravity.

Question 41.
A man rowing a boat upstream is at rest w.r.t. the shore. Is any work being done in this case? Why?
Yes, as there is a relative displacement between the boat and the stream.

Question 42.
Why friction is a non-conservative force?
It is because the work done against friction along a closed path is non-zero.

Question 43.
Does the work done in raising a load onto a platform depend upon how fast it is raised?
No, work done does not depend upon the time taken in doing it.

Question 44.
On what factors, does the work done depend?
It depends upon the applied force and the displacement of the body.

Question 45.
How will the kinetic energy of a body change if its momentum is tripled? Why?
K.E. becomes nine (9) times its initial value as K.E. ∝ p2.

Question 46.
Which physical quantity is conserved during both the elastic and inelastic collision?
Linear momentum is conserved in both collisions.

Question 47.
What is the most common feature of all types of collisions?
Linear momentum is conserved in all types of collisions.

Question 48.
Define variable force.
It is defined as the force which either changes in magnitude or direction.

Question 49.
Give an example of variable force.
When an object is falling towards the Earth, the gravitational force acting on it goes on changing, so it is a variable force.

Question 50.
Define an electron volt (eV).
It is defined as the K.E. acquired by an electron when a potential difference of 1 volt is applied across it.

### Work, Energy and Power Important Extra Questions Short Answer Type

Question 1.
An airplane’s velocity is doubled,
(a) What happens to its momentum? Is the law of conservation of momentum obeyed?
The momentum of the airplane will be doubled. Yes, the law of conservation of momentum will also be obeyed
because the increase in momentum of the airplane is simultaneously accompanied by an increase in momentum of exhaust gases.

(b) What happens to its kinetic energy? Is the law of conservation of energy obeyed?
K.E. becomes four times. Yes, the law of conservation of energy is obeyed with the increase in K.E. coming from the chemical energy of fuel i. e. from the burning of its fuel.

Question 2.
In a thermal station, coal is used for the generation of electricity. Mention how energy changes from one form to the other. before it is transformed into electrical energy?
When coal is burnt, heat energy is produced which converts water into steam. This steam rotates the turbine and thus heat energy is converted into mechanical energy of rotation. The generator converts this mechanical energy into electrical energy.

Question 3.
Chemical, gravitational and nuclear energies are nothing but potential energies for different types of forces in nature. Explain this statement clearly with examples.
A system of particles has potential energy when these particles are held a certain distance apart against some force. For example, chemical energy is due to the chemical bonding between the atoms. Gravitational energy arises when the objects are held at some distance against the gravitational attraction.

Nuclear energy arises due to the nuclear force acting between the nuclear particles.

Question 4.
What went wrong at the Soviet atomic power station at Chernobyl?
In this reactor, graphite was used as a moderator. The fuel elements were cooled by water and steam was produced from within the reactor. Both water and the steam came in contact with hot graphite. Due to this hydrogen and carbon-monoxide (CO) were released. When they came in contact with air, there was a big explosion.

Question 5.
A man can jump higher on the moon than on Earth. With the same effort can a runner improve his timing for a 100 m race on the moon as compared to that on Earth?
Man can jump higher on the moon because the acceleration due to gravity on the moon is less than that on the Earth. But acceleration due to gravity does not affect the horizontal motion. Hence the runner can’t improve his timing on the moon for the 100 m race.

Question 6.
How many MeV are there in a 1-watt hour?
We know that 1 watt hour = 1 JS-1 × 3600 s = 3600 J

Question 7.
What is Newton’s experimental law of impact?
The ratio of the relative speed of separation after a collision to the relative speed of approach before the collision is always constant. This constant is known as the coefficient of restitution. It is denoted by e.
∴ e = $$\frac{\mathbf{V}_{2 \mathrm{f}}-\mathbf{v}_{1 \mathrm{f}}}{\mathbf{u}_{1 \mathrm{i}}-\mathbf{u}_{2 \mathrm{i}}}$$

where u1i and u2i, are the velocities of the bodies before collision and v2f, v1f are the velocities of the bodies after the collision.

Question 8.
Two masses one n times as heavy as the other have the same K.E. What is the ratio of their momenta?

Question 9.
Two bodies A and B having masses mA and mB respectively have equal K.E. If pA and pB be their respective momenta, then prove that the ratio of momenta is equal to the square root of the ratio of respective masses. fc.
Let vA and vB be the velocities of A and B respectively.
Since their kinetic energies are equal,

Hence Proved

Question 10.
How fast-moving neutrons can be quickly slowed down by passing through heavy water?
Heavy water (D2O) contains deuterium atoms i.e. hydrogen nuclei. They have nearly the same mass as those of neutrons. So when neutrons strike against deuterium atoms, most of the K.E. of the former is transferred to the deuterium atoms i.e. K.E. is exchanged between them.

Question 11.
Will water at the foot of the waterfall be at a different temperature from that at the top? If yes, explain.
Yes, when water reaches the ground, its gravitational potential energy is converted into kinetic energy which is further converted into heat energy. This raises the temperature of the water. So water at the foot of the waterfall is at a higher temperature as compared to the temperature of water at the top of the waterfall.

Question 12.
How is the kinetic energy of a particle related to the direction of motion of the particle? Can K.E. be negative?
The kinetic energy of a particle is not related to the direction of motion of the particle. K.E. is always positive. It cannot be negative.

Question 13.
An automobile jack is employed to lift a heavyweight. The applied force is much smaller than the weight of the automobile. Can it be said that the work is done in less than the work done in? lifting the automobile directly through a height.
No, in the case of the automobile jack, the work done is not calculated from the equation W = FS cos θ, but the work done is calculated from an equation (W = τ θ = FRθ) of rotational motion. In both cases, the same amount of work is done.

Question 14.
What would be the effect on the potential energy of the system of two electrons brought closer?
Work has to be done to overcome the force of repulsion. This work done will be stored in the form of P.E. so, it increases the potential energy of the system.

Question 15.
Can the kinetic energy be increased without the application of an external force? If yes, give an example.
Yes, this is possible. If work is done by the internal force, then K.E. will be increased. As an example, when a bomb explodes, the combined K.E. of all the fragments is greater than the initial K.E.

Question 16.
Mountain roads rarely go straight up the slope, but wind up gradually why?
If roads were to go straight up, the slope (θ) would have been large, the frictional force (µmg cos θ) would be small. The wheels of the vehicle would slip. Also for going up a large slope, a greater power shall be required.

Question 17.
A truck and a car moving with the same K.E. are stopped by applying the same retarding force by means of brakes. Which one will stop at a smaller distance?
Both will stop at the same distance which follows from the work-energy theorem.
K.E. = work done in stopping= retarding force × distance
As K.E. and force for both are equal, so the distance covered must be equal.

Question 18.
A truck and a car are moving with the same K.E. on a straight road. Their engines are simultaneously switched off. Which one will stop at a lesser distance?
The vehicle stops when its K.E. is spent working against the force of friction between the tires and the road. This force of friction varies directly with the weight of the vehicle.

As K.E. = work done
= force of friction × distance
or
E = F × S
or
S = E/F.
Forgiven K.E., S, will be smaller where F is larger such as in the; case of a truck.

Question 19.
A man rowing a boat upstream is at rest w.r.t. the shore. Is any work being done in this case?
No, because the man is applying the force and there is no relative motion between the boat and the shore.

Question 20.
A body is heated by giving Q an amount of heat energy. Will its mass increase or decrease or remain constant? If it increases or decreases, then by how much?
The variation in mass depends on the nature of the body. For most solids, there is no variation in mass on heating. However, if the solid is volatile, heating will change the mass of the solid. Heating of solids, in general, bring about a change in state, and during this process, some loss of mass occurs due to evaporation.

Question 21.
A football kicked by a player leaves the ground and after traveling in the air reaches the ground and comes to stop at some other position on the ground. Identify the energy transformation in this process. Is the energy of the ball conserved?
The kinetic energy imparted to the ball is converted into potential energy at the highest point and a small part remains kinetic. Then it is converted into kinetic energy when it hits the ground. This energy is used up in overcoming the friction against air and the ground. Due to the dissipation of energy in overcoming friction against air and ground, the kinetic energy of the ball is not conserved.

Question 22.
A child enjoys going high and high in a swing by pumping energy. Explain the large amplitude from the rest position using the work-energy idea.
The child pumps energy to the swing at the appropriate time and place from his muscular work. In each swing friction at hinges and air, friction reduces the velocity which he compensates by providing energy to the swing at the appropriate time which partly increases the amplitude of the swing as well. So, periodic muscular work provides the energy to the swing.

Question 23.
How fast neutrons can be slowed down by heavy water? Where does the energy go?
The fast neutrons collide with water molecules and transfer a good part of their kinetic energy to the water molecules. As a result, the kinetic energy and hence the speed of the neutrons is decreased i.e. they are slowed down. The energy received by the water molecules increases their speed and there is an increase in the thermal energy of the mass of water.

Question 24.
Show that the work done by an elastic force on a spring is stored as the potential energy of the spring.
Consider an elastic spring that is extended or compressed. The force does not remain constant in this case. The elastic force Fel according to Newton’s Third law of motion is
Fel = – Fext
when Fext = external force applied
The Fel is always proportional to the compression or extension
i.e. Fel = – kx
∴ Fext = kx
If Fav be the average force, then

If x1 = O (choosing x1 as origin), then x2 = x
The work done by an elastic force is stored as potential energy in the spring and is given by
W = P.E. = $$\frac{1}{2}$$ kx2.

Question 23.
What ¡s transformation of energy? Give at least five exam pies.
The change of energy from one form to another is known as energy transformation. For example:

1. Diesel on burning produces hot gases and heat which is transformed into mechanical energy and is used to run trains and motor vehicles or pumps.
2. Electric energy is transformed into light in a tube or bulb.
3. The electric energy is changed into heat in an oven or heater.
4. Light is converted into electricity with the help of a photovoltaic cell.
5. In a microphone, sound energy is converted into electrical energy.

Question 26.
Which of the following does work: the hammer or the nail, a cricket bat or a ball? Explain.
Hammer does work on nails. The force is applied by a hammer on the nail (object) and moves it through some distance. A bat does work on the ball and pushes. Initially, the bowler does work on the ball which init turn exerts impulse on the bat. The bat reacts with the added force of the batsman and thus, the bat does work on the ball.

Question 27.
A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to $$\frac{\mathbf{k}}{\mathbf{r}^{2}}$$ where k is a constant. What is the total energy of the particle?

Question 28.
What are the relations between
(a) kilowatt hour and kilocalorie?

(b) metric horsepower and kilocalorie per hour?

Question 29.
Prove that instantaneous power is given, by the dot product of force and velocity i.e. P = F.v.
Let AW be the amount of work done in a small time interval Δt. If Pav be the average power, then

where F constant force producing a displacement S
∴ From (ii) and (iii), we get

Question 30.
Define spring constant or force constant. What is its S.I. unit?
It is defined as the extension per unit displacement produced in the spring or wire.
According to Hook’s law, the extension is always directly proportional to the tension or force applied on a spring i.e.
F ∝ x
or
F = k x
or
k = $$\frac{F}{x}$$

where k is the force constant
S.I unit of k is Newton metre-1 (Nm-1).

### Work, Energy and Power Important Extra Questions Long Answer Type

Question 1.
(a) State work-energy theorem or principle.
It states that the work done on a body is equal to the change in its kinetic energy.
i.e. W = change in kinetic energy
Proof: Let m = mass of a body moving in a straight line with a constant initial velocity u.

Let F = force applied on it at point A to B so that its velocity is V at B.
If dx = small displacement from P to Q
and a = acceleration produced in the body, then
F = ma

If dw be the work done from P to Q, then

If W = total work done from A to B, then

(b) State and prove the law of conservation of energy.
It states that energy can neither be created nor can be destroyed but it can be changed from one form of energy into another i.e. total energy = constant.

Proof: Let a body of mass m be lying at rest at point A at a height h above the ground. Let it be allowed to fall freely and reaches a point B after falling through a distance x and it finally hits the ground at point C. Let v and V be its velocities at points B and C respectively.
∴ AB = x and BC = h – x

At point A: u = 0
∴ K.E. = 0
P.E. = mgh

If E be the total energy of the body, then
E = K.E. + P.E. = 0 + mgh
or
E = mgh …. (i)

At point B: using the relation,

At point C: Here, v = vc, a = g, s = h

Thus, from (i), (ii), and (iii), it is clear that total energy at points A, B, and C is the same. It is purely P.E. at A and purely K.E. at point C.

Numerical Problems:

Question 1.
An engine draws a train up an incline of 1 in 100 at the rate of 36 km h-1. If the resistance due to friction is 5 kg wt per ton, calculate the power of the engine. Mass of train and engine is 100

Here, m = 100 metric ton = 100 × 1000 kg
Total force of friction,
f1 = 100 × 5 = 500 kg wt
= 500 × 9.8 N = 4900 N
sin θ = $$\frac{1}{100}$$

Let f2 = Downward force on the train
= component of its weight acting in downward direction parallel to the inclined plane = mg sin θ
= 100 × 1000 × 9.8 × $$\frac{1}{100}$$ = 9800 N

If F be the total force against which engine has to work, then

Question 2.
Calculate the work done and power of an engine that can maintain a speed of SO ms-1 for a train of mass 3 × 106 kg on a rough level track for 5 km. The coefficient of friction is 0.05. Given g = 10 ms-2.
Here,

Question 3.
A 10 kg block slides without acceleration down a rough inclined plane making an angle of 20° with the horizontal. Calculate the work done over a distance of 1.2 m when the inclination of the plane is increased to 30°.
Here, angle of sliding = 20°
μ = coefficient of friction
= tan 20° = 0.3647
S = 1.2 m

m = mass of block = 10 kg.

Let a = acceleration when the inclination is increased to 30° and F be the value of limiting friction then

Question 4.
A bullet of mass 0.03 kg moving with a speed of 400 ms-1 penetrates 12 cms into a fixed block of wood. Calculate the average force exerted by the wood on the bullet.
Here,
mass of bullet, m = 0.03 kg
initial K.E. = $$\frac{1}{2}$$mv2 = $$\frac{1}{2}$$ × 0.03 × (400)2 = 2400 J
Final K.E. = $$\frac{1}{2}$$mv2 = 0

∴ ΔK = Loss in kinetic energy = 2400 J
Let F = average force applied by block on bullet
S = 0.12 m

∴According to work energy principle
W = ΔK
or
FS = ΔK
or
F × 0.12 = 2400
or
F = 2 × 104 N

Question 5.
A boy of mass 50 kg sits in a swing suspended, by a rope 5 m long. A person pulls the swing to one side so that the rope makes an angle of 30° with the vertical. What is the gain in gravitational P.E. of the boy?
Here, θ = 30°
OA = OB = 5 m …. (i)
m = 50 kg

Vertical height through which boy at B rises up is given by

Question 6.
(a) bullet of mass 20 g moving with a velocity of 500 ms-1 strikes a tree and goes out from the other end. with a velocity of 400 ms-1. Calculate the work done in passing through the tree.
Here m = 0.02 kg
u = 500 ms-1
v = 400 ms-1
w = ?
from work energy theorem, work done loss in K.E.

(b) A bullet of mass 0.01 kg is fired horizontally into a 4 kg wooden block at rest on a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.25. The bullet remains embedded in the block and the combination moves 20 m before coming to rest. With what speed did the bullet strike the block?
Here S = 20 m
μ = 0.25
M = mass of block = 4 kg
m = mass of bullet = 0.01 kg

∴ R = normal reaction = Mg = 4 × 9.8
F = μR = 0.25 × 4 × 9.8 = 9.8 N
v = 0
u = ?
or
loss in KE. = work done by the force of friction against the block.

Question 7.
A stone of mass 5 kg falls from the top of a cliff 30 m high and buries itself one meter deep into the sand. Find the average resistance offered and time taken to penetrate into the sand.
Here, m = 5 kg
h = 30 m
S = 1 m
∴ P.E. of stone = mgh = 5 × 9.8 × 30 = 1470 J

Let F = average force of resistance offered by the sand and F’ = net upward force.

Now momentum of stone before penetration
mu = mv = 5 × 24.25 (∵ here u = v)

Final momentum of stone when it becomes stationary (i.e. v’ = 0)
= mv’ = m(0) = 0
∴ Ft = – mv’ + mu

Question 8.
An automobile moving at a speed of 72 km h-1 reaches the foot of a smooth incline when the engine is switched off. How much distance does the automobile go up the incline before coming to rest? Take angle of incline 30°, g = 9.8 ms-2.
Here, θ = 30°
Initial velocity of the automobile at point A when it is switched off = 72 kmh-1
= 72 × $$\frac{5}{18}$$ = 20 ms-1

Let m = mass of automobile.
Initial K.E. of the automobile at A = $$\frac{1}{2}$$ m(20)2 = 200 m

F = mg sin θ = force opposing the motion of the automobile
= m × 9.8 sin 30° = (4.9m)N

Let it stop at B after travelling AB = S
∴ Final K.E. = 0
∴ change in K.E. = 200m – 0 = 200m J
∴ W = work done by the automobile
= FS = (mg sin θ) × S = 4.9 m × S

∴ According to work-energy theorem,
W = change in K.E.
or
4.9m S = 200m
or
S = $$\frac{200}{4.9}$$ = 40.8m

Question 9.
A pump-set is used to lift water to a reservoir of 6000 liters capacity over an average height of 20m. If it takes 1 hour to All the reservoir completely, calculate the work done and the power supplied to the set if its efficiency is 60%, g = 9.8 ms-2.
Mass of 1 litre of water = 1 kg .
∴ m = 6000 kg
h = 20 m

∴ W = work done = P.E. = mgh
= 6000 × 9.8 × 20= 11.76 × 10s J
t = 1 h = 3600 s

Question 10.
A wooden ball is dropped from a height of 2 m. What is the height up to which the ball will rebound if the coefficient of restitution is 0.5?
Here, e =0.5
Let the ball fall from a height h1 from point A and rebound to B at a height of h2.
∴ h1 = 2 m

∴ If u1 be the velocity on reaching the ground = velocity of approach and u2 = Velocity on leaving the ground = velocity of separation

Question 11.
What percentage of K.E. of a moving particle is transferred to the stationary particle of
(a) 9 times it’s mass
Let m1 = mass of moving particle = m
and m2 = mass of stationary particle = 9 m
and their velocities are u1 and u2 = 0 (given)
v2 = velocity of the stationary particle after the collision.

So, % of K.E. transferred to the stationary particle

(b) equal mass,

(c) $$\frac{1}{9}$$ th of its mass.

Question 12.
Calculate the work done in raising a stone of mass 6 kg of specific gravity 2 immersed in water from a depth of 4 m to 1 m below the surface of the water. Take g = 10 ms-2.
Here, m = 6 kg
∴ weight of stone = 6 × 10 = 60 N
specific gravity of stone = 2
∴ d = density of stone
= 2 × lg cm-3
= 2g cm-3.

∴ volume of stone = $$\frac{\mathrm{m}}{\mathrm{d}}=\frac{6 \times 1000 \mathrm{~g}}{2 \mathrm{gcm}^{-3}}$$ = 3000 cm3
= 3000 × 10-6 m3
= 3 × 10-3 m3
= volume of displaced water.

Weight of water displaced or upward thrust on the stone
= volume × density of water × g
= 3 × 10-3 × 1000 × 10 = 30 N

∴ The net force acting on the stone = weight of stone – upward thrust on the stone
or F = 60 – 30 = 30 N
Distance through which stone is raised = (4 – 1 )m = 3m
∴ work done, W = FS = 30 × 3 = 90 J.

Question 13.
An antenna radiates energy for 24 hours at the rate of 1 kW. Calculate the equivalent mass for radiated energy.

Question 14.
A particle moves from position r1 = 3î + 2ĵ – 6k̂ to position r2 = 14î + 13ĵ – 9k̂ under the action of a force (4î + ĵ + 3k̂)N. Calculate the work done.
Here, F = (4î + ĵ + 3k̂)N
Displacement of the particle,

Question 15.
The displacement x of a particle moving in one dimension under the action of a constant force is related to the time by the equation t = $$\sqrt{x}$$ + 3.
where x is in meters and t is in seconds.
Find (i) the displacement of the particle when its velocity is zero and
(ii) work is done by force in the first six seconds.

Question 16.
A 1 kg mass on a floor is connected to a 2 kg mass by a string passing over a pulley as shown in the figure. Obtain the speed of the masses (after they are released) when the 2 kg mass just touches the floor. Show that the gain in kinetic energy of the system equals the loss in its potential energy. The 2 kg mass is initially at a height 3 m above the ground.
Here, m2 = 1kg
m1 = 2kg
u = 0
h = 3m
The system is shown in the figure. According to Newton’s Second law of motion, the equation of motion for 2 kg mass is
mg – T = ma
or
2 × 9.8 – T = 2a …(i)

(Free body diagrams of m1 and m2)

Since both the masses are initially at rest, therefore initial K.E. of the system = 0
Final K.E. of the system

∴ Gain in K.E. = 29.4 – 0 = 29.4 J
Initially P.E. of the system = m1gh1 + m2gh2
= 2 × 9.8 × 3 + 1 × 9.8 × 0
= 58.8 J ( ∵ here h2 = 0)

Finally, the 2 kg reaches the floor and mass 1 kg is at a height of 3m.
∴ Final P.E. of the system = m1gh1 + m2gh2
= m1g × 0 + 1 × 9.8 × 3
= 29.4 J

∴ Loss of P.E. = 58.8 – 29.4
= 29.4 J
Thus Gain in K.E. = Loss of P.E. = 29.4 J.

Question 17.
A homogeneous and inextensible chain of length 2 m and mass 100 g lies on a smooth table, A small portion of the chain of length 0.5 m hangs from the table. Initially, the part of the chain lying on the table held and then released. Calculate the velocity with which the chain leaves the table, g = 10 ms’2.
Mass of chain, M = 100 g = 0.1 kg
length of chain, l = 2 m

∴ Mass per unit length of the chain, m = $$\frac{M}{l}$$

Let v = velocity of the chain when it leaves the table.

Question 18.
The K.E. of a body increases by 300%. How much linear momentum of the body will increase?

Question 19.
A ball falls under gravity from a height of 10m with an initial downward velocity of u. It collides with the ground, loses 50% of its energy in the collision, and then rises back to the same height. Find the initial velocity u.
Here, u = initial velocity of the ball
h = its height

Let M = mass of the ball
∴ The total initial energy of the ball at height h,

The ball rises to height h due to this energy.
∴ P.E. of ball = mgh
∴ According to the law of conservation of energy

Question 20.
The heart of a man pumps 4-liter blood per minute at a pressure of 130 mm of mercury. If the density of mercury is 13.6 gm cm3, then calculate the power of the heart.
Here, the height of the mercury column,

Question 21.
A body is allowed to fall from a height of 100 m. Find the increase in temperature of the body on reaching the ground if the whole of heat produced remains in the body. Given that the specific heat of the body = 0.2, J = 4.2 J cal-1.
Loss in P.E. of the body in falling through 100 m = m × 9.8 × 100 J = $$\frac{980 \mathrm{~m}}{4.2}$$ (∵ P.E = mgh)

This loss is transformed into the heat state of the body = ms θ
where θ = rise in temperature.

Question 22.
A truck of mass 2000 kg has a velocity of 8 ms-1 when it starts from a point to descend a slope 200 m long. The vertical height of the slope is 18 m and the truck arrives at the bottom with a velocity of 20 ms-1. Calculate the resistance offered.
Take g = 10 ms-2.

Here, l = XY = length of slope.
= 200 m
OX = h = 18 m
Let u and v be the velocities of the .truck at points X and Y respectively.

Question 23.
A body of mass m slowly hauled up the hill by a force F which at each point was directed along a tangent to the trajectory. Find the work performed by this force if the height of the hill is h, the length of its base is l, and the coefficient of friction is μ.

Let PQ be the plane patch of the hill
dz = its length inclined at an angle θ with the horizontal

Force required to move the mass from P to Q
= mg sin θ + μmg cos θ

∴ Work done from P to Q = mg (sin θ + μcos θ).

∴ W = total work done in moving the mass m from A to B is given by

Question 24.
A chain of mass m and length l lying on a rough table starts sliding off the table all by itself till the hanging part equals n. Calculate the total work performed by the friction forces acting on the chain by the moment it slides off the table completely.
Weight of suspended chain = n mg
weight of the part of the chain on the table = (1 – n) mg

Let μ = coefficient of friction of tabletop.
For sliding motion of the chain on the tabletop under the weight of’ the hanging chain,
μ(l – n) mg = n mg
or
μ = $$\frac{n}{1 – n}$$ …(i)

The initial force of friction when the chain is pulled on the tabletop
= μ(1 – n)mg

Final force, when all the chain has been pulled = 0
mean friction = $$\frac{\mu(1-n) m g}{2}$$

Distance for which the chain is pulled = – (1 – n)l
∴ W = mean force × distance

Question 25.
A flexible chain of length l and mass m is slowly pulled up at a constant speed over the edge of a table by a force F(x) parallel to the edge of the tabletop. Calculate the work done by F(x).
Here the force required to pull the chain over the table varies with distance.
It is equal to the weight of the fraction of the chain $$\frac{(l-\mathrm{x})}{l}$$still hanging over the edge.

Value-Based Type:

Question 1.
Rajesh and Rakesh are two friends who participated in a quiz contest. The teacher asked Ramesh to lift a 20 kg weight vertically upwards to a height of 10 m and Rekcsh asked to pull 20 kg of weight by a rope over a pulley from a depth of 10 m from a well. There was a prize for one who will do more work. Rajesh claimed that he has done more work than Rakesh because he has not used any kind of lever. The teacher told Rajesh that he has done zero work but Rakesh has done the work. Hence, the prizes were given to Rakesh
(i) Why the teacher told that Rajesh did not do any work? justify your answer.

1. The teacher wanted to check the knowledge of physics and awareness in a playful manner.
He used the demonstration method to create interest.
2. We know that:
W = F d cos θ
In the case of Rajesh:
θ = 90° and cos 90°=o
∴ work done(w) =Fd × o = o
Hence, no work was done.

Question 2.
Shikha is doing her Engineering from IIT Delhi. Her mother was also an Engineer who wanted to check the knowledge of her daughter. She asked a question as to under. “If we assume that the moon orbits around the earth is perfectly circular then the earth’s gravitational force does no work. Is it correct?”
She explained that the moon’s displacement is tangential Whereas the earth’s force is inwards and θ = $$\frac{π}{2}$$
So, W = F.d. cos $$\frac{π}{2}$$ = o
(i) Which values are displayed by Shikha?
Values displayed are:
Explanatory, intelligent, cooperative.

(ii) Why her mother asked such a question?
Her mother wanted to test the knowledge and presence of mind of her daughter to ask such type of tricky questions.

Question 3.
Suraj went to Big Bazaar to purchase certain goods. There he has noticed an old lady struggling with her shopping. Immediately he showed her the lift and explained to her how it carries the load from one floor to the next. Even then the old lady was not convinced. Then Suraj took her in the lift and showed her how to operate it..That old lady was very happy.
(a) What values does Suraj possess?
Suraj is sympathetic and also has the attitude of helping others. He has patience.

(b) An elevator can carry a maximum load if 1800 kg is mov¬ing up with a constant speed of 2 m/s, The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horsepower.
The downward force on the elevator is
F = mg + Ff = (1800 × 10) + 4000 = 22000 N

The motor must supply enough power to balance this force.
Hence P = F.V = 22000 × 2 = 44000 = 59 hp

## Mechanical Properties of Solids Class 11 Important Extra Questions Physics Chapter 9

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 9 Mechanical Properties of Solids. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

## Class 11 Physics Chapter 9 Important Extra Questions Mechanical Properties of Solids

### Mechanical Properties of Solids Important Extra Questions Very Short Answer Type

Question 1.
Give an example of pure shear.
The twisting of a cylinder produces pure shear.

Question 2.
What is an elastomer?
It is a substance that can be elastically stretched to large values t of strain.

Question 3.
What is breaking stress?
It is defined as the ratio of maximum load to which the wire is < subjected to the original cross-sectional area.

Question 4.
What is the:
(a) value of modulus of rigidity of a liquid?
zero.

(b) order of strain within the elastic limit?
10-3 cm per cm = 10-3 cm/cm.

Question 5.
A wire is stretched to double its length. What is the value of longitudinal strain?
Unity.

Question 6.
Mention a situation where the restoring force is not equal and opposite to the applied force.
This happens when the body is deformed beyond the elastic limit.

Question 7.
What is a Cantilever?
It is a beam loaded at one end and free at the other end.

Question 8.
A wire is suspended from a roof but no weight is attached to the wire. Is the wire under stress?
Yes, the weight of the wire itself acts as the deforming force.

Question 9.
Why strain has no units?
As it is the ratio of two similar quantities.

Question 10.
What is Poisson’s ratio?
It is the ratio of lateral strain to linear strain.

Question 11.
What is the:
(a) the bulk modulus of a perfectly rigid body?
infinite.

(b) value of Y for a perfectly rigid body?
infinite.

(c) bulk modulus for an incompressible liquid?
infinite.

Question 12.
Why does spring balance shows wrong readings after they have been used for a long time?
Because of elastic fatigue.

Question 13.
Name three physical properties which can have different values in different directions.
Thermal conductivity, compressibility, and electrical conductivity.

Question 14.
What will happen to the potential energy if a wire is
(a) compressed,
(b) stretched?
In both cases, the potential energy of the wire increases as work has to be done on the wire.

Question 15.
Which of the two materials (see figure here) would you choose for a car tire? Why?

Rubber (A) would be used to avoid excessive heating of the car tire.

Question 16.
How are depression and load interrelated in the case of a beam
(a) having rectangular cross-section and loaded in the middle?
δ = $$\frac{\mathrm{W} l^{3}}{4 \mathrm{bd}^{3} \mathrm{Y}}$$

(b) having circular cross-section and loaded in the middle?
δ = $$\frac{\mathrm{W} l^{3}}{12 \mathrm{Y} \pi \mathrm{r}^{4}}$$

Question 17.
Write copper, steel, glass, and rubber in the order of increasing coefficient of elasticity.
Rubber, glass, copper, and steel.

Question 18.
How does Young’s modulus change with rising in temperature?
Young’s modulus of a material decreases with rising in temperature.

Question 19.
The length of a wire is cut in half. What will be the effect on the increase in its length under a given load?
An increase in length will be reduced to half as Δl ∝ l.

Question 20.
A wire is replaced by another wire of the same length and material but of twice diameter. What will be the effect on the:
(a) increase in its length under a given load?
Increase in length will be reduced to one fourth as Δl ∝ $$\frac{1}{\mathrm{r}^{2}}$$.

(b) maximum load which it can bear?
The maximum bearable load becomes four times as breaking force ∝ area. (= πr²)

Question 21.
Sand does not possess any definite shape and volume, still, it is solid?
Sand is a divided rock. It has all properties of solids and even its particles have indefinite volume like amorphous solids.

Question 22.
Name one material that is famous for a large elastic after effect.
Glass.

Question 23.
Why some solids have high thermal and electrical conductivity?
Because such solids have a large number of free electrons available in them (e.g. metals).

Question 24.
When a wire is bent back and forth, it becomes hot. Why?
When a wire is bent back and forth, the deformations are beyond the elastic limit.

Question 25.
Two identical solid balls, one of ivory and the other of wet- clay are dropped from the same height on the floor. Which one will rise to a greater height after striking the floor and why?
The ivory ball will rise to a greater height as compared to a clay ball because ivory is more elastic than wet clay.

Question 26.
Define ductile materials.
They are defined as materials whose plastic range is relatively large.

Question 27.
Define brittle materials.
They are defined as materials whose plastic range is relatively small.

Question 28.
Define elastic fatigue.
It is defined as the loss of strength of the material caused due to repeated alternating strains to which the material is subjected.

Question 29.
Define elastic after effect.
It is defined as the delay in regaining the original state by a body after the removal of the deforming force.

Question 30.
Define elasticity.
It is the property of matter due to which it regains its original shape and size when the deforming forces have been removed.

Question 31.
Define hydrostatic stress.
When a body is subjected to a uniform and equal force from all sides, then the corresponding stress is called hydrostatic stress.

Question 32.
Define plasticity.
It is defined as the property of matter due to which it does not regain its original shape and size after the removal of deforming forces.

Question 33.
Define tangential or shearing stress.
It is defined as the deforming force acting per unit area tangential to the surface.

Question 34.
Define deforming force.
It is defined as the external force which when applied to a body changes its configuration (i.e. shape and size).

Question 35.
Define elastic limit.
Maximum stress is called the elastic limit.

Question 36.
Define Young’s modulus.
Within elastic limits, it is defined as the ratio of normal stress to longitudinal strain.

Question 37.
Define compressibility.
It is defined as reciprocal of bulk modulus
i. e. compressibility = $$\frac{1}{\mathrm{~K}}=\frac{\Delta \mathrm{V}}{\mathrm{PV}}$$

Question 38.
Define modulus of rigidity.
Within elastic limits, it is the ratio of tangential stress (T) to shear strain (θ)

Question 39.
What is the limitation of Hook’s law?
It holds good when the wire is loaded within its elastic limit.

Question 40.
What are the factors on which the modulus of elasticity of material depends?
Nature of the material and the manner of deforming the wire.

Question 41.
Name the property of a body that opposes its deformation?
Elasticity.

Question 42.
Which of the three types of elasticity (Y, K, and t) is possessed by all the three states of the matter?
The volume elasticity (K) is possessed by all three states of the matter.

Question 43.
Explain why liquids don’t possess rigidity?
Liquids don’t possess rigidity because they have no shape of their own.

Question 44.
A wire of length L and cross-sectional area A is made of a material of Young’s modulus Y. If the wire is stretched by an amount x, then what is the work done?

Question 45.
What is the importance of the stress-strain curve?
Its slope gives the modulus of elasticity.

Question 46.
A bar is subjected to equal and opposite forces PQ is a plane ‘ making angle θ with the cross-section ‘a’ of the bar. Calculate the tensile stress on PQ.

Area PQ = $$\frac{a}{\cos \theta}$$

Tensile stress = $$\frac{\mathrm{F} \cos \theta}{\mathrm{a} / \cos \theta}=\frac{\mathrm{F} \cos ^{2} \theta}{\mathrm{a}}$$

Question 47.
Calculate shearing stress on PQ in Fig.

Shearing stress = $$\frac{F \sin \theta}{a / \cos \theta}=\frac{F}{a} \sin \theta \cos \theta=\frac{F}{2 a}(2 \sin \theta \cos \theta)$$
= $$\frac{F}{2a}$$ sin 2θ

Question 48.
In a bar is subjected to equal and opposite forces PQ is a plane making angle θ with the cross-section ‘a’ of the bar. Calculate the tensile stress on PQ.when the tensile stress can be maximum?
We know that, Tensile stress = $$\frac{\mathrm{F} \cos ^{2} \theta}{\mathrm{a}}$$
For tensile stress to be maximum, cos20 must be maximum – i.e. cos θ = 1 or θ = 0.

Question 49.
In Calculate shearing stress on PQ, when the shearing stress can be maximum?

Answer:Shearing stress = $$\frac{F}{2a}$$ sin2θ
For shearing stress to be maximum, sin20 must be maximum

Question 50.
The length of a rod is doubled, under the action of a constant force. If the initial stress is S, then what will be the final stress?
We know that stress = $$\frac{F}{A}$$ . Also AL = constant for a given rod. When L is doubled, A becomes half, so stress becomes double.

### Mechanical Properties of Solids Important Extra Questions Short Answer Type

Question 1.
What are the factors due to which three states of matter differ from one’s Other?
Three states of-matter differ from each other due to the following two factors:
(a) The different magnitudes of tester atomic and intermolecular forces.
(b) The degree of random thermal motion of the atoms and molecules of a substance depends upon the temperature.

Question 2.
When we stretch a wire, we have to perform work Why? What happens to the energy given to the wire in this process?
In a normal situation, the atoms of a solid are at the locations of minimum potential energy. When we stretch a wire, the work has to be done against interatomic forces. This work is stored in the wire in the form of elastic potential energy.

Question 3.
Why are the bridges declared unsafe after long use?
A bridge during its use undergoes alternative strains a large number of times each day, depending upon the movement of vehicles on it. When a bridge is used for a long time it loses its elastic strength, due to which the number of strains in the bridge for given stress will become large and ultimately the bridge may collapse. Thus, !» to avoid this, the bridges are declared unsafe after long use.

Question 4.
Why are the springs made of steel and not of copper?
Spring will be a better one if a large restoring force is set up in it on being deformed, which in turn depends upon the elasticity of the material of the spring. Since Young’s modulus of elasticity of steel is more than that of copper, hence steel is preferred in making the springs.

Question 5.
A heavy machine is to be installed in a factory. To absorb vibrations of the machine, a block of rubber is placed between the machinery and the floor. Which of the two rubbers (A) and (B) of Figure would you prefer to use for this purpose? Why?

The area of this hysteresis loop measures the amount of heat energy dissipated by the material. Since the area of the loop B is more than that of A, therefore B can absorb more vibrations than that of Av Hence B is preferred.

Question 6.
Metal wires after being heavily loaded dop’\ regain their lengths completely explain why?
A material regains its original Configuration (length, shape dr volume) only when the deforming force is within the elastic limit. Beyond the elastic limit, the bodies lose the property of elasticity and hence don’t completely regain the length of being heavily loaded.

Question 7.
Explain. Why spring balances show wrong readings after they have been, Used for a long time?
When spring balances are used for a long time, they get fatigued. So the springs of such balances will take time to recover their original configuration. Hence the readings shown by such spring balances will be wrong.

Question 8.
Elasticity is said to be the internal property of matter. Explain.
When a deforming force acts on a body, the atoms of the substances get displaced from their original positions. Due to this the configuration of the matter (substance) changes. The moment, the deforming force is removed, the atoms return to their original positions and hence the substance or matter regains its original configuration. Hence elasticity is said to be the internal property of matter.

Question 9.
Define tensor physical quantities. Give an example.
They are defined as the physical quantities having different values in different directions e.g. stress.

Question 10.
Define compressional stress.
It is defined as the restoring force developed per unit area of cross-section of a body when it is compressed i.e. when its length decreases under the action of deforming force.

Question 11.
Define longitudinal or tensile stress.
It is defined as the restoring force developed per unit area of cross-section of a body when the length of the body increases in the direction of the deforming force.

Question 12.

and tell in which significant ways do these curves differ from the stress-strain curve.

1. Hook’s law is not obeyed at all.
2. The elastic region is large here.

Question 13.
Define restoring force.
It is defined as the internal force which comes into play from within the body due to which it regains or tends to regain its original configuration.

For a perfectly elastic body restoring force = Deforming force.

Question 14.
Define longitudinal strain.
It is defined as the ratio of change in length to the original length of an object when deformed by an external force

i.e. Longitudinal strain = $$\frac{l}{\mathrm{~L}}=\frac{\text { change in length }}{\text { original length }}$$

Question 15.
Define volumetric strain.
It is defined as the ratio of change in volume per unit original volume of the body when deformed by an external force change in volume AV

i.e. Volumetric strain = $$\frac{\text { change in volume }}{\text { original volume }}=\frac{\Delta \mathrm{V}}{\mathrm{V}}$$

Question 16.
Define shear strain.
It is defined as the ratio of the lateral displacement of a layer to its perpendicular distance from a fixed layer.

Aliter: It is defined as the angle through which a line originally perpendicular to the fixed face gets turned on applying tangential deforming force.

Question 17.
State Hook’s law.
Hook’s law: States that within elastic limits, stress is directly proportional to the strain.
i. e. stress ∝ strain
or
$$\frac{\text { Stress }}{\text { Strain }}$$ = constant = E.
where E is called coefficient or modulus of elasticity of the material.

Question 18.
Define bulk modulus.
Within elastic limits, it is defined as the ratio of normal stress to volumetric strain.

Question 19.
On what factors does the value of the coefficient of elasticity depend? Why it is of it three types?
Its value depends upon the:

1. nature of the material of the body.
2. the way in which the body is deformed.

It is of three types as strain is of three types.

Question 20.
Why a hard wire is broken by bending it repeatedly in opposite direction?
It is because of the loss of strength of the material due to repeated alternating strains to which the wire is subjected.

Question 21.
When a cable is cut to half its original length, the maximum load it can withstand does not change. Why?
The breaking stress is constant for a given material. Now breaking lord = breaking stress × area. When we cut the cable to half its length, its area of cross-section does not change. Hence there is no effect on the maximum load, the cable can support.

Question 22.
What causes restoring stress when a wire is stretched and when a body is compressed?
When a wife is stretched, the restoring stress is caused by inter atomic-attraction and when a body is compressed, the restoring stress is caused by interatomic repulsion.

Question 23.
Are elastic restoring forces conservative?
The elastic restoring forces are conservative when the loading and unloading curves coincide, but when these curves are different, the elastic restoring forces are non-conservative.

Question 24.
When a cable is cut to half its original length, how does this affect the elongation under the given load?

Now as L is made $$\frac{L}{2}$$ l will become $$\frac{L}{2}$$, so elongation will become half of its value before cutting.

Question 25.
Why do solids have well-defined and reproducible external shapes?
It is because, the atoms and molecules are arranged in a definite and regular way throughout the body of the solids, so they have well-defined and reproducible external shapes.

Question 26.
Why any metallic part of the machinery is never subjected to stress beyond the elastic limit of the material?
If it is subjected beyond the elastic limit, then a permanent deformation will be set in that metallic part of the machine.

Question 27.
The braking force for a wire is F. What will be the breaking force for
(a) two parallel wires of the same size?
When two wires of the same size are suspended in parallel, force F equal to the braking force for a wire will acton each wire if a force 2F is applied on the parallel combination.

(b) for a single wire of double the thickness?

If the thickness is doubled, then the braking force will be four times the braking force.

Question 28.
Elasticity has a different meaning in Physics and in our daily life. Why?
In daily life, a body is said to be more elastic if large deformation or strain is produced on subjecting the material to a given stress. But in Physics, it is exactly the opposite. A body is said to be more elastic if a small strain is produced on applying the given stress.

Question 29.
The length of a Wire is increased by 16 cm when a weight of 5 kg is hung. If all conditions are the same, what will be the increase in its length when the diameter is doubled?
We know that,

when d = diameter of wire

∴ l ∝ $$\frac{1}{\mathrm{~d}^{2}}$$
Thus when d is doubled, l reduces to $$\frac{1}{4}$$th
i.e. l’= $$\frac{l}{4}=\frac{8}{4}$$ = 2 cm. 4 4

Question 30.
When stress is equal to Young’s modulus of elasticity, then calculate the extension of a wire of length l.
Here, L = original length of wire.
Let Δl = extension of wire = ?
Stress = Y (given)

### Mechanical Properties of Solids Important Extra Questions Long Answer Type

Question 1.
(a) What are the factors affecting elasticity?
The following factors affect the elasticity of a material:

1. Effect of hammering and rolling: It causes a decrease in the plasticity of the material due to break¬up of crystal grains into smaller units and hence – the elasticity of the material increases.
2. Effect of Annealing: Annealing results in the increases in the plasticity of the material due to, the formation of large crystal grains. Hence the elasticity of the material decreases.
3. Effect of the presence of impurities: The effect of the presence of impurities in a material can be both ways i.e. it can increase as well as decrease the elasticity r of the material. The type of effect depends upon the
nature of the impurity present in the material.
4. Effect of temperature: The increase in the temperature of the material in most cases causes a decrease in the elasticity of the material. The elasticity of invar does not change with the change of temperature.

(b) Define Poisson’s ratio.
Poisson’s Ratio (σ): Within elastic limits, it is defined; as the ratio of lateral strain (β) to the linear strain i.e.
σ = $$\frac{β}{α}$$

Breaking Load: It is defined as the product of the breaking stress and area of cross-section of the given object. It is also called maximum load a body (cable/ wire) can support

i.e. breaking load = Breaking stress × area of cross-section. It should be noted that breaking stress is a constant for the given material.

Question 2.
The length of a metallic wire is L1 when tension is T1 and L2 when tension is T2. Find the original length of the wire.
Let L, A be the length and area of the cross-section of the wire.
Also, let l be the extension produced on applying a force F, then

where Y = Young’s modulus.
Now when F = T1 and l = L1 – L.

Numerical Problems:

Question 1.
A mass of 5 kg is hung from a copper wire of 1 mm diameter and 2 m in length. Calculate the extension produced. What should be the minimum diameter of the wire so that its elastic limit is not exceeded? Elastic limit for copper = 1.5 × 109 dyne cm-2, Y for copper = 1.1 × 1012 dyne cm-2.

Let d1 be the minimum diameter, then maximum stress

Question 2.
A cube of aluminum of each side 4 cm is subjected to a tangential (shearing) force. The top of the cube is sheared through 0.012 cm w.r.t. the bottom face.
Find (a) shearing strain,
(b) shearing stress,
(c) shearing force. Given η = 2.08 × 1011 dyne cm-2.
Here,
length of each side, L = 4 cm
Lateral displacement, x = 0.012 cm ,
η = 2.08 × 1011 dyne cm-2.

Question 3.
The breaking stress of aluminum is 7.5 × 108 dyne cm-2. Find the greatest length of aluminum wire that can be hung vertically without breaking. The density of aluminum is 2.7 g cm-3. Take g = 980 cm s-2.
Here, breaking stress = 7.5 × 108 dyne cm-2, ρ = 2.7 g cm-3

Let l be the greatest length of the wire that can be hung vertically without breaking.
mass of wire, m = area of cross-section × length × density = a /ρ.
∴ Weight of wire, W = mg = a / ρ g.

This is equal to the maximum force that the wire can withstand.

Question 4.
Calculate the % increase in the length of a wire of diameter 2,5 unstretched by a force of 100 kg. Y for the wire = 12,5 × 1011 dyne cm-2.

Question 5.
Compare the densities of water at the surface and bottom of a lake 1oo m deep, given that the compressibility is $$\frac{10^{3}}{22}$$ per atm and 1 atm = 1.015 × 105 Pa.

Let V = Volume of 1 kg water at the surface.
V’ = Volume of 1 kg water at the bottom of lake 100 m deep
= V – ΔV, where ΔV = decrease in volume, increase in pressure, P = hρg = 100 × 103 × 9.8 Nm2

If ρs and ρb be the densities of water at the surface and at the bottom of the lake respectively, then

Question 6.
A steel wire 2 mm in diameter is stretched between two clamps, when its temperature is 40° C. Calculate the tension in the wire, when its temperature falls to 30° C. Given, coefficient of linear

If Δl be the change n length of the wire, then

Question 7.
When a weight W is hung from one end of a wire of length L (other end being fixed), the length of the wire increases by l fig. (a). If the wire is passed over a pulley and two weights W each is hung at the two ends fig. (b), what will be the total elongation in the wire?

(a) Let Y = Young’s modulus of the material of the wire. If ‘a’ be its area of cross-section, then

(b) When the wire is passed over the pulley, let l’ be the increase in the length of each segment. Since $$\frac{L}{2}$$ = length of each segment.

∴ Total increase in the length of the wire is given by
= l’ + l’ = 2l’ = 2 × $$\frac{l}{2}$$ = l.

Question 8.
A uniform cylindrical wire is subjected to longitudinal tensile stress of 5 × 107 Nm-2. The Young’s Modulus of the material of the wire is 2 × 1011 Nm-2. The volume change in the wire is 0.02%. Calculate the fractional change in the radius of the wire.

Question 9.
A wire loaded by the weight of density 7.6 g cm-3 is found to measure 90 cm. On immersing the weight in water, the length decreases by 0.18 cm. Find the original length of the wire.
Let L = original length of the wire =?
A = be its area of cross-section.
W = load attached to the wire.

Then Young’s Modulus of the wire is given by

Here, ΔL = 90 – L = Change in the length of wire.

Volume of weight attached,

∴ Mass of water displaced = V × density of water
= $$\frac{W}{7.6}$$ × l = $$\frac{W}{7.6}$$

∴ Net weight after immersing in water ¡s
W’ = W – $$\frac{W}{7.6}$$ = $$\frac{6.6}{7.6}$$ W

Length of wire after immersing ¡n water
= 90 – 0.18 = 89.82 cm.

∴ Increase in length on immersing in water,

Question 10.
Two exactly similar wires of steel and copper are stretched by equal forces. If the total elongation is 1 cm, find by how much each wire is elongated. Given Y for steel = 20 × 1011 dyne cm-2, Y for copper = 12 × 1011 dyne cm-2.
Let Δls and Δlc be the elongation produced in steel and copper wires respectively.
Ls, Lc be their respective lengths,
Ls = Lc (∵ wires are similar)
Ys = 20 × 1011 dyne cm-2
Yc = 12 × 1011 dyne cm-2
Δls + Δlc = 1 cm
A = area of cross-section of each wire.
F = equal force applied.

∴ Using the relation,

Dividing (ii) by (i), we get

Question 11.
A rubber rope of length 8 m is hung from the ceiling of a room. What is the increase in the length of the rope due to its own weight? Given Young’s modulus of elasticity of rubber is 5 × 106 Nm-2 and density of rubber = 1.5 × 103 kg m-3. Take g = 10 ms-2.
Here, ρ = 1.5 × 103 kg m-3
L = 8m
Y = 5 × 106 Nm-2

Let Mg and A be the weight and area of the cross-section of the rubber rope.
As weight acts at C. G. so effective length = $$\frac{L}{2}$$
∴ Mg = A L ρ g (∵ M is distributed over L)

Let ΔL = increase in the length of the rope =?

Question 12.
A sample of bone in the form of a cylinder of cross-sectional area 1.5 cm2 is loaded on its upper end by a mass of 10 kg. By careful measurements with a traveling microscope, the length of the cylindrical sample is observed to decrease by 0.0065%. Calculate the value of Young’s Modulus for the specimen. Given g = 9.8 ms-2.

Question 13.
Estimate the maximum height of a mountain.
The elastic behavior of the earth helps us to calculate the maximum height of mountains on earth.

Let h be the height of a mountain.
ρ = density of rocks of the mountain

∴ The pressure at the base of the mountain h ρ g = Stress
The elastic limit of a typical rock 3 × 108 Nm-2

The stress must be less than the elastic limits, otherwise, the rock-begins to sink.

It may be noted that the height of Mount Everest is nearly 9 km.

Question 14.
A uniform pressure P is exerted on all sides of a solid cube. It is heated through t°C in order to bring its volume back to the value it had before the application of pressure. Find the value of t.
Let γ = coefficient of cubical expansion of the cube.
Let K be the bulk modulus of elasticity of its material.
V = initial volume.
P = pressure applied.
ΔV = Decrease in its volume.

where t = rise in its temperature so as to increase the volume by ΔV s.t. it is brought back to its initial volume.
∴ From (i) and (ii), we get

Question 15.
If K be the bulk modulus of a metal and a pressure P is applied uniformly on all its sides. If p be the density of metal, then find the fractional increase in its density.
Let M = mass of metal
V = mass of volume
∴ ρ = $$\frac{M}{V}$$ ….(i)

Let ΔV = Decrease in its volume when a pressure P is applied.
If ρ’ be its new density, then

Question 16.
The rubber cord catapult has a cross-sectional area of 1 mm2 and a total unstretched length of 10 cm. It is stretched to 12 cm and then released through a projectile of the mass of 2 gm. Taking Y =5 × 108 Nm-2, find tension ¡n the cord and velocity of projection.
Here, Y = 5 × 108 Nm-2
A = 1mm2 = 10-6 m2
L = 10 cm = 0.1 m
Δl = 12 – 10 = 2cm = 0.02m
m = 2 × 10-3 kg
T =?
V =?

This is converted into K.E. of the projectile.

Question 17.
What should be the greatest length of a steel wire which when fixed at one end can hang freely without breaking? Density (ρ) of steel = 7.8 g cm3, breaking stress for steel = 7.8 × 109 dynes/cm2.
Here, breaking stress = 7.8 × 109 dyne cm2
Le.t A = area of cross-section of the wire
ρ = 7.8 g cm-3
g = 980 cms-2

Let l = length required = ?
∴ V = volume of wire = lA

∴ Stretching force = weight of the wire = ρ V g
= ρlAg

∴Maximum stress = Breaking stress

Question 18.
A copper wire of negligible mass, area of cross-section 10-6 m2, and length 1 m are kept on a smooth horizontal table with one end fixed. A ball of mass 1 kg is attached to the other end. The wire and ball are rotating with an angular velocity of 20 rad s-1. If the elongation in the wire is 10-3 m, then find Young’s modulus.
Here, L = original length of the wire = 1 m
l = elongation length of the wire = 10-3 m
A = area of cross-section of wire = 10-6 m2
M = mass of ball attached to the wire = 1 kg
ω = angular velocity of rotation = 20 rad s-1
Y = ?
Force on wire = centripetal force

Question 19.
If the angular velocity of the wire and the ball in the above question is increased to Loo rad s, then the wire breaks down. Calculate the breaking stress.
As per above question M = 1 kg
L = 1m .
A = 10-6 m2
Here, ω = angular velocity = 100 rad s-1
Breaking stress = ?

Using the formula,

Question 20.
A wire of radius r stretched without tension along a straight line is tightly fixed at points A and B as shown in the figure. What is the tension in the wire when it is pulled into the shape XYZ?

Let T = Tension in the wire = ?
r = radius of the wire
∴ A = area of cross-section of the wire = πr²
θ = ∠XYC = ∠ZYC
l = X C = CZ

If Δl be the extension of the wire, then

Question 21.
A wire of density 9 g cm-3 is stretched between two clamps 100 cm apart, while subjected to an extension of 0.05 cm. What is the lowest frequency of transverse vibrations in the wire, assuming Y = 9 × 1011 dyne/cm2?
Here, ρ = density of wire = 9 g cm3
Y = 9 × 1011 dyne/cm2
L = distance between two clamps = 100 cm
l = extension of wire = 0.05 cm

Let v = lowest frequency of transverse vibrations = ?
If F and A be the stretching force and area of the cross-section of the wire, then

Also, the lowest frequency or fundamental frequency of transverse vibrations is given by

where L’ = 100 + 0.05 = 100.05 cm
m = mass/length = Aρ = area × density

Question 22.
A steel wire with a cross-sectional area of 0.5 mm2 is held between two fixed supports. If the tension in the wire is negligible and it is just taut at a temperature of 20°C. Determine the tension when the temperature falls to 0°C. Y = 21 × 1011 dyne/cm2 and coefficient of linear expansion is 12 × 10-6 per degree centigrade. Assume distance between two supports. remains same.
Here, let L = length of the wire
A = its area of cross-section = 0.5mm2
= 0.005 cm2
Y = 21 × 1011 dyne/cm2

α = coefficient of linear expansion
= 12 × 10-6 °C-1

Let T = tension in the wire = ?
If Δl be extension, then
Δl = α L Δt
where Δt = 20 – 0 = 20°C

This strain is tensile as the wire is kept taut and not allowed to contract.

∴ Tensile stress in the wire = Y × strain
= Y α Δt

∴ T = Tensile stress × A
= Y α Δt A
= 21 × 1011 × 12 × 10-6 × 20 × 0.005
= 25.2 × 105 dyne.

Question 23.
Two rods of different metals having the same area of cross-section A are placed between the two fixed points and they have the parameters as l1, α1, Y1, and l2, α2, Y2 respectively where l stands for their lengths, a for coefficients of linear expansion. The temperature of both is increased by T°C. Find the force with which the rods act on each other.

Due to the rise in temperature by T°C, let Δl1 and Δl2, be the increase in lengths of the two rods.

As the two fixed points don’t allow the length to increase, hence thermal stress $$\frac{F}{A}$$ is developed.

∴ Decrease in length due to thermal stress = increase in length due to thermal linear expansion

Question 24.
Two rods of different materials having coefficients of linear expansion α1 and α2 and Young’s modulus Y1 and Y2 respectively are fixed between two massive points and are heated such that they undergo the same increase in temperature. There is no bending in rods of α1: α2 = 2: 3 and thermal stresses produced in two rods are equal. Calculate the ratio Y1: Y2.
It is a common cause of thermal expansion and elastic compression.

For the same rise in temperature (Δt) and same stress,
Yα = constant
or
Y1α1 = Y2α2

Question 25.
What force is required to stretch a copper wire 1 cm2 in cross¬section to double its length? Y for copper is 1.26 × 1012 dyne cm-2.
Here, A = 1 cm2
Let L = length of the wire.

Also, Let L’ be the new length when a stretching force F is applied.
L’ = 2L
Then l = extension = L’ – L = 2L – L = L
F = ?
Y = 1.26 × 1012 dyne cm-2

Value-Based Type:

Question 1.
Raman and his friend Sbhan while going to the school on a motorcycle noticed that a bridge had collapsed. Immediately they went to their physics teacher and enquired about the reasons for falling off the bridge. After knowing the reasons that very interesting; they decided to pursue their career as civil engineers and vowed to construct 100% quality dams and bridges.
(a)Comment upon the values possessed by them.
Sympathy, determination, and concern for society, honesty, and integrity.

(b) Name the property that helps in constructing bridges. Also, define the property.
Elasticity: The property of the body to regain its original configuration (i .e length, volume, or shape) when the deforming force is removed, is called elasticity.

Question 2.
Ram and his friend Ramesh are the students of class XI. They are discussing “elastic fatigue”. Ram told that just as a tired person elastic body also relieved of the fatigue or regains its original degree of elasticity when allowed to rest for some time. Raman did not convince Ram. They went to solve this problem before their physics teacher.
(i) What values are displayed by Ram and Ramesh?
The values displayed by them are:
(a) Curiosity
(b) Awareness
(c) Attitude to find the solution by discussion
(d) Interested in learning.

(ii) Whether Ram’s ideas and reasons were correct or not?
Yes, Ram’s ideas and reasons were correct and were recommended by their teachers.

(iii) Why do spring balance shows wrong readings after they have been used for a long time?
This is due to elastic fatigue.

Question 3.
Kamal read a topic in a newspaper that a bridge declared unsafe by the Government. The bridge was 2 km away from his house and he used to cross it almost daily. The condition of the bridge does not look so bad. However, it was built 60 years ago. The next day Kamal asked the reason for this question to his friends.
(i) What value is displayed by Kanal here?
Values displayed are:
Curiosity, awareness, group discussion, and keenness to know the scientific reasons.

(ii) Why the bridge declared unsafe after long use?
Due to the repeated stress and strain, the material used in the bridges loses elastic strength and ultimately may collapse. Hence, bridges are declared unsafe after long use.

Question 4.
Two friends Mohan and Dinesh arc discussing elasticity. Mohan told that steel is more elastic than rubber. Dinesh was surprised and asked his teacher the-Fcascm-Jachind it.
(i) What value is displayed by Dinesh?
The values displayed by Dinesh are:
Curiosity, Interested in learning.

(ii) How his teacher explained it?
Let the two pieces of wire, one of steal and the other of rubber having an equal length (L) and equal area of cross-section (A). Let each be stretched by equal force (F).

Then, Young’s modulus of steel and rubber are:

Hence, Steal is more elastic than rubber.

## System of Particles and Rotational Motion Class 11 Important Extra Questions Physics Chapter 7

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 7 System of Particles and Rotational Motion. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

## Class 11 Physics Chapter 7 Important Extra Questions System of Particles and Rotational Motion

### System of Particles and Rotational Motion Important Extra Questions Very Short Answer Type

Question 1.
Can the geometrical centre and C.M. of a body coincide? Give examples.
Yes, C.M. and geometrical centre may coincide when the body has a uniform mass density, e.g. C.M. and geometrical centre are the same in case of a sphere, cube and cylinder etc.

Question 2.
How does the M.I. change with the speed of rotation?
M.I. is not affected by the speed of rotation of the body.

Question 3.
Under what conditions, the torque due to an applied force is zero?
We know that τ = rF sin θ. If θ = 0 or 180,
or
r = 0, then τ = 0, r = 0 means the applied force passes through the axis of rotation.

Question 4.
Is it correct to say that the C.M. of a system of n-particles is always given by average position vectors of the constituent particles? If not, when the statement is true?
No, this statement is true when all the particles of the system are of the same mass.

Question 5.
A cat is able to land on her feet after a fall. Which principle of Physics is being used by her?
Principle of conservation of angular momentum.

Question 6.
What is conserved when a planet revolves around a star?
Angular momentum.

Question 7.
If no external torque acts on a body, will its angular velocity remain conserved?
No, it is the angular momentum that will be conserved.

Question 8.
A body is rotating at a steady rate. Is a torque acting on the body?
No, torque is required only for producing angular acceleration.

Question 9.
What is the other name for angular momentum?
Moment of momentum.

Question 10.
Out of two spheres of equal masses, one rolls down a smooth inclined plane of height h and the other is falling freely through height h. In which case, the work done is more?
The same work is done in both cases.

Question 11.
Many great rivers flow towards the equator. What effect does the sediment they carry to the sea have on the rotation of the Earth?
This will increase the M.I. of the Earth about its own axis and hence the time of rotation of Earth will increase.

Question 12.
Can the mass of a body be considered concentrated at its centre of mass for purposes of computing its rotational inertia?
No, the distribution of mass is extremely important in the calculation of M.I.

Question 13.
On what factors does the M.I. of a body depend?
It depends on:

1. Position of the axis of rotation.
2. Way of distribution of mass about the axis of rotation.
3. Mass of the body.

Question 14.
Should there exist mass at the location of C.M. of a system? Give an example.
No, e.g. the C.M. of a ring is at the centre where there is no mass.

Question 15.
Should the C.M. of a body necessarily, lie inside the body? Explain.
No, it may lie outside the body. In the case of the semicircular ring, it is at the centre which is outside the ring.

Question 16.
What is the position vector of C.M. of two particles of equal masses?
It is the average of the position vectors of the two particles.

Question 17.
If one of the particles is heavier than the other, to which side will their C.M. shift?
It will shift closer to the heavier particle.

Question 18.
What is an isolated system?
An isolated system is that on which no external force is acting.

Question 19.
Why do we prefer to use a wrench with a long arm?
The turning effect of a force is τ = r × F. When the arm of the wrench is long, r is larger. So smaller force is required to produce the same turning effect.

Question 20.
What is the rotational analogue of mass and force?
M.I. and torque are the respective rotational analogue.

Question 21.
Is the radius of the gyration of a body a constant quantity? Why?
No. It depends upon the position of the axis of rotation and also on the distribution of mass of the body about this axis.

Question 22.
Two solid spheres of the same mass are made of metals of different densities. Which of them has a larger M.I. about a diameter?
The sphere of metal with smaller density shall be bigger in size and hence it will have a larger M.I.

Question 23.
A solid cylinder is rolling without slipping on an inclined plane. What is the torque provided by the weight of the cylinder, about the C.M. of the cylinder?
Zero, This is because the line of action of the weight passes through the C.M.

Question 24.
A solid cylinder is rolling without slipping on an inclined plane. What is the torque provided by the weight of the cylinder, about the C.M. of the cylinder?
(a) What is the torque provided by the normal reaction? Why?
Zero as the line of action of the force passes through the C.M.

(b) Which force causes the cylinder to rotate about its C.M.?
Force of friction

(c) Through which point the axis of rotation passes?
Point of contact.

(d) What is the instantaneous velocity of the point of contact?
Zero.

(e) What is the acceleration of the cylinder rolling down the inclined planed.
a = $$\frac{2}{3}$$ g sin θ

(f) What is the condition for rolling without slipping?
$$\frac{1}{3}$$tan θ < μs.

Question 25.
Give an example of rigid body motion in which the centre of mass of the body is in motion.
The motion of a cylinder rolling down an inclined plane is-the required example.

Question 26.
Give an example of rigid body motion in which the centre of mass of the body remains at rest.
The motion of a point mass tied to a string and wound on a cylinder capable of rotating without friction about its own axis is the required example.

Question 27.
The angular speed of a body changes from to, of oo2 without applying a torque but due to changes in M.I. Find the ratio of the radii of gyration in the two cases.

Question 28.
A disc is recast into a hollow and thin cylinder of the same radius. Which will have larger M.I.? Explain.
M.I. of the cylinder will be larger because most of its mass is located away from the axis of rotation as compared to that of the disc.

Question 29.
Angular momentum of a system is conserved if its M.I. is changed. Is its angular K.E. also conserved?
No, as I change, K.E. also changes and hence it does not remain conserved.

Question 30.
If a body is rotating, is it necessarily being acted upon by an external torque.
No. Torque is required only for producing angular acceleration.

Question 31.
Can a torque be balanced by a single force?
No.

Question 32.
To cut a tree, why we often make a cut on the side facing the direction in which we wish the tree to fall?
It is done so that the weight of the tree applies torque about the cut and thus the tree falls in that direction.

Question 33.
About which axis the moment of inertia of a body is minimum?
The moment of inertia of a body is minimum about the axis passing through the centre of mass of the body.

Question 34.
Is the moment of inertia a vector quantity?
No, it is a scalar quantity.

Question 35.
What are the dimensions of a moment of inertia?
The dimensions of the moment of inertia are one in mass, two in length and zero in time as its dimensional formula is [M L2 To].

Question 36.
Can all magnitudes of angular displacement be treated as vectors?
No, only infinitesimally small magnitudes of the angular displacement can be treated as vectors.

Question 37.
Can all magnitudes of angular velocity be treated as vectors?
Yes, all magnitudes of instantaneous angular velocity can be treated as vectors, but it may not be possible to treat all magnitudes of average angular velocity as vectors.

Question 38.
Does the moment of inertia of a rigid body change with the speed of rotation?
No, as it is independent of the speed of rotation of the body.

Question 39.
Can θ, ω and α be expressed in degrees instead of radians in the rotational kinematic equations?
Yes, as it makes no difference whether we express them in radians or in degrees.

Question 40.
Does the radius of gyration depend on the angular speed of rotation of the body?
No, it is independent of the angular speed of rotation of the body.

Question 41.
How does the ice skater or an acrobat takes the advantage of the law of conservation of the angular momentum?
They do so by folding or extending their arms and legs and thus change their moment of inertia which causes a decrease or an increase in their speed of .pinning.

Question 42.
Why there are two propellers in the helicopter?
The two propellors help in making the helicopter fly at the same level. If there is only one propeller in it, then it may begin to rotate in a direction opposite to that of the propeller so as to conserve the angular momentum.

Question 43.
A homogenous disc is revolving about its axis which is stationary? Is the disc possessing linear momentum? Why?
No, it is zero as the momentum of one half of the disc is equal and opposite to that of the other half.

Question 44.
Does the disc in A homogenous disc is revolving about its axis which is stationary? Is the disc possessing linear momentum? Why possesses angular momentum?
Yes, it possesses angular momentum.

Question 45.
No torque acts on a body. Should its angular velocity be conserved? Why?
No, in absence of torque, angular momentum remains conserved and not the angular velocity.

Question 46.
A particle revolves uniformly along a circle (on a smooth horizontal table) by means of a string connected to it. Does its angular momentum changes from its initial value if the string is cut suddenly?
No, the particle flies away tangentially, keeping angular momentum conserved as no external torque acts on the particle.

Question 47.
Is the angular momentum of a comet revolving around a massive star in a highly elliptical orbit constant over the entire orbit?
Yes, the force acting on the comet is radial and hence there is no change in its angular momentum.

Question 48.
What is the significance of the C.M. of the system of particles?
It helps us in describing the behaviour of a macroscopic body in terms of the laws developed for microscopic bodies.

Question 49.
Does the acceleration of a cylinder rolling down an inclined plane depend upon the mass of the cylinder?
No.

Question 50.
When there is no external torque acting on a rotating body which of the following quantities can change?
(i) angular acceleration
(ii) angular momentum
(iii) angular speed.
Both angular acceleration and angular momentum do not change, but the angular speed may change if the M.I. is allowed to change.

### System of Particles and Rotational Motion Important Extra Questions Short Answer Type

Question 1.
What is the difference between the centre of gravity and C.M.?
C.G.: It is the point where the whole of the weight of the body is supposed to be concentrated i.e. on this point, the resultant of the gravitational force on all the particles of the body acts.
C.M.: It is the point where the whole of the mass of the body may be supposed to be concentrated to describe its motion as a particle.

Question 2.
There are two spheres of the same mass and radius, one is solid and the other is hollow. Which of them has a larger moment of inertia about its diameter?
The hollow sphere shall have greater M.I., as its entire mass is concentrated at the boundary of the sphere which is at maximum distance from the axis.

Question 3.
What shall be the effect on the length of the day if the polar ice caps of Earth melt?
Melting of polar ice caps will produce water spread around the Earth going farther away from the axis of rotation that will increase the radius of gyration and hence M.I. In order to conserve angular momentum, the angular velocity ω shall decrease. So the length of the day (T = $$\frac{2 \pi}{\omega}$$) shall increase.

Question 4.
If only an external force can change the state of motion of the C.M. of a body, how does it happen that the internal force of brakes can bring a vehicle to rest?
The internal force of brakes converts the rolling friction into sliding friction. When brakes are applied, wheels stop rotating. When they slide, the force of friction comes into play and stops the vehicle. It is an external force.

Question 5.
What do you understand by a rigid body?
A rigid body is that in which the distance between all the constituting particles remains fixed under the influence of external force. A rigid body thus conserves its shape during its motion.

Question 6.
Distinguish between internal and external forces.

1. The mutual forces between the particles of a system are called internal forces.
2. The forces exerted by some external source on the particles of the system are called external forces.

Question 7.
Two equal and opposite forces act on a rigid body. Under what conditions will the body (a) rotate, (Z>) not rotate?
Two equal and opposite forces acting on a rigid body such that their lines of action don’t coincide constitute a couple. This couple produces a turning effect on the body. Hence the rigid body will rotate. If the two equal and opposite forces act in such a way that their lines of action coincide, then the body will not rotate.

Question 8.
(a) Why is it easier to balance a bicycle in motion?
The rotating wheels of a bicycle possess angular momentum. In the absence of an external torque, neither the magnitude nor the direction of angular momentum can change. The direction of angular momentum is along the axis of the wheel. So the bicycle does not get tilted.

(b) Why spokes are fitted in the cycle wheel?
The cycle wheel ¡s constructed in such a way so as to increase the M.I. of the wheel with minimum possible mass, which can be achieved by using spokes and the M.I. is increased to ensure the uniform speed.

Question 9.
A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The C.M. of the rod is at a distance x from A. Find the normal reactions at knife edges A and B.
Let RA and RB be the normal reactions at points A and B respectively.

As the rod in is an equilibrium position, so the sum of moments of the forces about the knife edges must be equal to zero.

∴ Taking moments of forces about point A, we have

Question 10.
What are the essential features of angular momentum?
Following are the essential features of angular momentum:

1. The angular momentum of a particle with respect to a point gives an idea of the strength of its rotational tendency about that point.
2. The magnitude of the angular momentum is defined in terms of mass and velocity of the particle and its distance from the reference point i. e. L = m v r.
3. The vector concept of the angular momentum is useful. Its direction is the axial direction given by the right-hand rule. The direction of L is ⊥ to the plane containing r and v.

Question 11.
(a) What is the physical significance of M.I.?
The inability of a body to change its state of uniform rotation about an axis is called rotational inertia or M.I. of the body.
It plays the same roll in rotatory motion as is being played by the mass in translational motion i.e. it is a rotational analogue of mass.

(b) Define M.I. of a body rotating about an axis.
M.I. of a body about a fixed axis of rotation is defined as the sum of the products of the masses and square of the perpendicular distances of various constituent particles from the axis of rotation.

Question 12.
Define the radius of gyration of a body rotating about an axis.
It is defined as the ⊥ar distance of the point from the axis of rotation at which if the whole mass of the body is supposed to be concentrated then its M.I. about the given axis remains the same as with the actual distribution of mass. It is denoted by K.

Question 13.
Derive the expression for the radius of gyration.

Let M = total mass of a rigid body rotating about ZZ’ axis.
Let it is made up of n particles each of mass m and situated at Tar
distances r1, r2, …. and rn from the axis of rotation.
∴ M = mn …. (i)

If I am its M.I. about ZZ’ axis,
Then by def. of M.I.,

Also by definition of K,
I = MK2 ….(iii)
∴ from (ii) and (iii), we get

Question 14.
Show that in the absence of an external force the velocity of the C.M. of a system remains constant.
Let us consider a body of mass m whose C.M. is moving with a velocity v and external force is zero. The equation of motion of the C.M. can be written as

i.e. in the absence of external force, the C.M. moves with a constant speed along a straight line. If R0 be the position vector of the C.M. at time t = 0, then the position vector of the C.M. after the time t is given by
R(t) = R0 + vt.

Question 15.
Define Torque. What is its physical significance?
It is defined as the turning effect of a force about the axis of rotation or it is the moment of a force about the axis of rotation.
i.e τ = Fd
where F = force applied on a body.
d is the ⊥ar distance of the line of action of the force from the axis of rotation.

Mathematically in vector form, τ may be expressed as
τ = r × F

i.e. it is the cross product of the position vector r and Force F

Note: d is called lever arm of the force
i.e. Torque = Force × lever arm

Question 16.
There is a stick half of which is wooden and half is of steel. It is pivoted at the wooden end and force is applied at the steel end at right angles to its length. Next, it is pivoted at the steel end and the same force is applied at the wooden end. In which case is the angular speed greater and why?
We know that the greater the angular acceleration greater is the angular speed. Also α = $$\frac{τ}{I}$$
In the former case, M.I. is greater than in the latter case (steel being heavier than wood).

Thus angular acceleration in the 2nd case is greater than in the 1st case and hence the angular speed.

Question 17.
How would you distinguish between a hard-boiled egg and a raw egg each spinning on a tabletop?
The raw egg will have greater M.I. as compared to the boiled egg because the mass of the raw egg tries to move away from the axis of rotation.

We know that α = $$\frac{τ}{I}$$, so the angular acceleration is inversely proportional to the M.I. as the boiled egg will spin with greater speed than the raw egg.

Question 18.
Using the expression for power and K.E. of rotational motion, derive the relation τ = Iα.
We know that power is given by
P = τ ω
Also, we know that,

Question 19.
Equal torques are applied on a hollow cylinder and hollow sphere, both having the same mass and radius. The cylinder rotates about its axis and the sphere rotates about its diameter. Which one will acquire greater speed and why?
The speed of rotation depends upon the angular acceleration produced.

where α1 and α2 are angular accelerations of hollow-cylinder and hollow sphere respectively.

The sphere acquires greater speed than the cylinder as α is more for it and ω is directly proportional to α.

Question 20.
A thin wheel can stay upright on its rim for a considerable time when rolled with a considerable velocity, while it falls from its upright position with the slightest disturbance, when stationary. Explain.
When the wheel is still and standing upright on the rim, the equilibrium is unstable. But when the wheel is rolling it possesses an angular momentum in the horizontal direction which is a conserved quantity of the motion. The angular momentum does not change unless the force of friction slows down the angular velocity.

Greater is the velocity, greater is the angular momentum and hence greater internal torque is needed to change the angular momentum. When the angular velocity tends to zero, the angular momentum also tends to zero, so even with little disturbance, the angular momentum can be changed.

Question 21.
Three homogeneous rigid bodies: a solid sphere, a solid cylinder and a hollow cylinder are placed at the top of an inclined plane. If they all are released from rest at the same elevation and roll without slipping, which one reaches the bottom first and which one reaches last?
Let a1, a2, a3 be their respective linear accelerations respectively.

∴ solid sphere will reach first and hollow cylinder reaches last.

Question 22.
Three uniform and thin rods each of mass m and length l are placed along the three co-ordinate axes with one end at the origin. Find the M.I. of the system about the x-axis.
We know that the M.I. of’a rod about its end is given by
= $$\frac{\mathrm{m} l^{2}}{12}+\frac{\mathrm{m} l^{2}}{4}=\frac{\mathrm{m} l^{2}}{3}$$

The M.I. of two rods lying along y and z axes about x-axis will be
Ix = $$\frac{\mathrm{m} l^{2}}{3}+\frac{\mathrm{m} l^{2}}{3}=\frac{2 \mathrm{~m} l^{2}}{3}$$

Question 23.
A mass m is supported by a massless string wound round a uniform cylinder of mass m and radius r. Find with what acceleration will it fall when released from rest?
Let it falls through a height h on releasing.

According to the law of conservation of energy,

Question 24.
The speed of the inner layers of the whirlwind in a tornado is alarmingly high. Explain why?
The inner layers of the whirlwind in a tornado are close to the axis of rotation. It means the M.I. of the air molecules of inner layers is small. Hence according to the law of conservation of angular momentum, co of the inner layers of the whirlwind in a tornado is very high.

Question 25.
State the two theorems of the moment of inertia. Give an example of the application of each.
There are two theorems of the moment of inertia of a body namely,
(a) the theorem of perpendicular axes and
(b) the theorem of parallel axes.

(a) Theorem of perpendicular axes: It states that the moment of inertia of a body about an axis passing through a point in the body is equal to the sum of the moment of inertia about two mutually perpendicular axes lying in a plane perpendicular to the given axis and passing through the chosen point.
i.e. Iz = Ix + Iy.

(b) Theorem of parallel axes: It states that the moment of inertia of a body about an axis is equal to the sum of the moment of inertia of the body about an axis parallel to the given axis and passing through its centre of mass plus the product of mass and the square of the distance between the two axes
i.e. I = Ic + mh2.

Question 26.
Why cannot a single force balance the torque?
The effect of torque is to produce the angular acceleration and its effect is entirely different than that of the force which causes linear acceleration. Thus a single force cannot balance the torque.

Question 27.
Equal torques are applied on a cylinder and a sphere, both having the same mass and radius. The cylinder rotates about its axis and the sphere rotates about one of its diameters. Which one will acquire greater speed and why?
The speed of rotation acquired will depend upon the angular acceleration produced and is given by
α = $$\frac{τ}{I}$$

where τ = is the torque, I am the moment of inertia,
As τ is the same in both the cases, so

Thus, the sphere acquires greater speed than the cylinder as αs > αc.

Question 28.
Explain why small angular displacements are treated as vectors?
Let a particle undergoing pure rotational motion describes a circular path as shown here.

Let r = radius of its path in XY plane (say).
Let its position changes from P to Q in time At so that radial line OP describes an ∠POQ = Δθ.
Δθ is called angular displacement of the particle.

We know that Δθ = $$\frac{\text { Length of arc } \mathrm{PQ}}{\text { Radius of the circle }(=\mathrm{r})}$$
or
arc PQ = (Δθ)r …. (1)

If Δθ is very small, then Δθ → 0, so arc PQ ≅ chord PQ i.e. the particle may be assumed to be displaced along the chord PQ. In that case, the displacement vector of the particle P is given by
PQ = Δr(say)
i.e. arc PQ = |Δr| ……(2)

The position vector of the particle at P may be written as:
OP = r
Also OQ = r
∴ |OP| = |OQ| = r

∴ from (1) and (2), we get
|Δr| = (Δθ) | r | ….(3)

If we represent Δθ as a vector along the axis of rotation, then from (2), we get
Δr = Δθ × r.

Question 29.
The same solid sphere is made to roll down from the same height on two inclined planes having different angles of inclination. In which case will it take less time to reach the bottom?
Let L be the length of the inclined planes having angels of inclination θ1, and θ2 respectively from which the ball begins to slide down

Let r = radius of the sphere.
m = mass of the sphere.
v = speed with which the sphere will reach the bottom,

let ω = $$\frac{v}{r}$$ be its angular speed.
If I be a M.I. of the sphere, then

this show that in both cases, the sphere will reach the bottom with the same velocity. Let t be the time taken to reach the ground
∴ using v = u + at, we get
v = a . t
or
t = $$\frac{v}{a}$$ …. (3)
Also for an inclined plane,
a = g sin θ
or
a ∝ sin θ …..(4)

From (3) and (4), we get
t ∝ $$\frac{v}{sin θ}$$
or
t ∝ $$\frac{1}{sin θ}$$ as v is same in both the cases. sm0
i.e. t will be less for the inclined plane with a larger angle of inclination.

Question 30.
The angular momentum of a particle relative to certain I points Ovaries with time as L = a + bt2 where a and b are constant vectors with a ⊥ b. Find the force moment $$\vec{τ}$$ relative to the point O acting on the particle when the angle between the vectors $$\vec{τ}$$ and L equal 45°.
Here, let a and b act along X and Y axis respectively. L = a + bt2 = aî + bĵt2

i.e. τ acts along Y-axis.
when L is at 45° with τ i.e. with Y-axis,

Question 31.
What are the important characteristics of C.M.?

1. Its position remains unchanged in rotational motion about itself but changes in linear motion.
2. Its position is independent of the coordinate system chosen.
3. The sum of the moments of the masses of the system about the C.M. is zero.
4. Its position depends upon the shape of the body.
5. Its position depends upon the mass distribution of the body.
6. C.M. coincides with C.G. for a symmetrical body.
7. The C.M. of a circular ring lies outside the material body.
8. The C.M. of a circular disc lies within the material of the body.

### System of Particles and Rotational Motion Important Extra Questions Long Answer Type

Question 1.
Discuss the rolling of a cylinder (without slipping) down a rough inclined plane and obtain an expression for the necessary coefficient of friction between the cylinder and the surface.
Consider a solid cylinder of mass m, radius R and MJ. I rolling down an inclined plane without slipping as shown in the figure. The condition of rolling down without slipping means that at each instant of time, the point of contact P of the cylinder with the inclined plane is momentarily at rest and the cylinder is rotating about that as the axis.

Let θ = angle of inclination of the plane. The forces acting on the cylinder are:

1. The weight mg of the cylinder acting vertically downward.
2. The force of friction F between the cylinder and the surface of the inclined plane and acts opposite to the direction of motion.
3. The normal reaction N due to the inclined plane acting normally to the plane at the point of contact. The weight W of the cylinder can be resolved into two rectangular components:
(a) mg cos θ along ⊥ to the inclined plane.
(b) mg sin θ along the inclined plane and in the downward direction. It makes the body move downward.

Let a = linear acceleration produced in the cylinder,
Then according to Newton’s 2nd law of motion,
ma = mg sin θ – F …. (1)
and N = mg cos θ …. (2)

If α = angular acceleration of the cylinder about the axis of rotation, then
τ = I α …. (3)

Here, τ is provided by F i.e.
τ = F.R …. (4)

∴ from (3) and (4), we get

If μs be the coefficient of static friction between the cylinder and the surface,
Then

For rolling without slipping

equation (9) is the required condition for rolling without slipping i.e. $$\frac{1}{3}$$ tan θ should be less than equal to μs i.e. the maximum allowed inclination of the plane with the horizontal is given by
θmax = tan-1 (3 μs)

Question 2.
Prove that
(a) Δω = τ Δθ
Let F = force applied on a body moving in XY plane.
Δr = linear displacement produced in the body by the force F in moving from P to Q.
If Δω is the small work done by the force, then by definition of work.

ΔW = F . Δr ….(1)

In component form,

Let PN ⊥ on X-axis & PON = θ
∴ in rt ∠d ΔPNO,

As Δθ is very small, i.e. Δθ → 0, cos Δθ → 1 and sin Δθ → Δθ

(b) P = τ ω.

Numerical Problems:

Question 1.
Three particles of masses 1g, 2g and 3g have their centre of mass at the point (2,2,2). What must be the position of the fourth particle of mass 4g so that the combined centre of mass may be at the point (O, O, O).
Let r1, r2, r3 are the position vectors of the three mass particles w.r.t. origin. Then their centre of mass is given by

If r’ be the position vector of the new C.M. then
r’ =0
and r4 be the position vector of the fourth particle from the origin, then

∴ the new particle should be at (- 3, – 3, – 3).

Question 2.
Two balls of masses 3m and m are separated by a distance of l.
(a) Find the position of the C.M.
(b) If these two balls are attached to a vertical axis by means of two threads whose combined length is l and rotated in a horizontal plane with uniform angular velocity ω about an axis of rotation passing through C.M. Prove that the tensions in the two strings will become equal.
Here, m1 = 3m, m2 = m; AB = l
(a) Let x = distance of the C.M. from the ball A of mass 3m
∴ l – x = distance of the C.M. from the ball B.

∴ From the definition of C.M.
3m . x = m (l – x)
or
x = $$\frac{l}{4}$$

(b) Let T1 and T2 be the tensions in the strings connecting 3m and m to the axis through C.M. respectively and rotating in a horizontal plane with angular velocity coω.
∴ radii of their circular paths = $$\frac{l}{4}$$ and $$\frac{3l}{4}$$ respectively.

Hence proved.

Question 3.
Calculate the M.I. about a transverse axis through the centre of a disc whose radius is 20 cm. Its density is 9g cm-3 and its thickness is 7 cm.
Here, R = 20 cm, t = thickness = 7 cm density of material of disc,

Question 4.
A cylinder of mass 5 kg and radius 30 cm and free ω rotate about its axis, receives an angular impulse of 3 kg mV initially followed by a similar impulse after every 4s. What is the angular speed of the cylinder 30s after the initial impulse? The cylinder is at rest initially.
Here, M = 5 kg
R = 30 cm = 0.3 m
ω = ? after 30s
I = M.I. of the cylinder about its axis
= $$\frac{1}{2}$$mR2 = $$\frac{1}{2}$$ × 5 × (0.3)2
= 0.225 kg m2

∴ The initial angular speed of the-cylinder, ωo = 0
Initial angular momentum of the cylinder, Lo = Iωo = 0
Initial impulse received by the cylinder, L’o = 3 kg m2 s-1

The angular impulse received by the cylinder will produce a change in its angular momentum. If L is the angular momentum of the cylinder after the initial impulse, then
L’o = L – Lo
or
L = L’o + Lo = 3 + 0 = 3 kg m2 s-1

As the cylinder receives angular impulse after every 4s.
∴ Angular momentum of the cylinder after 4s = 3 + 3 = 6kg m2 s-1
Angular momentum of the cylinder after 8s = 3 × 2 + 3 = 9kg m2 s-1
Angular momentum of the cylinder after 12s = 3 × 3 + 3 = 12kg m2 s-1
Angular momentum of the cylinder after 28s = 3 × 7 + 3 = 24kg m2 s-1

Next angular impulse is to be received by the cylinder at t = 32s
∴ between 28s and 30s no impulse is received
∴ angular momentum of the cylinder after 30s = 24kg m2 s-1

If ω = angular speed of the cylinder after 30s, then

Question 5.
A cord is wound around the circumference of a wheel of diameter 0.3m. The axis of the wheel is horizontal. A mass of 0.5kg is attached at the end of the cord and it is allowed to fall from rest. If the weight falls 1.5m in 4s, what is the angular acceleration of the wheel? Also, find out the M.I. of the wheel.
Here, mass attached, m = 0.5kg
radius of wheel, R = $$\frac{0.3}{2}$$ = 0.15m
initial linear velocity of the mass attached, u = 0
Distance covered, s = 1.5m, t = 4s

let a = linear acce. of the mass after 4s

If τ = torque applied by the mass attached, then

Question 6.
One end of a uniform rod of mass m and length L is supported by a frictionless hinge which can withstand a tension of 1.75mg. The rod is free to rotate in a vertical plane. To what maximum angle should the rod be rotated from the vertical position so that when left, the hinge does not break?
Let θ = maximum angle = ?
T = 1.75 mg
When the rod is in a vertical position, then the net force acting on the rod
F= 1.75mg – mg = 0.75mg
when the rod rotates, centripetal force,

When the rod is in the displaced position A, then K.E. of rotation is converted in RE.
∴ P.E. of the rod at displaced position = mg h

For rod I = $$\frac{1}{3}$$ mL2

∴ According to the law of conservation of energy,

Question 7.
A uniform cylinder of radius r is rotating about its axis at the angular velocity ωo. It is now placed into a corner as shown in the figure. The coefficient of friction between the wall and the cylinder as well as the ground and cylinder is μ. Then And the number of turns n, the cylinder completes before it stops?
The forces acting on the system are shown in the figure.

Let m = mass of the cylinder
The initial K.E. = $$\frac{1}{2}$$ Iωo2

where I = M.I. of the cylinder about its axis of rotation.

work done by the frictional force before the cylinder comes to rest is

Question 8.
Two masses m1 = 15kg, m2 = 10kg are attached to the ends of a cord which passes over the pulley of an Atwood’s machine. The mass of the pulley is m = 10kg and its radius is R = 0.1m. Calculate the tension in the cord, the acceleration ‘a’ of the system and the number of revolutions made by the pulley at the end of 2s from start.
Here, m1 = 15kg, m2 = 10kg and tensions are shown in the figure.
As the pulley has a finite mass m = 10kg, so the two tensions are not equal.

Let a = linear acceleration of the system,
∴ according to Newton’s law of gravitation,
m1g – T1 = m1a …. (i)
and
T2 – m2g = mo1a ….(ii)

Here tension in the cord exerts torques on the pulley assumed to be a solid disc.
∴ Torque on the pulley, τ = (T1 – T2)R …. (iii)
(in the clockwise direction)

Question 9.
A wheel of radius 6cm is mounted so as to rotate about a horizontal axis through its centre. A string of negligible mass wrapped around its circumference carries a mass of 0.2kg attached to its free end as shown in the figure. When it falls, the mass descends through one metre in 5s. Calculate the angular acceleration of the wheel, its M.I. and tension in the cord.
Let a = acce. of the falling body.
Here, s = 1m
t = 5s
r = 6 cm = 6 × 10-2 m
m = 0.2kg
u = 0

If v = velocity of the wheel after 5 s, then using the relation

P.E. lost by falling mass = mgh
Increase in translational KE. = $$\frac{1}{2}$$mv2
Increase in rotational K.E. = $$\frac{1}{2}$$Iω2
or
$$\frac{1}{2}$$Iω2 + $$\frac{1}{2}$$mv2 = mgh

Question 10.
A rod of length l, whose lower end is resting along the horizontal plane starts to topple from the vertical position. What is the velocity of the upper end when it hits the ground?
Let m = mass of the rod
when the rod is in the vertical position, it possesses P.E. given by
P.E = mg ($$\frac{l}{2}$$)
Here, h = $$\frac{l}{2}$$ because C.G. of the rod is at a height $$\frac{l}{2}$$ above the ground. When the rod topples, it gains K.E. of rotation.

Question 11.
A thin wire of length l and uniform linear mass density ρ is bent into a circular loop with a centre at O as shown in the figure. What is the moment of inertia of the loop about the axis XX’?

Here, ρ = mass per unit length,
m = mass of the loop,
∴ m = ρl
R = radius of the ring or loop.
ID = M.I. of the ring about its diameter.
ID = $$\frac{1}{2}$$mR2 = $$\frac{1}{2}$$ρlR2

Since XX’ is parallel to the diameter of the loop,
∴ According to the theorem of parallel axes, M.I. about XX’ is given by

Question 12.
A long horizontal rod has a bead that can slide along its length and initially placed at a distance l from one end A of the rod. The rod is set in angular motion about A with constant angular acre. α. If the coefficient of friction between the rod and the bead be μ and gravity is neglected, then find the time after which the bead starts slipping?
Bead starts slipping if centripetal force = force of friction

Question 13.
Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length l and negligible mass. The rod is to be set rotating about an axis ⊥ to it. Find the position on this rod through which the axis should pass in order that the work required to set the rod rotating with angular velocity ω0 should be minimum.
Let the axis of rotation be at a distance x from m1.
∴ the distance of the axis from m2 = l – x
When the rod is set into rotation, the increase in the rotational K.E. is given by

According to the work-energy theorem,
work done = increase in rotational K.E.

Question 14.
An isolated particle of mass m is moving in a horizontal plane (x – y) along the X-axis at a certain height above the ground. It suddenly explodes into 2 fragments of masses $$\frac{m}{4}$$ and $$\frac{3}{4}$$ m. An
instant later, the smaller fragment is at y = + 15 cm. What is the position of the larger fragment at this instant?
There is no motion along the y-axis as the isolated particle moves along the x-axis at a certain height. The C.M. will remain stationary along the y-axis even after the collision.

∴ larger fragment will be at – 5cm.

Question 15.
What would be the duration of the day if:
(a) Earth suddenly shrinks to $$\frac{1}{64}$$ th of its original volume, mass remaining uncharged. (I of earth = $$\frac{2}{5}$$ mR2)?
(b) Earth suddenly contracts to half of its present radius (without any central torque acting on it). By how much would the day be decreased?
Let T1 and ω1, be the time period and angular velocity of revolution of Earth before contraction.
T2, ω2 = angular velocity after contraction.
I1, I2 = its M.I. before and after contraction.
R1, R2 radius of Earth before and after contraction.
Since no external torque acts, so

(a) Volume after contraction = $$\frac{1}{64}$$ (volume before contraction)

Question 16.
The energy of 484J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. Find the M.I. of the flywheel.

Let I = M.I. of the flywheel

Question 17.
What constant torque must be applied to a disc of mass 10kg and diameter 50cm so that it acquires an angular velocity of 2π rads-1 in 4s? The disc is initially at rest and rotates about an axis through the centre of the disc and in a plane ⊥ to the disc. Also, find the tangential force to be applied.
Here, m = 10 kg, r = $$\frac{50}{2}$$ = 25 cm = 0.25 m

Question 18.
On the application of constant torque, a wheel is turned from rest through an angle of 200 rads in 8s.
(a) What is its angular acceleration?
(b) If the same torque continues to act what will be the angular velocity of the wheel after 16s from start?

Question 19.
A cat weighing lO kg is standing on a flatboat so that she is 20 m from the shore. She walks 8m on the boat towards the shore and then stops. The boat weighs 40 kg and we can assume that there is no friction between it and the water. How far is she from the shore at the end of this time?
Here, initially, the C.M. of the cat-boat system is at the centre of the boat at x = 20m away from shore.

M = mass ofboat4O kg
m = mass of cat 10 kg
Taking mass moments about the shore.

Total initial mass moment = (M + m)x
= (40+ 10)20= 1000 kgm.

Let the final distance of the cat be x, from the shore, then C.M. of the boat will be at a distance (x1 + 8)m from the shore.
Total final mass moment = mx1 + M(x1 + 8)
= 10x1+ 40(x1 + 8) = 50x1 + 320

according to mass moment conservation,
1000 = 50 x1 + 320
or
50x1 = 1000 – 320 = 680
∴ x1 = $$\frac{680}{50}$$
= 13.6m

Aliter: If the walking of a cat does not move the boat, then the distance of the cat from the shore would have been = 20 – 8 = 12 m.

But here, as no external force acts in this situation, so the C.M. of the boat-cat system must not move and it causes the boat to move behind. C.M. has shifted towards the shore due to the 8 m walk of cat to a position dividing the 8 m distance in the inverse ratio of its own and cat’s mass.

∴ Shift of C.M = $$\frac{8 \times 10}{40+10}=\frac{80}{50}$$ = 1.6 m

Since the C.M. is not to shift, the boat must move behind by a distance 1.6 m
∴ distance of cat from the shore = 12 + 1.6= 13.6 m.

Question 20.
A stationary pulley carries a rope whose one end supports a ladder with a man and the other end carries a weight of mass M. The man climbs up a distance w.r.t the ladder and then stops. Find the displacement x of the centre of inertia of the system.

Let m = mass of man
Here, M = mass attached to one end
l = distance by which the man climbs up w.r.t. ladder.

In the ladder-climbing process, the man is being away from the C.M.
∴ increase in the moment of the mass of man about C.M. = ml
= decrease of the moment about C.M.

It must be shared by the whole mass 2M of the system
If x be the shift in the C.M. in the same direction,
Then ml = 2Mx
or
x = $$\frac{\mathrm{ml}}{2 \mathrm{M}}$$ in the same direction.

Question 21.
A small ball is suspended from a point O by a light thread of length l. Then the ball is drawn aside so that the thread deviates through an angle θ from the vertical and set in motion in a horizontal direction at right angles to the vertical plane in which the thread is located. What is the initial velocity that has to be given to the ball so that it could deviate by a maximum angle $$\frac{π}{2}$$ during the motion?
Let r be the radius of the circle in which the ball rotates.

T = tension in the thread,
w = angular velocity of the ball in the circle,
mg = weight of the ball.

∴ T is resolved in two rectangular components.
∴ for vertical equilibrium,

Let v be the linear velocity required to start the motion of the ball and is given by
∴ v = l ω = l. $$\sqrt{\frac{\mathrm{g}}{l \cos \theta}}=\sqrt{\frac{\mathrm{g} l}{\cos \theta}}$$

During motion of the ball producing deviation of $$\frac{π}{2}$$, velocity vector rotates by $$\frac{π}{2}$$ causing a change in velocity = v$$\sqrt{2}$$

Let v0 = initial imparted velocity then v0 must provide v$$\sqrt{2}$$
or
vo = $$\sqrt{2}$$v
= $$\sqrt{2}$$. $$\sqrt{\frac{\mathrm{g} l}{\cos \theta}}=\sqrt{\frac{2 \mathrm{~g} l}{\cos \theta}}$$

Question 22.
A solid cylinder
(i) slides
(ii) rolls from rest down an inclined plane.
Compare the velocities when the cylinder reaches the bottom of the plane.
Let l = length of an inclined plane having an angle of inclination θ.
(i) When the cylinder slides down the plane,
a = g sin θ
Let v1 = velocity with which it reaches the bottom
∴ using the relation,

(ii) When the cylinder rolls, then let v2 be its velocity on reaching the bottom.
when r = radius of the cylinder, for a solid cylinder,

Question 23.
A flexible chain of weight W hangs between two fixed points A and B at the same level as shown here.
Find (i) force applied by a chain on each endpoint.
(ii) the tension in the chain at the lowest point.
Let W = weight of the chain
∴ $$\frac{w}{2}$$ = reaction at each endpoint A and B vertically upward

$$\frac{w}{2}$$ balances downward
(i) Component F sin 0 of force F applied by a chain on each point

(ii) At lowest point C, the tension T is horizontal and equals the horizontal component of force F

Question 24.
The moment of inertia of a body about a given axis is 1.2 kg m2. Initially, the body is at rest. In order to produce a rotational K.E. of 1500J, for how much duration, an acceleration of 25 rads-2 must be applied about that axis.

Question 25.
A thin bar XY of negligible weight is suspended by strings R and S shown in fig. The bar carries masses of 10 kg and 5 kg. Find the tensions in the strings and the angle θ if the system is in the static equilibrium.
T1, T2 = ?
θ = ?

Taking moments about point X, we get

Again taking moment about point Y, we get

Also for horizontal equilibrium

Value-Based Type:

Question 1.
A physics teacher explained about conservation of Angular momentum in the class. After the completion of her explanation, she wants to test how far the students are able to understand the topic.
In the process. she selected two students to name Babita and Ram. Both could explain the topic with examples.
(a) What questions of them impress you?
Both were doing a group study, discussing together they have given answers.

(b) A physics teacher sits on a stool that is free to rotate nearly without friction about a vertical axis her outstretched hands each hold a large, mass so that the rotational inertia is 12 kg m2 by pulling her arms close to her body she is able to reduce her rotational inertia to 6 kg m2. If her student starts spinning at 0.5 rad/s, what is her speed after she draws her arms in?
Solution:
In the absence of external torque, her angular momentum stays constant so that

When her rotational inertia halves, her angular velocity doubles.

Question 2.
Radha found the wheel getting detached from her uncle’s car. She took it to the workshop and got it repaired. She informed her uncle, who is a mechanical engineer, about this matter.
(a) What according to you the values displayed by Radha?
Radha takes care of things and has a concern for others. Prac deal with finding the solutions to problems.

(b) A thin wheel can stay up-right on its rim for a considerable length of time when rolled with considerable velocity, while it falls from its upright position at the slightest disturbance, when stationary, Explain.
When the wheel is rolling, the angular momentum is conserved. However, due to the frictional force, it continues to decrease. Thus, the wheel can stay upright on its rim only for a certain interval of time. In the stationary position, the wheel falls due to unstable equilibrium.

Question 3.
Two friends Raman and Sohan are discussing the speed and rotation of a wheel. One of them said that it is due to ‘ rotational motion about a fixed axis. The next day Raman brought a problem which is as under:
“The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds.”
(i) What values are displayed by Raman?
Values displayed are:
(a) Curiosity.
(b) Keen observation of the surroundings.
(e) Keen to find a solution.

(ii) What is the angular acceleration, assuming the accelera¬tion to be uniform?
Here, Wo = Initial angular speed (in rad/s)
W = Final angular speed (in rad/s)

Question 4.
A 10-year boy was sitting stationary at one end of a long trolly. The trolly is moving on a smooth horizontal floor at a definite speed. He was a naughty boy and enjoying it very much. Suddenly he stood up and ran about the trolly in any manner. He was surprised to see that trolly did not become unbalanced at all. He was curious to know the reason for it, he asked this question to his science teacher. The teacher explained to him about the centre of mass of the trolly and child together.
(i) what values of the boy are depicted here?
Values depicted here in the above problem are:
Curiosity, awareness, willing to know the scientific reason, presence of mind.

(ii) What is the centre of mass of the body?
For a system of particles, the centre of mass is defined as that point where the entire mass of the system is imagined to
be concentrated.

(iii) How did the teacher explain the reason?
The teacher explained that the body of the boy and trolly is a single system and the forces involved are totally internal. Since there is no external force. So, there is no change in the momentum of the system and hence the velocity of the system remains unchanged when the child gets up and runs about on the trolly in any manner.

## Laws of Motion Class 11 Important Extra Questions Physics Chapter 5

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 5 Laws of Motion. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

## Class 11 Physics Chapter 5 Important Extra Questions Laws of Motion

### Laws of Motion Important Extra Questions Very Short Answer Type

Question 1.
(a) Why do we beat dusty blankets with a stick to remove dust particles?
It is done due to inertia of rest.

(b) A stone when thrown on a glass window smashes the window pane to pieces. But a bullet fired from a gun passes through it making a hole. Why?
This is due to the inertia of rest.

Question 2.
(a) If you jerk a piece of paper from under a book quick enough, the book will not move, why?
It is due to the inertia of rest.

(b) Passengers sitting or standing in a moving bus fall in forward direction when the bus suddenly stops. Why?
This is due to the inertia of motion.

Question 3.
(a) Why passengers are thrown outward when a bus in which they are travelling suddenly takes a turn around a circular road?
This is due to the inertia of direction.

(b) Is any force required to move a body with constant velocity?
No.

Question 4.
(a) Why a one rupee coin placed on a revolving table flies off tangentially?
This is due to the inertia of direction.

(b) Why mud flies off tangentially to the wheel of a cycle?
This is due to the inertia of direction.

Question 5.
(a) When the electric current is switched off, why the blades of a fan keep on moving for some time?
This is due to the inertia of motion.

(b) Why the passengers fall backward when a bus starts moving suddenly?
This is due to the inertia of rest.

Question 6.
(a) A body of mass m is moving on a horizontal table with constant velocity. What is the force on the table?
mg i.e. equal to the weight of the body.

(b) Name a factor on which the inertia of a body depends.
Mass.

Question 7.
(a) Rocket works on which principle of conservation?
Law of conservation of linear momentum.

(b) Is the relation $$\overrightarrow{\mathbf{F}}=\mathbf{m} \overrightarrow{\mathbf{a}}$$ applicable to the motion of a rocket?
No.

Question 8.
(a) Will a person while firing a bullet from a gun experience a backward jerk? Why?
Yes, it is due to the law of conservation of linear momentum.

(b) A bomb explodes in mid-air into two equal fragments. What is the relation between the directions of their motion? Answer:
The two fragments will fly off in two opposite directions.

Question 9.
(a) What happens to the acceleration of an object if the net force on it is doubled?
As a = $$\frac{F}{m}$$ i.e. a ∝ F, so acceleration will be doubled when m the force is doubled.

(b) An electron moving with a certain velocity collides against a stationary proton and sticks to it. Is the law of conservation of linear momentum true in this case?
Yes, it is true.

Question 10.
(a) According to Newton’s third law of motion, every force is accompanied by an equal (in magnitude) and opposite (in direction) force called reaction, then how can a movement take place?
As the action and reaction never act on the same body, so the motion is possible.

(b) You can move a brick easily by pushing it with your foot on a smooth floor, but, if you kick it, then your foot is hurt. Why?
As Ft remains constant, so if t is reduced, then F will be increased and hence hurt our foot.

Question 11.
(a) Why does a swimmer push the water backward?
So as to get forward push according to Newton’s third law of motion.

(b) Why does not a heavy gun kick so strongly as a light gun using the same bullets (i.e. cartridges)?
The recoil speed of the gun is inversely proportional to its mass. So the recoil speed of the heavy gun is lesser than that of the light gun.

Question 12.
(a) Can a rocket operate in free space?
Yes.

(b) In a game of tug of war, two opposing teams are pulling the rope with equal (in magnitude) but the opposite force of 1000 kg wt at each end of the rope. What is the tension in the rope if a condition of equilibrium exists?
1000 kg wt.

Question 13.
(a) Which of Newton’s laws of motion is involved in rocket propulsion?
Newton’s third law of motion.

(b) Action and reaction are equal in magnitude and opposite in direction, then why do not they cancel/balance each other?
They don’t balance each other as they act on different bodies.

Question 14.
(a) A passenger sitting in a bus at rest pushes it from within. Will it move? Why?
No, internal forces are unable to produce motion in a system.

(b) How would you explain the motion of a motorcyclist in a globe of death in a circus?
It is a case of motion in a vertical circle.

Question 15.
Can a body in linear motion be in equilibrium? why?
Yes, it will be in equilibrium if the vector sum of the forces acting upon the body is zero.

Question 16.
Is the law of conservation of momentum valid for a system consisting of more than two particles?
Yes, the law of conservation of momentum is a general law that is applicable to all systems.

Question 17.
Which is greater out of the following
(i) The attraction of 1 kg lead for Earth,
(ii) the attraction of Earth for 1 kg of lead? Why?
Both are equal, the forces of action and reaction are always equal and opposite in direction according to Newton’s third law of motion.

Question 18.
Mention the conditions for the maximum and minimum pull of a lift on a supporting cable.

1. The pull of the cable is minimum (zero) when the lift is falling freely.
2. The pull of the cable is maximum when the lift is moving up with the same acceleration.

Question 19.
A man is at rest in the middle of a pond on perfectly frictionless ice. How can he get himself to the shore of the pond?
He can get himself to the shore if he throws away his shirt or anything in his possession in a direction opposite to the desired direction of motion or by spitting in the forward direction or by blowing air from his mouth.

Question 20.
Suppose you are seated in a cabin that has no doors, no windows, etc., and is also soundproof. Shall it be possible to detect the uniform velocity with which this cabin is moving? Why?
No, this is because when the cabin is. moving with uniform velocity, there will be no net unbalanced force.

Question 21.
(a) Why do we pull the rope downwards for climbing up?
When we pull the rope downwards, an upward reaction helps us to rise up.

(b) Why is it easier to roll than to pull a barrel along a road?
It is due to the fact that rolling friction is less than sliding friction.

Question 22.
(a) Why are the lubricants used in machines?
Lubricants are used in machines so as to reduce friction.

(b) Friction is independent of the area, but brakes of a very small contact area are not used. Why?
This is done so as to avoid wear and tear.

Question 23.
Mention a factor on which coefficient of friction depends.
The coefficient of friction depends upon the nature of the surfaces in contact.

Question 24.
Carts with rubber tires are easier to ply than those with iron tires. Why?
The coefficient of friction between the rubber tires and the road is lesser than the coefficient of friction between iron and steel.

Question 25.
Why are wheels made circular? Explain.
Circular wheels roll without sliding on the road. Since rolling friction is less than sliding friction, so they move easily.

Question 26.
(a) What do you mean by dry friction?
When both the bodies in contact are solids, then the force of friction is called dry friction.

(b) A soda water bottle is falling freely. Will the bubbles of the gas rise in the water of the bottle?
A freely falling soda water bottle is in a state of weightlessness. Thus the bubbles of the gas will not rise in the water of the bottle rather they remain floating.

Question 27.
What do you mean by liquid or fluid or wet friction?
It is defined as the friction which conies into play between a surface and a liquid or fluid.

Question 28.
“Friction is a self-adjusting force.” Is this statement correct?
This statement is correct only so long as the friction is static friction. Up to the limiting friction, the force of friction is equal (in magnitude) and opposite to the applied force.

Question 29.
(a) Several forces act simultaneously on a body. In which direction will it move?
It will move in the direction of the net force.

(b) Name the physical quantity which is a measure of inertia of a body?
Inertial mass.

Question 30.
Can a force change only the direction of the velocity of an object keeping its magnitude constant?
Yes, a force can only change the direction of the velocity of an object keeping its magnitude constant.

Question 31.
(a) Two objects having different masses have some momentum. Which one of them will move faster?
The object with a smaller mass will move faster.

(b) A table is lying on the floor of a room. Is some force of friction acting on it?
No.

Question 32.
(a) A book is lying on an inclined plane. Is some force of friction acting on the book?
Yes

(b) Name the physical quantity which can be found from the area under the force-time graph.
Impulse

(c) Will the body be in equilibrium under the action of three non-coplanar forces.
No.

Question 33.
(a) Write the S.I. units of force, momentum, and impulse.
S.I. unis of force, momentum, and impulse are newton (N), kg ms-1, and Ns (Newton-second).

(b) Why should the hammer be heavier to push the nail deeper into the wooden blocks?
It should be heavier so as to increase the impact of force i.e. more force applied for a shorter time.

Question 34.
(a) Why rockets are given conical shapes?
The rockets are given conical shapes so as to reduce atmospheric friction.

(b) How does air friction affect the maximum height of a projectile?
The maximum height of a projectile is reduced due to air friction.

Question 35.
(a) Explain why jet planes cannot move in air-free space but rockets can move?
Jet planes use atmospheric oxygen to foil their fuel but rockets carry their own fuel and don’t depend on atmospheric oxygen.

(b) Is it correct to state that a body always moves in the direction of the net force acting on it?
The statement is true only for bodies at rest before the application of force.

Question 36.
At which place on Earth, the centripetal force is maximum?
The centripetal force is maximum at the equator.

Question 37.
What provides the centripetal force in the following cases?
(i) Electron revolving around the nucleus.
It is provided by
The electrostatic force of attraction between the electron and the nucleus.

(ii) Earth revolving around the sun.
The gravitational force of attraction between Earth and Sun.

(iii) Car taking a turn on a banked road.
A component of the reaction of the road.

Question 38.
(a) What is the direction of the angular velocity of the minute hand of a wall-clock?
The direction of the angular velocity of the minute hand of a wall- clock is perpendicular to the wall and directed inwards.

(b) What is the difference between ‘Newton’ and ‘newton’?
Newton is the name of a scientist but newton is the S.l. unit of force named after the scientist Newton.

Question 39.
(a) Is it possible that a particle moving with a constant velocity may not have a constant speed?
No.

(b) For uniform circular motion, does the direction of centripetal force depend on the sense of rotation {i.e. clockwise or anti-clockwise)?
No. It is always radial irrespective of the sense of rotation.

Question 40.
Why chinaware crockery is wrapped in paper or straws?
Paper or straw provides a cushion between th° pieces of crockery.
In case of any jerk (impulse), these will prolong the time of impact and reduce its effect, the crockery will thus be saved.

Question 41.
AH, vehicles are provided with springs and shockers, why?
To reduce the impact of force on the vehicle as shockers and springs break the impact of force and increase the time of action of force thus reduce the impulse and save the vehicle from shock.

Question 42.
A body falls from a single-story building roof on a muddy floor and another boy falls from the same height on a stone, who is likely to survive out of them and why?
The boy who falls on a muddy floor will survive due to the reduction of the impact of force by prolonging the time of reduction of force from maximum to zero.

Question 43.
What is the effect on the direction of centripetal force when the revolving body reverses its direction of motion?
The centripetal force will be directed towards the center of the circle. This fact does not depend upon the sense of the rotation of the circle.

Question 44.
Is it correct to say that the banking of roads reduces the wear and tear of the tires of automobiles? If yes explain.
Yes, if the road is not banked, then the necessary centripetal force will be provided by the force of friction between tires and the road. On the other hand; when the road is banked, a component of the normal reaction provides the necessary centripetal force, which reduces wear and tear.

Question 45.
The linear velocity of a particle moving on the circumference of the circle is equal to the velocity acquired by a freely falling body through a distance equal to one-fourth the diameter of the circle. What is the centripetal acceleration of the particle moving along the circle?

Question 46.
(a) A body is moving with uniform velocity. Can it be said to be in equilibrium? Why?
Yes, it can be said to be in equilibrium when it moves with uniform velocity as no acceleration i.e. no net force acts on the body.

(b) Why Newton’s second law of motion is not applicable to the motion of a rocket?
Newton’s second law i.e. F = ma is applicable only if the mass (m) of the body remains constant. In the case of the rocket, the mass continuously decreases and hence F = ma is not applicable.

Question 47.
(a) A thief jumps from the upper story of a house with a load on his back. What is the force of the load on his back when the thief is in the air?
When the thief is in the air, he is in the state of free fall, and hence in the state of weightlessness. So the force of the load on his back is zero.

(b) When a body falls to the Earth, the Earth also moves up to meet it. But the motion of Earth is not noticeable. Why? Answer:
We know that acceleration is the ratio of the applied force to the mass of the body. As the mass of Earth is very large, so its acceleration is very small.

Question 48.
What are the different effects a force is capable of producing?
Force can cause the following

1. Circular motion
2. Translatory motion
3. Deformations in the body on which it is applied.

Question 49.
A ball is thrown up at a speed of 36 ms-1 by a thrower. If the ball returns to the thrower with the same speed. Will, there be any change in:
(a) Momentum of the ball?
There will be a change in the direction of the momentum of the ball.

(b) Magnitude of the momentum of the ball?
There is no change in the magnitude of the momentum of the ball.

Question 50.
When a ball falls from a height its momentum increases. What causes the increase in the momentum of the ball?
The gravitational force acting in the direction of motion increases the velocity and hence momentum of the ball.

Question 51.
When a high jumper leaves the ground, where does the force which accelerates the jumper upward comes from?
The high jumper after taking a short run presses the ground hard, the ground, in turn, reacts on him and provides the necessary upward accelerating force to the jumper. Thus, the reaction of the ground on the jumper is the required force.

Question 52.
Name the forces which are in equilibrium in each of the following situations:
(a) a book resting on a table.
The gravitational force on the book and a force of reaction of the table.

(b) a cork floating on water.
The gravitational force on the cork and an upward thrust or buoyant force of water.

(c) a pendulum bob suspended from the ceiling with the help of a string.
The gravitational force on the bob and the tension in the string.

### Laws of Motion Important Extra Questions Short Answer Type

Question 1.
(a) A learner shooter fired a shot from his rifle and his shoulder got injured ¡n the process. What mistake did he commit?
We know that a gun recoils i.e. moves back after firing. To avoid injury to the shoulder, the gun must he held tightly against the shoulder. The learner shooter might have not held it tightly against his shoulder and hence the gun must have injured his shoulder after firing.

(b) When the horse suddenly stops, the rider falls in the forward direction. Why? Explain it.
When the horse suddenly stops, the rider falls in forwarding direction due to the inertia of motion.

Explanation: The lower portion of the rider comes to rest along with the horse while the upper portion of the rider continues to move forward. Hence, he falls forward.

Question 2.
(a) Newton’s first law of motion is the law of Inertia. Explain.
According to Newton’s first law of motion, a body can’t change its state of rest or of uniform motion along a straight line unless an external force acts on it. It means that the natural tendency of the material body is to continue in the state of rest or that of uniform motion which is termed as inertia. Thus Newton’s first law is the law of inertia.

(b) What happens to a stone tied to the end of a string and whirled in a circle if the string suddenly breaks? Explain why?
The stoneflies off tangentially to the circle along a straight line at the point where the string breaks. It is due to the inertia of direction. When the string breaks, the force acting on the stone ceases. In the absence of force, the stoneflies away in the direction of instantaneous velocity which is along the tangent to the circular path.

Question 3.
(a) An astronaut accidentally gets separated out of his small spaceship accelerating in inter-steller space at a constant rate of 100 ms-2. What is the acceleration of the astronaut at the instant after he is outside the spaceship?
According to Newton’s first law of motion, the moment he is out of the spaceship, there is no external force on the astronaut, thus his acceleration is zero. Here we are assuming that he is out of the gravitational field of heavenly bodies i.e. there are no nearby stars to exert a gravitational force on him and the small spaceship exerts a negligible gravitational attraction on him.

(b) How is it that a stone dropped from a certain height falls much more rapidly as compared to a parachute under similar conditions?
As the surface area of a parachute is much larger as compared to the surface area of a stone, so the air resistance, i. e. fluid friction in the case of the parachute is much larger than in the case of stone. Hence the parachute falls slowly.

Question 4.
(a) When a man jumps out of a boat, then it is pushed away. Why?
This is due to Newton’s third law of motion. When the man jumps out of the boat, he applies a force on it in the backward direction, and in turn, the reaction of the boat on the man pushes him out of the boat.

(b) Explain how lubricants reduce friction?
The lubricants spread as a thin layer between the two surfaces. The motion now is between the surface and the lubricant layer which changes the dry friction into wet friction. As wet friction is less than dry friction, hence lubricants reduce friction.

Question 5.
Two hoys on ice-skates hold a rope between them. One boy is much heavier than the other. The lightweight boy pulls on the rope. How will they move?
The light-weight boy is doing the action on the heavy boy by pulling the rope. According to Newton’s third law, equal and opposite force (reaction) also acts on the light boy. As the mass of the boy pulling the rope is lesser, so the acceleration produced in him will be more. Thus both the boys move tow; rds each other and the lighter boy will move faster.

Question 6.
Explain why ball bearings are used in machinery?
We know that rolling friction is much lesser than sliding friction, so we convert the sliding friction into rolling friction which is done using ball bearings that are placed in between the axle and the hub of the wheel. The ball bearings tend to roll around the axle as the wheel turns and as such the frictional force is reduced.

Question 7.
Why a horse has to apply more force to start a cart than to keep it moving? Explain.
Static friction comes into play when the horse applies force to start the motion in the cart. On the other hand, kinetic friction comes into play when the cart is moving.

Also, we know that the static friction is greater than the kinetic friction, so the horse has to apply more force to start a cart than to keep it moving.

Question 8.
Sand is thrown on tracks or roads covered with snow. Explain why?
When the roads (or tracks) are covered with snow, then there is a considerable reduction of frictional force between the tires of the vehicles and the road (or between the track and the wheels of the vehicle or train) which leads to the skidding of the vehicles (or trains). Thus, driving is not safe. When sand is thrown on the snow-covered roads (or tracks), then the force of friction increases, so safe driving is possible.

Question 9.
Explain why is it difficult to move a cycle along a road with its brakes on?
When we move the cycle with its brakes on, then its wheels can only skid i.e. slide along the road as they can’t rotate. So the friction is sliding in nature. As the sliding friction is greater than fr,e rolling friction, therefore, it is difficult to move a cycle with its brakes on.

Question 10.
Explain how proper inflation of tires saves fuel?
When the tires are properly inflated, the area of contact between the tires and the ground is reduced which in turn reduces the rolling friction. As a result of this, there is less dissipation of energy against friction. So the cover of the automobile/eater distance for the same quantity of fuel consumed. Hence proper inflation of tires leads to saving fuel.

Question 11.
Explain how the man at rest in the middle of a pond of perfectly frictionless ice comes out by blowing air or splitting etc.?
By blowing air from b’s mouth in a forwarding direction, the man applies some force in that direction and hence imparts some momentum to the air blown out from his mouth. According to the law of conservation of linear momentum, an equal and opposite momentum will be imparted to his body. Since there is no force of friction and hence no energy is wasted. Finally, by doing the same action again and again he gets himself on the shore.

Question 12.
Give one argument in favor of the fact that frictional force is a non-conservative force.
The direction of the frictional force is opposite to the direction of motion. When a body is moved, say from point A to B and then back to A, then work is required to be done both during forward and backward motion. So the network done in a round trip is not zero. Hence the frictional force is a non-conservative force.

Question 13.
Why is it more dangerous to fall on frozen ice than on fresh snow?
The fresh snow is softer and frozen ice is harder. So the impulse caused by falling on fresh snow is much lesser than frozen snow. Thus, falling on frozen ice may cause serious injury or maybe fatal (the reaction will be much greater in this case).

Question 14.
(a) An astronaut in open space is away from his spaceship. How can he return to his ship?
Using Newton’s third law of motion, he can reach his spaceship. He must throw objects with appropriate force opposite to the direction of the spaceship. The recoil will send him-to his spaceship.

(b) Why mudguards are used in vehicles?
The mud stuck to the tires fly off tangentially to the tires due to the inertia of direction. The mudguards are i so placed that they come in the way of mud and it is stopped by them from spilling all over.

Question 15.
What is the difference between absolute and gravitational units of force?
The absolute units of force don’t depend upon the value of acceleration due to gravity, so they remain the same throughout the universe. But the gravitational units of force depend upon the value of ‘g’ which is different at different places.

Question 16.
A disc of mass m is placed on a table. A stiff spring is attached to it and is vertical. To the other end of the spring is attached a disc of negligible mass. What minimum force should be applied to the upper disc to press the spring such that the lower disc is lifted off the table when the external force is suddenly removed?
The minimum force should be mg. When a force mg is applied vertically downwards on the upper disc, the lower disc will be pressed against the floor with a force mg. The floor will exert an upward reaction mg. When the external force is suddenly removed, this reaction will just lift the lower disc.

Question 17.
A force acting on a material particle of mass m first grows to a maximum value Fm and then decreases to zero. The force varies with time according to a linear law, and the total time of motion is tm. What will be the velocity of the particle at the end of this time interval if the initial velocity is zero?
In the given problem,

= area under F – t graph
= area of ΔOAB ..(1)

Also by definition of impulse, we know that

Where m = mass of the particle,
v = its velocity after time interval
i.e. after time tm

∴ From (1) and (2), we get

Question 18.
Sometimes we need to increase friction. Why? Given an example.
Sometimes friction between two surfaces decreases to such an extent that it is difficult to move on that surface. So friction needs to be increased. For example, vehicles can’t move on a road covered with snow. In such cases, we have to throw sand on the road to increase the friction.

Question 19.
Vehicles stop applying brakes. Does this phenomenon violate the principle of conservation of momentum?
The law of conservation of momentum is not violated by stopping vehicles from applying brakes. Some retarding force is being applied due to brakes and the vehicle comes to rest such that the total loss of its momentum is equal to the impulse of the applied force. Thus the law of conservation of momentum is not violated.

Question 20.
“Two surfaces if made extremely smooth, will have a very low value of friction between them.” Is the statement true? Justify your answer with two illustrations.
The given statement is wrong. This is because the force of frictions increased when the surfaces in contact are made highly smooth.

Examples:

1. In factories, the conveyer belt is made smooth by rubbing it with wax or resin. The wheel over which the belt is to move is also made extremely smooth. Due to the increased force of friction between the belt and the wheel, slipping does not take place.
2. The wheels of the railway train and the surfaces of the railway tracks are also made very smooth.

Question 21.
Why force of friction increases when the two surfaces in contact are made extremely smooth i.e. polished beyond a certain limit?
When the two surfaces in contact are made extremely smooth, then they come in intimate contact with each other. So the force of adhesion comes into play- Due to this force, the motion of one surface over the other surface becomes retarding and hence causes an increase in the friction.

Question 22.
(a) A small amount of water spread on a marble floor causes slipping. Why?
Water fills the grooves in the floor and makes it more smooth thus reducing friction and it causes slipping.

(b) Why automobile tires have generally irregular projections over their surface?
The automobile tires have generally irregular projections over their surface so as to

1. Increase friction.
2. Increase the grip with the ground and thus avoiding their skidding.

Question 23.
A body moving over the surface of another body suddenly comes to rest. What happens to friction between the two surfaces?
The force of friction will be present only so long as there is relative motion between the two bodies and it automatically disappears as soon as the relative motion ceases.

Question 24.
Explain why one should take short steps rather than long steps while walking on ice?
There is a danger of slipping if one walks on ice by taking long steps. This can be explained as –

Explanation: Let R represents the reaction offered by ice while walking on it.
Also, let mg = weight of the person walking on the ice.

Let f = force of friction between ice and the foot of the person walking on it.
Let θ = angle made by R with the vertical.

The rectangular components of R are shown in the figure. The vertical component i.e. R cos θ will balance the weight of the person and the horizontal component i.e. R sin θ will help the person to walk forward by balancing the f.
i. e. R cos θ = mg …(i)
and R sin θ = f …(ii)

Dividing (ii) by (i), we get

where μ = coefficient of friction and has a fixed value.

If steps are longer, then θ is more, so tan θ will be more. But as p. is fixed, so f is to be increased which is not possible, hence there is a danger of slipping in a long step.

Question 25.
Why a cricket player lowers his hands while catching a cricket ball?
While taking a catch, a cricket player moves his hands backward. He has to apply retarding force to stop the moving ball in his hands. If he catches the ball abruptly, then he has to apply a large retarding force for a short time. So he gets hurt.

On the other hand, if he moves his hands backward then the player applies the force for a longer time to bring the ball to rest. In this case, he has to apply less retarding force and thus will not get hurt.

Thus to avoid injuries to his hands, he lowers his hands while catching a cricket ball.

Question 26.
Why are buffers provided between the bogies of a railway train?
Buffers are provided between the bogies of a train to reduce shocks or jerks by reducing the impulse or impact of a force. Springs of buffers increase the time of contact in the collision of bogies and thus the force acting on the bogies will be small and hence the passengers sitting inside the train will feel fewer jerks.

Question 27.
It is more difficult to catch a cricket ball than to catch a tennis ball moving with the same velocity. Explain why?
The mass of a cricket ball is greater than the mass of a tennis ball, so the change in momentum of the cricket ball is more than that of the tennis ball. Hence greater force is required to catch the cricket ball than that of the tennis ball.

Question 28.
(a) A body of mass 25 g is moving with a constant velocity of 5 ms’1 on a horizontal frictionless surface in a vacuum, what is the force acting on the body?
F = ma, where a = acceleration of the body.
For constant velocity, a = 0, so F = 0.
Thus when a body moves with constant velocity then no force acts on it.

(b) A bird is sitting on the floor of a wire cage and the cage is in the hand of a boy. The bird starts flying in the cage. Will the boy experience any change in the weight of the cage? Why?
Lighter. Since the cage is made of wire, the air inside the cage is not bound with the cage, rather it is in contact with the atmospheric air. Therefore the boy will not experience the weight of the bird when it flies in the cage. Thus the cage will appear lighter when the bird starts flying in the cage.

Question 29.
A woman stands on a spring scale on an elevator. In which case will the scale record the minimum reading and the maximum reading?
(i) elevator stationery,
When the elevator is stationary, then reading of the scale = actual weight of woman = mg.

(ii) elevator cable breaks, free fall,
When the elevator falls freely, it is the case of weightlessness.
∴ Reading of the scale = m(g – g) = 0 (∴ a = g)

(iii) elevator accelerating upward and
When the elevator accelerates upwards, the reading of the scale = m(g + a), where a = acceleration of the elevator.

(iv) elevator accelerating downward.
When the elevator accelerates downwards, then the reading of the scale = m(g – a).
Thus, the reading of the scale is minimum when the elevator falls freely and the reading of the scale is maximum when the elevator accelerates upwards.

Question 30.
Two bodies of different masses m1 and m2 are falling from the same height. If resistance offered by the air be the same for both the bodies, then will they reach the Earth simultaneously? Assume m1 > m2?
Let f be the resistance offered by the air for both the bodies.
If F = Net force acting on the body of mass m, i.e. weight W1 = m1g,

Then F = W1 – f
= m1g – f …(1)

If a1 = acceleration produced in the body of mass m, then

Similarly if a2 be the acceleration produced in the body of mass m2,
then
a2 = g – $$\frac{\mathrm{f}}{\mathrm{m}_{2}}$$ ….(3)

Now as m1 > m2, so it is clear from equation (2) and (3) that
a2 < a1 or a1 > a2.

Thus, we conclude that the body having larger acceleration will reach earlier, so the body of larger mass will reach the earth earlier than the body of smaller mass.

Question 31.
How does air friction affect the maximum height of a projectile?
The air friction reduces both vertical and horizontal components of velocity. The air friction depends on the volume of the body or the air displaced. So the height attained by the body in projectile motion is severely restricted by the air friction. The height reached is reduced compared to one which is attained in the absence of air.

Question 32.
Why is it difficult to climb a greasy pole or rope?
For climbing any pole or rope, the climber presses the pole or rope with his feet which in turn pushes the feet and the person climbs up. But the man’s feet push the pole only if there is friction between the pole and the feet so that feet can hold on.

When grease is applied to the pole, the friction is sharply reduced and the foot cannot press it in the absence of friction. Thus, man does not get the reaction required to climb, hence it is difficult to climb a greasy pole or rope.

Question 33.
Does F = ma represent Newton’s second law of motion under all conditions? Give a reason for your answer.
No, F = ma does not represent Newton’s second law of motion under all conditions. In fact the statement of Newton’s second law of motion relates force with a change in momentum of the body i. e.

In a reference frame where the velocity of the body is too high i.e., it approaches the speed of light then mass does not remain the same. Thus, in that case,

holds goods which are different from F = ma.

Question 34.
Several passengers are standing in a running bus, it is said to be dangerous. How will you justify the statement?
There are three reasons for such situations to be dangerous –

1. Standing passengers raise the center of gravity of the bus and thus the bus comes in unstable equilibrium. This becomes dangerous.
2. A sudden brake applied by the driver will cause passengers to fall forward one over the other. Thus, they may be injured and there may be stamped.
3. If the bus is suddenly stopped, the passenger’s forward momentum may push it slightly ahead which may cause the accident in a congested traffic situation.

Question 35.
In a circus in the game of swing, the man falls on a net after leaving the swing but he is not injured, why?
When the man falls on the net it is depressed where a man falls on it and thus the time of contact is increased. Due to this, the force of reaction on the man is reduced to a great extent. Because the increase in time reduces the impulse which is equal to the changes in momentum (FΔt = m Δv). So, F is quite less and the man is not injured by the net. In fact, F pushes him up once and again he falls on the net.

Question 36.
(a) A meteorite burns in the atmosphere before it reaches the Earth’s surface. What happens to its momentum?
A meteorite while traveling towards earth shares its momentum with atmospheric particles and the remaining momentum is imparted to the Earth.

(b) By accident a person fell on a floating ice slab in a pond. He has nothing with him to get out and the ice slab is big enough so that he cannot put his hands or feet in the water. How can he save himself by coming out of the ice slab?
The person should know how to generate a force so that the ice slab is thrown away opposite to the direction of the force along with. This, he can do by spitting forcefully or sneezing, etc. again and again till the slab reaches the shore of the lake from where he can get out from the slab by the same technique.

Question 37.
In pushing a box up an inclined plane, is it better to push horizontally or to push parallel to the inclined plane? Why?
The box should be pushed by a force parallel to the plane. This will neither increase the reaction nor the force of friction. If the force is applied horizontally, only a component of the force will push the box up but the second component acting normally to the inclined plane will increase the reaction of the plane and thereby increase the force of friction. This will make the pushing of the box up the inclined harder.

Question 38.
Explain the role of friction in the case of bicycle brakes. What will happen if a few drops of oil are put on the rim?
When the brakes are applied, the rubber padding creates

1. a normal force on the rim, and the brakes and
2. force of friction are created in the direction opposite to the rotation of the wheel.

Thus, brakes stop the rotation of the wheels and the cycle is stopped.

If a few drops of oil are put on the rim, the friction is reduced but brakes apply a force on the rim which increases the normal reaction and thereby the force of friction. No doubt that now bicycle will take a longer time to stop and a small amount of oil will be thrown out due to centrifugal force.

Question 39.
A man cannot run faster on the sandy ground but a snake can. Why?
The greater friction of sandy ground stops the man from running faster but the snake gets support as its points of contact with the ground help in holding and pushing its body. The normal reaction of man is much more as compared to that of snake, that too is a factor in the relative force of friction acting on the two.

### Laws of Motion Important Extra Questions Long Answer Type

Question 1.
(a) State and prove impulse-momentum Theorem.
It states that the impulse of force on a body is equal to the change in momentum of the body.
i.e. J = Ft = P2 – P1

Proof: From Newton’s Second law of motion, we know that

Let P1 and P2 be the linear momenta of the body at time t = 0 and t respectively.
∴ integrating equation (i) within these limits, we get

Hence proved.

(b) Prove that Newton’s Second law is the real law of motion.
Proof: If we can show that Newton’s first and third laws are contained in the second law, then we can say that it is the real law of motion.
1. First law is contained in second law: According to Newton’s second law of motion,
F = ma …(i)
where m = mass of the body on which an external force F is applied and a = acceleration produced in it.

If F = 0, then from equation (1), we get
ma = 0, but as m ≠ 0
∴ a = 0

which means that there will be no acceleration in the body if no external force is applied. This shows that a body at rest will remain at rest and a body in uniform motion will continue to move along the same straight line in the absence of an external force. This is the statement of Newton’s first law of motion. Hence, the First law of motion is contained in the Second law of motion.

2. Third law is contained in second law: Consider an isolated system of two bodies A and B. Let them act and react internally.
Let FAB = force applied on body A by body B
and FBA = force applied on body B by body A

It $$\frac{\mathrm{d} \mathbf{p}_{\boldsymbol{A}}}{\mathrm{dt}}$$ = rate of change of momentum of body A
and
$$\frac{\mathrm{d} \mathbf{p}_{\boldsymbol{B}}}{\mathrm{dt}}$$ = rate of change of momentum of body B

Then according to Newton’s second law of motion,

(2) and (3) gives

As no external force acts on the system (∵ it is isolated), therefore according to Newton’s second law of motion,

or
Action = – Reaction,
which means that action and reaction are equal and opposite. It is the statement of Newton’s 3rd law of motion. Thus 3rd law is contained in the second law of motion.

As both First and Third Law is contained in Second law, so Second law is the real law of motion.

Question 2.
Derive the general expression for the velocity of a rocket in flight and obtain the expression for the thrust acting on it.
The working of a rocket is based upon the principle of conservation of momentum. Consider the flight of the rocket in outer space where no external forces act on it.
Let mo = initial mass of rocket with fuel.
Vu = initial velocity of the rocket,
m = mass of the rocket at any instant t.
v = velocity of the rocket at that instant.
dm = mass of the gases ejected by the rocket, in a small-time it.
u =H velocity of exhaust gases,
DV = increase in the velocity of the rocket in a time dt.

∴ Change in the momentum of exhaust gases = dm. u
Change in momentum of rocket = – (m – dm) dv.

A negative sign shows that the rocket is moving in a direction opposite to the motion of exhaust gases.

Applying the law of conservation of linear momentum,
dm.u = – (m – dm) dv …(1)

As dm being very small as compared to m, so it can be neglected, Thus, eqn. (1) reduces to
dm.u = – m dv
or
dv = – u $$\frac{dm}{m}$$ …(2)

Instantaneous velocity of the rocket:
At t = 0, mass of rocket = m0, velocity of rocket = vo.
At t = t, mass of rocket = m, velocity of rocket = v.

∴ Integrating Eqn. (1) within these limits, we get

In actual practice, the velocity of exhaust gases nearly remains constant.

equation (3) gives the instantaneous velocity of the rocket. In general vo = 0 at t = 0,

∴ Eqn. (3) reduces to

From Eqn. (4), we conclude that the velocity of the rocket at any instant depends upon:

1. speed (u) of the exhaust gases.
2. Log of the ratio of initial mass (m0) of the rocket to its mass (m) at that instant of time.

Upthrust on the rocket (F): It is the upward force exerted on the rocket by the expulsion of exhaust gases. It is obtained as follows:
Dividing Eqn. (2) by dt, we get

where F = ma is the instantaneous force (thrust).

From Eqn. (5), we conclude that the thrust (F) on the rocket at any instant is the product of the velocity of exhaust gases and the rate of combustion of fuel at that instant. Here negative sign shows that the thrust and velocity of exhaust gases are in opposite direction.

Question 3.
(a) Define inertia. What are its different types? Give examples.
The tendency of bodies to remain in their state of rest or uniform motion along a straight line in the absence of an external force is called inertia.

Inertia is of the following three types:
1. The inertia of rest: When a body continues to lie at the same position with respect to its surrounding, it is said to possess inertia of rest. This situation may be changed only by the application of external force. For example, if a cot or sofa is lying in a particular place in the house, it will remain there even after days or years unless someone removes (by applying force) the same from its position. This is an example of the inertia of rest.

2. The inertia of motion: When a body is moved on a frictionless surface or a body is thrown in a vacuum, it will continue to move along its original path unless acted upon by an external force. In actual situations, air or floor etc. exert friction on the moving bodies so we are unable to visualize a force-free motion. This type of inertia when a body continues to move is called the inertia of direction.

3. In the above examples it is found that the direction of motion of the body or particle also does not change unless an external force acts on it. This tendency to preserve the direction of motion is called the inertia of direction.

(b) Explain Newton’s First law of motion. Why do we call it the law of inertia?
According to the First law of motion, “Everybody continues to be in the state of rest or of uniform motion along a straight line until it is acted upon by an external force.”

It means that if a book lying on a table,-it will remain there for days or years together unless force is applied on it from outside to pick it.

Similarly, if a body is moving along a straight line with some speed, it will continue to do so until some external force is applied to it to change its direction of motion.

Thus First law tells us the following:

1. It tells us about the tendency of bodies to remain in the state of rest or of motion and the bodies by themselves can neither change the state of rest nor of uniform motion. This tendency is called inertia. To break the inertia of rest or motion or direction, we need an external force. Thus the definition of the first law matches with the definition of inertia and hence first law is called the law of inertia.
2. The first law of motion also provides the definition of another important physical quantity called force. Thus force is that agency which changes or tends to change the state of rest or of uniform motion of a body along a straight line.

(c) State Newton’s Second law of motion. How does it help to measure force? Also, state the units of force.
It states that the time rate of change of momentum of a body is directly proportional to the force applied to it.

where a = $$\frac{dv}{dt}$$ = acceleration produced in the body of mass m.
k = proportionality constant which depends on the system of units chosen to measure F, m, and a.

In the S.I. system, k = l,
∴ F = ma

The magnitude of the force is given by
F = ma …. (2)

Note: We have assumed that the magnitude of velocity is smaller and much less than the speed of light. Only under this condition Eqns. (1) and (2) hold good.

The definition of the Second law and its mathematical form is given in Eqn. (2) provide us a mean of measuring force.

One can easily find the change in velocity of a body in a certain interval of time. Both velocity and time can be easily measured. Thus by knowing the mass of the body one can determine both change in momentum as well as the acceleration of the body produced by an external force. If the force is increased, the rate of change of momentum is also found to increase. So also is the acceleration. Now with known values of m and we can find F.

Units of force: Force in S.I. units is measured in newton or N. From Eqn. (1) or (2) we can see that a newton of force is that fore? which produces 1 ms-2 acceleration in the body of mass 1 kg.
1 newton = 1 kilogram × 1 metre/(second)2
or
1 N = 1 kg × 1 ms-2 = 1 kg ms-2

In CGS system force is measured in dyne
1 dyne = 1 gram × 1 cm/s2 = 1 g cm s-2
Since 1 N = 1 kgm s-2= 1000 g × 100 cm s-2
= 105 g cm s-2 = 105 dyne
1 N = 105 dyne
or
1 dyne = 105 N

Gravitational Unit: If a falling mass of 1 kg is accelerated towards the Earth with 9.8 ms-2, then the force generated is called 1 kg wt (1-kilogram weight) force. It is the S.I. gravitational unit of force.

We know that the earth accelerates the mass with g = 9.8 ms-2
1 Kg wt = 9.8 N [1 kg × 9.8 ms 2 = 9.8 N]

C.G.S. gravitational unit is gf or g wt.
1 gf = 1g × 980 cms-2
= 980 dyne

Question 4.
(a) State Newton’s Third law of motion. Discuss its consequences.
Newton’s Third law of motion states that “to every action, there is always an equal and opposite reaction.”’
So, if a body 1 applies a force F12 on body 2 (action), then body 2 also applies a force F2] on body 1 but in opposite direction, then
F21 = – F12

In terms of magnitude
|F21| = |-F12|

It is very important to note that F12 and F21 though are equal in magnitude and opposite in direction yet act on different points or else no motion will be possible.

For example, hands pull up a chest expander (spring), and spring in turn exerts a force on the arms. A football when pressed reacts on the foot with the same force and so on. The most important consequence of the third law of motion is the law of conservation of linear momentum and its application in collision problems.

Here Δt is the time for which the bodies come in contact during impact. This is the same for the two bodies of masses m1 and m2 and having velocity changes Δv1 and Δv2 respectively. Therefore,
m1 Δv1 = – m2 Δv2
or
m1 Δv1 + m2Δv2 = 0

Let u1, u2 and v1 and v2 be initial and final velocities of the two masses before and after collision, then
m1(v1 – u1) = – m2(v2 – u2)
or
m1u1 +m2u2 = m1v1 + m2v2

Momentum before impact = momentum after impact
This is the law of conservation of momentum.

(b) State the law of conservation of linear momentum and illustrate it with examples.
‘The linear momentum of an isolated system always remains the same provided no external force is applied on it.’ This is the law of conservation of linear momentum.

The linear momentum of a body = mass × velocity
p = mv
If a system has several bodies initially at rest then initial momentum = 0.
The final momentum = p1 + p2 + p3 + ……

According to law of conservation of linear momentum
p1 + p2 + p3 + …. = 0

Linear momentum is a vector quantity and is measured. in kg ms-1or Ns.

For example, a gun and a bullet make a system in which both are initially at rest. When the bullet of mass m is
fired with muzzle velocity v, the gun of mass M gets a recoil velocity V. Since the initial linear momentum of the system is zero.
MV + mv = 0
or
MV = – mv

Thus gun moves in the opposite direction to that of the bullet.

(c) Define the terms – momentum and impulse. What are their units in the S.I. system?
The total quantity of motion possessed by the body is called is momentum. Mathematically, it is equal to the product of the mass of the body and the velocity of the body.

In linear motion, this term is called linear momentum P.
It is a vector quantity.
p = mv
The units of linear momentum are kg ms-1 or NS in S.I. units.

Impulse: The action or impact of force is called the impulse of force. Mathematically, impulse J is equal to the product of the force F acting on the body and the time for which the force acts on it. Thus
J = F × t = Ft

J is a vector quantity and is measured in Ns or kg ms-1
The action of force or impulse is increased if the force acts for a smaller interval.

Question 5.
(a) State and explain the laws of friction.
Following are the laws of friction:
1. The direction of the force of friction is always opposite to the direction of relative motion i.e. motion of the body over the surface of another body.

It can be seen easily that if a ball is rolled on the floor, it will be stopped after traveling through some distance if no other force is applied to it. This is because the force of friction that comes into play opposes the motion and slows down the ball which finally comes to rest.

2. The force of friction always acts tangentially along the surfaces of contact of the two bodies.

3. The magnitude of the limiting friction (F) is always directly proportional to the normal reaction (R) between the two surfaces i.e.
F ∝ R
or
F = μR

If we take two bodies of masses m1 and m2 s.t. m1 > m2. We apply forces F1 and F2 which just move these bodies on a surface. We find that F1 > F2. The body of mass m presses the surface by a force m1g, and the surface exerts normal reaction R1 on it equal to m1g. Thus F1 ∝ R1 or F ∝ m1g.
Similarly F2 ∝ R2

4. The force of friction depends on the nature of smoothness or the state of polishing of the surfaces in contact.

Let us have a wooden plank on which blocks of wood, copper, and glass of the same mass are placed. These are pulled by a spring balance. We find that spring balance shows different readings for just starting the motion.
Hence we conclude that friction depends on the nature of the surfaces in contact.

5. The force of friction is independent of the area of the surface of contact of two bodies as long as the normal reaction is constant.

The force applied to slide the body is the same but the contact area is different for the wooden block placed on a wooden plank.

1. It helps in walking, talking, writing, sleeping, etc.
2. The brakes of vehicles can’t work without friction.
3. Moving belts remain on the rim of wheels due to friction.
4. Adhesives work due to friction.
5. Cleaning with sandpaper is due to friction.
6. Nuts and bolts are held together due to friction.

1. A significant amount of energy of a moving object is wasted in the form of heat energy to overcome friction.
2. It causes a lot of wear and tear of the parts of machinery in contact, thus reducing their lifetime.
3. It restricts the speed of moving vehicles like buses, airplanes, trains, etc.
4. The efficiency of machines decreases due to the presence of a force of friction.
5. A machine gets burnt due to the force of friction between its different moving parts due to friction.

(c) What are various methods of reducing friction?
Reduction of friction saves a lot of energy. The various methods of reducing friction are listed below:
1. Making the surfaces smooth by various methods such as:

• Polishing the surfaces and rubbing surfaces.
• Covering the surface with smooth materials such as metal foils or sun mica etc.
• Lubricating the surfaces by lubricating oils and other materials such as granite.

2. With the help of these above methods, the irregularities, grooves, etc. are filled up and the surface becomes smooth. For example in many machines grease is used to lower friction.

3. Streamlining the shape of bodies: The shape of bodies is made streamline to reduce friction. For example, the shape of high-speed buses, railway engines, ships, boats, airplanes, etc. is made of streamline to reduce fluid friction. It not only reduces friction but also helps in providing the pushing force.

4. Making changes to convert sliding into rolling friction: Since the rolling friction is much less than the sliding friction, several machine parts are designed in such a manner that they have rolling friction. For example, ball bearings are used in bicycle and machine shafts. The vehicles have wheels instead of sliding arrangements for the same reason.

Question 6.
(a) Discuss the motion of a vehicle on a level road having a circular turn.
Consider a vehicle (say a car) of mass m moving with a constant speed v on a circular level (flat) road of radius r. While taking the round, the tires of the car tend to leave the road and move away from the center of the curve. So the forces of friction f1 and f2 act inward on the two tires 1 and 2 respectively shown in fig. If R1 and R2 be the normal reactions of the ground on the tires, then
f1 = μR1 and
f2 = μR2

If f be the total force of friction, then
f = f1 + f2
= μ(R1 + R2) = μR ….(1)

where R = R1 + R2 is the total reaction of the ground on the car.
Also R = mg (weight of the car) …. (2)

∴ From (1) and (2), we get
f = μmg …. (3)

This total- force of friction provides the necessary centripetal force $$\frac{m v^{2}}{r}$$ to the vehicle to move in the circular path.

Equation (4) gives the maximum speed with which the vehicle can move on the given circular road without skidding.

If its speed becomes more than the speed given by (4), then the centripetal force needed by the car will not be provided by the force of friction and hence the vehicle will skid and go off the road.

(b) What is the need for the banking of circular roads/tracks? Describe the motion of a car on a banked road.
We know that a vehicle cannot remain on a circular road if its speed is more than the speed given v = $$\sqrt{µrg}$$. In such a case, the force of friction is not sufficient to provide the necessary centripetal force. The additionally required centripetal force is obtained by banking the circular road i.e. its outer edge is raised a little above the inner edge by a suitable angle θ with the horizontal. This process is called banking and the angle is called the angle of banking.

Case I:
The motion of the car without taking into account the force of friction between the tires of car and the road:
Let OX = horizontal line
m = mass of the car
V = its velocity

The following force acts on the car:

1. Its weight mg acts vertically downwards.
2. The normal reaction of the road acts at an angle θ to the vertical.

Let us resolve R into two rectangular components:
R cos θ which is equal and opposite to mg
i.e. R cos θ = mg …(1)
R sin θ which acts towards the centre of the circular path provides the necessary centripetal force $$\left(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\right)$$ to the car i.e.

eqn. (3) gives the maximum safe speed of the car.

If it negotiates the curve on the banked road with a speed greater than that given by equation (3), then it will skid away.

Case II:
The motion of a car on a banked road taking into account the friction between the tires and the road.
The following forces act on the car:

1. Its weight mg acting vertically downwards.
2. Reaction R of the road acts at an angle θ to the verticle.
3. The force of friction between the tires and the road acts along the road inwardly.

Let us resolve f and R into rectangular components along with horizontal and vertical directions as shown in the figure.

Necessary centripetal force required for motion around the circular path i.e.

R sin θ + f cos θ = $$\frac{\mathrm{mv}^{2}}{\mathrm{r}}$$ ….(5)

Dividing (5) by (4), we get

Dividing on L.H.S. by R cos θ in numerator and denominator, we get

Equation (6) gives the maximum speed for safely negotiating a curve on a banked road.

(c) Describe the motion of a body in a vertical circle.
When a body moves in a vertical circle, its velocity decreases from lowest point L to highest point H. Then it again increases as it moves from H to L. The.reasOn is that gravity opposes the motion of the body as it rises and helps in motion as it falls. Thus, motion is not uniform. Let m = mass of a body tied to one end of a string of length r and is rotated in a vertical circle of center O. L, H are lowest and highest points respectively.

Let v1 and v2 be the velocities of the body at L and H respectively which act along the tangent to the circle. mg= weight of the body which acts vertically downward both at L and H.

Also, let T1 and T2 be the tensions in the string at L and H respectively and both act towards O. Thus, the net force at L towards O is T1 – mg and provides the necessary centripetal force i.e.

Similarly, T2 + mg at H provides the necessary centripetal force at H i.e.

Subtracting Egn. (2) from (1), we get

According to the law of conservation of energy,
Total energy at L = Total energy at H

i. e. in a vertical circle, the difference in tensions at the lowest and highest points is six times the weight of the body.
For completing the vertical circle successfully, the string must remain tight even at H i.e. it should not slack at H.

For this, T2 must be zero i.e. T2 = 0.

∴ From eqn. (2), mg = $$\frac{\mathrm{mv}_{2}^{2}}{\mathrm{r}}$$
or
v2 = $$\sqrt{gr}$$ ….(6)

Eqn. (6) gives the minimum velocity of the body at the highest point i.e. vmin = (rg)1/2

If v2 < $$\sqrt{gr}$$, then the string will slack at H and the vertical circle will not be completed.
From Eqn. (4) and (6), we get

Equation (7) gives the minimum velocity of projection required at the lowest point L so that the body just completes the vertical circle successfully. This phenomenon is called looping the vertical loop.

Numerical Problems:

Question 1.
A machine gun placed horizontally in front of a target fires 600 bullets per minute into the target. The average mass of the bullet is 50 gm and the bullets are fired at a speed of 800 ms-1. Calculate the average force applied to the target.
Mass of each bullet, m = 50 gm = $$\frac{1}{20}$$ kg.
v = velocity of each bullet = 800 ms-1
No. of bullet fired, n = 600,
t = 60s,
average foce exerted on the target = fav = ?
∴ p1 = initial momentum of each bullet
= mv = $$\frac{1}{20}$$ × 800 = 40 kg ms-1

∴ Total momentum of 600 bullets per min.,
Pi = nP1
= 600 × 40 = 24000 kg ms-1

As the bullets come to rest after striking the target,
∴pf = final momentum of bullets = 0

Thus change in momentum ΔP = pi – pf = 24000 kg ms
But by definition,

Impulse = average force × time
= change in momentum
or
Fav × t = Δp

∴ Fav = $$\frac{\Delta \mathrm{p}}{\mathrm{t}}=\frac{24000}{60}$$

Question 2.
A rubber ball of mass 50 gm falls from a height of 1 m and rebounds to a height of 0.5 in. Finds the impulse and average force between the ball and the ground if the time for which they are in contact is 0.1 s.
m = mass of ball = 50gm = $$\frac{50}{1000}=\frac{1}{20}$$ kg, J =?. F =?

Case I:
u = initial velocity of ball = O
h1 = height through which it falls 1m
y = final velocity of the ball =?
a = g = 9.8 ms-2
∴ using the relation,

If p1 be the momentum of the ball when it is about to reach the ground or before the impact,
then
p1 = mv = $$\frac{1}{20}$$ × 4.43 = 0.221 kg ms-1

Case II: u’ initial velocity = velocity after rebounding = ?
v = final velocity = 0
S = h = height up to which it rises = 0.5 m = $$\frac{1}{2}$$ m
a = – g = – 9.8 ms-2

∴ using equn. (i), we get

If p1 = momentum of the ball after impact, then
p2 = mu = $$\frac{1}{20}$$ × 3.13 = 0.156 kg ms

∴ impulse is given by

Question 3.
A bus moving with a velocity of 60 kmh-1 has a weight of 50 tonnes. Calculate the force required to stop it in 10s.
m = mass of bus = 50 tonnes = 50 × 1000 kg
u = initial velocity of bus = 60 kmh-1 = 60 × $$\frac{5}{18}$$ ms-1 = $$\frac{50}{3}$$ ms-1
v = final velocity of bus = 0
t = 10 s, a = ? and retarding force, F = ?

Question 4.
A hunter has a machine gun that can fire 50 g bullets with a velocity of 150 ms-1. A 60 kg tiger springs at him with a velocity of 10 ms-1. How many bullets must the hunter fire into the tiger so as to stop him in his track?
m = mass of bullet = 50 gm = 0.050 kg
M = mass of tiger = 60 kg
v = velocity of bullet = 150 ms-1
V = velocity of tiger = – 10 ms-1
(∵ it is coming from opposite direction) .

Let n = no. of bullets fired per second at the tiger so as to stop it.
∴ pi = 0 before firing …. (i)
pf = n(mv) + MV …. (ii)

∴ According to the law of conservation of momentum,

Question 5.
A mass of 200 kg rests on a rough inclined plane of angle 300. If the coefficient of limiting friction is $$\frac{1}{\sqrt{3}}$$ find the greatest and the least forces in newton, acting parallel to the plane to keep the mass ¡n equilibrium.
Here, m = mass of body = 200 kg
Let angle of inclination = θ
μs = coefficient of limiting friction = $$\frac{1}{\sqrt{3}}$$

The rectangular components of mg are as shown in fig.

Here mg sin θ acts along the plane in the downward direction and is given by

(i) the least forces in newton, acting parallel to the plane to keep the mass in equilibrium. given by
f21 =mg sinθ – F = 980 – 9800 = 0

(ii) The greatest force to be applied to keep the mass in equilibrium is given by
f2 = mg sin θ + F = 980 + 980 = 1960 N.

Question 6.
Find the force required to move a train of 2000 quintals up an incline of 1 in 50, with an acceleration of 2 ms-2, the force of friction being 0.5 newtons per quintal.
Here, m = 2000 quintals
= 2000 × 100 kg (v 1 quintal = 100 kg)
sin θ = $$\frac{1}{50}$$ , acceleration, a = 2 ms-2

F = force of friction
= 0.5 N per quintal
= 0.5 × 2000 = 1000 N

In moving up an inclined plane, force required against gravity
= mg sin 0
= 2000 × 100 × 9.8 × $$\frac{1}{50}$$
= 39200N.
Also if f = force required to produce acce. = 2 ms-2.
Then f = ma = 200000 × 2 = 400000 N

∴Total force required = F + mg sin θ + f
= 1000 + 39200 + 400000 = 440200 N.

Question 7.
A bullet of mass 0.01 kg is fired horizontally into a 4 kg wooden block at rest on a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.25. The bullet remains embedded in the block and the combination moves 20 m before coming to rest. With what speed did the bullet strike the block?
Here, m1 = mass of the bullet = 0.01 kg
m2 = mass of the wooden block = 4 kg
μ2 = coefficient of kinetic friction = 0.25
initial velocity ofblock, u2 = 0, s = distance moved by combination=20 m

Let u1 = initial velocity of the bullet
If v = velocity of the combination, then according to the principle of conservation of linear momentum,

If F = kinetic force of friction,
Then

Then retardation ‘a’ produced is given by

Question 8.
A force of 100 N gives a mass m1 an acceleration of 10 ms-2, and of 20 ms-2 to a mass m2. What acceleration would it give if both the masses are tied together?
Let a = acceleration produced if m1 and m2 are tied together.
F = 100 N, Let a1 and a2 be the acceleration produced in m1 and m2 respectively.
∴ a1 = 10 ms-2, a2 = 20 ms-2 (given)

Question 9.
A balloon with mass M is descending down with an acceleration ‘a’ < g. What mass m of its contents must be removed so that it starts moving up with an acceleration ‘a’?
Let F = retarding force acting on the balloon in the vertically upward direction.

When the balloon is descending down with an acceleration ‘a’, then the net force acting on the balloon in the downward direction is given
by
Ma = Mg – F
or
F = Mg – Ma ….(i)
When the mass m is taken out of the balloon, then its weight is
= (M – m)g

Now as the balloon is moving upward with acceleration ‘a’, so the net force acting on the balloon in the upward direction is given by:

Question 10.
Three blocks are connected as shown below and are on a horizontal frictionless table. They are pulled to right with a force F = 50 N. If m1 = 5 kg, m2 = 10 kg and m3 = 15 kg, find tensions T3 and T2.
Here, F = 50 N, m1 = 5 kg, m2 = 10 kg m3 = 15 kg.

As the three blocks move with an acceleration ‘a’

To determine T2: Consider the free body diagram (1). Here F and T2 act towards the right and left respectively.

As the motion is towards the right side, so according to Newton’s Second law of motion:

To determine T3: Consider the free body diagram (2)

Question 11.
A monkey is descending from the branch of a tree with constant acceleration. If the breaking strength is 75% of the weight of the monkey, then find the maximum acceleration with which the monkey can slide down without breaking the branch.
Let a = constant acceleration of the monkey descending from the branch of tree = T

∴ According to Newton’s Second law of motion,

∴ from (1) and (2), we get

Question 12.
A body of mass 1 kg initially at rest explodes and breaks into three fragments of masses in the ratio 1:1:3. The two pieces of equal mass fly off perpendicular to each other with a speed of 30 ms-1 each. What is the velocity of the heavier fragment?
Let m1, m2, and m3 be the masses of the three fragments

Also let v and v2 be the velocities of the two fragments. ..
∴ v1 = v2 = 30 ms-1
If v3 be the velocity of the heaviest fragment, then according to the law of conservation of linear momentum in the horizontal direction,

Question 13.
A mass of 4 kg is suspended by a rope of length 30 m from the ceiling. A force of 30 N in the horizontal direction is applied at the mid-point of the rope. What is the angle, the rope makes with the vertical in equilibrium? Take g = 10 ms-2 and neglect the mass of the rope.
OA = OB 1.5 m
Let ∠AOC = 0
F = horizontal force = 30 N
m = mass of the body = 4 kg
mg = 4 × 10 = 40 N

Let T1 and T2 be the tensions in parts OA and OB respectively, where O = midpoint of the rope
∴ T2 = mg = 40 N
At point O, three forces T2, T1, and 30 N are acting.

Let us resolve T1 into rectangular components.
Horizontal component of T1 = T1 sin θ
Vertical component of T1 = T1 cos θ
Now T1 cos θ =40 ….(i)
and T1 sin θ = 30 …. (ii)

Dividinig (ii) by (i), we get

tan θ = $$\frac{3}{4}$$ = 0.75 = tan 36.87°
θ = 36.87°.

Question 14.
An elevator weighs 4000 kg. when the upward tension in the supporting cable is 48000 N, what is the upward acceleration? Starting from rest how far does it rise in 3s?
Weight of elevator,
W = Mg = 4000 kgf = 4000 × 9.8 N
= 39200 N
mass of elevator, M = 4000 kg

F = Force on elevator in upward direction = upward tension in the cable = 48000 N.
If F’ = Net force on the elevator in the upward direction
Then F’ = F – W = 48000 – 39200 = 8800 N

If a = upward acceleration,
Then a = $$\frac{\mathrm{F}^{1}}{\mathrm{M}}=\frac{8800}{4000}=\frac{11}{5}$$ = 2.2 ms-2

Let S be the distance covered by the elevator in 3s. For upward motion, we have
u = 0,a = 2.2ms-2, t = 3s

Question 15.
A disc of mass 10 gm is kept floating horizontally by throwing 10 marbles per second against-it from below. If the mass of each marble is 5 gm, calculate the velocity with which the marbles are striking the disc. Assume that the marbles strike the disc normally and rebound downward with the same speed.
When each marble hits the disc, it produces an impulsive force and acts against the weight of the disc. This force keeps the disc floating for $$\frac{1}{10}$$ s.

Let v = velocity with which each marble strikes the disc and rebounds downward.
Impulse imparted by each marble = charge in momentum
= mv – (- mv) = 2 mv

Since m = 5g = 5 × 10-3 kg
∴ Impulse = 2 × 5 × 10-3v
= 10-2 v kg ms-1.

Now average force between the disc and each marble = weight of the disc.
The resultant downward force produces the change in momentum.

Question 16.
A box is placed on a horizontal measuring scale that reads zero when the box is empty. A stream of pebbles is then dropped in to the box from a height h above its bottom at the rate of n pebbles per second. Each pebble has a mass m. If the pebbles collide with the box such that they immediately come to rest after collision. Find the scale reading at time t after the bubble begins to fill the box.
Let v = speed the pebble when it strikes the box.
Then using v2 – u2 = 2as, we get
v2 – 0 = 2gh (∵ here a = g, S = h, u = 0)
or
v = $$\sqrt{2gh}$$

Mass of each pebble = m
Momentum of each pebble = mv = m $$\sqrt{2gh}$$
Each pebble comes to rest after colliding the box

∴ Change in momentum during each collision = m $$\sqrt{2gh}$$
Δt = Time for each colbsion = $$\frac{1}{n}$$s

Force exerted by pebbles on the box,

Weight of pebbles after time t is

Question 17.
A stone of mass 5 kg falls from the top of a cliff 40 m high and buries itself 2 m deep in the sand. Find the average resistance offered by the sand and the time it takes to penetrate (g = 9.8 ms-2).
Let the acceleration of the stone in the sand be “a”.
The velocity it gains by falling through a height of 40 m is
v = $$\sqrt{2gh}$$ = $$\sqrt{2 \times 9.8 \mathrm{~ms}^{-2} \times 40 \mathrm{~m}}$$
= 28 ms-1 = u = initial velocity

When it buries in the sand its final velocity is zero so the acceleration
a = $$\frac{0^{2}-\left(28 \mathrm{~ms}^{-1}\right)^{2}}{2 \times 2 \mathrm{~m}}$$ = -196 ms-1

(i) The average resistance offered by the sand is
F = ma = 5 kg × 196 ms-2 = 980 N

(ii) Let t be the time of penetration in the sand
using equation

Question 18.
An airplane required a speed of 80 km h-1 for takeoff on a 100 m long runway. The coefficient of friction between the tires of the plane and the runway is 0.2. Assuming that the plane accelerates uniformly, what is the force required by the engine to take off the plane? The mass of the airplane is 10,000 kg.
Take off speed of plane, v = 80 km h-1
Run-on the ground, s = 100 m
u = initial velocity = 0
Acceleration ‘a’ of the plane is obtained from

The force to produce this acceleration = ma= 10,000 × $$\frac{200}{81}$$ N
Force needed to overcome friction = μm = μmg
= 0.2 × 10,000 × 9.8 N

The force required to be given to the engine is

Question 19.
A 4 m long ladder of aluminum having 20 kg mass rests against a smooth wall making an angle of 30° with it. Find the coefficient of friction between the ladder and the ground so that ladder does not slip. Take g = 10 ms4.
Let the ladder AB of length 4 m rest against smooth wall BC. The reaction of the ground is say R1 and that of the wall on the ladder is R2. The weight of the ladder acts at the midpoint O in a vertically downward direction (assuming the ladder to be uniform).

Let F be the force of friction.
Taking moments of mg and R2 about A, we get
mg × AD = R2 × BC

Question 20.
Prove that a motor car moving over a convex bridge is lighter than the same car resting on the same bridge.
The motion of the motor car over the convex bridge is the motion along the segment of a circle. The centripetal force is provided by the difference of weight mg of the car and the normal reaction R of the bridge.

∴R = apparent weight of the moving car
Clearly R < mg i.e. the weight of the moving car is less than the weight of the stationary car.

Question 21.
A hammer weighing 1 kg moving with a speed of 10 ms-1 strikes the head of a nail driving it 10 cm into a wall. Neglecting the mass of the nail, calculate:
(a) the acceleration during the impact.
(b) the time interval during the impact,
(c) the impulse.
Here, the initial velocity of nail = velocity of the hammer
i. e. u = 10 ms-1
Final velocity of nail, v = 0, m = mass of hammer = 1 kg
Distance covered, s = 10 cm = 0.1 m

Let t be the time interval for which impact lasts and let a be the acceleration produced =?
(а) Using the relation,

(b) Also using the relation,

(c) Impulse
Ft = (ma)t
= 1 × (-500) × 0.02
= – 10 Ns.

Question 22.
A rocket with a lift of mass 20,000 kg is blasted upward with an initial acceleration of 5 ms2. Calculate internal thrust of the blast
Here, m = 20,000 kg
a = 5 ms-1
∴ F = ma = 20,000 × 5N = 105 N

Let T = upward of thrust and
W = weight of rocket.

Question 23.
A car starts from rest and accelerates uniformly with 2ms2. At t = 10 s, a stone is dropped out of the window (1m high) of the car. What are the : (a)velocity, (b) acceleration of the stone at t = 10.ls. Neglect air resistance and take g = 9.8 ms2.
(a) Here, the stone will possess horizontal velocity due to the motion of the car and vertical downward velocity due to gravity.
Here u = 0, a = 2 ms-2, t = 10s, g = 9.8 ms-2

If vH and vy be its horiìontal and vertical velocities respectively, then using relation,
v = u + at, we get
vH = u + at = 0 + 2 × 10 = 20 ms-1
and
vy = u + gt = 0 + 9.8 × 0.1 = 0.98 ms-1

If $$\overrightarrow{\mathbf{v}}$$ be its resultant velocity at < θ with horizontal direction, then

The vertical component of acceleration,
g = 9.8 ms-2

Since the velocity is constant as no force acts on the stone at the time of dropping, so the horizontal component of acce. is zero.

If a’ be its acce. at 10.1 s, then
a’ = $$\sqrt{\mathrm{g}^{2}+0^{2}}$$ = g = 9.8 ms-2.

Question 24.
A stream of water flowing horizontally with a speed of 15 ms-1 gushes out of a tube of cross-sectional area 10-2 m2 and hits at a vertical wall nearby. What is the force applied on the wall by the impact of water, assuming it does not rebound? The density of water is 1000 kg m-3.
Here, A = area of cross-section of tube
= 10-2 m2
v = speed of water = 15 ms-1
ρ = density of water = 103 kg m-3

If V be the volume of water striking the wall per second, then
V = Av = 10-2 × 15 = 0.15 m3 s-1

Let m be the mass of water striking/sec.
∴ m = Vρ = 0.15 × 1000 = 150 kg s-1

Let pi = initial momentum of water striking the wall s-1,
∴ Pi = mv
= 150 × 15 = 2250 kg ms-2

pf = final momentum of water/sec = 0 as it does not rebound
∴ Δp = change in momentum/s = pf – Pi
= 0 – 2250 = – 2250 kg ms-2
or
F = 2250 N.

Question 25.
A body of mass 1 kg lies on a rough horizontal plane. A horizontal force of 15 N produces in the body an acceleration of 10 ms”2. Find the force of friction and coefficient of friction between the body and the table.
Here, m = mass of body = 1 kg
F = force = 15 N
a = acceleration = 10 ms-2

Let f be the force of friction and μ, be the coefficient of friction
∴ the net force on the body = F – f

Value-Based Type:

Question 1.
Anil is a student of the science stream who lived on the first floor of a building. One day he and his grandfather were enjoying the holiday. Suddenly he observed that dense smokes are coming from his neighbor’s flat. People were crying and panic was there. There was a rush on the staircase. He was thinking about how he could save his grandfather. Suddenly an idea came to his mind. He took a turban cloth and suspended it on the ground and asked a person to hold the other end of the cloth tightly; Then he asked his grandfather to gently sit on the turban cloth. He was now able to comfortably reach the ground:
(i) What values and qualities displayed by Anil?
The values displayed by Anil are Caring, Presence of mind, Courageous, and affection to elderly people.

(ii) Which concept is being used here?
Motion on a Rough inclined plane

(iii) What will be the tension in the cloth inclined at an angle of 30° from horizontal when a person of mass 60 kg falls through it with an acceleration Of 2 m/s2

Here, m = 60 kg, θ = 30°, a = 2m/s2, g = 10 m/s2
∴ Tension T =mg sin θ – ma
= m(g sin θ – a)
= 60(10 × sin30°- 2)
= 60 × (10 × $$\frac{1}{2}$$ – 2)
= 60 × 3 = 180N

Question 2.
Two friends Vipul and Mohan are cycling at 18 km/h on a level road. Vipul told that there is a sharp circular turn of the radius of 3 m. So, reduce your speed to avoid slipping. But Mohan was a careless boy and he did not follow his advice. Suddenly he slipped and felt the importance of his friend’s idea.
(i) Which values are depicted by Vipul?
Cooperative, caring, awareness, and helping nature.

(ii) Explain why the cyclist Mohan slip while taking the turn?
[Take us = 0.1 ]
If the speed is too large or if the turn is too sharp (i.e. of too small a radius) or both, the frictional force is not sufficient to provide the necessary centripetal force, and the cyclist slips. The condition for the cyclist not to slip is given by:

Hence, the condition is not obeyed. So, the cyclist will slip while taking the circular turn.

Question 3.
Mahesh was driving a car with his old grandfather. After some time when the destination was about to come, he did not apply the brakes. But he stopped the engine. Even then the car was running on the road for some time. His grandfather surprised and asked his grandson the reason for the car running without the engine on. Mahesh was a student of the science stream of class XI. He explained that it is due to the momentum of the car.
(i) Which values are displayed by Mahesh here?
The values displayed by Mahesh are:
Intelligence, awareness, helping nature, and willingness to explain the scientific reason, (i.e explanation).

(ii) What is momentum and on which factors it depends?
The momentum of a body is defined as the product of its mass and velocity.
i.e Momentum = mass × velocity
or
P = m × v
Momentum is directly proportional to mass as well as its velocity.
So, Momentum depends on the mass and the velocity of a moving body.

Question 4.
Suresh noticed a big Granite Rock in his locality. He thought that if they worked upon it they could earn money. He took permission from the Government, completed all the formalities. He broke the Rock using a bomb. The rock was made into slices. They established a Granite industry. Many of the people in the surroundings started to earn and live comfortably.
(a) What values of Suresh impress you?
Suresh knows how to utilize the natural resources, has got concerned for others. Also, he knows how to complete all legal formalities before taking up any work.

(b) A bomb is thrown in a horizontal direction with a velocity of 50 m/s. It explodes into two parts of masses 6 kg and 3 kg. the heavier fragment continues to move in the hori¬zontal direction with a velocity of 80 m/s. Calculate the velocity of the lighter fragment.
Solution: r
According to low of conservation of momentum
Total momentum of fragments = Momentum of the Bob
m1 v1 + m2 v2 = M V
⇒ 6 × 80 + 3 × V2 = 9 × 50
⇒ V2 = -10 m/s.

Question 5.
Rakesh with an intention to win in the interschool sports practiced the high jump every day for about a month. He participated and won I position in the interschool sports.
(a) Comment upon the values Rakesh possesses.
Rakesh has the determination, he plans and executes his plan accordingly.

(b) Why does an athlete run some steps before taking a jump?
An Athlete runs some steps before taking a jump to gain some initial momentum, which helps him to jump more?

## Motion in a Plane Class 11 Important Extra Questions Physics Chapter 4

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 4 Motion in a Plane. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

## Class 11 Physics Chapter 4 Important Extra Questions Motion in a Plane

### Motion in a Plane Important Extra Questions Very Short Answer Type

Question 1.
Under what condition |a + b| = |a| + |b| holds good?
When a and b act in the same direction i. e. when 0 = 0 between • them, then |a + b|=|a| + |b|.

Question 2.
Under what condition |a – b| = |a| – |b| holds good?
The condition |a – b|=|a| – |b| holds goods when a and b act in the opposite direction.

Question 3.
The sum and difference of the two vectors are equal in magnitude
i. e. |a + b|=|a – b|. What conclusion do you draw from this?
The two vectors are equal in magnitude and are perpendicular to each other.

Question 4.
What is the angle between $$\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}$$ and $$\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}}$$?
The given vectors act along two parallel lines in opposite directions i.e. they are anti-parallel, so the angle between them is 180°.

Question 5.
What is the minimum number of coplanar vectors of different magnitudes which can give zero resultant?
3, If three vectors can be represented completely by the three sides of a triangle taken in the same order, then their resultant is zero.

Question 6.
When a – b = a + b condition holds good than what can you say about b?
For a – b = a + b condition to hold good, b must be a null vector.

Question 7.
What is the magnitude of the component of the 9î – 9ĵ + 19k̂ vector along the x-axis?
9.

Question 8.
Can displacement vector be added to force vector?
No.

Question 9.
What is the effect on the dimensions of a vector if it is multiplied by a non-dimensional scalar?
There is no effect on the dimensions of a vector if it is multiplied by a non-dimensional scalar.

Question 10.
(a) What is the angle between î + ĵ and î vectors?
45°

(h) What is the angle between î – ĵ and the x-axis?
45°

(c) What is the angle between î + ĵ and î – ĵ?
90°

Question 11.
What is the dot product of 2î + 4ĵ + 5k̂ and 3î + 2ĵ + k̂?
19.

Question 12.
What must be the value of ‘a’ in 2î + 2ĵ – ak̂ so that it is perpendicular to 5î + 7ĵ – 3k̂?
8.

Question 13.
Is finite rotation a vector quantity? Why?
No. This is because the addition of two finite rotations does not obey commutative law.

Question 14.
Is infinitesimally small rotation a vector quantity? Why?
Yes. This is because the addition of two infinitesimally small rotations obeys commutative law.

Question 15.
(a) Can the resultant of two vectors of different magnitudes be zero?
No

(b) Can the magnitude of the rectangular component of a vector be greater than the magnitude of the vector itself?
No.

Question 16.
A quantity has both magnitude and direction. Is it necessarily a vector? Why? Give an example.
No. The given quantity will be a vector only if it obeys the laws of vector addition. Electric current.

Question 17.
In a vector equation, all the quantities are of similar nature but their directions are different. Does it mean that the vector equation is necessarily incorrect? Electric current.
No. In a vector equation, the vectors need not be in the same direction.

Question 18.
Why vectors cannot be added algebraically?
The vectors cannot be added algebraically because they possess both magnitude and direction.

Question 19.
Fifty vectors each of magnitude 10 units are completely represented by the sides of a polygon in the same order. What will be the resultant?
Their resultant will be zero. This is because the vector sum of all the vectors represented by the sides of a closed polygon taken in the same order- is zero.

Question 20.
What will be the angle between $$\vec{A}$$ and $$\vec{B}$$ if $$\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}=\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{A}}$$?
The angle between A and B may be of any value if A + B = B + A.

Question 21.
How will you prove that the given vectors are neither parallel nor perpendicular?
The given vectors will neither be parallel nor perpendicular if neither their cross product nor their dot product is zero.

Question 22.
How will you prove that the two vectors are parallel?
If their cross product is zero, then the two vectors are said to be parallel.

Question 23.
How will you prove that two vectors are perpendicular?
Two vectors will be perpendicular to each other if their dot product is zero.

Question 24.
(a) What is the minimum possible resultant of two forces 2N and 1N?
1N

(b) What is the maximum possible resultant of two forces 2N and 1N?
3N.

Question 25.
P2 is scalar because it is the dot product of P and P.

Question 26.
Is it necessary to mention the direction of a vector having zero magnitudes? Why?
No. A vector having zero magnitudes is called a null vector. In this case, there is no sense of direction as the null vector is represented by a point.

Question 27.
Can a vector vary with time? Give example.
Yes. For example, when an object moves, its position vector continuously changes with time.

Question 28.
Define a projectile.
It is defined as an object thrown with some initial velocity moves under the effect of gravity alone.

Question 29.
Define trajectory.
It is defined as the path followed by the projectile during its flight.

Question 30.
Define maximum height attained by a projectile.
It is defined as the maximum vertical distance traveled by the projectile during its journey.

Question 31.
Define the horizontal range of the projectile.
It is defined as the maximum horizontal distance between the point of projection and the point in the horizontal plane where the projectile hits it back after the journey.

Question 32.
At what angle a ball must be thrown to get maximum horizontal range?
It must be thrown at an angle of 45° to the horizontal.

Question 33.
At what point in its trajectory does a projectile have its minimum velocity?
At the highest point of the trajectory, a projectile has its minimum velocity.

Question 34.
How much is the velocity of the projectile at its highest point along the vertical direction?
It is zero.

Question 35.
When a projectile is fired at an angle with the horizontal, then which of the components of its velocity remains constant throughout the trajectory?
The horizontal component of its velocity remains constant throughout the trajectory.

Question 36.
At what point of the trajectory of a ball thrown upward is the acceleration perpendicular to the velocity?
The acceleration is perpendicular to the velocity at the highest point of the trajectory.

Question 37.
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2m long is set into oscillation. The speed of the bob at its mean position is 1 ms-1. What is the trajectory of the bob if its string is cut when the bob is: (a) at one of its extreme position (b) at its mean position.
(a) straight line,
(b) parabolic path.

Question 38.
A cannon on a level plane is aimed at an angle P above the horizontal and a shell is fired with a muzzle speed v towards a vertical cliff a distance x away. At what height y from the bottom, the shell would hit the cliff?
y = x tan β – $$\frac{\mathrm{gx}^{2}}{2 \mathrm{v}^{2} \cos ^{2} \beta}$$

Question 39.
A body is projected so that it has maximum range R. What is .the maximum height reached during the flight?
Maximum height is $$\frac{1}{4}$$ of maximum range i.e. h = $$\frac{R}{4}$$.

Question 40.
A bullet x is fired from a gun when the angle of elevation of the gun is 30°. Another bullet y is fired from the gun at an angle of elevation 60°. Tell which of the two bullets would have a greater horizontal range?
Both will have the same range.

Question 41.
A particle moves in a plane with uniform acceleration ¡n a direction different from its initial velocity. What is the nature of the path followed by it?
It is a parabolic path in nature.

Question 42.
A projectile of mass m is fired with initial velocity u at angle 6 with the horizontal. What is the change in momentum as it rises to the highest point of the trajectory?
Change in momentum = Force × time = mg × $$\frac{\mathrm{u} \sin \theta}{\mathrm{g}}$$ = mu sinθ

Question 43.
Name the five physical quantities which change during the motion of an oblique projectile.
Five physical quantities are velocity, the vertical component of velocity, momentum, kinetic energy, potential energy.

Question 44.
Name the two quantities which would be reduced if air resistance is taken into account in the study of the motion of the oblique projectile which quantity would be increased?
The quantities which are reduced:

1. maximum height
2. horizontal range.

The angle at which the projectile hits the ground would be increased.

Question 45.
At which point of the trajectory of the projectile, the speed is maximum?
The speed is maximum at two points which are:

1. The point from where the projectile is thrown.
2. The point where the projectile returns back to the plane of projection.

Question 46.
What is the condition for the following formula to be valid?
Average velocity = \frac{\text { Initial velocity + final velocity }}{2}
It is valid only if acceleration is constant.

Question 47.
A body travels with uniform acceleration ai for time t, and with uniform acceleration a2 for time t2. What is the average acceleration?
aaverage = $$\frac{a_{1} t_{1}+a_{2} t_{2}}{t_{1}+t_{2}}$$.

Question 48.
Even when rain is falling vertically downwards the front screen of a moving car gets wet. On the other hand, the back screen remains dry. Why?
The rain strikes the car in the direction of the relative velocity of the rain with respect to the car.

Question 49.
Can a body have a constant velocity and still have a varying speed?
No. If the velocity of a body is constant then its speed cannot vary.

Question 50.
(a) What are the units of angular speed?

(b) What is the relation between time period (T) and frequency (v)?
T= $$\frac{1}{v}$$

Question 51.
(a) Give an example of a body moving with uniform speed but having a variable velocity and an acceleration that remains constant in magnitude but changes in direction.
A body having uniform circular motion.

(b) What is the direction of a centripetal acceleration with reference to the position vector of a particle moving in a circular path?
It is always along the position vector but directed towards the center of the circle.

### Motion in a Plane Important Extra Questions Short Answer Type

Question 1.
Name two quantities that are the largest when the maximum height attained by the projectile is largest.
Time of flight and the vertical component of velocity are the two quantities that are the largest when the maximum height attained by the projectile is the largest.

Question 2.
A stone dropped from the window of a stationary railway carriage takes 2 seconds to reach the ground. At what time the stone will reach the ground when the carriage is moving with
(a) the constant velocity of 80kmh-1
(b) constant acceleration of 2ms-2?
The time taken by the freely falling stone to reach the ground is given by
t = $$\sqrt{\frac{2 h}{g}}$$

In both cases, the stone will fall through the same height as it is falling when the railway carriage is stationary. Hence the stone will reach the ground after 2 seconds.

Question 3.
Can a particle accelerate when its speed is constant? Explain.
Yes. A particle can be accelerated if its velocity changes. A particle having uniform circular motion has constant speed but its direction of motion changes continuously. Due to this, there is a change in velocity and hence the particle is moving with variable velocity. Thus particle is accelerating.

Question 4.
(a) Is circular motion possible at a constant speed or at constant velocity? Explain.
Circular motion is possible at a constant speed because, in a circular motion, the magnitude of the velocity i.e. speed remains constant while the direction of motion changes continuously.

(b) Define frequency and time period.
Frequency is defined as the number of rotations completed by a body in one second and the time period is defined as the time taken by an object to complete one rotation.

Question 5.
When the component of a vector A along the direction of vector B is zero, what can you conclude about the two vectors?
The two vectors A and B are perpendicular to each other.
Explanation: Let θ = angle between the two vectors A and B component of vector A along the direction of B is obtained by resolving A i.e. A cos θ.

Now according to the statement
A cos θ = 0
or
cos θ = 0 = cos 90°
θ = 90°
i.e. A ⊥ B

Hence proved.

Question 6.
Comment on the statement whether it is true or false “Displacement vector is fundamentally a position vector.’’ Why?
The given statement is true. The displacement vector gives the position of a point just like the position vector. The only difference between the displacement and the position vector is that the displacement vector gives the position of a point with reference to a point other than the origin, while the position vector gives the position of a point with reference to the origin. Since the choice of origin is quite arbitrary, so the given statement.

Question 7.
Does the nature of a vector changes when it is multiplied by a scalar?
The nature of a vector may or may not be changed when it is multiplied.

For example, when a vector is multiplied by a pure number like 1, 2, 3,…. etc., then the nature of the vector does not change. On the other hand, when a vector is multiplied by a scalar physical quantity, then the nature of the vector changes.

For example, when acceleration (vector) is multiplied by a mass (scalar) of a body, then it gives force (a vector quantity) whose nature is different than acceleration.

Question 8.
Can the walk of a man be an example of the resolution of vectors? Explain.
Yes, when a man walks, he pushes the ground with his foot. In return, an equal and opposite reaction acts on his foot. The reaction is resolved into two components: horizontal and vertical components. The horizontal component of the reaction helps the man to move forward while the vertical component balances the weight of the man.

Question 9.
Explain under what conditions, the resultant of two vectors will be equal to either of them.
The resultant of two vectors will be equal to either of them if:

1. The two vectors are of the same magnitude.
2. The two vectors are inclined to each other at 120°.

Explanation: Let x be the magnitude of each of the two vectors say P and Q. If 9 = angle between P and Q, then the magnitude of their resultant vector (R) is given by the relation

Question 10.
Why the magnitude of the rectangular components of a vector can’t be greater than the magnitude of the vector itself?
The magnitude of rectangular components of a vector itself cannot be greater than the magnitude of the vector itself because the rectangular components of a vector A are Ax = A cos θ and Ay = A sin 9. As sin θ and cos θ both are ≤ 1, so both Ax and Ay cannot be greater than A.

Question 11.
Can a flight of a bird be an example of the composition of vectors? Explain.
Yes, the flight of a bird can be an example of the composition of vector i.e. addition of vectors as is shown in the figure here. As it flies, it strikes the air with its wings W, W along with WO. So, according to Newton’s third law of motion, airstrikes the wings in opposite directions with the same force in reaction represented by OA and OB. Thus according to the parallelogram law of vectors, the resultant of OA and OB is OC. This resultant upward force OC is responsible for the flight of the bird.

Question 12.
Can commutative law be applied to vector subtraction?
Let P and Q be the two vectors.
∴ As P – Q ≠ Q – P, so commutative law is not applied to vector subtraction.

Question 13.
Write down the vector whose head is at (4, 3, 2) and whose tail is at (3, 2, 1).
Let r, and r2 be the position vectors of the points P (4, 3, 2) and Q (3, 2, 1) respectively
∴ r1 = 4î + 3ĵ + 2k̂
and r2 = 3î + 2ĵ + 1k̂

Let QP = Δr be the required vector
Now applying triangle law of vector addition of ΔOQP, we get
r2 + Δr = r1
or
Δr = r1 – r2
= (4î + 3ĵ + 2k̂) – (3î + 2ĵ + 1k̂)
= (î + ĵ + k̂ ).

Question 14.
If A.B = A.C, is it safe to conclude that B = C?
Let θ1, θ2 be the angler between A and B, B and C respectively

So we conclude that it is not safe to conclude that B = C until θ1 = θ2.

Question 15.
Define Tensor. Give example.
It is defined as a physical quantity that has no direction but has different values in different directions. It is neither a vector nor a scalar, e.g. moment of Inertia has no direction but its values are different in different directions. Hence moment of inertia is neither a scalar nor a vector but a tensor. Other examples of the tensor are refractive index, stress, strain, density.

Question 16.
How does a sling work?
It works on the principle of parallelogram law of vectors. A rubber sling is attached to the two ends of a sling. The stone to be thrown is held at point O. When the rubber string is pulled, the tensions are produced along with OA and OB. When the string is released, the stone moves under the effect of resultant force OC.

Question 17.
Why does a tennis ball bounce higher on hills than in plains?
We know that the maximum height of a projectile is given by the relation

where θ = angle of projection
and u = initial velocity of projection.

From (i), we see that the maximum height is inversely proportional to the acceleration due to gravity (g). So smaller the g, greater shall be the maximum height. Since the value of g is lesser on the hills than on the plains, so a tennis ball will bounce more on hills than on plains.

Question 18.
Two particles are moving with equal and opposite velocities in such a way that they are always at a constant distance apart. Calculate the time after which the particles return to their initial positions.
Clearly, the two particles are at the two ends of the diameter of a circular path. It is further clear that each particle will return to its initial position after describing one circle.

If t = required time, then
t = $$\frac{2 \pi r}{v}=\frac{\pi d}{v}$$
where v = speed of particles.
r = radius of the circular track,
d = constant distance between particles.
= diameter of the circular path = 2r.

Question 19.
A ball P is projected directly towards a second ball Q. The horizontal distance x of the second ball is lesser than the horizontal range R of the first ball. The second ball is released from the rest at the instant the first is projected, will the two balls collide.
Yes, the initial elevation of 2nd ball = x tan θ
During time t, the second ball covers a distance = $$\frac{1}{2}$$ gt2

If y = elevation at the instant of collision, then

Thus from (i) and (iv), it is clear that both the balls attain the same elevation at time t. So that two must collide, whatever may be the initial velocity of the first ball.

Question 20.
Which one of the following is greater?
(a) The angular velocity of the hour hand of a watch.
(b) The angular velocity of the Earth around its own axis. Why?
The angular velocity of the hour hand of a watch.is greater than the angular velocity of the earth around its own axis.

Explanation: We know that the angular velocity (w) of an object has
Time period (T) is given by
ω = $$\frac{2 \pi}{\mathrm{T}}$$ ….(i)
(a) T for hour hand of a watch is 12h

Thus clearly angular velocity of the hour hand is more than the angular velocity of the earth around its own axis because of the Tearth > Thour hands.

Question 21.
A bomber in the horizontal flight drops a bomb when it is just above the target. Explain whether the bomb hits the target or misses it?
The bomb will miss the target. This is due to the fact that at the time of dropping, the bomb possesses a horizontal speed equal to the speed of the bomber. So bomb will drop and hit the ground some distance away (= horizontal speed of bomber x time) from the place of dropping.

Question 22.
(a) What is the direction of the area of the vector?
It acts perpendicular to the plane containing the area i.e. it acts normal to the surface area.

(b) What are the characteristics of the motion of an Earth satellite revolving around it in a fixed orbit?
The satellite motion in a fixed orbit is characterized by the following factors:

1. The Earth’s gravitational pull at the height is equal to the centripetal force at that height.
2. The angular velocity of the satellite about the Earth’s center remains constant.
3. The linear speed of the satellite is also constant.

Question 23.
A ball is thrown horizontally and at the same time, another ball is dropped from the top of a tower with the same velocity.
(i) Will both the balls hit the ground with the same velocity?
When the balls hit the ground, their vertical velocities are equal. However, the horizontal velocities will be different. Hence the resultant velocities of the two balls are different, so the balls would hit the ground with different velocities.

(ii) Will both the balls reach the ground at the same time?
Both the balls would reach the ground at the same time because their initial vertical velocities, acceleration, and distances covered are all equal.

Question 24.
Three balls thrown at different angles reach the same maximum height (fig. given), then answer the following:

(a) Are the vertical components of the initial velocity the same for all the balls? If not, which one has the least vertical velocity component?
Since maximum height is directly proportional to the vertical component of initial velocity hence the vertical components of velocity in the three cases are the same as the maximum height attained is the same.

(b) Will they all have the same time of flight?
Yes, all will have same time of flight T = $$\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}$$ as sin θ is same for all.

(c) Which one has the greatest horizontal velocity component?
The projectile having maximum range has the greatest horizontal component of the initial velocity.

Question 25.
When a car is driven too fast around a curve it skids outward. How would a passenger sitting inside explain the car’s motion? How would an observer standing by the roadside explain the event?
The passenger inside the car is in the reference frame of the car so he experiences centrifugal force $$\frac{\mathrm{mv}^{2}}{\mathrm{r}}$$ acting on the car. So to remain in a circular path force should be larger as it is proportional to v12.

An observer on the roadside finds that the centripetal force $$\frac{\mathrm{mv}^{2}}{\mathrm{r}}$$ acting on the car becomes inadequate with an increase in v. Since again v2 r and r is insufficient so the car skids.

Question 26.
Show that the horizontal range of projectile for two angles of projection α and β is same when α + β = 90°.
The horizontal ranges for two angles of projections are

assuming that projection velocity is the same in both cases

Question 27.
Why is it easier to pull a lawn roller than to push it? Explain.
(a) Consider the case of the pull of roller as shown in Fig. (a):
Here the force of pull (F) has two components:
(i) F cos θ which moves the roller horizontally.
(ii) F sin θ which acts vertically upward and opposite to the weight (mg) of the roller.
∴ Net weight of the roller = mg – F sin θ ….(i)

(b) Case of the push of roller: In this case, the components of F are:
(i) F cos θ which helps to move the roller horizontally.
(ii) F sin θ which acts vertically downward in the direction of the weight (mg) of the roller.
∴ Net weight of the roller = mg + F sin θ ….(ii)

Now from equations (i) and (ii), it is clear that the net weight (i.e. apparent weight) of the roller becomes greater in the case of pushing than in the case of pulling. But due to friction, the force required to move a body is directly proportional to the effective weight of the body. So a smaller force is required to move the roller in case of pulling than in case of pushing. Hence it is easier to pull than to push a lawn roller.

Question 28.
(a) Is the maximum height attained by a projectile largest when the maximum range is maximum?
No. the angle for the maximum horizontal range is 45° but for the maximum height the largest range is 41°:

(b) A body in uniform horizontal circular motion possesses variable velocity. Does it mean that the kinetic energy of the body is also variable?
No, since the magnitude of velocity in uniform motion is constant, hence the kinetic energy will also not change.

(c) Why vector cannot be added algebraically?
The algebraic sum does not take into account the direction of a physical quantity. It adds only magnitudes, hence it is not suitable for vector sum.

Question 29.
(a) How can a vector be tripled?
By multiplying the vector by number 3 (a scalar) or by adding two more similar vectors i.e. vectors of the same magnitude and direction to the given vector.

(b) State Triangle law of vector addition.
It states that if two vectors are represented completely (i.e. both in magnitude and direction) by the two sides of a triangle taken in the same order then their resultant is represented completely by the third side of the triangle taken in the opposite order.

Question 30.
State polygon law of vector addition. Show it graphically.
It states that if a number of vectors are represented completely by the sides of a polygon taken in the same order, then their resultant is represented completely by its closing side taken in the opposite order. If P, Q, S, T, and U be the vectors represented completely by the sides OA, AB, BC, CD, and DE of the polygon taken in the same order, then their resultant (R) is represented by OE taken in opposite order s.t.
R = P + Q + S + T + U

Question 31.
Define null vector. What are its properties? What is its physical significance?
It is defined as a vector having zero magnitudes and acting in an arbitrary direction. It is denoted by O.
Properties of null vector:

1. The addition or subtraction of zero vector from a given vector is again the same vector.
i.e. A + O = A
A – O = A
2. The multiplication of zero vector by a non-zero real number is again the zero vector
i.e. n.O = O
3. If n1 A = n2B where n1 and n2 are non-zero real numbers, then the relation will hold good
if A = B = O
i.e. both A and B are null vectors.

The physical significance of null vector: It is useful in describing the physical situation involving vector quantities.
e.g. A – A = O
0 × A = O.

### Motion in a Plane Important Extra Questions Long Answer Type

Question 1.
Discuss the problem of a swimmer who wants to cross the river in the shortest time.
Let vs and vr be the velocities of swimmer and river respectively.
Let v = resultant velocity of vs and vr

1. Let the swimmer begins to swim at an angle θ with the line OA where OA is ⊥ to the flow of the river.
If t = time taken to cross the river, then
t = $$\frac{l}{v_{s} \cos \theta}$$ …(i)

where l = breadth of the river
For t to be minimum, cos 0 should be maximum.
i.e. cos θ = 1

This is possible if θ = 0
Thus, we conclude that the swimmer should swim in a direction perpendicular to the direction of the flow of the river.

2. v = $$\sqrt{v_{s}^{2}+v_{r}^{2}}$$
where v$$\vec{v}$$ is the resultant velocity of vs and vr.

3. tan θ = $$\frac{v_{\mathrm{r}}}{\mathrm{v}_{\mathrm{s}}}=\frac{\mathrm{x}}{l}$$
or
x = l$$\frac{\mathbf{v}_{\mathrm{r}}}{\mathbf{V}_{\mathrm{s}}}$$

4. t = $$\frac{l}{\mathrm{~V}_{\mathrm{s}}}$$

Question 2.
State and prove parallelogram law of vector addition. Discuss some special cases.
It states that if two vectors can be represented completely (i.e. both in magnitude and direction) by the two adjacent sides of a parallelogram drawn from a point then their resultant is represented completely by its diagonal drawn from the same point.

Proof: Let P and Q be the two vectors represented completely by the adjacent sides OA and OB of the parallelogram OACB s.t.
$$\overrightarrow{\mathrm{OA}}$$ = P, $$\overrightarrow{\mathrm{OB}}$$ = Q
or
| $$\overrightarrow{\mathrm{OA}}$$ | = |P|, | $$\overrightarrow{\mathrm{OB}}$$ | = |Q|

θ = angle between them = ∠AOB

If R be their resultant, then it will be represented completely by the diagonal OC through point O s.t. OC = R
The magnitude of R: Draw CD ⊥ to OA produced,

eqn. (vii) gives the magnitude of R.

The direction of R: Let β be the angle made by R with P
∴ in rt. ∠d ΔODC,

Special cases: (a) When two vectors are acting in the same direction:
Then θ = 0°

Thus, the magnitude of the resultant vector is equal to the sum of the magnitudes of the two vectors acting in the same direction, and their resultant acts in the direction of P and Q.

(b) When two vectors act in the opposite directions:
Then θ = 180°

Thus, the magnitude of the resultant of two vectors acting in the opposite direction is equal to the difference of the magnitude of two vectors and it acts in the direction of the bigger vector.

(c) If θ = 90° i.e. if P ⊥ Q,
then cos 90° = 0
and
sin 90° = 1

R = $$\sqrt{P^{2}+Q^{2}}$$
and
tan β = $$\frac{O}{P}$$.

Question 3.
Derive the relation between linear velocity and angular velocity. Also, deduce its direction.
Let R be the radius of the circular path of center O on which an object is moving with uniform angular velocity co. Let v = its linear velocity. Let the object move from point P at time t to point Q at time t + Δt. If r and r + Δr be its position vectors at point P and Q respectively, then

∴ Linear displacement of the particle from P to Q in small time interval Δt = Δr.
Let Δθ = its angular displacement
∴ ω = $$\frac{Δθ}{Δt}$$
or
Δθ = ωΔt ….(1)

Also we know that Δθ = $$\frac{\widehat{\mathrm{PQ}}}{\mathrm{R}}$$ …(2)

∴ from (1) and (2), we get

Now when Δt → 0, then from eqn. (1) Δθ → 0
so arc PQ = $$\widehat{\mathrm{PQ}}$$ = chord PQ

Thus eqn. (3) reduces to

where v = $$\frac{PQ}{Δt}$$ is the linear velocity of the object.

Direction of velocity vector: In isosceles ΔOPQ,

when Δt → 0, ∠QPO → $$\frac{π}{2}$$
i.e. $$\overrightarrow{\mathrm{OP}}$$ tends to become ⊥ to $$\overrightarrow{\mathrm{OP}}$$
or
$$\overrightarrow{\mathrm{OP}}$$ tends to lie along the tangent at P. Hence velocity vector at P is directed along the tangent to the circle in the direction of motion.

Question 4.
What do you understand by the rectangular resolution of a vector? Resolve it into its two rectangular components.
It is defined as the process of splitting a given vector into two or three-component vectors at right angles to each other. The component vectors are called rectangular components of the given vector. Let R be the given vector acting in the X – Y plane at an angle θ with the x-axis. Let $$\overrightarrow{\mathrm{OC}}$$ = R. From point C, draw perpendiculars CA and CB on X and Y axes respectively. If P and Q be the rectangular components of R along the X and Y axis respectively, then

Also according to the triangle law of vector addition,
R = P + Q
= Pî + Qĵ
or
R= (R cosθ)î+ (R sinθ)ĵ
Thus if Ax and Ay be the two rectangular components of A along the X and Y axes respectively.
Then A = Ax + Ay
A = Axî + Ayĵ

Note: Similarly in three dimensions,
A = Axî +Ayĵ + Az

Position vector in two and three dimensions can be expressed as:
r = xî + yĵ
and
r = xî + yĵ + zk̂ .

Question 5.
Define centripetal acceleration and derive its expression. Also, deduce its direction.
Centripetal acceleration: It is defined as the acceleration which always acts towards the center along the radius of the circular path.
“Centripetal” comes from a Greek term that means “centre¬seeking”.

The expression for centripetal acceleration: Let a body be moving wi^h constant angular velocity 0) and constant speed v in a circular path of radius R and center O.

Let the body be at points P and Q at time t and t + Δt respectively.
Also, let Δθ = ∠POQ be the angular displacement in a small-time inverted Δt, thus

If v and v + Δv be the velocity vectors of the body at points P and Q acting along PL and QM respectively. As the body is moving with uniform speed, so lengths PL and PM are equal i.e. PL = PM. The change in velocity Δv from P to Q is due to the change in the direction of the velocity vector.

Hence the magnitude of the acceleration of the body is given by

From point ‘a’ draw $$\overrightarrow{\mathrm{ab}}$$ parallel and equal to $$\overrightarrow{\mathrm{PL}}$$ and $$\overrightarrow{\mathrm{ac}}$$ parallel and equal to $$\overrightarrow{\mathrm{QM}}$$, thus

Clearly, the angle between ab and ac is Δθ. As Δt is quite small so
b and c lie on very close to each other, thus be can be taken to be an arc $$\widehat{\mathrm{bc}}$$ of a circle of radius ab = |v|.

Also when Δt → 0, then L.H.S. of Eqn. (4) represents acceleration. Thus Eqn. (4) can be written as

eqn. (6) gives the magnitude of the acceleration produced in the body.

The direction of acceleration: Acceleration always acts in the direction of Δv which acts from b to c. When Δt is decreased, Δθ also decreases. If Δt → 0, then Δθ → 0, so Δv tends to be perpendicular to ab. As ab ∥ PL, so Δv tends to act perpendicular to PL i.e. along PO i.e. along the radius towards the center of the circular path.

Since v and R are always constant, so the magnitude of centripetal acceleration is also constant. But the direction changes pointing always towards the center. So centripetal acceleration is not a constant vector.

Question 6.
A body is projected with some initial velocity making an angle θ with the horizontal. Show that its path is a parabola. Find the maximum height attained, time for maximum height, horizontal range, maximum horizontal range, and the time of flight.
Let the body be projected with velocity u inclined at angle θ with the horizontal. The horizontal and vertical components of velocity and acceleration are
Ux, ax, and Uy, ay where
ux = u cos θ, Uy = u sin θ, ax = 0, ay = -g
where g is the acceleration due to gravity.

The coordinates of O are (0, 0). Considering horizontal motion, The position of the body after time t has coordinates (x, y) where
x (t) = Xo + ux t + 1/2 ax × t2

Substituting for various factors
x(t) = 0 + u cos θ . t + 0
x(t) = u cos θ t

Equation of trajectory:
Substituting for t from eq. (1) in eq. (2), we get

This is an equation of a parabola. Thus, the path of a projectile is a parabola.

Maximum height attained: At the maximum height, the vertical componènt of velocity becomes O (zero). Now using the equation of motion

Time for maximum height: Using equation of motion, v = u + at or vy = Uy + ay t, we have

Horizontal range: Let the horizontal range be R. Since there is no acceleration in the horizontal direction, so
x = x(0) + uxt + $$\frac{1}{2}$$ax t2
Here x(0) = 0, ux = u cos θ, ax = 0, and t is the total time of flight which is twice the time for maximum height because the body takes the same time in rising to and falling from the highest point.

Maximum horizontal range: For R to be maximum, sin 2θ in Eqn. (6) must be maximum i.e. sin 2θ = 1,

Thus, for R to be maximum, 0 must be 45°. It is independent of the mass of the body.

Time of flight of the projectile: The projectile after completing its flight returns back to the same horizontal level from which it was projected. Therefore, the vertical displacement in the whole flight is zero. Considering vertical motion.

Therefore, either T = 0 or u sin θ – $$\frac{1}{2}$$ g T = 0
T = 0 corresponds to the initial (starting) position,

Equation (8) gives the total time of flight This is twice the time for maximum height.

Question 7.
A projectile is projected at an angle a with the vertical with initial velocity u. Find the maximum height attained, time of flight, and the horizontal range.

Let the particle be projected from O with velocity u. The magnitude of horizontal and vertical components of the velocity are u sinα and u cosα respectively. Coordinates of O are (0, 0).

Maximum height: Consider the vertical motion. The initial speed Uy = u cosα and final speed vy = 0, acceleration ay = – g, then using the
equation of motion s = $$\frac{v^{2}-u^{2}}{2 a}$$ we have

Time of flight: It is the total time taken by the projectile in going from O to A. The vertical displacement = 0. Considering vertical motion and using the equation of motion

Horizontal range: Considering horizontal motion ux = u sinα,
ax = 0,
∴ using equation of motion

Numerical Problems:

Question 1.
What is the angle between the following pair of vectors?
A = î + ĵ + k̂,B = – 2î – 2ĵ – 2k̂
We know that
A.B = AB cos θ

Question 2.
Prove that 2î – 3ĵ – k̂ and – 6î + 9ĵ + 3k̂ are parallel.
Let A = 2î – 3ĵ – k̂
and B = – 6î + 9ĵ + 3k̂

These two vectors will be parallel if their cross-product is zero.
i.e. if A × B = 0

Hence proved.

Question 3.
Calculate the area of the triangle determined by two vectors
A = 3î + 4ĵ and B = – 3î + 7ĵ
We know that area of a triangle is given by

Question 4.
A man wants to cross the river to an exactly opposite point on the other bank. If he can row his boat with twice the velocity of the current, then at what angle to the current, he must keep the boat pointed.

Question 5.
A body is acted upon by the following, velocities:
(i) 7 ms-1 due to E,
(ii) 10 ms-1 due S,
(iii) 5$$\sqrt{2}$$ ms-1 due N.E.
Find the magnitude and direction of the resultant velocity.

Let OA, OB and OC represent the velocities given in the statement i.e.
OA = 7 ms-1
OB = 10 ms-1
and OC = 5$$\sqrt{2}$$ ms-1
To find their resultant velocity, resolve OC into two rectangular components along east and north.

Hence resultant velocity along east = 7 + 5 = 12 ms-1 and resultant velocity along south = OB – OF = 10 – 5 = 5 ms-1.

If R be the resultant velocity, then the magnitude of R is obtained by applying the parallelogram law of vector addition as

When OG = 12ms-1 and OH 5ms-1.

The direction of R: Let θ be the angle made by R with the east.
∴ in rt. ∠d ΔOGI,

Question 6.
If $$\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}=\overrightarrow{\mathbf{C}}$$, prove that C = (A2 + B2 + 2AB cosθ)1/2 where θ ¡s the angle between $$\overrightarrow{\mathbf{A}}$$ and $$\overrightarrow{\mathbf{B}}$$.

Question 7.
Determine the value of k so that the vectors $$\overrightarrow{\mathbf{A}}$$ = 6î + kĵ – 4k̂ and $$\overrightarrow{\mathbf{B}}$$ = 3î + 4ĵ + 5k̂ are perpendicular.

Question 8.
Determine a unit vector which is perpendicular to both $$\overrightarrow{\mathbf{A}}$$ = 2î + ĵ + k̂ and $$\overrightarrow{\mathbf{B}}$$ = î – ĵ + 2k̂
Let n̂ be the unit vector perpendicular to both A and B.
Also, Let C be the vector that acts along n̂
∴ By definition of the cross product,

Question 9.
A projectile is fired horizontally with a velocity of 98 ms-1 from the top of a hill 490 m high. Find:
(i) the velocity with which it strikes the ground.
(ii) the time is taken to reach the ground.
(iii) the distance of the target from the hill.

(i) h = 490 m, a = g = 9.8 ms2
Uy = initial velocity along the y-axis at the top of the tower = 0

(ii) Let v be the velocity along the y-axis with which the projectile hits the ground.

If V be the resultant velocity of hitting the ground

Let θ be the angle made by V with the horizontal

(iii) Let x, be the distance of the target from the hill.
∴ x = horizontal distance covered with u in a time t.
ut = 98 × 10 = 980 m.

Question 10.
A boy stands at 78.4 m from a building and throws a ball which just enters a window 39.2 m above the ground. Calculate the velocity of the projection of the ball.
Let the boy standing at A throw a ball with initial velocity u.
θ = angle of the projection made with the horizontal.

As the boy is at 78.4 m from the building and the ball just enters above the ground.

Question 11.
The equation of trajectory of an oblique projectile is
y = $$\sqrt{3}$$x – $$\frac{1}{2}$$gx2
What is the initial velocity in ms’ and the angle of projection of the projectile in degree?
Comparing the given equation with the standard equation of the trajectory of an oblique projectile.

Question 12.
Two particles located at a point begin to move with velocities 4 ms-1 and 1 ms-1 horizontally in opposite directions. Determine the time when their velocity vectors become perpendicular. Assuming that the motion takes place in a uniform gravitational field of strength g.
Let v1 and v2 be the velocities of first and 2nd particles respectively after a time t.
∴ v1 = 4î – gt ĵ
v2 = – î – gt ĵ
For v1 and v2 to be ⊥ to each other, then their dot product must be zero.

Question 13.
In the previous question, determine the distance between particles at t = $$\frac{2}{g}$$.
The separation between the particles is to take place only along the horizontal.
∴ the separation between the particles
= relative speed of the particles along horizontal × time

Question 14.
The maximum range of a projectile is range. What is the angle of projection for the actual range?
We know that the horizontal range of a projectile is given by
R = $$\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}$$ …(i)

Where u = initial velocity of projection
θ = angle of projection with horizontal

Also, we know that the maximum horizontal range (Rmax) is given by

Question 15.
A body is projected with a velocity of 40 ms-1. After two seconds, it crosses a verticle pole of 20.4 m. Find the angle of projection and the horizontal range.
Here, u = 40 ms-1
height of verticle pole, h = 20.4 m
t = 2 seconds

Let us take vertical motion

∴ The horizontal range is given by the relation,

Question 16.
The greatest and the least resultant of two forces acting at a point are 29 N and 5 N respectively. If each force is increased by 3 N, find the resultant of two new forces when acting at a point at an angle of 90° with each other.
Let A and B be the two forces.
∴ Greatest Resultant = A + B = 29 N ….(1)
least Resultant = A – B = 5 N ….(2)

Let A and B be the new forces such that
A’ = A + 3 = 17 + 3 = 20N and
B’ = B + 3 = 12 + 3 = 15 N

Here, θ = angle between A’ and B’ = 90°
Let R be the resultant of A’ and B’.
∴ according to parallelogram law of vector addition

The direction of R:
Let β be the angle made by R with A’

Question 17.
An aircraft is trying to fly due north at a speed of 100 ms-1 but is subjected to a crosswind blowing from west to east at 50 ms-1. What is the actual velocity of the aircraft relative to the surface of the earth?
Let Va and Vw be the velocities of aircraft and wind respectively.
∴ Va = 100 ms-1 along N direction
Vw = 50 ms-1 along E direction

If V be the resultant velocity of the aircraft, then these may be represented as in the figure given below. So the magnitude of V is given by,

Let ∠AOB = θ be the angle which the resultant makes with the north direction.

Question 18.
Calculate the total linear acceleration of a particle moving in a circle of radius 0.4 m at the instant when its angular velocity is 2 rad s-1 and angular acceleration is 5 rad s-2.
Since the particle possesses angular acceleration, so its total linear acceleration (a) is the vector sum of the tangential acceleration (a,) and the centripetal acceleration (ac). a1 and ac, are at right angles to each other.
a = $$\sqrt{a_{t}^{2}+a_{c}^{2}}$$ …. (1)

Now r = radius of the circular path = 0.4 m
ω = angular velocity of the particle = 2 rad s-1
α = angular acceleration = 5 rad s-2

Let θ = angle made by a with a,

Question 19.
Calculate the greatest number of rotations per minute which can be given to a body of mass 500 gms tied to a string of length 1.5 m, if the string can withstand a maximum tension of 40 N.
Here, m = mass of body = 500 gms = 0.5 kg
r = radius of circle = 1.5 m
Fc = Centripetal force = 40 N
Now we know that

Hence the body can be given 70 rotations per minute without breaking the string.

Question 20.
An airplane flies 400 km west from city A to city B then 300 km north-east to city C and finally 100 km north to city D. How far is it from city A to city D? In what direction must the airplane go to return directly to the city A from city D?
Given, AB = 400 km
BC = 300 km
CD = 100 km

Let N1, N2 represent north directions.
∠ABC = 45°
Draw CC’ ⊥ AB, And CB’ ⊥ BN2
Now in ΔBC’ C

From AAC’D, AD is given by

Question 21.
Which of the following quantities are independent of the choice of the orientation of the coordinates axes:
a + b, 3ax + 2by, [a + b – c], angle between b and c, a?
a + b, |a + b – c|, angle between b and c, a are the quantities that are independent of the choice of the orientation of the coordinate axes.

But the value of 3ax + 2by depends on the orientation of the axes.

Question 22.
A machine gun is mounted on the top of a tower 100 m high. At what angle should the gun be inclined to cover a maximum range of firing on the ground below? The muzzle speed of the bullet is 150 ms-1. (take g = 10 ms-2)
Here, u = 150 ms-1
h = 100 m
g = 10 ms-2

Let R = horizontal range of the projectile.
If t = time during which the bullet covers, horizontal range R = time during which the bullet falls through a height h (= 100 m).
Let θ = angle of projection with horizontal, then
ux = u cos θ = 150 cos θ = horizontal component of speed of the projection,
uy = 150 sin θ = vertical component of the speed of projection.

Considering the upward motion positive, then using the equation

∴ The horizontal range covered by the (using positive sign only) projectile is given by
x = ux × t

To maximize this function graphically, a graph between R and θ by taking arbitrary values of θ is plotted. We find from the graph that R is maximum for θ = 43.8°.

Question 23.
Two vectors of magnitude A and $$\sqrt{3}$$ A are perpendicular to each other. What is the angle which their resultant makes with A?
Let θ = angle made by R (resultant of A and $$\sqrt{3}$$ A) with A
∴ In ΔOPQ

Question 24.
If |a + b| = |a – b|, then find the angle between a and b using properties of the dot product of vectors.
Here, |a + b| = |a – b| …. (i) given
Squaring on both sides of eqn. (i), we get

Question 25.
The express cross product of two vectors in cartesian coordinates.
Let A and B be the two vectors s.t.

It can be written in a determinant form as

Value-Based Type:

Question 1.
Two friends Raman and Madhav arc playing on the bank of a river. They are throwing stones in the river. Raman is a student of the science stream whereas Madhav from the commerce stream. Every time Raman throws the stone far away than Madhav. Madhav was surprised and could not understand the reason. Finally, Raman explained the fact that it is relating to physics.
(i) What values are shown here by Roman?
Intelligence, social, cooperative, and willing to share his knowledge,

(ii) Which concept is being used here?
Horizontal range of a projectile

(iii) What is the minimum angle to get the maximum horizontal range?
We know that: Maximum range: R = $$\frac{u^{2} \sin 2 \theta}{\mathrm{g}}$$

For maximum range
Sin 2θ = maximum value
i.e sin 2θ =1
⇒ sin 2θ = sin 90°
⇒ 2θ = 90° or θ =45°
Hence, if the body is projected at an angle of 45°, it will cause maximum horizontal range.

Question 2.
Himanshu and Shika are the two students of class XI. Himanshu fired a bullet at an angle of 30° with the horizontal which hits the ground 3 km away. Shika told that I could hit a bird that is 4 km away by adjusting its angle of projection. Himanshu immediately betted, He told if you can I would give you Its. 1000 otherwise you have to pay the same to me. Himanshu was sure that you can never hit that bird. However, killing a bird is not good for the environment. Assume the muzzle speed to be fixed, and neglect air resistance.
(i) Is betting a good practice?
No

(ii) How Himanshu is sure that the bullet could not hit the target?
Max horizonatal range = $$\frac{\mathrm{u}^{2}}{\mathrm{~g}}$$ [∵ muzzle velocityin fixed]
ie Rmax = $$\frac{\mathrm{u}^{2}}{\mathrm{~g}}$$ = 2$$\sqrt{3}$$ = 3.464 km g
So, the bullet can not hit the target which is 4 km away.

(iii) Which values are depicted in the above question?
Values depicted are:
(a) Concern for environment and living creature
(b) Avoid betting
(c) Intelligent

Question 3.
Rain is falling vertically with a speed of 35 m s-1. A girl of class XI rides a bicycle with a speed of 12 m s-1 in the east to west direction. She could not understand that in which direction she should half her umbrella. When she came back to her home she asked her mother who was a physics teacher in a reputed school. She explained as under:
(i) Which value is displayed by the girl?
Willing to know the scientific reason, curiosity

(ii) What justification was given by her mother?
Vf = The velocity of rain
Vb = The velocity of bicycle
Both these velocities are with respect to the ground.
The velocity of rain relative to the velocity of the bicycle is given by
Vrb = Vr – Vb

This relative velocity vector as shown above makes an angle θ with the vertical.
It is given by:

∴ The girl should hold her umbrella at an angle of about 19° with the vertical towards the west.

Question 4.
Sita a student of class XI was suffering from malaria. The area is full of mosquitoes. She was not having a mosquito net. Her friend Geeta has an extranet. She gave it to Sita. Also, she took Gita to Doctor, got her medicines. After a week Sita became normal.
(a) Comment upon the qualities of Sita.
Sita has a caring attitude, and concern for others.

(b) The mosquito net over a 7 m × 4 m bed is 3 m high. The net has a hole at one corner of the bed through which a mosquito enters the net. It files and Sita at the diagonally opposite upper corner of the net
(i) Find the magnitude of the displacement of the mosquito
We know that
A = $$\sqrt{\left(A_{x}\right)^{2}+\left(A_{y}\right)^{2}+\left(A_{x}\right)^{2}}=\sqrt{7^{2}+4^{2}+3^{2}}$$
= $$\sqrt{74}$$ m

(ii) Taking the hole as the origin, the length of the bed as the X-axis, its width as the Y-axis and vertically up as the Z-axis, write the components of the displacement vector.
The components of the vector are 7 m, 4 m, and 3 m

## Physical World Class 11 Important Extra Questions Physics Chapter 1

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 1 Physical World. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

## Class 11 Physics Chapter 1 Important Extra Questions Physical World

### Physical World Important Extra Questions Very Short Answer Type

Question 1.
Name that branch of science that deals with the study of Earth.
Geology.

Question 2.
Name that branch of science that deals with the study of stars.
Astronomy.

Question 3.
Name the scientist and the country of his origin whose field of work was elasticity.
Robert Hook, England.

Question 4.
The word “Physics” comes from a Greek word. Name the word.
The word is ‘fuses meaning ‘Nature’.

Question 5.
The word science has come from a Latin verb. Name the verb.
The name of the Latin verb is ‘Scientia’.

Question 6.
What is the meaning of the verb ‘Scientia’?
To ‘know’

Question 7.
Name the scientist and the country of his origin who received the Nobel Prize for his work on molecular spectra.
C.V. Raman, India.

Question 8.
What is the most incomprehensible thing about the world?
It is comprehensible.

Question 9.
Name a great scientist who gave the following comment on science.
“Science is not just a collection of laws, a catalog of unrelated facts. It is a creation of the human mind, with its freely invented ideas and concepts.”
Albert Einstein.

Question 10.
Which famous philosopher gave the following comments on science?
“We know very little and yet it is astonishing that we know so much, and still more astonishing that so little knowledge of science can give so much power.”
Bertrand Russel.

Question 11.
Who discovered the electron?
J.J. Thomson.

Question 12.
Who discovered neutron?

Question 13.
Who gave the general theory of relativity?
Albert Einstein.

Question 14.
Who proposed the wave theory of light?
Huygen.

Question 15.
Name four physics devices widely used in medical diagnosis.

1. X-rays,
2. Ultrasound,
3. Stethoscope,
4. Microscope.

Question 16.
Name Indian-born scientist who received Nobel Prize for his discoveries in astronomy.
S. Chandra Shekhar.

Question 17.
Metaphysics is a science that is concerned with what?
Supernatural .events.

Question 18.
Which science is considered to be the mother of all sciences?
Physics.

Question 19.
Name the discovery made by S.N. Bose.
Bose-Einstein Statistics.

Question 20.
Name the scientist and the country of his origin whose field of work was ‘cosmic rays’.
Hess, Austria.

Question 21.
What are the meaning of the Sanskrit word ‘Vijnan’ and the Arabic word ‘Ilm’?
Knowledge.

Question 22.
Name the Sanskrit equivalent word of Physics.
Bhautiki.

Question 23.
Name the field of Physics in which India was a leading country in the sixties.
Cosmic rays.

Question 24.
Who discovered X-rays?
W. Roentgen.

Question 25.
Which electronic media can help in eradicating illiteracy in India?
Television.

Question 26.
Name the technology based on the amplification of light by population inversion?
Laser.

Question 27.
Who discovered nuclear forces?
H. Yukawa.

Question 28.
To which country he belonged?
japan.

Question 29.
Pierre Curie and Marie Curie.

Question 30.
Name the discovery made by W. Roentgen.
X-rays.

Question 31.
What has been said by P.A.M. Dirac regarding physics in relation to society?
P.A.M. Dirac said, “It is more important to have beauty in the equations of physics than to have them agree with experiments.”

Question 32.
What did Issac Newton say to measure the degree of impact of science on society?
He said “Nature is pleased with simplicity, and affects not the pomp of superfluous causes.

Question 33.
What Neils Bohr said regarding science in relation to society?
He said “The task of science is both to extend the range of our experience and to reduce it to order.

Question 34.
Name a few Indian physicists who have made significant contributions in the field of physics.
C.V. Raifiaq, S. Chandra Shekhar, S.N. Bose, Homi J. Bhabha, and Meghnath Saha.

Question 35.
Name the scientific principle on which airplane works.
Bernoulli’s theorem.

Question 36.
Name the scientific principle on which radio and T.V. works.
Propagation of electromagnetic waves.

Question 37.
Name the scientific principle upon which laser works.
Amplification by a process called population inversion.

Question 38.
Name the technology which works on the scientific principle “Newton’s second and third laws of motion”.
Rocket propulsion.

Question 39.
Name the forces which are of nuclear origin.
Strong forces.

Question 40.
What is Physics?
It is that branch of science which deals with nature and natural phenomena.
Or
It is that branch of physical science that is to seek out and understands the basic laws of nature upon which all physical phenomena depend. It has brought to us deeper and deeper levels of understanding nature.

Question 41.
What is Science?
It is defined as the systematic study of physical phenomena.

Question 42.
What are Biological Sciences? Give three examples.
Those sciences which deal with living things are called Biological Sciences, e.g. Zoology, Botany, Ornithology.

Question 43.
What are Physical Sciences? Give a few examples.
They ate defined as the sciences which deal with non-living things,
e.g. Physics, Chemistry, Astronomy, Astrology, Geology, Geography, Oceanology.

Question 44.
Define Theory.
It is defined as the behavior of physical systems explained in terms of a set of a minimum number of laws.

Question 45.
What do you understand by the term scientific method?
The systematic observations, logical reasoning, model-making, and theoretical prediction form the scientific method.

Question 46.
Name the scientific principle on which electric generator works.
Electromagnetic induction (E.M.I.).

Question 47.
Name the technology which works on the scientific principle ‘Nuclear Fission’.
Nuclear Reactor.

Question 48.
Name the technology which works on the scientific principle “Digital logic of electronic circuits”.
Calculators and computers.

Question 49.
Name the scientific principle upon which the working of cyclotron defends.
The motion of charged particles under electric and magnetic fields.

Question 50.
Name the. scientist and his country who discovered wireless1 telegraphy.
G. Marconi, Italy.

### Physical World Important Extra Questions Short Answer Type

Question 1.
Differentiate between Biological and Physical sciences?

 Biological Sciences Physical Sciences (i) They deal with living things. (i) They deal with non-living things. (ii) The study of the biological specimens is conducted at the molecular level. (ii) The study of matter is conducted at atomic or ionic levels i.e. at much smaller levels.

Question 2.
What is the relation between Physics and Technology?
Broadly speaking, physics and technology both constitute science. Physics is the heart and technology is the body of science.

The application of the principles of physics for practical purposes becomes technology, e.g.

1. Airplanes fly on the basis of Bernoulli’s theorem.
2. Rockets propulsion is based on Newton’s second and third laws of motion.
3. The generation of pow%r from the nuclear reactor is based on the phenomenon of controlled nuclear fission.
4. Lasers are based on the population inversion of electrons and so on. Thus, we can say that to some extent technology is applied to Physics.

Question 3.
What is the relation between Physics and society?
Most of the development made in Physics has a direct impact on society, e.g.

1. Exploration of new sources of energy is of great importance to society.
2. Rapid means of transport are no less important for society.
3. society has-been enriched due to the advances in electronics, lasers, and computers.
4. The development of T.V., radio, satellites, telephone, the telegraph has revolutionized the means of communications which have a direct impact on society and so on.

Question 4.
Is Science on speaking terms with humanities?
Yes, there is a deep relation between the development of humanity on account of science. Many socio-economic, political, and ethical problems are being tackled and solved by science. Science has greatly helped in developing art and culture. Many musical instruments have been developed due to the theories in Physics. The steam engine is inseparable from the industrial revolution which had a great impact on human civilization.

Question 5.
What is the relation between Physics and Technology?
The interplay between physics and technology is the basic to the progress of science which is ever dynamic. Laws in waves and oscillation opened several technological fields which include telescopy, ultrasounds, microscopy, X-rays, and laser. Powerhouses, big cranes, healing devices, etc. work on the principle of electromagnetism. Atomic energy and nuclear weapons are on account of fission. Similarly, Radar, television, the internet, etc. are all based on simple laws of physics. So until there is no theory i.e. physics, there can be no experiment i.e. technology. Hence both are deeply related.

Question 6.
Is Physics more of a philosophy or more of a mathematical science?
Physics is not a purely abstract science devoid of philosophy. Physicists are natural philosophers and Einstein is an example to quote. So Philosophy has provided the backbone to Physics.

Question 7.
Define Biophysics.
It is defined as the understanding of biological processes based upon the principles of Physics. For example, spectroscopic techniques are used to study the constitution of biological molecules and disorders in them. Laws of thermodynamics are used to explain various biological activities of predators and also the activities of molecules.

Hence the application of Physics to bioscience is now well known to all of us.

Question 8.
Define Technology?
It is defined as the study of newer techniques of producing machines, gadgets, etc. by using scientific discoveries and advancements. It is largely dependent on Physics.

Question 9.
Has imagination any role in Physics?
One of the definitions of Physics says that “It is the science-based on imagination and intuition which can be tested experimentally and mathematically.” Thus, imagination has a great role in the development of physics. Schrodinger, De-Broglie, Heisenberg, and most of the other scientists who were physicists were great imaginers.

Question 10.
Name a few aspects of your daily life in which you rely on the simplicity of nature.
Laws of Physics represent the nature in simplest form. We face nature in many ways in our daily life. For example, we work, walk, write, talk and stand on our feet, and so on. The natural way of taking bath, chewing food, etc. can easily be understood in terms of simple laws of science. Even though actions like swimming, running, and playing may be complex but the underlying laws of nature are quite simple such as Newton’s laws, friction, etc.

Question 11.
The physicists think at a level far higher than a normal individual. Explain.
For everyone to become a leader in his field, he has to think for a higher level than an ordinary person. This is more so for the case of physicists as the technological development meant for uplifting the living condition of mankind is highly dependent on the farsightedness of the physicists in particular. He must think at a level that is philosophical and mathematically quantifying so that they can visualize the requirement of people is quite advance.

Question 12.
Name a few wartime and maritime applications of Physics.
(a) Wartime: The wartime applications are Bombs, nuclear weapons, jet fighter bombers, missiles, ships, radar, sonar, wireless communications, transportation, and electronics.

(b) Maritime: The maritime application of physics is Navigation of ships, tankers, airplanes, T.V., radio, and music system, etc.

Question 13.
Name five Indian scientists and the field of their work.
Following are the five Indian Scientists and the field of their work:

 Name Field of work (1) C.V. Raman Raman effect (Scattering of light by molecules) (2) S. Chandrashekar Theory of Black Hole (i.e. structure and evolution of stars) (3) J.C. Bose E. M. Waves (4) S.N. Bose Bose-Einstein statistics (5) H.J. Bhabha Cosmic rays

Question 14.
Who invented:
(i) Computer
Charles Babbage

(ii) Transistor,
J. Bardeen

(iii) Electric bulb and Telegraphy,
A. Edison

Appleton

(v) Wireless telegraphy,
Marconi

(vi) Telephone.
Graham Bell.

Question 15.
Name the theories given by the following:
(i) Neil Bohr
Theory of atomic structure

(ii) Lawrence,
Cyclotron

(iii) Henry Becquerel,

(iv) Galileo,
Principle of Inertia

(v) Bragg,
Crystal structure by X- rays

(vi) Abdus Salam,
Unified Field Theory

(vii) Millikan.
Charge on an electron.

Question 16.
Give the nationality of the following scientists:
(i) Van der Waals,
Dutch

(ii) Curie,
French

(iii) Yukawa,
Japanese

(iv) Galileo,
Italian

(v) Michelson,
U.S.A. (American)

(vi) Heisenberg,
German

(vii) Archimedes,
Greek

(viii) Maxwell,
Scottish

(ix) Cavendish,
English

(x) Hubble.
Austrian.

Question 17.
The following are the gadgets commonly used in our house:

1. Pressure Cooker
2. Electric light
3. Tube light
4. Electric fan
5. Water cooler
6. Refrigerator
7. Washing machine
8. Gas stove
9. Electric iron
10. Mixi
11. Geyser
12. Electric motor.

Question 18.
Write the physical principle upon which the working of the gadgets mentioned in the above question is based.
These are based on the following physical principles:

1. The boiling point rises with the increase in pressure.
2. Light is produced when the current is passed through a given resistor.
3. Light is emitted when an electric discharge is passed through the gas.
4. A rotating magnetic field is produced on passing current which notates the motor.
5. Due to evaporation of water, cooling in the air which is being forced out by the fan is produced.
6. On absorbing heat from the surroundings, compressed volatile liquid on sudden expansion causes cooling.
7. Current produces a rotating magnetic field that operates the motor.
8. Heat is produced due to the burning of L.P.G.
9. It works on the principle of heating effect of electric current.
10. Torque is produced on the coil due to the electric current passed through it, hence it rotates.
11. Current shows the heating effect when passed through the conductor.
12. It rotates due to the torque produced on the coil on passing an electric current through it.

Question 19.
Name one Scientist each from the following countries who have won Nobel Prize.
(a) Japan
H. Yukawa

(b) England

(c) India
C.V. Raman

(d) The U.S.A.
K. Feynman

(e) Germany.
Max. Plank.

Question 20.
How Darwin showed that scientific themes are at once simple even though phenomena in nature may be complex.
Darwin found a simple basis for the origin of species and descent of man which is “Living things change producing descendants with different characteristics in a process that has been going on for as long as there has been life” by taking a large number of observations on the theory of evolution while onboard ship.

Question 21.
Illustrate by an example the beauty of a Scientific Theory.
The theory proposed by Darwin was opposed by the church and now we have new discoveries such as selfish genes and punctuated equilibria but Darwin’s basic theory still holds. This is the beauty of Davin’s theory of evolution.

Question 22.
In science sometimes we observed certain phenomena experimentally but are unable to give a logical equation or theory for that sometimes, it also happens that we have a scientific theory supported by’ mathematical formulation yet are unable to test it immediately. Site one such example.
Einstein worked to establish a relation between the energy and mass of the body. He was of the view that these are the two sides of the same coin or two facts of the same physical quantity. He succeeded when he gave his mass-energy equation E = mc2. But its experimental verification came 40 years later in 1945 when the atomic bomb was exploded over Japan.

Question 23.
Why do we call physics an exact science? What is the aim of science?
Physics is called exact science because it is based on the measurement of fundamental quantities.
The main aim of science is to find the truth behind the various processes taking place in the universe.

Question 24.
How science has helped in solving the food problem in several countries?
Science has helped in solving food problem in the following ways:
(a) It has given improved and new agricultural implements.
(b) Science has improved the quality of seeds by genetic engineering.
(c) High-yielding hybrid varieties of grains have been developed. Some easily reaping varieties have also been developed and grown.
(d) Use of pesticides and insecticides has saved crops from being destroyed by insects and pests.
(e) Some new types of crops are also developed and are being developed to meet the requirement of society.

Question 25.
What is a scientific temperament and scientific way of doing things?
A mindset molded in a particular set of thinking called the scientific way is known as scientific temperament. It is not only based on logic, facts but on reliable observations. The ultimate test of truth in science is experimental verification.

A scientific way of doing things involves the following steps:
(a) Identifying the problem or aim.
(b) Collecting all relevant information or data related to the problem.
(c) Hypothesising or proposing a possible theory.
(d) Taking experimental observation yielding consistent results.
(e) Predicting or making statements.

Question 26.
What is the scope of Physics?
The scope of Physics is very wide i.e. the domain of Physics covers a very wide variety of natural phenomena.

For example, the range of distances we study in Physics varies from 1014 m (size of the nucleus) to 1025 m (size of the universe).

Similarly, the range of masses included in the study of Physics varies from 10-30 kg (mass of an electron) to 1055 kg (mass of the universe). Also, the range of time i.e. time intervals of events we come across in the study of Physics varies from 1022 seconds (time taken by light to cross a nuclear distance) to 10-8 seconds (lifetime of the sun).

Thus we see that the scope of Physics is really very wide. It includes; optics, electricity waves, and oscillations, heat and thermodynamics, magnetism, atomic and nuclear physics, computers, and electronics.

Question 27.
Physics is an exciting subject! Comment.
The study of Physics is exciting in many ways, e g.:

1. Journey to the moon with controls from the grounds.
2. Lasers and their ever-increasing applications.
3. Live transmission of events thousands of kilometers away on the T: V.
4. The speed and memory of the fifth generation of computers.
5. Study of various types of forces in nature.
6. Technological advances in health science.
7. The use of robots is quite exciting.
8. Telephone calls over long distances and so on. Thus, Physics is exciting not only to the scientist but also to a layman, children, women, etc. The musical instruments, toy guns, toy trains, etc. all are constructed using simple principles of physics like collision, potential energy, and vibration, etc. Today the situation is that even our thought process and social values are affected by Physics. Thus, it is quite amazing.

Question 28.
Write a short note on the origin and development of Physics.
Physics as a science took roots from the days of Copernicus, i. e., nearly four centuries ago when it was not well understood and it was considered as a part of philosophy, i.e., knowledge. Later on, with the development of knowledge about nature and its various activities, the knowledge was divided into physical and biological sciences.

Some important developments like Newton’s law of gravitation, ideas about light were developed in the 18th century. The 19th century saw some of the great discoveries in Physics and at the end f the century i.e. 1889, the electromagnetic theory was developed, Fouriuatun of Einstein’s and Plank’s ideas were laid down apart from laying the basis for the industrial revolution. Physics progressed very fast in the first quarter of the 20th century.

Atomic structure, the theory of relativity, quantum theory, nuclear physics, basics of laser theory and most of the other developments took place in this period. Then came transistors, semiconductors, television, radar, and few important discoveries during World War II.

Further development in quantum mechanics, thin-film technology, computers, lasers was developed from 1950 onward. Today we have no theoretical development beyond quantum mechanics. A unified theory is not being tried yet. This is the present status with achievements in applied fields.

Question 29.
Explain the role of science in the entertainment industry.
Progress in science especially in physics and technology has enriched no other field as the field of entertainment. We see scientific toys like robots for children and merry-go-rounds of all sorts in amusement and entertainment parks which are not only highly entertaining but also test the endurance of an individual.

These are based on the laws of forces and the critical stages are those when a man is pitted against gravity. T.V. has invaded a large number of houses as a source of entertainment and so are the music blaring CD players. Computers have become another source of individual indoor entertainment. The use of laser beams in music and drama shows and disco dances is highly rewarding.

With the improvement in physics and technology, circus entertainment to have changed with the application of science. Thus, we conclude that the role of science and technology in the entertainment industry has increased tremendously.

Question 30.
Give some of the uses and applications of artificial satellites.
The following are the fields where satellites are being used –
(a) Remote sensing
(b) Communication
(c) Spying
(d) Weather forecasting

(a) Remote sensing: In remote sensing, infrared photography s used from a high altitude. The technique has improved a lot and the resolution has gone down to about 5m in an area. This technique has helped in mineral and oil exploration. It has also helped in the study of forest living and crop patterns.

(b) Communication: In the field of communication, satellites have brought revolution during the last 20 years. Now new items are flashed all around the globe. Cricket matches can be seen anywhere on the globe which is played in one small part of a country. Internet, E-mail, etc. have brought people much more closer and the world has become a unified entity.

(c) Spying: In spying also use IR technology.

(d) Weather forecasting: Weather forecasting has become more reliable with the use of satellites. The rains, cyclones can now be predicted with greater accuracy 36 hours in advance or even earlier. The movement of glaciers, the position of ice and snow deposition, and the resulting flow of water in rivers is known well in advance.

### Physical World Important Extra Questions Long Answer Type

Question 1.
How Physics is related to other sciences?
Physics is so important to a branch of science that without the knowledge of Physics, other branches of science cannot make any progress.

This can be seen from the following:
(a) Physics in relation to Mathematics: The theories and concepts of Physics lead to the development of various mathematical tools like differential equations, equations of motion, etc.

(b) Physics in n .ation to Chemistry: The concept of iñteraction between various particles leads to understanding the bonding and the chemical structure of a substance. The concept of X-ray diffraction and radioactivity has helped to distinguish between the various solids and to modify the periodic table.

(c) Physics in relation to Biology: The concept of pressure and its measurement has helped us to know the blood pressure of a human being, which in turn is helpful to know the working of the heart. The discovery of X-rays has made it possible to diagnose the various diseases in the body and fracture in bones.

The optical and electron microscopes are helpful in the studies of various organisms. Skin diseases and cancer can be cured with the help of high-energy radiation like x-rays, ultraviolet rays.

(d) Physics in relation to Geology: The internal structure of various rocks can be known with the study of the crystal structure. The age of rocks and fossils can be known easily with the help of radioactivity i.e. with the help of carbon dating.

(e) Physics in relation to Astronomy: Optical telescope has made it possible to study the motion of various planets and satellites in our solar system.

The radio telescope has helped to study the structure of our galaxy and to discover pulsars and quasars (heavenly bodies having star-like structures). Pulsars are rapidly rotating neutron stars. Doppler’s effect predicted the expansion of the universe. Kepler’s laws are responsible to understand the nature of the orbits of the planets around the sun.

(f) Physics in the relation to Meteorology: The variation of pressure with temperature leads to the forecast of the weather.

(g) Physics in relation to Seismology: The movement of the earth’s crust and the types of waves produced help us in studying the earthquake and its effect.

Question 2.
Write a short note on origin and Fundamental forces in nature.
These are the following four basic forces in nature:
(a) Gravitational forces
(b) Electromagnetic forces
(c) Weak forces
(d) Strong force or nuclear forces.

Some of the important features of these forces are discussed below:
(a) Gravitational forces: These are the forces of attraction between any two bodies in the universe due to their masses separated by a definite distance. These are governed by Newton’s law of gravitation given by
F = G $$\frac{m_{1} m_{2}}{r^{2}}$$

where m1, m2 are the masses of two bodies
r = distance between them
G = universal gravitational constant
= 6.67 × 10-11 Nm2kg-2

Characteristics of gravitational forces:

1. They are always attractive. They are never repulsive. They exist between macroscopic as well as microscopic bodies.
2. They are the weakest forces in nature.
3. They are central forces in nature i.e. they set along the line joining the centers of two bodies.
4. They are conservative forces.
5. They obey inverse square law i.e. F ∝$$\frac{1}{r^{2}}$$ they vary inversely as the square of the distance between the two bodies.
6. They are long-range forces i.e. gravitational forces between any two bodies exist even when their distance of separation is quite large.
7. The field particles of gravitational forces are called gravitons. The concept of the exchange of field particles between two bodies explains how the two bodies interact from a distance.

(b) Electromagnetic forces: They include the electrostatic and magnetic forces. The electrostatic forces are the forces between two static charges while magnetic forces are the forces between two magnetic poles. The moving charges give rise to the magnetic force. The combined action of these forces is called electromagnetic forces.

Characteristics of electromagnetic forces:

1. These forces are both attractive as well as repulsive.
2. They are central forces in nature.
3. They obey inverse square law.
4. They are conservative forces in nature.
5. These forces are due to the exchange of particles known as photons which carry no charge and have zero rest mass.
6. They are 1036 times stronger as compared to gravitational forces and 1011 times stronger than weak forces.

(c) Strong forces: They are the forces of nuclear origin. The particles inside the nucleus are charged particles (protons) and neutral particles (neutrons) which are bonded to each other by a strong interaction called nuclear force or strong force. Hence they may be defined as the forces binding the nucleons (protons and neutrons) together in a nucleus. They are responsible for the stability of the atomic nucleus.

They are of three types:

1. n-n forces are the forces of attraction between two neutrons.
2. p-p forces are the forces of attraction between two protons.
3. n-p forces are the forces of attraction between a proton and a neutron.

Characteristics of Nuclear forces:

1. They are basically attractive in nature and become repulsive when the distance between nucleons is less than O.S fermi.
2. They obey inverse square law.
(a) and
(b) types are the forces that we encounter in the macroscopic world while
(c) and
(d) types are the forces that we encountered in the microscopic world.

(c) Weak forces: They are defined as the interactions which take, place between elementary particles during radioactive decay of a radioactive substance. In P-decay, the nucleus changes into a proton, an electron, and a particle called anti-neutrino (which is uncharged). The interaction between the electron and the anti-neutrino is known as weak interaction or weak force.

Characteristics of Weak forces:

• They are 1025 times stronger than the gravitational forces.
• They exist between leptons and leptons, leptons and mesons. etc.

Question 3.
Distinguish between the studies in the fields of science, engineering, and technology. Give an outline of the two or three industrial revolutions brought about by advancements in technology over the last twenty-five years or so.
Science is concerned with the unfolding of the basic aspects of nature. It formulates simple laws and finds the rhythm in nature, materials, and energy. Using basic principles of science, the ways to use them for the production of different kinds of articles is called technology, i.e., it is the application of science.

The execution of the application of technology in engineering. The production of articles using machines and implements in engineering. This involves the design, development, and manufacturing of articles.

The most notable technology development in the last 25 years is in the field of information technology, computers, and electronic media. The revolution in information technology has opened up fields on the internet, satellite linking of information systems and services other peripheral developments in the industry.

Computers have changed the face of society and made life easy in several fields. It has improved work efficiency in many segments of the industry and public life. Computers have touched the lives of children playing video games and adults alike. It has helped big organizations like railways, banks, and financial institutions like the insurance sector.

India has become one of the biggest centers of software exports and a big foreign exchange earner. Advance scientific research and industrial designing are being done by computers. TV has entered most Indian houses and community centers-courtesy revolution in electronic media.

The younger generation is mad after the stereo music with CD facilities. The transistors and tape recorders are left far behind. Electronic media has changed the face of the entertainment industry as well as information dissemination. Quick transmission of news, views, and comments are accepted as natural ones by listeners and viewers.

Value-Based Type:

Question 1.
A debate was organized by a school on woman’s innate nature, capacity, and intelligence. The arguments given by the two groups were as under:
Team A: Nature makes little difference in men and women in their anatomy and feeling. So, women are not on par with men. Hence, she is inferior to men in spheres of activity like sports, scaling of mountains like the Himalayas, etc.
Team B: The students of team ‘B’ have demolished this view using scientific arguments, and by quoting examples of great women, in science and other spheres; and persuade yourself and others that, give equal opportunity, women are on par with men.
(i) Which values are displayed by team ‘B’?
The values displayed by team B are:
(a) Gender equality.
(b) Justice and support to women.
(c) Equal opportunities to women.

(ii) Do you think team B was correct? Give proper justification with examples.
Yes, there is no difference in the capacity of women and men as far as working capacity, intelligence, decision making is concerned. It is a biological fact that the development of the human brain does not depend upon sex but on nutrition contents and heredity. If equal opportunities are afforded to both men and women, then the female mind will be as efficient as the male mind.

The list of successful women from various fields is as under Kalpana Chawala, Madame Curie, Indira Gandhi, Mother Teresa, Margaret Thatcher, etc.

Question 2.
Two friends of class XI Mohan and Raghav were discussing the role of physics in society. Mohan Said to Raghav that physics does not have any impact on our society. Since he was a student of Arts Stream. But Mohan who is a science student explained that physics and Society are directly linked because whatever is discovered in physics, it immediately affects society.

The latest technology has played an important role in the fields of communication such as radios, computers, TVs, mobile phones and connects the people with each other. Nuclear energy has brought a profound change in the thinking and living style of human beings.
(i) Which values are displayed here by Mohan?
The values displayed by Mohan are :
(a) Intelligence
(b) Awareness
(c) Logical

(ii) Give some discoveries which have really affected society more.
(a) Theory of relativity
(b) X-ray
(d) Photoelectric effect
(e) Steam engine (based on laws of thermodynamics)
(f) Computer (Based on Digital logic)
(g) Super Conductivity

Question 3.
India has had a long and unbroken tradition of great Scholarship in mathematic, astronomy, linguistics, logic, and ethics. Yet, in parallel with this, several superstitious and obscurantist attitudes and practices flourished in our society and unfortunately continue even today-among many educated people too.

A student of Class XI gave an idea to overcome this evil of our society by using scientific explanations through mass media, radio television, and newspapers.
(i) Which values are displayed by the student?
(a) Awareness
(b) Scientific and logical ideas
(c) Concern for the upliftment of our society to eradicate the evils of society.
(d) Intelligence, Sharp mind.

(ii) How will you use your knowledge of science to develop strategies to counter these attitudes?
Solution:
Educating the common man is the only way to get rid of superstitions and obscurantist attitudes. The mass media like newspapers, magazines, radio, television, etc. Can play a vital role. School and college curriculum can be suitably developed so that there is an emphasis on these topics. Teachers can play an important role to organize the seminar and motivate them.

Question 4.
A seminar was organized to know the views of students on whether the particular application is good, bad or something that cannot be so clearly categorized:
(a) Prenatal sex determination.
(b) Development of nuclear weapons.
(c) Mass vaccination
(d) Cloning
(e) Purification of water for drinking.

(i) Which values arc depicted here?
To create awareness through group activities, to develop problem-solving ability, scientific knowledge can be put to good or bad use, depending on the user.

(ii) Do you think such type of group activities are important in day-to-day life?
Yes.

(iii) Give justification for each of the above topics from (a-e):
(a) Bad, it leads to the practice of abortion in the case of the female fetus.
(b) Bad, because nuclear weapons may cause mass destruc¬tion of mankind.
(c) Good, because it helped to eradicate the disease.
(d) Bad, because it can destroy the normal family life.
(e) Good, because it will help to improve good health of the citizens.

## Units and Measurements Class 11 Important Extra Questions Physics Chapter 2

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 2 Units and Measurements Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

## Class 11 Physics Chapter 2 Important Extra Questions Units and Measurements

### Units and Measurements Important Extra Questions Very Short Answer Type

Question 1.
If the size of the atom were enlarged to the tip of the sharp pin, how large would the height of Mount Everest be?
1010 m.

Question 2.
What does the LASER mean?
It stands for Light Amplification by Stimulated Emission of Radiation.

Question 3.
If the Universe were shrunk to the size of the Earth, how large would the Earth be on this scale?
1o-11 m (size of an atom.).

Question 4.
A research worker takes 100 careful readings in an experiment. If he repeats the same experiment by taking 400 readings, then by what factor will the probable error be reduced?
By a factor of 4.

Question 5.
What is the number of significant figures in 0.06070?
4.

Question 6.
Which of the following reading is most accurate?
(a) 7000m,
(b) 7 × 102 m,
(c) 7 × 103 m
(a) i.e. 7000 m.

Question 7.
The density of a cube is calculated by measuring the length of one side and its mass. If the maximum errors in the measurement of mass and length are 3% and 2% respectively, then what is the maximum possible error in the measurement of density?
3% + 3 × 2% = 9%.

Question 8.
The mass of a body as measured by two students is given as 1.2 kg and 1.23 kg. Which of the two is more accurate and why?
The second measurement is more accurate as it has been made to the second decimal point.

Question 9.
Do the inertial and gravitational masses of ordinary objects differ in magnitude?
No.

Question 10.
Are S.I. units Coherent? Why?
Yes, because all the derived units in this system can be obtained by multiplying or dividing a certain set of basic units.

Question 11.
Do A.U. And Å represents the same magnitudes of distance?
No, 1 A.U. = 1.496 × 1011 m and 1 Å = 1010 m.

Question 12.
What does SONAR stand for?
It stands for Sound Navigation and Ranging.

Question 13.
What is the atomic mass unit (a.m.u.)?
It is defined as $$\frac{1}{12}$$th of the mass of one 6C12 atom
i.e. 1 a.m.u. = $$\frac{1}{12} \times \frac{12}{6.023 \times 10^{23}}$$ = 1.66 × 10-27 kg

Question 14.
Which is the most accurate clock?
Cesium atomic clock.

Question 15.
Write the S.I. units of the following physical quantities:
(a) Luminous intensity
Candela (cd)

(b) Temperature
Kelvin (K)

(c) Electric current
Ampere (A)

(d) Amount of substance
Mole (mol)

(e) Plane angle

(f) Solid angle

(g) Pressure.
Nm-2 = pascal (pa).

Question 16.
What is the difference between mN, Nm, and nm?

• mN means.milli-newton, 1 mN = 10-3N.
• Nm means newton-meter, 1 Nm = 1 J
• nm means namometer, 1 nm = 10-9 m.

Question 17.
If x = a + bt + ct2 where x is in meter and t in seconds, what is the unit of c?
The unit on the left-hand side is a meter so the units of ct2 should also be a meter. Since t2 has units of s2, so the unit of c is ms-2

Question 18.
Will the dimensions of a physical quantity be the same, whatever be the units in which it is measured? Why?
Yes, the dimensions don’t depend on the system of units chosen.

Question 19.
Write the dimensions of:
(i) gravitational constant
[M-1 L3 T2]

(ii) Plank’s constant
[M L2 T-1]

(iii) torque
[M L2 T2]

(iv) surface tension
[M L0 T-2]

(v) angular momentum.
[M L2 T-1]

Question 20.
Name at least two physical quantities each having dimensions:
(a) [M L-1 T-2]
Pressure and stress,

(b) [M L2 T-1]
Plank’s constant and angular momentum.

Question 21.
State the principle of homogeneity of dimensions?
It states that the dimensions of each term on both sides of an equation are the same.

Question 22.
Which are the main types of errors in a physical measurement?
Main errors are systematic error, random error, gross error, relative error, and percentage error.

Question 23.
Which one is large, the number of microseconds in a second or the number of seconds in a year?
The number of seconds in a year = 107s and the number of microseconds in a second = 106μs. So the number of seconds in a year is larger than microseconds in a second.

Question 24.
Do significant figures change if the physical quantity is measured in different systems of units?
No, significant figures don’t depend on the system of units. e.s. 250 g = 2.50 × 10-1 kg.
Both have 3 significant figures.

Question 25.
Suggest a distance corresponding to each of the following order of length:
(a) 10-4 m
Size of the atomic nucleus

(b) 10-9 m
Size of the oil molecule

(c) 104 m
Height of Mount Everest

(d) 107 m

(e) 109 m.

Question 26.
What do you understand by the following?
(a) Century
It is the largést unit of time, 1 century = 100 years.

(b) Shake
It is the smallest unit of time, 1 shake = 10-8s.

(c) Lunar month
It is the time taken by the moon to complete one revolution around the Earth, 1 lunar month = 27.3 days.

(d) Leap year
A year that is divisible by four and in which the month of February is of 29 days is called a leap year.

(e) Tropical year.
The year in which the total solar eclipse takes place is called a tropical year.

Question 27.
What is the role of power (index) of a measurable in the formula used for calculation of a quantity in regard to the error in the quantity as determined in the given experiment?
The error in the quantity becomes power times the error,
i.e. if x = pa qb. Then
$$\frac{\Delta x}{x}=a \frac{\Delta p}{p}+b \frac{\Delta q}{q}$$

Question 28.
How will you find the size of a liquid molecule?
Using the formula, t = $$\frac{\mathrm{nV}}{500 \mathrm{~A}}$$ we can find the size of the liquid molecule.
where V = volume of liquid (say oleic acid),
n = no. of drops in the solution of $$\frac{1}{500}$$ concentration
A = area of the film of oleic acid left
t = thickness of the film.

Question 29.
What do you mean by the term measurement?
Measurement means the comparison of a physical quantity with its unit to find out how many times the unit is contained in the given physical quantity.

Question 30.
Sort out the incorrect representation of units and write them
(i) m/sec
ms-1

(ii) Newton
newton

(iii) kelvin
kelvin

(iv) m.m.
mm

(v) Jk-1
JK-1

(vi) kg/m3
kgm-3

(vii) wH
Wh

(viii) gms-2
gs2

(ix) length = 5M
length = 5 m

(x) B = 4g (B = magnetic field intensity).
B = 4G

Question 31.
Define light year.
It is defined as the distance traveled by light in one year.
1 L.Y. (ly) = 3 × 108 ms-1 × 365 × 24 × 60 × 60s ≈ 9.46 × 1015 m.

Question 32.
Define Astronomical distance.
It is defined as the distance between the Earth and Sun.
1 A.U. = 1.496 × 1011 m~ 1.5 × 1011 m.

Question 33.
What is the limit of
(i) accuracy
The least count of the measuring instrument is the limit of accuracy with which a physical quantity can be measured.

(ii) error?
The error in measurement is taken equal to half the least count.

Question 34.
What do you mean by ‘Order of magnitude’?
Order of magnitude is defined as the approximation to the nearest power of 10 used to express the magnitude of a physical quantity under consideration, e.g.

1. The order of magnitude of the time interval of 1.2 × 10-6 s is -6.
2. The order of magnitude of the distance of 4.5 × 106 is +6.

Question 35.
Find the order of magnitude of a light-year.
I light year = 9.46 × 1015 m ≈ 1016m
∴ The order of magnitude of light-year is +16.

Question 36.
Derive the dimensional formula of:
(a) Angular velocity
Angular velocity = $$\frac{\text { Angle }}{\text { Time }}=\frac{1}{\mathrm{~T}}$$ = [M0 L0 T-1]

(b) Angular momentum
Angular momentum = momentot inertia × Angular velocity
= mass x (radius of rotalion)2 (Time)-1
= [M L2 T-1].

Question 37.
Derive the dimensional formula of:
(a) Impulse
Impulse = Force x Time
= [M L T-2][T]
= [M L2 T-1].

(b) Surface energy
Surfäce energy = $$\frac{\text { Energy }}{\text { Surface area }}$$
= $$\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^{2}\right]}$$
= [M L0 T-2].

Question 38.
Derive the dimensional formula of
(a) Specific gravity
Specific gravity = $$\frac{\text { Density of substance }}{\text { Density of water at } 4^{\circ} \mathrm{C}}$$
= $$\frac{\left[\mathrm{ML}^{3} \mathrm{~T}^{0}\right]}{\left[\mathrm{ML}^{-3} \mathrm{~T}^{0}\right]}$$
= [M0 L0 T0]

(b) Coefficient of viscosity .
Coefficient of viscosity = $$\frac{\text { Force } \times \text { Distance }}{\text { Area } \times \text { velocity }}$$
= $$\frac{\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]}{\left[\mathrm{L}^{2}\right]\left[\mathrm{LT}^{-1}\right]}$$
= [M L-1 T-1]

Question 39.
Derive the dimensional formula of:
(a) Universal gas constant
Universal gas constant = $$\frac{\text { Pressure } \times \text { Volume }}{\text { Temperature }}$$
= $$\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^{3}\right]}{[\mathrm{K}]}$$
= [M L T2 K-1]

(b) Specific heat.
Specific heat = $$\frac{\text { Heat }}{\text { Mass } \times \text { Temperature }}$$
= $$\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{MK}]}$$
= [M0 L2 T-2 K-1].

Question 40.
Derive the dimensional formula of:
(a) Coefficient of elasticity
Coefficient of elasticity = $$\frac{\text { Stress }}{\text { Strain }}$$
= $$\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]}{\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]}$$
= [M L-1 T-2]

(B) Boltzmann’s constant
Boltzmann’s constant = $$\frac{\text { Energy }}{\text { Temperature }}$$
= $$\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{K}]}$$
= [M L2 T-2 K-1]

Question 41.
Define dimensions of a physical quantity.
They are defined as the powers to which the fundamental units of mass, length, and time have to be raised to obtain its units, e.g. derived unit of area is [M0 L2 T0]. Thus its dimensions are 1 zero in mass, 2 in length, and zero in time.

Question 42.
Define the dimensional formula of a physical quantity.
It is defined as an expression that shows which of the fundamental units and with what powers appear into the derived unit of a physical quantity.
e.g. dimensional formula of force is [M1 L1 T2].

Question 43.
Define dimensional equation of a physical quantity.
It is defined as the equation obtained by equating the symbol of a physical quantity with its dimensional formula, e.g. [F] = [M L T-2] is the dimensional equation of force.

Question 44.
Define one kilogram.
It is the mass of platinum-iridium cylinder (90% Pt + 10% Ir) having its diameter equal to its height (both equal to 3.9 cm) kept in the International Bureau of Weights and Measures of Sevres near Paris.

Question 45.
Define one second.
It is defined as the time interval occupied by 9, 192, 631, 770 vibrations corresponding to the transition between two hyperfine levels of cesium -133 (Cs133) atom in the ground state.

Question 46.
Define one ampere.
It is defined as that constant current which when flowing through two parallel, straight conductors of the infinite length of negligible cross-section held one meter apart in a vacuum produces a force of 2 × 10-7 N/m between them.

Question 47.
Define Kelvin.
It is defined as $$\frac{1}{273.16}$$ fraction of the thermodynamic temperature at the triple point of water.

Question 48.
It is defined as the angle made at the center of a circle by an arc of length equal to the radius of the circle.

Question 49.
It is defined as the solid angle made at the center of a sphere by an area cut from its surface whose area is equal to the square of the radius of the sphere.

Question 50.
Define one mole.
It is defined as the amount of substance that contains the same number of elementary units {i.e. atoms) as there are atoms in 0.012 kg of carbon-12.

Question 51.
Define standard meter.
It is defined to be equal to exactly 1650763.73 wavelengths of orange-red light emitted in vacuum by krypton-86 atom i.e. kr86.
i. e. 1 metre = 16,50,763.73 wavelengths.
Or
It is also defined as the distance traveled by light in $$\frac{1}{299792458}$$ second.

Here 299792458 ms-1 is the exact value of the velocity of light.

For all practical purposes, c = 2.9 × 108 ms-1 = 3.0 × 108 ms-1.

### Units and Measurements Important Extra Questions Short Answer Type

Question 1.
If the size of a nucleus is scaled up to the tip of a sharp pin, what roughly is the size of an atom?
The size of a nucleus is in the range of 10-15 m to 10-14 m. The tip of a sharp pain may be taken to be in the range of 1o-5 m to 10-4 m. Thus we are scaling up the size of the nucleus by a factor of 10-5/10-15 = 1010. An atom roughly of size 10-10 m will be scaled up to a rough size of 10-10 × 1010 = 1 m. Thusanucleus in an átom is as small in size as the tip of a sharp pin placed at the center of a sphere of radius about a meter.

Question 2.
(a) What do you mean by physical quantity?
It is defined as a quantity that can be measured, e.g. mass, length, time, etc.

(b) What do you understand by:
(i) Fundamental physical quantities?
They are defined as those quantities which cannot be expressed in terms of other quantities and are independent of each other, e.g. mass, length, time.

(ii) Derived physical quantities?
They are defined as the quantities which can be expressed in terms of fundamental quantities, e.g. velocity, acceleration, density, pressure, etc.

Question 3.
(a) Define the unit of a physical quantity.
It is defined as the reference standard used to measure a physical quantity.

(b) Define:
(i) Fundamental units.
They are defined as the units of fundamental quantities. They are independent of each other and are expressed by writing the letter of the fundamental quantity in a parenthesis.
e.g. Fundamental units of mass, length and time are [M], [L], [T] respectively.

(ii) Derived units.
They are defined as those units which can be derived from fundamental units. They are expressed by writing the symbol of a derived quantity in a parenthesis.
e.g. D.U. of velocity = [u]
acceleration = [a]
pressure = [P]
work = [W] and so on.

Question 4.
Define one Candela.
It is defined as the luminous intensity in a perpendicular direction of a surface of $$\frac{1}{600,000}$$ square meter area of a black body at a temperature of freezing platinum (1773°C) under a pressure of 101,325 N/m2.

It is the S.I. unit of luminous intensity.

Question 5.
What is the advantage of choosing wavelength of light radiation as standard of length?

1. It can be easily made available in any standard laboratory as Krypton is available everywhere.
2. It is well defined and does not change with temperature, time, place or pressure, etc.
3. It is invariable.
4. It increases the accuracy of the measurement of length (1 part in 109).

Question 6.
Which type of phenomenon can be used as a measure of time? Give two examples of it.
Any phenomenon that repeats itself regularly at equal intervals of time can be used to measure time.

The examples are:

1. Rotation of earth – the time interval for one complete rotation is called a day.
2. Oscillations of a pendulum.

Question 7.
Find the number of times the heart of a human being beats in 10 years. Assume that the heartbeats once in 0.8s.
In 0.8 s, the human heart makes one beat.

∴ In 1 s, the human heart makes = $$\frac{1}{0.8}=\frac{10}{8}$$ beats.

∴ In 10 years, the human heart makes
= $$\frac{10}{8}$$ × 365 × 24 × 60 × 60 beats.
= 3.942 × 108 beats.

Question 8.
Why it is not possible to establish a physical relation involving more than three variables using the method of dimensions?
The dimensional analysis fails to derive a relation involving more than three unknown variables. The reason is that there will be more than three unknown factors in that case whose values cannot be determined from the three relations which we get by comparing the powers of M, L, and T.

Question 9.
What is the difference between accurate and precise measurement?
A given measurement is said to be accurate in relation to other similar measurements if the error involved in it is least.

A given measurement is said to be precise in relation to other similar measurements if it is taken with an instrument with the minimum least count.

Question 10.
Pick up the most accurate and most precise measurement out of (a) 50.0 m, (b) s.oe m, (e) 5.00 cm, (f) 5.00 mm.
The following table gives the relative error in count of the given measurement:

 l Δl Δl/l 50.00 m 0.01 m 0.01/50 = 0.2 × 10-3 5.00 m 0.01 m 0.01/5 = 0.2 × 10-2 5.00 cm 0.01 cm 0.01/5 = 0.2 × 10-2 5.00 mm 0.01 mm 0.01/5 = 0.2 × 10-2

Clearly, the first measurement is most accurate because the relative error in it is minimum. The fourth measurement is most precise because it is taken with an instrument having the minimum least count among the v given measurements.

Question 11.
Define one parsec.
It is defined as the distance at which an arc of length equal to y one astronomical unit subtends an angle of one second at a point.

Note: 1 L.Y. = 6.3 × 104 A.U.
1 parsec = 3.26 L.Y.

Question 12.
Define annual parallax.
It is defined as the angle (θ) subtended by the basis at the object (0). It is also called the parallactic angle.

∴ Parallactic angle = $$\frac{\text { Length of arc }}{\text { Radius }}$$
Or
θ = $$\frac{b}{s}$$

∴ s = $$\frac{b}{θ}$$

The parallax method is used to measure the distance of stars.
Here, b = basis = E1E2 distance where E1 and E2 are the two observation points on earth. θ = angle made by the star at point O. To find θ, let us observe the star 0’ simultaneously and let Φ1 and Φ2 be the angles made at E1 and E2.
∴ θ = Φ1 + Φ2
∴ s = $$\frac{b}{\phi_{1}+\phi_{2}}$$

Question 13.
Give Avogadro’s method to measure distances of the order of 10-10 m.
It is the indirect method of measuring the distances of the order of 10-10 m which is the size of an atom i.e. small distances. An atom is a tiny sphere. When such atoms lie packed in any substance; empty spaces are left in between. According to Avogadro’s hypothesis, the volume of all atoms in one gram of substance is $$\frac{2}{3}$$ of the volume occupied by one gram of the substance.

i.e. V’ = $$\frac{2}{3}$$V ….(i)
Where V = actual volume of one gram mass.
V’ = volume occupied by atoms in I gram mass.
ρ = density of the substance.
∴ V = $$\frac{1}{ρ}$$ …(ii)

Let m = atomic weight of the substance.

∴Number of atoms in I gram of the substance = $$\frac{N}{M}$$
If r be the radius of each atom, then
V’ = no. of atoms in I gram x volume of each atom
Or
V’ = $$\frac{N}{M}$$ x $$\frac{4}{3}$$ 4πr3 …. (iii)

∴ From (i), (ii), and (iii), we get

Thus, r the radius of an atom can be calculated from equation (iv).

Question 14.
What arè the characteristics of a standard unit?
The characteristics of a standard unit are as follows:

1. It should be well defined.
2. It should be of the proper size.
3. It should be easily accessible.
4. It should be reproducible in all places.
5. It should not change with time, place, and physical conditions such as pressure, tèmperature, etc.
6. It should be widely accepted.

Question 15.
What are the advantages of the S.I. system?
Following are the main advantages of the S.I. system over other systems of units i.e. (C.G.S, FPS, and MKS).
1. It is a coherent system of units i.e. a system based on a certain set of fundamental units from which all derived units are obtained by multiplication or division without introducing numerical factors i.e. units of a given quantity are related to one another by the power of 10. So the conversions are easy.

2. S.I. is a rational system of units as it assigns only one unit to a particular physical quantity e.g. Joule is the S.I. unit for all types of energies while MKS units of mechanical energy, heat energy, and electrical energy are Joule, calorie, and watt-hour respectively.

3. It is an absolute system of units: There are no gravitational units on the system. The use of factor ‘g’ is thus eliminated.

4. It is a metric system i.e. the multiples and submultiples of units are expressed as powers of 10.

Question 16.
Point out the measurable likely to create the maximum error in the following experimental measurement.
Young’s modulus ‘Y’ of the material of the beam is calculated using the relation Y = mgl3/4bd3δ
When w = mg, δ = depression, I, b, d = length, breadth, thickness.

Thus clearly m, l, b, d and 8 introduce the maximum error in the measurement of Y.

Question 17.
Classify the physical quantities on basis of their dimensional formula.
They are divided into the following four categories:
1. Dimensional variables: They are defined as the physical quantities which possess dimensions and have variable values e.g. Area, velocity, force, etc.

2. Non-dimensional variables: They are defined as the physical quantities which have no dimensions but have variable values, e.g. Angle, specific gravity, strain, sin0, cos0 (i.e. trigonometric functions).

3. Dimensional constants: They are defined as the physical quantities which have both dimensions and constant values, e.g. Plank’s constant, speed of light in vacuum, gravitational constant.

4. Dimensionless constants: They are defined as the physical quantities which do not have dimensions but have constant values, e.g. n, e, pure numbers 1, 2, 3, …… etc.

Question 18.
What are the limitations of dimensional analysis?
Following are the limitations of dimensional analysis:

1. Dimensionless constants involved in the physical relationship can not be determined.
2. It fails to derive the relations involving plus or minus signs like s = ut + $$\frac{1}{2}$$at2,
v = u + at,
v2 — u2 = 2as etc.
3. It fails to derive a relation involving more than three physical quantities.
4. This method does not help to derive the relations containing exponential and trigonometrical functions.
5. This method gives no information on whether a quantity is a scalar or vector.
6. It does not tell about the physical correctness of a relation.

Question 19.
Define significant figures. State the rules for determining the number of significant figures in a given measurement.
They are defined as the number of digits up to which we are sure about their accuracy. In other words, they are defined as those digits that are known correctly in an experimental observation plus one more digit that is uncertain.

Following are the rules for determining significant figures:

1. All non-zero digits are significant.
2. All zeros occurring between non-zero digits are significant.
3. All zeros to the right of the last non-zero digit are not significant.
4. All zeros to the right of a decimal point and to the left of a non¬zero digit are not significant, e.g. 0.000879 has (3) significant figures.
5. All zeros to the right of a decimal point and to the right of a non-zero digit are significant, e.g. 0.2370 contains (4) sig-fig.
6. All zeros to the right of a non-zero digit and to the left of an expressed decimal are significant, e.g. 21900. has (5) sig-fig.

Question 20.
Mankind has existed for about 106 years whereas the s. the universe is about 1010 years old. If the age of the universe is taken to be one day, how many seconds has mankind existed?
Since

Mankind has existed for 8.64s on the new chosen scale.

Question 21.
State and explain the rule for finding the maximum possible error in a result.
The maximum possible error is found in different ways in different types of results as follows:
(a) When the result involves the sum of two quantities,
i. e. if x = p + q
then, maximum possible error = maximum absolute error in first quality + maximum absolute error in the second quality. i.e. Δx = Δp + Δq

(b) When the result involves the difference of the quantities
i.e. if x = p – q
then, maximum possible error in x = maximum absolute error in p + maximum possible error in q
i.e. Δx = Δp + Δq

(c) When the result involves the multiplication of two quantities i.e. x = pq; then maximum relative error in x is given by
$$\frac{\Delta x}{x}=\frac{\Delta p}{p}+\frac{\Delta q}{q}$$

(d) When the result involves the quotient of tw«j observed quantities i.e. when x = p/q, then Maximum relative error in x is given by
$$\frac{\Delta x}{x}=\frac{\Delta p}{p}+\frac{\Delta q}{q}$$

(e) When the result involves some power oFa physical quantity, i.e. when x = pa qb, then maximum absolute e> u r in x is given by
$$\frac{\Delta x}{x}=a \frac{\Delta p}{p}+b \frac{\Delta q}{q}$$

i.e. maximum possible relative error in x = power × relative . error in p + power × relative error in q.

Thus the error is always additive in nature and maximum, permissible error is equal to the sum of maximum possible error in the i. individual quantities of that quantity.

Question 22.
How do you represent very large and very small physical y quantities? Write the prefixes, multiple, submultiple, and their symbols in a tabular form.
The very long and very small quantities are written in powers of 10. The prefixes are tabled below:

 S.No. Power of 10 Prefix Symbol 1. 1012 Tera (Trillion) T 2. 109 Giga (billion) G 3. 106 mega(million) M 4. 103 kilo K 5. 102 Hecta h 6. 101 Deca da 7. 10-1 Deci d 8. 10-2 Centi c 9. 10-3 mull m 10. 10-6 micro u 11. 10-9 nano n 12. 10-12 pico p 13. 10-15 femto f 14. 10-18 Atto a

Question 23.
Explain the importance of reference frames in measurements.
(a) All measurements are made with reference to a point or portion i.e., a frame of reference.
(b) The number of time units contained in a physical quantity gets changed with a change in the reference frame.
(c) The same physical quantity may have different values in different reference frames.

Question 24.
Briefly describe the various techniques to measure time.
The various techniques to measure time are:
(a) The synchronous motor run on a.c. of frequency 50 hertz is used to measure time as its rotation provides a time scale.

(b) Electronic oscillations: The semiconductor oscillators produce high-frequency oscillations i.e. of a very small time period. These oscillations thus can be used to measure the small time intervals.

(c) Quartz-crystal clocks: These clocks work on the piezoelectric effect. The oscillations so produced can be used to measure time intervals.

(d) Atomic clock: These are usually Cs-133 atom clocks whose electrons make a definite number of repeated jumps between two energy levels. These are very precise and are used to define second, so time can be measured.

(e) Decay of elementary particles: The study of decay can provide a scale for the measurement of very small intervals of time as unstable elementary particles decay between 10-16 to 10-24 seconds.

(f) Radioactive dating: Very long intervals of time can be measured by comparing them with the techniques of radioactive dating. The half-life period of decay of carbon is a standard time frame to determine the age of fossils etc.

Question 25.
Explain the rules for finding significant figures in the sum, difference, product, and quotients of true numbers.
The rules for counting significant figures in algebraic operations are given below:
1. Addition and Subtraction: The sum or the difference of two numbers has significant figures only in those places where these are in the least precise amongst the given number. For example, if we subtract 45.7 from 46.9267 the result is 1.2267. But it should be written only 1.2 because the least precise of the two numbers is 45.7 and there is only one digit after the decimal.

Similarly in the sum of numbers 4205, 112.39, 77.93, and 213.2532, the correct result is 824.0 to the significant figures. So, in addition, or subtraction the same number of the decimal. places are retained in the result as are present in the number with the minimum amount of decimal places.

2. Multiplication and Division: The product or quotient of two numbers does not have more significant figures than are present in the least precise of the given numbers. e.g. in the product of two quantities 0.025 with 40, we get 1.000 but the answer is to be written as 1.0 because there are two significant figures in 40 the least of the two numbers. Similarly in a quotient when we divide 16.775 by 2.5, we get 6.71. The result of the significant figure will be 6.7.

Question 26.
Is it possible for an equation to be dimensionally correct still to be wrong? If so indicate the number of ways in which this might happen.
It is possible that an equation may be dimensionally correct but physically it is wrong. For example the displacement of a particle moving with velocity u and acceleration ‘a’ after time t is given by
s = ut + 1/2at2

However, s = 1/2at2 is dimensionally correct, as [L] = [LT2][T2]2 shows dimensions on LHS = dimensions on RHS. Therefore, in certain circumstances, an equation may be dimensionally correct but actually, it is not physically correct. This happens especially in equations involving the sum and difference of two or more terms and in functions involving trigonometric functions. For example, the displacement y maybe y = a where a is the amplitude. This is dimensionally correct but does not give the full picture of the physical phenomenon.

The correct equation is

y = a sin wt or y = a sin $$\frac{2 \pi}{\lambda}$$ (vt + x) etc.

Question 27.
Show that bigger is the unit smaller is the numerical value of physical quantity and vice-versa.
In any system of unit, the following relations hold good
n1u1 = n2u2

Where n1 and n2 are the numerical values, u1 and u2 represent the unit of a physical quantity. This relation is based on the simple thing that the magnitude of a physical quantity remains the same in different systems of units.
Thus nu = constant or n ∝ $$\frac{1}{u}$$

If larger the n smaller will be the unit (u) and smaller the numerical value (n) Larger will be u. E.g. if the length of a rod l be 1 m in the S.I. system. Then it is 100 cm in the C.G.S. system i.e.

l = 1 m = 100 cm. Here clearly 1 < 100 and m > cm. Hence proved.

Question 28.
The difference in the order of magnitude of the longest and shortest distances and most massive and least massive objects are quite different. Write the same and compare them.
The size of the nucleus is of the order 10-15 m and the intergalactic distance is of the order of 1022 m. So, the ratio between the longest and the shortest distance is:
$$\frac{10^{22} \mathrm{~m}}{10^{-15} \mathrm{~m}}$$ = 1037

The smallest mass is the mass of an electron of the order of 10-30 kg and the longest mass is the mass of the galaxy of the order of 1042 kg. So, the ratio of maximum to minimum mass is
$$\frac{10^{42} \mathrm{~kg}}{10^{-30} \mathrm{~kg}}$$ = 1072

Now the ratio between the mass ratio and distance ratio is
$$\frac{10^{72}}{10^{37}}$$ = 1035

The ratio betwe< n minimum mass and minimum distance is
$$\frac{10^{-30} \mathrm{~kg}}{10^{-15} \mathrm{~m}}$$ = 10-15

The ratio between maximum mass and minimum distance is
$$\frac{10^{42} \mathrm{~kg}}{10^{22} \mathrm{~m}}$$ = 1020

Question 29.
Define the terms
(i) mean absolute error
The arithmetic means of all the absolute errors is known as mean absolute error.

(ii) relative error and
The relative error is the ratio of mean absolute error to the mean or true value of the quantity measured.
Relative error = $$\frac{\Delta \mathrm{a}_{\text {mean }}}{\mathrm{a}_{\text {mean }}}$$

(iii) percentage error.
Percentage error is the relative error expressed in percent
Percentage error = $$\frac{\Delta \mathrm{a}_{\text {mean }}}{\mathrm{a}_{\text {mean }}}$$ × 100

Question 30.
State the rules applied in rounding off measurements.
While rounding off measurements, the following rules are applied:
1. If the digit to be dropped is smaller than 5, then the preceding digit should be left unchanged.
e.g. 7.93 is rounded off to 7.9.

2. If the digit to be dropped is greater than 5, then the preceding digit should be raised by 1.
e.g. 17.26 is rounded off to 17.3.

3. If the digit to be dropped in 5 followed by digits other than zero, then the preceding digit should be raised by 1.
e.g. 7.351 on being rounded off to first decimal becomes 7.4.

4. If the digit to be dropped in 5 or 5 followed by zeros, then the preceding digits are not changed if it is even.
e.g. (a) 3.45 on being rounded off becomes 3.4.
(b) 3.450 on being rounded off becomes 3.4.

5. If the digit to be dropped in 5 or 5 followed by zeros, then the preceding digit is raised by one if it is odd.
e.g. (a) 3.35 on being rounded off becomes 3.4.
(b) 3.50 on being rounded off becomes 3.4.

Note: Rules (iv) are based on the convention that the number is to be rounded off to the nearest even number.

### Units and Measurements Important Extra Questions Long Answer Type

Question 1.
State the rules for writing the units of physical quantities in the S.I. system.
While writing the units of physical quantities following rules are followed with S.L units:
(1) The S.l. units are written in the form of symbols after the number i.e., number of time, the unit is contained in the physical quantity so that physical quantity = nu

With symbols, certain rules are laid down:

• Units in symbols are never written in plural i.e., meters is only m and not ms, years is y.
• The units based on the name of the scientists are written beginning with small letters and with capital letters in symbolic form viz, weber (Wb), newton (N), etc.
• No full stop is used at the end of the symbol.
• Symbols of units not based on the name of scientists are written as small letters viz. kilogram (kg), second (s), etc.

(2) Bigger and smaller number of units are represented with symbols corresponding to the power of 10 viz. 106 is mega (M), 1012 is Tera (T), 10-3 is milli (m), 10-9 is nano (n), etc.

(3) All units are written in numerator viz. kg/m3 is kg m, Nm2c2.

(4) The units are written within parenthesis in graphs below the corresponding taxes viz. (ms-1) and (s) in the velocity-time graph.

(5) Units of a similar physical quantity can be added or subtracted.

Question 2.
Explain the Triangular method.
It is used to measure the distance of an accessible or inaccessible hill or a tower by measuring the angle which the object makes at point P (say)

Let x = distancy of point P from the foot of tower = PA .
∴ h = x tan θ

It is also used to measure the distance of an inaccessible object eg. a tree on the other bank of a river.

Let h = height of the inaccessible object.
Let θ1, θ2 = be the angle made at P and Q by the object.
Let PA = d, PQ = x.

∴ In ΔPAB and ΔQAB,

Question 3.
What are the uses of dimensional analysis? Explain each of them.
Dimensional analysis is used for:
(a) checking the dimensional correctness of the given physical equation or relation.
(b) converting one system of units to another system.
(c) deriving the relationship between various physical quantities.

(a) checking of the dimensional correctness of a physical relationship is done by using the principle of homogeneity of dimensions. If the dimensions of M, L, T of each term on R.H.S. are equal to the dimensions of M, L, T of each term on L.H.S., then the given- physical relation is dimensionally correct, otherwise wrong.

(b) conversion: It is based on the fact that the magnitude of a physical quantity remains the same whatever may be the system of units, i.e. n1u1 = n2u2.

are the units of M, L, T in the first and second system of units of a physical quantity having dimensions of M, L, T, and a, b, c respectively.

Thus if fundamental units of both systems, dimensions of the quantity, and its numerical value n1 in one system, are known then we can easily calculate n2 in another system.

(c) Derivation of a relationship between various physical quantities is based on the principle of homogeneity of dimensions.

Following are the steps used:

1. We must Know the physical quantities (say p, q, r) upon which a physical quantity say x depends.
2. We must know the dimensions of p, q, r say a, b, c respectively.
3. Now, write the dimensions of each physical quantity on both sides of the equation
4. and compare the powers of M, L, T to find a, b, c. Putting values of a, b, c in the equation
5. we get the required relation.

Numerical Problems:

Question 1.
The average distance from Earth to the Sun is 1.49 × 1011 m. Find out the value of 1 percent in m.
According to the definition of parsec,

Question 2.
The parallax of a heavenly body measured from two points: diametrically opposite on the Earth’s equator is 60 seconds. If the radius of the Earth is 6.4 × 106 m, determine the distance of the heavenly body from the center of Earth. Convert this distance in AU.
Here

∴ The distance of the heavenly body from Earth is given by

Question 3.
Find the height of a rock mountain if on moving 100 m towards the rock in the horizontal direction through the base of
the rock, the angle of elevation of itš top increases from 300 to 45°.
Here d = 100 m, θ1 = 30°, θ2 = 45°, h =?

Question 4.
Find the number of air molecules in a room of volume 12 m3. Given 1 mole of air at N.T.P. occupies a volume of 22.4 liters.
No. of atoms in I mole of air = N
= 6.023 × 1023

∴ Also at N.T.P., 1 mole of air occupies 22.4 liters of volume
= 22.4 × 10-3 m3

∴ No. of molectiles ir 22.4 × 10-3 m3 volume = 6.023 × 1023
∴ No. of molecules in 12 m3 volume = $$\frac{6.023 \times 10^{23}}{22.4 \times 10^{-3}}$$ × 12
= 3.23 × 1026

Question 5.
(a) Convert ION into dyne using dimensional analysis.
Newton (N) and dyne are the S.I. and C.G.S. units of force having dimensional formula [MLT-2]
∴ a = 1, b = 1, c = – 2

S.I. system
n1 = 10
M1 = 1 kg
L1 = 1 m
T1 = 1 s

C.G.S. system.
n2 = ?
m2 = 1 g
L2= 1 cm
T2 = 1 s

(b) Find the units of length, mass, and time of the unit of force, velocity, and energy are 100 dynes, 10 cms-1, and 500 erg respectively.

Question 6.
Check the dimensional correctnesïof the relation
V = $$\sqrt{\frac{2 G M}{R}}$$
It is done using the principle of homogeneity of dimensions i.e. if the dimensions of each term of both sides of the equation are the same, Then it is dimensionally correct.
Now dimensions of V = [LT-1].
G = [M-1 L3 T2]
M = [M]
R = [L]

∴ Dimensions of L.H.S. = [V] = [LT-1] …(f)
Dimensions of R.H.S = $$\sqrt{\frac{G M}{R}}$$

Now from equations (i) and (ii), it is clear that the dimensions of L.H.S = R.H.S, so the given relation is dimensionally correct.

Question 7.
Suppose that the oscillatións of a simple pendulum depend on
(i) mass of the bob (m),
(ii) length of the string (1),
(iii) acceleration due to gravity (g) and (iv) angular displacement
(iv) Dimensionally show which of the above factors here an influence upon the period and in what way?
Let t ∝ ma lb gc θd
or t = k’ ma lb gc qd ….(i)
here k’ is a dimensionless constant.

Since q is dimensionless, hence equation (i) reduces to
t = K ma lb gc …(ii)

where K. = k1d is another dimensionless constant.
Now writing the dimensional formula of each physical quantity on both sides of equation (ii), we get
[M0 L0 T-1] = [M]a [L]b [LT2]c = [Ma Lb-c T-2c]

Comparing dimensions of

Thus from equation (vii), we see that the period of the pendulum is directly proportional to the square root of the length of string, and inversely proportional to the square root of acceleration due to gravity and is independent of the mass of the bob.

Question 8.
Given that the period T of oscillation of a gas bubble from an explosion underwater depends on P, d, and E, where the symbols are pressure, density, and total energy of the explosion. Find dimensionally a relation for T.
Let T ∝pa db Ec
or
T = k pa db Ec …(i)
where k is a dimensionless constant.

Writing dimensional formula of each physical quantity on both sides of equation (i), we get

Question 9.
If the velocity of light c, the constant of gravitation G, and Plank’s constant h be chosen as fundamental units, find the value of a gram, a centimeter, and a second in terms of new units of mass, length, and time respectively.
Given c = 3 × 1010 cms-1
G = 6.67 × 10-8 dyne cm2 g-2
h = 6.6 × 10-27 ergs.

Question 10.
Assuming that the mass m of the largest stone that can be moved by a flowing river depends on the velocity y density p and acceleratión due to gravity g. Show that m varies as the sixth power of the velocity of flow.
Let, m ∝ va
m ∝ρbb

where k is a dimensionless constant.

Writing dimensions of each physical quantity on both sides of equation (i), we get
[M1 L0 T0] = [LT-1] [ML-3]b [LT-2]c
[Mb L a-3b+2c T-a-2c]

Comparing powers of:

Question 11.
The radius of Earth is 6.37 × 106 m and its average density is 5.517 × 103 kg m-3. Calculate the mass of Earth to correct significant figures.
M = pV
Here, R = 6.37 × 106 m,
ρ = 5.517 × 103 kg m3

Here R has 3 significant figures and the density has four. Thus the final result should be rounded off to 3 significant figures.

Hence, M = 5.97 × 1024 kg.

Question 12.
The length, breadth, and thickness of a block of metal were measured with the help of Vernier Callipers. The measurements are:
l = (5.250 ± 0.001) cm
b = (3.450 ± 0.001) cm,
t = (1.740 ± 0.001) cm.
Find the percentage error in the volume of the block.
The volume of the block is given by:
V = lbt

∴ The relative percentage error in V is given by:

Question 13.
An experiment measures quantities a, b, c, and x is calculated from the relation:
x = ab2/c3
The percentage errors in a, b, c are ± 1%, ± 3%, and ± 2% respectively what is the % error in x?

Question 14.
The radius of the proton is about 10-9 microns and the radius of the universe is about 1028 cm. Name a physical object whose size is approximately halfway between these two extremes on the logarithmic scale.
Radius of proton, x1 = 10-9 microns = 10-9 × 106m .
= 10-15m

Radius of universe, x2 = 1028 cm = 1026m.

If x be the size halfway between these two extremes on the logarithmic scale is:

Since 106m is the order of the size of the Earth. So the physical object is Earth.

Question 15.
The velocity v (cms-1) of a particle is given in terms of time t (s) by the equation y = at + $$\frac{b}{t+c}$$. What are the dimensions of a, b and c?
v = at + $$\frac{b}{t+c}$$.

L.H.S. is velocity, so according to principle of homogeneity t, of dimensions, at and must have the dimensions of $$\frac{b}{t+c}$$ velocity.

Question 16.
The young’s modulus (Y) of a material is given by the relation Y = $$\frac{\mathbf{M g L}}{\pi \mathbf{r}^{2} \boldsymbol{l}}$$. If the percentage error in W(= mg), L, r and l are 0.5%, 1%, 3% and 4% respectively, what is the percentage error in Y? If the calculated value of Y is 18.79 × 1011 dyne cm-2, to what value should we round off the result?
The total percentage error in Y is given by:

Since Y = 18.79 × 1011 dyne cm-2, it can be rounded off to
Y = 19 × 1011 dyne cm-2

The total percentage error in this result is:
$$\frac{19 \times 11.5}{100}$$ = 2.2

So, the result of Y can be quoted as (19 ± 2.2) × 1011 dyne cm-2

Question 17.
The refractive index as measured by relation:
μ = $$\frac{\text { real depth }}{\text { apparent depth }}$$ was found to have values 1.29, 1.33, 1.34, 1.35, 1.32, 1.36, 1.30, 1.33.
Find the mean value of μ, the mean value of the absolute error, the relative error, and the percentage error.
Mean value of μ

Question 18.
Derive the dimensions of a/b in the relation:
F = a$$\sqrt{x}$$ + bt2
where F is the force, x is the distance and t is the time.
Here, F= a$$\sqrt{x}$$ + bt2 …(i)

According to the principle of homogeneity of dimensions, a $$\sqrt{x}$$ and bt2 should have dimensions of F.

Question 19.
Calculate the number of seconds in a:
(i) day
(ii) year
and express them in orders of magnitude.
(i) No.of seconds in a day = 24 × 60 × 60s
= 86400s
= 8.64 × 104s
= 0.864 × 105s
∴ order of magnitude = 5

(ii) No.of seconds in a year = 365 days
= 365 × 0.864 × 105s
= 315360 × 105s
= 31536 × 107s
∴ order of magnitude = 7.

Question 20.
A stone is lying in a fluid stream. The force F acting on it depends on the density of the fluid δ, the velocity of flow v, and the maximum area of cross-section A perpendicular to the direction of flow. Find the relation between the force F and the velocity v.
Let, F ∝ ρa
∝ vb
∝ Ac
or
F ∝ pa vb Ac
or
F = k pa vb Ac …(i)

Now writing the dimensional formula of each physical quantity on both sides of equation (i), we get:
[MLT-2] = [ML-3]a [LT-1]b [L2]c
= [Ma L-3a+b+2c T-b] …(ii)

Comparing dimensions of M, L, T on both sides of equation (ii), we get:

Question 21.
If we suppose the velocity of light (c), acceleration due to gravity (g), and pressure (p) as the fundamental units, then And the dimensional formula of mass in this system of units.

Question 22.
The density p of a piece of metal of a mass m of and volume V is given by the formula, ρ = m/V.
If, m = 325.32 ± 0.0lg
V = 136.41 ± 0.01 cm3.
Find the percentage error in p:
Here, m = 325.32g, Δm = 0.01 g
V = 136.41 cm3, ΔV = 0.01 cm3

Question 23.
The specific resistance σ of a circular wire of radius r cm, resistance R Ω and and length L is given by:
σ = $$\frac{\pi r^{2} R}{L}$$
If, r = 0.20 ± 0.02 cm
R = 20 ± 1Ω
L = 80 ± 0.01 cm, then find the % error in σ.
% error in σ is given by:

Question 24.
If S2 = a t4, where S is in meters, t in second, then find the unit of a.
Here, S2 = a t4
According to the principal of homogeneity of dimensions,
at4 = m2
∴ a = $$\frac{\mathrm{m}^{2}}{\mathrm{t}^{4}}$$ = m2 s-4

Question 25.
Calculate the value of 600m. + 600mm with due regard to the significant digits.
Here, because there is no significant digit after the decimal point in 600m, so 600m + 600mm.
= 600m + 600 × 10-3m
= 600m + 0.600m
= 600.6m = 601m.

Value-Based Type:

Question 1.
The teacher of class XI asked Madan and Prathiva to find the distance of the moon from the Earth. Pratibha said it is impossible to find. But Madan was excited to know. He observed the moon from two diametrically opposite points A and B on Earth. The angle q subtended at the moon by the two directions of observation is 1°54′. Given the diameter of the Earth to be about (1.276 × 107 m).
(i) Which values arc depicted by Madan?
The values depicted by Madan are :
(a) Curiosity
(b) Sincerity
(c) Willing to know and implement the scientific ideas.

(ii) Which mathematical concept is used in the above problem?
Parallax method

(iii) Compute the distance of the moon from the Earth?

We have

Question 2.
Muesli went to London to his uncle who is a doctor over there. He found that the currency is quite different from India. He could not understand the pound and how it is converted into rupees. He asked an English man that how far is central London from here.

He replied that it is about 20 miles. Mukesh was again confused as he never used these units in India. When his uncle came back from his clinic he curiously inquired all about it. His uncle told him about the unit system used in England. He explained that here F.P.S system is used.

It means distance is measured in the foot, Mass in the pound, and time in seconds. But in India, it is an MKS system.
(i) What values are depicted by Mukesh?
(a) Curiosity
(b) Willing to know
(c) Intelligence

(ii) How many types of unit systems are there?
The unit system is:
(a) CGS (centimeter, gram and second) system
(b) FPS (foot, pound and second) system
(c) MKS (meter, kilogram, and second) system

Question 3.
Two friends Sachin and Dinesh are confused as a book with many printing errors contains four different formulas of the displacement ‘y’ of a particle undergoing a certain periodic motion.
(a) y=asin 2πt/T
(b) y = asinvt

[a = Maximum displacement of the particle,
v = Speed of the particle, T = time-period of motion]

Sachin told that he has no idea whereas Dinesh ruled out the wrong formulas on dimensional grounds.
(i) which values are displayed by Dinesh?
the values displayed by Dinesh are:
(a) Sincerity
(b) Curiosity
(c) application of knowledge

(ii) which one of the above is correct? Give justification to support your answer.

L.H.S = R.H.S ; Hence the equation is correct.

## International Business Class 11 Important Extra Questions Business Studies Chapter 11

Here we are providing Business Studies Class 11 Important Extra Questions and Answers Chapter 11 International Business. Business Studies Class 11 Important Questions with Answers are the best resource for students which helps in class 11 board exams.

## Class 11 Business Studies Chapter 11 Important Extra Questions International Business

Question 1.
Importance of External Trade: Due to unequal distribution of natural resources and skills of different countries, foreign trade is the only solution to specialise in the production of those goods for which a large number of resources and facilities available in a country and export the surplus production to other countries and simultaneously make imports other goods from some other country.

Foreign trade makes available the goods to the consumers of countries where they are not produced. Thus, it improves the standard of living of the people. Foreign trade is also important for the economic development of a nation. Capital equipment and scarce raw materials can be imported. Similarly, surplus commodities can be exported to other countries and foreign exchange may be earned.

1. Optimum use of Resources: Foreign trade leads to the international division of labour and specialisation. It reduces wastage of resources resulting from the production of uneconomic goods. The resources are also used efficiently.

2. Standard of Living: International trade helps the people living in different countries to raise their standard of living by providing goods and services which cannot be produced economically in a particular country.

3. International Relations: Foreign trade makes different countries dependent upon each other. A country having surplus products can sell its surplus stock to the deficient countries and a country having a deficiency of a product can import it from another country. This promotes goodwill and cordial relations among the nations of the world.

4. Stabilisation of Prices: Foreign trade leads to stabilisation of prices of commodities throughout the world by adjusting demand and supply. This would not have been possible in the absence of foreign trade.

5. Employment: Foreign trade helps in increasing employment opportunities in the export-oriented industries.

6. Economies of Large-scale: Foreign trade facilitates the specialisation of a country in the production of certain goods. This will help to carry on the production of some commodities not only for home consumption but also for external consumption. This will lead to several economies of large scale production. The resources will also be utilised in a better way.

7. Growth of Economy: Under-developed and developing countries can exploit their unutilised natural resources with the import of technical know-how, machinery and equipment from the advanced countries.

Question 2.

Question 3.
Explain the various terms used in foreign trade.
Some of the important terms used in export trade are given below –

1. Free on Board (FOB): The importer has to bear all costs and risks of loss or damage from the port of shipment.
2. Cost and Freight (C&F): Under this contract, the exporter is expected to deliver the goods at the port of shipment. The freight charges are payable by the exporter. The importer bears the risk of loss or damage to the goods after this destination. C&F price consists of the FOB price plus freight charges.
3. Cost, Insurance and Freight (CIF): Under CIF contract, the exporter bears the costs and freight for bringing the goods to the port of destination. It includes charges of insurance against the risks of loss or damage to the goods during transit.

Question 4.
Differentiate between Bill of Lading and Charter Party.
The distinction between Bill of Lading and Charter Party:

 Points of Distinction Bill of Lading Charter Party 1. Receipt It is a receipt of goods on board the ship. It is not a receipt of goods. 2. Document of title It is a document of title to the goods. It is not a document of title. 3. Transferability It can be transferred freely by endorsement and delivery. It cannot be transferred. 4. Lease It is not a lease of the ship. It is a lease of the ship. 5. Use It is used when a part of the ship is to be hired. It is used when the whole ship is to be hired. 6. Stamp It requires a 25 paise stamp. It requires a stamp of higher value. 7. Control Shipowner always retain control on the ship Shipowner loses control of the ship for a temporary’ period. 8. Crew Master and crew remain agents of the shipowner. Master and crew become agents of the exporter for a temporary’ period. 9. Types (a) Clean (b) Foul (a) Voyage charter (b) Time charter 10. Collateral security It can be used as collateral to borrow money. It can no he used as a collateral-dl to borrow none’. 11. Clauses and Warranties Does not mention loading and unloading day and lay days. Mentions days allowed for loading and unloading and lay days.

Question 5.
Explain the various methods of payment in External Trade.
External Trade payment can be made through various methods. The important methods for payment of international trade are as follows:

2. Open Account
3. Documentary Bills
4. Letter of Credit
5. Direct Remittance
6. International Credit Card.

1. Advance Payment: Advance payment means the payment made along with the order by the importer as the exporter is always interested in advance payment of the goods exported. The importer can send payment to the exporter by means of:
(a) Bank Draft.
(b) International Money Order.
(c) Telegraphic/Mail Transfer.
(d) Electronic Transfer.

2. Payment against Open Account: Exporter generally ship the goods and send the shipping documents to the importer and make debit his account for the payment due on goods sent. The importer may make periodic payments against his account. This is convenient for both the exporter and the importer. The exporter is relieved of the botheration of drawing and discounting the bills of exchange. The importer is relieved of the botheration of accepting bills of exchange or getting letters of credit in favour of the exporter.

3. Documentary Bill: The exporter draws a bill of exchange on the importer. The bill may be

• Sight Bill, or
• Usance Bill.

In case of a sight bill, the importer has to make the payment immediately and obtain the shipping documents/ This mode is known as Documents against Payment (D/P). In case of a usance bill, the exporter is given some time to make the payment. However, documents are passed on to him against acceptance of the bill. It is known as Documents against Acceptance (D/A) bill.

The exporter draws a bill of exchange on the importer and attaches to it shipping documents such as a bill of lading, insurance policy, invoice, consular invoice, certificate of origin, and certificate of quality. Such a bill is known as a documentary bill. It is sent through the exporter’s bank which will present it to the importer through his bank or agent. Documentary bills may be of two types:

1. D/A Bill: In case of documents against acceptance, the bank will hand over the shipping documents to the importer when the latter gives acceptance on the bill of exchange.
2. D/P Bill: In case of documents against payment bill, the documents are to be released by the bank only on the payment of the bill either at the time of presentation or within a specified period of time.

4. Letter of Credit: The exporter can request the importer to open a letter of credit with his bank in favour of the exporter. When such an arrangement is made, the importer’s bank will accept the bill of exchange drawn by the exporter under the terms of the letter of credit. After acceptance, the bill is returned to the exporter who can get it discounted with his bank. He can also wait till the period of maturity and instruct his bank to collect payment on its maturity from the importer’s bank.

5. Direct Remittance: Under this method, the exporter sends the goods to the importer and also passes on the shipping documents to him. After receipt of documents, the importer can remit the payment to the exporter through the banking channel or telegraphically. This method involves risk for the exporter as the importer may not send the payment in time.

The modes of direct remittance are as follows –
1. Bank Draft: Bank draft is a popular method of making payment in respect of foreign transactions. A bank draft is issued in favour of the exporter by a commercial bank on receipt of the necessary amount.

In fact, a bank draft is a sort of cheque drawn by a bank on its foreign branch, directing the foreign branch to pay the specified amount of money in a particular currency to the person named in the bank draft. The bank charges a nominal commission for providing this service.

2. Telegraphic Transfer: Under this method, there is a transfer by the telegraph or cable of bank deposits from one country to another. As in case of bank draft, the money is deposited with the importer’s bank which will in return send a telegram to its bank in the exporter’s country to pay a specified amount either to the exporter or his bank.

This method of making the transfer is adopted only when the traders are in a hurry to settle accounts. This method is costlier as compared to other methods. That is why it is not commonly used.

6. International Credit Card: Multinational banks issue international credit cards to importers whose financial position are very sound. They make payment through these cards to the exporter for the goods shipped.

Question 6.
What is special Economic Zones (SEZ’s) Explain their benefits in brief?
Special economic zones have been set up with a view to encouraging free trade for the pair of promotion of exports. A special Economic Zone is a duty-free enclave deemed to be foreign territory for the purpose of trade operations and duties and tariff. Goods going to SEZ area shall be treated as “deemed exports.” Goods coming from the SEZ area into DTA (Domestic Tariff Area) shall be treated as imported goods. An SEZ may be set up in the public, private or joint sector or by the state government as notified by the control government.

Special Economic Zones (SEZs) Meaning & Functions: A SEZ is a specifically delineated duty-free enclave and is deemed as a foreign territory for the purposes of trade operations, duties and tariffs. SEZ units may be set up for manufacture of goods and rendering of services, products, processing, assembling, trading, repair, remarking, reconditioning, re-engineering including making of gold/silver/platinum jewellery or tides thereof or in connection therewith.

The EPZs are being converted into SEZs. At present, there are four SEZs in India. The only difference between EPZs and SEZs is that in an EPZ customs permission is necessary for taking the raw material from one place to another, but it is not needed in an SEZ.

The SEZs are not subjected to any predetermined value additions, export-obligations, input/output wastage norms and are treated as outside the customer’s territory.

Under the Export-import (EXIM) Policy 2002-2007, SEZs would get income tax benefits. The advantage would be that the tax concession; would be available for the full tax concession period and not curtailed by the expiry of a notified date as happens now. Currently, income tax concessions for SEZs are for 10 years or 2009-10 whichever is earlier. Offshore banking has been presented in India for the first time.

This will provide units in the SEZ access to funds from abroad at international rates. This means that the cost of funds for the exporters from the SEZs would be less. Exporters in the SEZs wanting to raise ADRs/GDRs also need not go to the US and Europe. Sitting here, they will be able to do so using the overseas banking units (OBUs) set up in the SEZs. Units in the SEZs would be able to get world-class security in their backyard itself without having to go looking for it elsewhere. SEZs in India would become more attractive to foreign investors.

SEZs: described as the “best of our dream projects”: would also benefit in a big way with the government deciding to treat bank branches in these zones as overseas branches free of CRR, SLR and priority sector lending requirements. This would help SEZ units, as well as developers in bringing down the cost of funds as overseas branches of Indian banks, are in a position to lend at much lower rates than those prevailing in the Indian market.

SEZ units will also have the freedom to carry out hedging in commodities, make liberal overseas investments out of their export earnings and borrow overseas without being hindered by existing regulations. Ultimately, this could lead the banking sector to go in for global banking centres which can operate like offshore facilities despite their location on Indian soil.

The new exam policy has given a leg uH to banking sector reforms by permitting Indian banks to set up overseas banking units in SEZs. This means that exporters operating out of the SEZ units and developers would be permitted to hold dollar accounts and the OBUs operating out of the SEZs would be able to deal in multiple currencies. Additionally, through OBUs, exporters in SEZs would have access to financing at international costs.

This is because the OBUs would be exempt from CRR, SLR and priority’ sector lending requirements, which would permit them to operate on par with their overseas branches. Under the new scheme, foreign banks registered in India would also be permitted to set up OBUs in SEZs through the finer details of the scheme would be announced later.

Benefits available to Units in SEZs:

1. They can import goods without payment of duty.
2. Reimbursement of central sales tax.
3. Exemption from payment of central excise duty on all goods eligible for procurement.
4. Reimbursement of central excise duty, if any, paid on bulk-tea procured by SEZ units so long as a levy on bulk tea in this regard is in force.
5. Reimbursement of duty paid on fuels or any other^goods procured from DTA as per the rate of drawback notified by the Directorate General of Foreign Trade from the date of such notification.

Question 7.
Explain the various agreements of WTO.
WTO Agreements As against GATT which covered only rules relating to trade in goods, the WTO agreements cover trade in goods, services as well as intellectual property. WTO Agreements make the government responsible to formulate the policies and procedure and make them transparent in order to avoid disputes among the nations. Major WTO agreements are discussed below:

Agreements Forming Part of G ATT: The erstwhile General Agreement on Tariffs and Trade (GATT) after its substantial modification in 1994 (effected as part of the Uruguay Round of negotiations) is very much part of the WTO agreements. Besides the general principles of trade liberalisation, GATT also includes certain special agreements evolved to deal with specific non-tariff barriers. Some of the specific agreements contained in the GATT are listed in the bank on GATT 1994 major agreements.

Agreement on Textile and Clothing (ATC): This agreement was evolved under WTO to phase out the quote restrictions as imposed by the developed countries on exports of textiles and clothing from the developing countries. The developed countries were imposing various ends of quota restrictions under the Multi-Fibre Arrangement (MFA) that itself was a major departure from the GATT’s basic principle of free trade in goods.

Under the ATC, the developed countries agreed to remove quota restrictions in a phased manner during a period of ten years starting from 1995. ATC is considered a landmark achievement of the WTO. It is due to the ATC that the world trade in textile and clothing has become virtually quota-free since 1st January 2005, thus, benefiting immensely the developing countries to expand their textiles and clothing exports.

Agreement on Agriculture (AoA): It is an agreement to ensure free and fair trade in agriculture. Though original GATT rules were applicable to trade in agriculture, these suffered from certain loopholes such as exemption to member countries to use some non-tariff measures such as customs tariffs, import quotas and subsidies to protect interests of the farmers in the home country. Trade-in agriculture became highly distorted especially due to the use of subsidies by some of the developed countries.

AoA is a significant step towards an orderly and fair trade in agricultural products. The developed countries have agreed to lower down the customs duties on their imports and subsidies to the exports of agricultural products. Due to their higher dependence on agriculture, the developing countries have been exempted from making similar reciprocal offers.

General Agreement on Trade in Services (GATS): Agreements are applicable to services also like on goods or merchandise, although services are intangible and cannot be toughed like commodities. GATS is regarded as a landmark achievement of the Uruguay Rbiind as it extends the multilateral rules and disciplines to services. It is because of GATS that the basic rules governing ‘trade in goods’ have become applicable to trade in services.

Three major provisions of GATS governing trade in services are as follows –

• All member countries are required to remove restrictions on trade in services in a phased manner. The developing countries, however, have been given greater freedom to decide about the period by which they would liberalise and also the services they would like to liberalise by that period.
• GATS provides that trade in services is governed by ‘Most Favoured Nations’ (MFN) obligation that prevents countries from discriminating among foreign suppliers and services.
• Each member country shall promptly publish all its relevant laws
and regulations pertaining to services including international agreements pertaining to trade and services to which member is a signatory.,

Agreement on Trade-Related Aspects of Intellectual Property’ Rights (TRIPS): The WTO’s agreement on Trade-Related Aspects of Intellectual Property Rights (TRIPS) was negotiated in 1986-1994. It was the Uruguay Round of GATT negotiations where for the first time the rules relating to intellectual property rights were discussed and introduced as part of the multilateral trading system. Intellectual property means information with commercial values such as ideas, inventions, creative expression and others.

The agreement sets out the minimum standards of protection to be adopted by the parties in respect of seven intellectual properties, viz., copyrights and related rights, trademarks, geographical indication, industrial designs, patents, layout design of integrated circuits, and undisclosed information (trade secrets).

## Internal Trade Class 11 Important Extra Questions Business Studies Chapter 10

Here we are providing Business Studies Class 11 Important Extra Questions and Answers Chapter 10 Internal Trade. Business Studies Class 11 Important Questions with Answers are the best resource for students which helps in class 11 board exams.

## Class 11 Business Studies Chapter 10 Important Extra Questions Internal Trade

Question 1.
Differentiate between wholesalers and retailers.
Difference between Wholesaler and Retailer:

 Basis Wholesaler Retailer 1. Scale of Operations Deals in large quantities and on a large scale. Deals in small quantities and on a small scale. 2. Number of items Handles a small number of items. Handles a large number of items. 3. Trade Channel First outlet in the chain of distribution The second outlet in the chain of distribution 4. Customers Sells to retailers and industrial users. Sells directly to customers for final consumption. 5. Supplies Receives goods from manufacturers/producers. Receives goods from wholesalers and sometimes, from the manufacturers. 6. Location The location of a wholesaler’s shop is not very important. A wholesaler may have a godown at any place. Location of retailer’s shop near the residential areas is very important. It located in the heart of the city. 7. Window Display The window display is not very important. The window display is a must to attract customers. 8. Specialisation Specializes in the products he deals in Specialization is not possible as he deals in a large number of products produced by different producers. 9. Margin of Profit Sells at a very low margin of profit as turnover is very fast. Sells at a higher margin of profit as he has to spend on window dressing and pay higher rent for the shop at a central place. 10. After-Sale Service Does not provide after-sale service. Provides aftersale service. 11. Free Home Delivery Does not provide free home delivery of goods to customers. Provides free home delivery of goods to Customers. 12. Display No window dressing and shop decoration are important. Window-dressing and shop decoration are much more important to attract customers.

Question 2.
Mention the services provided by wholesalers to retailers.
Services to Retailers:
Wholesalers render the following services to retailers –
1. Wide Variety: A retailer has to stock a large variety of products to meet the individual tastes and needs of its customers. Since he can easily purchase the required goods of different varieties from a wholesaler, he is relieved of the botheration of collecting goods from several manufacturers.

2. Regular Supplies: A wholesaler is always well stocked with different types of goods. Therefore, the retailers are assured of a quick and regular supply of their requirements from time to time. They need not maintain a large stock of goods and have no fear of running out of stock. The wholesaler serves as their warehouse keeper from whom they can quickly replenish their stocks. The wholesaler saves the retailers from the trouble of searching out and assembling goods from several manufacturers.

3. Specialisation: The wholesalers generally specialize in one line of goods. They buy good quality goods at the minimum possible price from the manufacturers. Retailers receive the benefit of such specialization when they buy goods from wholesalers. Wholesalers advise retailers on matters like quality, price, and timing of purchase. They inform the retailers about the new products, their uses, quality, prices, etc. They may also advise on the decor of the retail outlet, allocation of shelf space, and demonstration of certain products.

4. Publicity: Wholesalers advertise their goods regularly. Such publicity helps retailers in increasing their sales. Some wholesalers also guide retailers’ in-store layout and selling techniques. The retailers are benefited as it helps them in increasing the demand for various new products.

5. Credit: Wholesalers grant liberal credit facilities to retailers. As a result, retailers can carry on a large volume of business even with a small amount of working capital. They would be placed in a difficult position if they were to buy goods on cash payment.

Question 3.
List the various types of retail trade organizations.

Question 4.
Explain the automatic vending machine as a source of retail trade.
Automatic Vending Machines:
A vending machine is a new and complementary form of retailing. It is a slot machine operated by coins or tokens. The buyer inserts the coin or a token into the machine and receives a specified quantity of a product from the machine. Vending machines are used to sell prepacked and low-cost products of mass consumption, soft drinks, hot beverages, cigarettes, tickets, etc. In Delhi, Mother Dairy sells milk through vending machines.

The vending machines have become popular due to convenience in the handling of products and in the collection of payment. The customer gets a fresh supply of goods with uniform weight and quality. Moreover, vending machines can sell goods at places and at times when other types of retailing are not convenient or economical. There is a saving of labor costs. But initial investment in the machine is quite high. The machine requires regular repairs and maintenance.

The merits of vending machines are as under –

1. It is convenient for the customers to buy the goods from the machine.
2. Machines provide quick service.
3. Machine develop the habit of self-service among the customers.
4. Special peeks are to be developed by manufacturers that suit the machine.
5. Care has to be taken about replenishing the stock of the machine regularly.
6. Machines are useful in selling only consumer goods that are usually edible in nature i.e. candies, chocolate, soft drinks, coffee, etc.

Question 5.
Write an essay on the organization of wholesale trade.
Organization of Whole Trade Board of Directors

Wholesale trade is generally carried on a large scale and a large amount of capital investment is required for it. Therefore, a wholesale firm is generally organized in the form of a joint-stock company or a partnership. The company works under the overall supervision of the board of directors and the managing director. The organization is divided into sections. Every section is managed by a sectional head. All sectional heads are responsible to the managing director. Each section is divided into various departments. The head of each department is accountable to the sectional head concerned.

The organization of wholesale trade may be divided into the following sections –
This section looks after the overall planning and control of the wholesale trade. It is usually divided into several departments.
(a) Records and filing department: This department handles the firm’s records and files. Filing relates to having records of the business correspondence for future reference. Proper binding is also necessary for an easy and quick location of files, whenever required.

(b) Correspondence department: This department is responsible for the receipt, typing, and despatch of all letters. Timely and prompt reply of all incoming letters is essential for the success of a business. A copy of every letter sent by the firm is sent to the filing department for ready reference.

(c) Accounts and finance department: This department is concerned with the proper maintenance of the firm’s accounts. These accounts are related to the firm’s purchases, sales, receipts, payments, debtors, creditors, etc. This department is also responsible for preparing budgets and raising the necessary funds for a business. This department works under the supervision and control of the chief accountant and clerks working under him.

(d) Labour department: This department is responsible for the recruitment, selection, training, remuneration, promotion, etc. of employees.

2. Cash Section:
This section is concerned with the receipt and payment of cash. It is the responsibility of this section to ensure that all payments are made promptly and on the due dates so that the firm enjoys a good credit standing in the market. Similarly, this department takes steps for prompt collection of debts from the firm’s debtors in order to minimize bad debt losses. For handling small payments, there is a petty cashier who is provided a small amount of impress cash. When he has spent the entire amount, the head cashier advances him some more important cash.

In some wholesale firms, there is a separate credit and collection department which keeps a check on the credit sales.

3. Planning and Executive Section: This section consists of the various functional departments.
(a) Buying department: This department is concerned with buying goods in bulk from different producers. Before buying goods it invites Quotations from the producers. After comparing different quotations with regard to price, quality of goods, delivery period, etc. it places its orders.

When the goods are received, they are compared with the order. If there is any discrepancy or damage to goods, the matter is duly settled with the supplier. After receiving the goods, arrangements are made for their storage. The buying department is headed by an expert buyer who has complete knowledge of the various producers and the market conditions.

(b) Sales department: This department is responsible for selling the goods to retailers. It conducts market surveys to find out the tastes, fashions, etc. of the customers and other market conditions. This information is passed on to the buying department. The sales department also handles the complaints of retailers. The sales manager is the head of this department. Several salespersons work under him. Their recruitment, training, remuneration, etc. is also the responsibility of the sales manager.

(c) Publicity department: This department is concerned with advertising goods in order to create demand. It also arranges fairs and exhibitions of products.

(d) Despatch department: The function of this department is to despatch goods to various retailers according to the instructions it has been given. It handles packing, marking, and labeling of goods and arranges for the delivery of goods to the retailers. Many wholesale firms have a separate warehousing department also.

Question 6.
What are multiple shops or chain stores? Explain its features, merits, and limitations.
Chain-Stores (Meaning): Chain stores or multiple shops are a group of branch shops dealing in the same line of goods under single ownership and centralized management. A chain store is a chain of identical retail stores situated in different localities. Such a chain may be established by a manufacturer or by a merchant. It is known as chain stores in the United States and multiple shops in Europe. Bata Stores and DCM Stores are examples of chain stores in India. These normally deal in standardized and branded consumer products that have rapid sales turnover.

Distinctive Features:

1. Large Size: Chain stores are an example of large-scale retail establishments. These are located popularly in the area where a sufficient number of customers can be approached.
2. Company Form: These stores are organized as a joint-stock company. All stores are owned and controlled by the same company. There are centralized management and control.
3. Specialization: All stores in a chain deal in the same line of products, usually necessities.
4. Centralized Purchasing: Goods for all chain stores are purchased by the head office. Through centralized purchasing. These shops enjoy economies of scale.
5. Decentralized Selling: Chain stores are situated in different parts of the city and country. These shops are run by the same organization and have identical merchandising strategies.
6. Elimination of Middlemen: A chain store is a form of direct selling in which middlemen are eliminated.
7. Uniform Price: Goods are sold in all the stores at a fixed price.
8. Standardization: Decoration of stores and window displays follow a uniform style or pattern.
9. Cash Sales: Goods are sold on a cash and carry basis. There is no loss on account of bad debts.

Merits:
1. Economics of Scale: Goods for all chain stores are purchased by the head office. Such bulk buying results hr several economies like heavy discounts, saving in transport costs, etc. Benefits of specialization and centralized management are also available. Large capital permits expansion and growth.

2. Convenient Location: Chain stores are located to suit the conveniencFoftheeustometa. This helps in increasing sales turnover and in retaining contact with customers. These shops me located in fairly populous localities where a sufficient number of customers can be approached.

3. Low Operating Costs: Chain stores sell goods on a cash basis H so that there is no loss due to bad debts. There is an economy in advertising because one advertisement is enough for all the stores. Large and rapid turnover and common advertising are possible.

4. Low Price: Due to low operating costs and the elimination of middlemen, goods are sold at relatively cheaper rates. A manufacturer can establish direct contact with customers through chain stores.

5. Flexibility: If one store out of stock, supplies can be easily transferred from a nearby store belonging to the same chain. Such inter-branch transfers help to avoid loss due to shortage or surplus of stock. If a branch is not doing well it can easily be closed down and a new one can be opened in another place without really affecting the profitability of the organization as a whole.

6. Public Confidence: Fixed prices arid standard quality help to increase confidence among consumers. Customers can easily identify the chain stores on account of uniform decoration and design.

7. Diffusion of Risk: Lack of demand in one area 4oes affect the
sales in other stores. But the loss incurred by one store can easily be absorbed by profits made in other stores reducing the risk of an organization.

8. Simplicity of Control: Goods are sent to the different stores by the head, office and cash receipts are sent by each branch to the head office. Price and the other policies are uniformly laid down for all the stores. In this way, the office can exercise effective supervision and control over all the branches.

Limitations:
1. Limited Choice f Ghain deal in a limited range of products and do not offer a wide variety of choice to customers,

2. Lack of Personal Touch: The paid employees of chain stores do not take a personal interest in each and every customer Tltey adopt a ‘take it or leave it’ attitude towards the customers. The owner loses all personal contracts with the customer Lack of initiative in the employees some times leads to indifference and lack of personal touch in them

3. Lack of initiative: The employees of chain stores have no freedom to make decisions. Rigid control and uniform policies discourage initiative on their part. They cannot exhibit business opportunities or adapt to local needs. This makes them habitual of looking up to the head office for guidance on all matters and takes away creative skills of their own.

4. Heavy Overheads: The head office has to incur heavy expenditure on rent, wages, salaries, furniture, fixtures, etc.

5. No Facilities: Customers do not get credit, home delivery, or other facilities. Therefore, chain stores do not attract rich customers. This discourages certain types of people to visit multiple shops.

6. Local Competition: Chain stores are considered a threat to small independent retail stores. Therefore, local retailers are hostile towards these stores and offer tough competition.

7. Remote Control: The head office is usually far removed from the stores. Therefore, there is a lack of close contact between them.

Question 7.
Super Market is the most famous retail trade-in city. Mention its merits and demerits.
Super Market (Self-Service Store):
Meaning:
A supermarket of Bazar is a large retailing business unit selling a wide variety of consumer goods such as food and grocery items on the basis of the low-margin appeal, wide variety, and assortments, self-service, and heavy emphasis on merchandising appeal.

A supermarket deals mostly in food and grocery items and convenience goods like household goods, hosiery items, cosmetics, medicines and drugs, electronic appliances, etc. It is generally situated at the main shopping center. Goods are kept in open racks, and the price and quality are clearly labeled on the goods.

A consumer can make a selection of goods moving from counter to counter and pick up the selected goods and place them in a trolly. After he has completed his selection, the trolly will be carried to the exit where a person computes the total charge and the buyer makes payment to the cashier and then takes delivery of goods. Thus, supermarket follows the policy of self-help’ by the customers. The customers are not pressurized by the salesmen. That is why many people are attracted to the supermarket.

The supermarket is organized on a departmental basis and a customer can buy various types of goods under one roof. A supermarket can be differentiated from the departmental store on the main ground that there are no salesmen at the super Bazar to deal with the customers. The customers are free to choose the commodities of their choice. Moreover, a supermarket does not offer certain services which are usually provided by a departmental store. For instance, a supermarket does not allow credit sales and does not provide free home delivery service.

Features:
The main features that distinguish a supermarket from other retail institutions are discussed below –

1. A supermarket generally carries a complete line of food items and groceries in addition to non-food convenience goods including drugs, cosmetics, household goods, etc.
2. Its organization resembles a departmental store. A customer can buy his requirements under one roof.
3. A supermarket operates on the principle of self-service. There are no salesmen or shop assistants to help or pressurise the customers in a supermarket. That is why it is known as a self-service store. The distribution cost is, therefore, lower.
4. A supermarket is a low-cost retail institution in comparison with Other types of retail stores. The prices of the products are generally lower than other types because of bulk purchasing, lower operational cost, and low-profit margins.

Merits:
The following are the merits of a supermarket –
1. A supermarket is a large-scale retailing store. It enjoys all the benefits of large-scale buying and selling. Because of this reason and because of large turnover, its operating costs are lower and it can sell goods at cheaper rates. These outlets are not only convenient but also economical to buyers for making their purchases.

2. Considerable attention is paid to the package of the products since there are no salesmen to convince and pressurize the customers. Many people like this distinct feature of self-service. They enjoy the freedom to ccunpar£-iii£fereat4minda of Q.product- and making a selection of goods without pressure from anybody.

3. The customers can make all their purchases under one roof. A supermarket provides goods to customers at cheaper rates because of large turnover and absence of salesforce. The administrative and distribution overheads per unit of a product are also lower.

4. Since supermarket sells only on a cash basis, there is no chance of bad debts.

Demerits:
A supermarket suffers from the following drawbacks –
1. There are certain people who give greater weightage to personal attention. Such people do not like shopping through the supermarket as there are no salesmen.

2. Supermarkets cannot handle commodities that require personal explanation by the salesmen. It works on the self-service principle.

3. Some customers handle the goods carelessly and misuse the ‘ opportunity of self-service and selection. This may cause loss to the supermarket.

4. In practice, supermarkets have not been able to create low price appeal among the customers because of higher overhead expenses.

5. Establishment and, running off a supermarket requires huge; investment, and its turnover should be higher to keep the overhead expenses under reasonable limits. Thus, a supermarket cannot be established if the necessary capital is not available and a Jarge turnover is not expected. In other words, supermarkets are not suitable for smaller towns.

Question 8.
Write in brief about Telemarketing or Teleshops in the modern world.
Telemarketing (Teleshops): Telemarketing means a form of non-store retailing in which the seller initiates contact with a shopper ‘ and closes the sale over the phone. A telemarketer may procure the names and telephone numbers of his prospective customers from the telephone directory and other sources.

Popular products such as electrical appliances, health products, and educational aids and services such as magazine subscriptions, credit-card membership, etc. can be promoted through telemarketing. ICICI and Citibank follow this technique to popularise their credit cards among people.

A telemarketer can advertise the product, its features, uses, and price through a TV network, say Doordarshan, Metro, Zee TV, or Sony TV channel. The interested customer can place an order directly over telephone, fax, e-mail, or by post to the advertiser, say, Asian Sky Shop. The delivery may be effected through courier, or post office, or the manufacturer’s distribution van. Payment is to be made at the time of delivery. Telemarketing is a very convenient method of shopping. It is becoming popular in India.

From the point of view of the customer, buying or ‘dial-n-order’ is a very convenient method of shopping. One need not visit the store for shopping. One can place an order over the phone and make payment through a credit card. Teleshopping and getting free home delivery at residence is not only convenient but also cheaper as no middlemen are involved. The popularity of telemarketing would be attributed to the growing focus on customers.

From the point of view of the seller, telemarketing is a cheaper method of retailing. It saves expenditure on retail showrooms and salesforce. Even a firm with retail outlets or stores in major cities can reach customers at far off places through telemarketing. The growing satellite networks have created brand awareness and facilitated Telebuying and selling. The industry grew rapidly over 5 years to reach a size of2000crores with over one million consultants.

Telemarketing is a major tool of direct marketing in the USA and is gradually gaining acceptance in India. But telemarketing has the disadvantage of lack of personal touch with the buyers. Moreover, the buyer can’t inspect the goods personally before placing an order. Some people don’t like teleshopping. They get irritated when they receive unsolicited calls from the call centers, say, for the marketing of credit cards or personal loans.

Question 9.
What are Internet or Online marketing, its benefits, and the difference between traditional marketing and online marketing?
Internet Marketing:
Internet marketing is emerging as an important form of e-commerce, In this form of marketing, orders are received and processed on the internet. The internet is the world’s largest computer network. In fact, it is a ‘network of networks’ of computers throughout the world. A computer network is basically a bunch of computers hooked together for receiving and transferring information.

The facility of linking millions of computers is provided through the internet. The use of the internet by marketers or producers for the purpose of selling their products is known as internet marketing. Internet marketing can take several forms e.g. Online services, worldwide webs (www), and CD ROMs.

Internet:
The internet is a global web or computer network that makes instantaneous global communication possible. Internet usage has surged with the development of the user-friendly World Wide Web (www) and Web browser software such as Microsoft Internet Explorer or Netscape Navigator. Users can surf the internet and send e-mails, shop for products, and can get news and other information.

The internet itself is free though the user has to pay some fee to the Internet Service Provider to be hooked up to it. Thus, the internet is changing the way marketing is done. Internet technology provides marketers with faster ‘ more efficient and much more powerful methods of designing, promoting, and distributing products, conducting research, and gathering market information.

Online Marketing:
The users of the internet are younger, educated, and more affluent. Business firms can use the internet to reach such users by sending catalogs, price lists, etc. through e-mail. Besides advertisement, the sellers can attend to the queries of the buyers and clinch the deal on-line. The buyers can on their own also log on the t computer to know about the products of different manufacturers and decide to buy the products that suit them.

The use of electronic channels for the direct marketing is on the rise these days. The term e-commerce describes a wide variety of electronic platforms such as the sending of purchase orders to suppliers via electronic data interchange (EDI), the use of fax and e-mail to conduct transactions, the use of ATMs (automated teller machines) and smart cards to facilitate payments and obtain digital cash and the use of internet an online > services.

Thousands of business firms have established their presence on the internet. A new firm, big or small, can do so in two ways:
(a) It can buy space on a commercial on-line service; or (b) It can open its own website. In the case of online service, the firm has to pay an annual fee and also a small commission as a percentage of sales.

But these days, companies prefer to set up their own websites. Such a website offers information * about the company’s history, products, and services. The company can interact with anyone visiting the website and explore the possibility of concluding sales.

Benefits of Online Marketing:
The marketers can enjoy the following benefits of internet marketing –

1. The cost of digital catalogs is much less than the cost of printing and mailing paper catalogs;
2. Digital catalogs can be revised quickly and without much difficulty.
3. The marketers achieve economy in operations. They have to maintain low inventories and incur low costs on storage and insurance.
4. Internet marketing helps in relationship building with the customers. Online marketers can interact with customers and learn from them and improve their products.
5. Marketers can contact the people who have visited their website and make them attractive offers to affect sales.
6. Orders can be received quickly.
7. The marketers can approach customers living anywhere in the world. Thus, there is no distance gap between the sellers and the buyers.
8. Marketers can size up the audience, know how many people visited their online site. This information helps in improving sales offers and advertisements.
9. Online marketers can build relationships with customers by interacting with them.

Comparison of Online and Traditional System of Marketing

 Online Shopping Traditional Shopping Advantages Convenience Saves time Reduces impulse buying Five senses influence buying Memory trigger Product sampling Exposure to new items Social interaction Disadvantages Less price and selection control Forget items Reliance on computer Delivery fee Time-consuming Waiting for lines and parking Carrying groceries home Impulse buying Safety

Question 10.
A franchise is a buzzard in the modern marketing world. What are its merits and limitations?
Franchise: Franchise is a commercial concession by which a company or person grants a retailer the right of selling its products or services in a specified area. The owner of a product (known as a franchiser) permits another business firm (called franchisee) to sell the product in exchange for royalty payments.

The franchise is the right or privilege to use an established business system is “a continuing relationship in which franchisee provides licensed privilege to do business plus assistance in organizing, training, merchandising and management in return for a consideration from the franchise.” A franchising operation is a contractual relationship between a franchiser and franchisee.

Thus, the franchise is a system under which the owner of a product or service grants the franchisee the exclusive right to distribute the product or service in a specific geographical area on specified terms and conditions. The owner of the product or service who grants the right to distribute is known as the franchiser. The person or firm who acquires the right or franchise is called the franchisee.

A franchise system is one in which a manufacturer grants selected retailers the exclusive right to sell their products or services in specified areas. Such retailers are required to promote and sell the product in a specific manner. There is a written agreement between the franchiser (supplier) and the independent franchisee (retailers) on the terms and conditions of the franchisee.

The franchise is found in several types of businesses. Consumer items such as cosmetics, readymade garments, television sets, V.C.R., music system, computers, machinery and equipment, automobiles, servicing of consumer durables, computer training, real estate are some of the examples where the franchise is popular. Wimpy, Nirulas, Essex farms, Snowhite Drycleaners, etc. are notable examples of franchises in India.

The franchiser receives either a fixed sum or periodical royalties for allowing the use of trademark and providing training. The franchisee pays for a reliable and proven business. He gets professional advice and national sales promotion support from the franchiser. Generally, all franchised outlets of a product or service have an identical trademark, standard symbols, standardized products, and uniform business policies. The franchisee has to raise his own finances.

Franchise arrangements may be of the following types –
1. Product and Trade Name Franchise: In this arrangement, the franchisee acquires the right to use the product and trade name of the franchiser. The franchisee can also use window-display, standardized operating procedures,s and a prescribed territory to the franchisee.

2. Exclusive Dealership: Under this system, a manufacturer signs an exclusive agency contract, with a distributor. The distributor gets the exclusive right to sell the product within a specified geographical area. The distributor agrees to certain conditions of the manufacturer, e.g. adequate stock, prices to be charged, the services to be provided, etc. The exclusive dealership is popular in automobiles.

3. Conversion Franchising: Herein an established businessman gets affiliated with a franchiser. The two share the benefits of a franchising relationship.

4. Combination Franchising: In this arrangement, two franchises share a location and management, site selection, training of the staff, financing, and marketing, record keeping, and production business also arranged.

5. This is a fully integrated and continuous relationship between a franchiser and franchisee. The relationship covers total operations of the franchise including product or service, trademark quality control, strategy, etc. Fast food restaurants such as McDonald’s are an example of such a franchise.

Merits:
The main advantages of the franchise are as follows –
1. Availability of Established Brands: The franchisee acquires the right to use the popular brand name or trademark of the franchiser. Association with an established name provides a ready market. A franchise gives a quick and easy start in business. There are greater chances of success of the franchisee because the products are well known.

2. Standardised Goods and Services: The reputation of the franchiser depends largely on the quality of products and services supplied by the franchisee. The franchiser takes steps to ensure that products and services in all the franchised outlets are uniforms. He provides the raw materials and keeps close control of the quality of goods. The quality of products helps the franchisee to satisfy his customers by offering quality products.

3. Advertising Support: The franchiser carries on advertising. The franchisee gets the benefit of such advertising and the reputation or goodwill of the franchiser. The products are well advertised in various media and are known to the people. It is easier for the franchisee to promote the sale of products.

4. Financial Assistance: Franchiser offers a wide range of financial assistance to the franchisee in the form of short-term credit, lower down payment, flexible repayment terms, etc. Financial assistance is available for plant and equipment, accounts receivable, etc.

5. Managerial Training: Franchisers provide technical and managerial training to franchisees and their staff. Prior to opening a franchise, counseling and training are provided in the professional and profitable operation of the business and in the fields of inventory management, accounting, sales promotion, advertising, etc. Assistance is also available in site selection, marketing research, in addition to ongoing business assistance.

6. Established Business Methods: The franchisee can capitalize on the accumulated knowledge, experience, and skill of the franchiser. He does not have to build a business from scratch. The franchisee buys a business that has proved its success and can, therefore, avoid many of the pitfalls faced by small business owners. The franchiser makes huge investments in the innovation of products and research and development.

7. Economies of Scale: Due to the group or cooperative purchasing,
costs of products are reduced. Mass buying provides economies of scale. The distribution system between franchisor and franchisee in the shortest possible time. Marketing costs are lowered.

8. Uniform Control System: All franchising outlets are subject to the uniform control system. Standardized inventory control enables the franchiser to have more accurate information about the merchandise available and needed. Standardized reporting procedures are also helpful. It enables the franchiser to increases his goodwill and reputation.

9. Higher Success Rate: On average, franchises survive better than other business start-ups. The success rate, of franchises, is higher than that of independently owned businesses, Therefore, franchises are more attractive to middle-aged people who are less willing to take full risk of starting their own business. Potential income can be higher than independent small businesses.

10. Benefits to Franchiser: Franchise enables the franchiser to enter a new business territory at a low cost. It is a relatively quick way to raise cash and expand business operations. Owner-operators (franchises) are highly motivated. Franchises can be used as outlets for goods and services manufactured or supplied by the franchiser. This provides economies of scale in manufacturing and purchases. The franchiser can exercise control over products, services, and processes. The franchiser gets feedback about the product’s popularity from the franchisee.

Limitations:
The franchise system suffers from the following disadvantages –
1. Fees and Royalties: Costs of a franchise include license fees and fees for the initial processing of the application. It is payable when the franchise agreement is signed and is not refundable. Other costs include down payment on equipment, decoration of the outlet, office furniture, publicity on opening, etc. The franchise has to bear travel and living expenses while undergoing training. In addition, He has to pay a royalty on a continuing basis.

2. Lack of Freedom: The franchisee does not have the freedom to run a business like an independent owner. He has to conform to the controls exercised by the franchiser to ensure quality and uniformity of standards of product or service. Quality standards and specifications for all items used in the franchise are established. The freedom of purchasing is also restricted.

3. Limited Product Line: The franchiser controls the products or services sold at the franchisee’s outlet. The franchisee cannot introduce other products except those permitted by the franchiser.

4. Restriction on Sale of Franchise: Sale, transfer, or assignment of ownership interest requires the franchiser’s approval. Even when the sale is approved, the new franchisee is required to conform to the terms and conditions of the franchise.

5. Disadvantages to the Franchiser: The franchiser cannot treat or control the franchisees like his employees. Franchisees tend to become quite vocal and demanding if they feel they are not getting fair treatment or do not see benefits in the franchise network. Extensive communications are necessary and the costs of visiting the franchisees at distant places can be high. The franchiser has to bear the expenses of administration, training, advertising, legal services, supervision, etc.

Question 11.
Wholesalers are parasites for society and should be eliminated as soon as possible. Give arguments in favor and against of elimination of wholesalers.
Elimination of Wholesalers: Wholesalers perform several useful functions and render many vital services. They provide a ready sales outlet to manufacturers and serve as a source of a steady supply of goods for retailers. Therefore, they provide a valuable link between producers and retailers. But in recent years there has been a trend towards the elimination of wholesalers.

Growth of large-scale retailing institutions, development of quick means of transport and communication, growth of cooperative movement among consumers, the rise of specialized advertising agencies, etc. have made it possible, in certain cases, to establish a direct link between producers and retailers.

Arguments in Favour of Elimination: Some people insist that wholesalers exploit producers and retailers. Therefore, they should be eliminated and their functions should be taken over by producers and retailers.

The following arguments are given in favor of the elimination of wholesalers –
1. Reduction in Prices: Wholesalers charge a substantial margin of profit and add to the cost of distribution. This results in higher prices payable by the ultimate consumers. Prices payable by consumers can be reduced by eliminating the wholesalers.

This would increase the sales volume which would benefit the producers. Many producers want to reduce prices to maximize sales in conditions of cut-throat competition but wholesalers discourage reduction in prices. Wholesalers are the parasites of society. The customers have to bear the cost of wholesalers.

2. Faster Distribution: Wholesalers are mere transfer agents who set up unnecessary road-blocks in the process of distribution. Their, interference in the distribution channel obstructs the smooth and quick flow of goods from manufacturers to consumers. By eliminating the wholesaler’s goods can be supplied to consumers more quickly. Modem means of transportation and communication do not favor unnecessary middlemen in the channels of distribution.

3. Manipulations: Many wholesalers indulge in malpractices such as hoarding and adulteration. They push up prices by creating an artificial scarcity of goods. Such malpractices can be avoided by eliminating the wholesalers. Wholesalers do not render services correspondingly for the profits they earn while handling goods.

4. No Risk-bearing: Wholesalers assume the little risk and make no improvement in the distribution techniques. Therefore, there is little justification for the existence of wholesalers. Sometimes, even the existence of wholesalers hinders the smooth flow of goods and services.

5. Unreliable: A manufacturer faces a sudden decline in sales when wholesalers discontinue their line of goods and switch over to a different competitive line. Many wholesalers sell so many products that they are unable to give equal attention to pushing the sale of all the products.

6. Better Alternatives: Large-scale retailers such as departmental stores, chain stores, and supermarkets have adequate funds and space to buy goods in bulk directly from manufacturers. They can bear risks and promote sales on a large scale. Therefore, they do not need the services of wholesalers.

Arguments against Elimination:
Those who believe that wholesale trade is essential, give the following arguments –
1. Undivided Attention: If wholesalers are eliminated, the producer will also have to undertake the distribution of his goods in small lots to a large number of widely scattered retailers. As a result, he will not be able to concentrate fully on production. The economies of large-scale production will have to be sacrificed. Wholesalers perform many marketing functions like buying and assembling, selling, market research, advertisement, transport, etc.

2. Necessary for Small Producers: Small manufacturers are not in a position to distribute their goods. Due to a large number of small producers and retailers, wholesalers still dominate the field of distribution of goods. Wholesalers bear the risk of fluctuations in prices, purchase of raw material, etc. Above all, they will have to find out means of financing which small producers are unable to do.

3. Storage of Goods: If wholesalers are eliminated, retailers will have to maintain large stocks and bear the risk of price fluctuations. Which they are generally unable to bear due to limited capital and lack of enough space.

4. Savings in Costs: Wholesalers are experts in the task of distribution. Therefore, they save expenses through more efficient marketing of goods. They maintain a sales force that calls upon retailers regularly. Their expenses are lower than those of manufacturers’ who sell directly to retailers. By eliminating the wholesaler, costs of distribution may increase. The manufacturers are often solely dependent upon distribution by the wholesalers as they can’t handle distribution themselves.

5. Seasonal Products: Some products are produced throughout the year but their demand arises only during a particular season. If wholesalers are eliminated, producers will have to keep huge stocks of such products. They will require a large amount of capital during the off-season.

6. Financing: In case wholesalers are eliminated, the manufacturers and retailers will have to invest more capital in the business.

Conclusion: By eliminating the wholesaler, his functions cannot be eliminated. These functions will have to be performed by either producers or retailers. Therefore, the elimination of the wholesaler is desirable only in those cases where manufacturers or retailers can perform these functions more efficiently than the wholesaler. Only large and well-established producers or retailers may be in a position to do so. Therefore, a wholesaler is an essential link in the distribution of all such commodities wherein producers and retailers are small and unable to assume the burden of wholesale trade.

In brief, wholesalers neither can be nor should be eliminated because of their services to the manufacturers, retailers, consumers, and society as a whole.

## Formation of a Company Class 11 Important Extra Questions Business Studies Chapter 7

Here we are providing Business Studies Class 11 Important Extra Questions and Answers Chapter 7 Formation of a Company. Business Studies Class 11 Important Questions with Answers are the best resource for students which helps in class 11 board exams.

## Class 11 Business Studies Chapter 7 Important Extra Questions Formation of a Company

### Formation of a Company Important Extra Questions Short Answer Type

Question 1.
Explain briefly the meaning of promotion of a Company9.
Meaning of Promotion: The term promotion is used as the sum total of activities by which a business enterprise is brought into existence. It is a term of business, not of law. Promotion consists of the. business operations by which a company is established.

It is the process of planning and organising the finances and other resources of a business enterprise in the corporate form. According to C.W. Gerstenberg, “Promotion is the discovery of business opportunities and the subsequent organisation of funds, property and management ability into business concern for the purpose of making profit therefrom.”

Promotion is the process of the discovery of a business idea, its investigation and assembling of necessary resources to set-up a business as a profitable concern. The person or group of persons who perform the work of promotion and form a company as a going concern are known as a promoter. A promoter is an entrepreneur or businessman who gives birth to a business concern. A promoter may be an individual, a firm or a company.

According to S. Francis Palmer, “Promoter means a person who originates the scheme of promotion of the company, has the Memorandum and Articles prepared, executed and registered and finds the first directors, settles the terms of preliminary contracts and prospectus (if any) and makes arrangements for advertising and circulating the prospectus and the capital.”

Thus a promoter is a person or a group of persons who conceive the idea of the formation of a company, and takes necessary steps for its incorporation, raising of capital and making it a going concern. In order to perform the task of promotion successfully, a promoter must have several essential qualities. Fertile imagination, sound judgement, initiative, resourcefulness and organising ability are the main qualities of a successful promoter.

Promotion may be done for several objectives, e.g. to start a new business, to expand an existing business or to take over an existing business. Out of this starting, an altogether new business is the most difficult task.

Question 2.
Draw up the various stages of formation of the company.

Question 3.
Explain briefly the various types of promoters.
Types of Promoters: Promoters can be of the following kinds:
1. Entrepreneur Promoters: An entrepreneur conceives the idea of a new business and performs all the work for establishing it as a going concern. He continues to manage and control the business promoted by him. Small-scale enterprises, such as sole proprietorships and partnerships are promoted by entrepreneurs. He undertakes risk and takes initiative in promoting the company.

2. Professional Promoters: These promoters are specialists in promoting new business ventures. Large-scale enterprises are generally promoted by experts. These experts possess the necessary skills and knowledge in the promotion. They promote a business as a going concern and then sell the proposition or hand-over its management and control others.

These promoters are interested only in looking out tor business reality. He needs to be action-oriented. He will assemble resources, prepare necessary document give a name to the company opportunities and converting them into business units in return for handsome remuneration. There are very few professional promoters in India. They initiate new enterprises and find out the persons who can supply capital.

3. Occasional Promoters: This type of promoters promote a business once a while rather than on a regular basis. Promotion is not their main job and after promoting a company they go back to their original occupation. For example, an engineer or a technical expert may promote a business to commercially exploit a patent or invention discovered by him. They manage the company’s affairs even after incorporation of the company.

4. Financial Promoters: These promoters float new companies during favourable conditions in the securities market. Banks and other financial institutions also perform the work of promotion owing to their experience in the financial sector. Investment bankers become active in the field of promotion when the securities market is able to absorb new issues of equity shares. In India, the Industrial Development Bank of India and other financial institutions carry out the work of promoting industrial concerns.

5. Government: Nowadays, the government has become the biggest promoter. For example, the Government of India has established several basic, strategic and defence industries to speed up the process of economic development in the country. It has promoted large-scale enterprises in iron and steel, coal, shipping, fertilisers, electronics, engineering, insurance, tourism, hotels, etc.

Question 4.
A promoter has various qualities. Explain in brief.
Qualities of a Promoter: A promoter should possess the following qualities.

1. Vision: A promoter requires a sound imagination and a clear but realistic view of the future. He analysis the prospects of a company and brings together the men, materials and machinery.
2. Alert mind: The promoter should be alert enough to notice business opportunities that can be used to advantage. An invention, a patent, a natural resource, an unsatisfied or poorly satisfied need are examples of such opportunities.
3. Resourcefulness: A promoter should be able to mobilise financial, human and physical resources so a
4. Risk-taking ability: The promoter should be able to bear the calculated risks of the business. Sometimes, an idea may be good but technically not possible to execute.
5. Tact: A promoter needs to be tactful so as to persuade people to invest money in the new venture.
6. Patience: Considerable patience is necessary to wait till the business idea takes a practical shape. Sometimes, it so happens that a project is technically feasible and viable, but chances of it being profitable are very little, the idea may have to be adopted later by investigating into details which require patience.
7. Analytical ability- The promoter should be able to carefully examine each idea and opportunity in terms of costs and benefits.

Question 5.
Define ‘Memorandum of Association’ and ‘Articles of Association’ as these are public documents.
Memorandum of Association: The Memorandum of Association is the principal or most important document of a company. According to Lord Macmillan, “The Memorandum of Association sets out the constitution of the company. It is, so to speak, the charter of the company and provides the foundation on which the structure of the company is built. It enables persons, who deal with the company, to know its permitted range of activities.” In the words of Lord Cains, “the Memorandum of Association of a company is its charter and defines the limitations of the power of the company established under the Act.

The Memorandum contains the fundamental conditions upon which alone the company is allowed to be incorporated.” It also lays down the scope of operations of the company beyond which it cannot operate. The purpose of the Memorandum of Association is to enable the shareholders, creditors and others who deal with the company to know its permitted range of activities. In fact, it can be considered as the .foundation on which the structure of a company is based. Its primary importance lies in the fact that a company cannot undertake such operations which are not mentioned in its memorandum.

Great care should be taken in preparing the Memorandum of Association because a company cannot go beyond the limits laid down in the Memorandum.

The Memorandum of Association must be
(a) printed,
(b) divided into suitable paragraphs numbered in sequence, and
(c) signed by the required number of persons in the presence of at least one witness.

The Memorandum must be published in sufficient numbers because it is a public document and a copy has to be given on demand and at a nominal charge. The facsimile and common seal of the company should be filed on the Memorandum. The Companies Act contains different forms of Memorandum (in Schedule 1), one each for companies limited by shares, companies limited by guarantee without shares capital, and unlimited company.

In brief Lord Justice Browen, “a Memorandum of Association is a fundamental document of a company which is also known as the charter of the company. It lays down the object and scope of activities and limitations on the power of a company beyond which the company cannot go. It is a document which contains all conditions upon which a company is allowed to be incorporated.

Articles of Association: Articles of Association are the bylaws of a company. They contain rules and regulations for the management of the internal affairs of a company. They define the mode and manner in which the company’s business is to be carried on. Articles of Association are public document. Outsiders dealing with the company are supposed to have read the Memorandum and Articles of the company. They are entitled to believe that the company conduct its business according to the rules and regulations. This is known as the Doctrine of Indoor Management.

A private company, a company limited by guarantee and an unlimited company, must prepare and file their own Articles of Association. But a public company limited by shares may adopt Table A in the First Schedule of the Companies Act, in case it does not want to prepare and file its own Articles of Association. While preparing the Articles great care should be exercised.

Anything contained in the Articles which is against the Memorandum of Association or against The Companies Act shall be null and void. The Articles of Association should be printed, properly divided into paragraphs, consecutively numbered and signed by the signatories to the Memorandum in the presence of at least one witness. The Articles of Association are subordinate to the Memorandum of Association.

Question 6.
What is ‘Statement in lieu of Prospectus*?
Statement in lieu of Prospectus: A public company having a share capital may sometimes decide not to approach the public for securing the necessary capital because it may be confident of obtaining the required capital privately. In such a case, it will have to file a ‘Statement in lieu of Prospectus’ with the Registrar. A ‘Statement in lieu of Prospectus’ is drafted in accordance with the particulars set out in “Schedule III” of the Companies Act. It .contains information very much similar to a prospectus:

It must be duly signed by all the directors and a copy thereof must be filed with the Registrar at least three days before the allotment of the shares. However, a private company is a riot required to either file “Prospectus” or a “Statement in the lie of Prospectus” with the Registrar. ‘Statement in lieu of Prospectus’ must be dated and signed by each director. It should not contain any untrue or misleading statement. Provisions regarding the penalty for issuing a misleading prospectus are also applicable to untrue details given in a statement in lieu of prospectus. A private company is not required to file either a prospectus or a statement in lieu of prospectus as it is not permitted to raise funds.

### Formation of a Company Important Extra Questions Long Answer Type

Question 1.
What is the procedure adopted for the alteration of various clauses of Memorandum of Association?
Alternation of the Memorandum: The Memorandum of Association can be altered in accordance with the procedure laid down in the Companies Act.

The provisions of the Act are as follows:
1. Alteration of Name Clause: A company can change its name in the following manner.
(a) If the name registered is identical with or similar to the name of an existing company, bypassing an ordinary resolution and obtain ing written approval of the Central Government;

(b) Bypassing a special resolution and obtaining the written consent of the Central Government in other cases.
It may be pointed out and delete that the word ‘private’ to the name of the company does not require the approval of Central Government. But where the addition of the word ‘private’ becomes essential due to the conversion of a public company into a private company, such conversion | is effective only with the approval of the Central Government.

2. Alternation of Registered Office Clause: This clause can be altered in the following ways.
(a) If the registered office is to be shifted from one state to another, bypassing a special resolution and obtaining the sanction of the Registrars of both the states;
(b) If the office is to be shifted from one town to another in the same state, bypassing a special resolution;
(c) If the office is to be shifted from one locality to another in the same town, bypassing an ordinary resolution.

The Registrar of Companies shall register the confirmation of the change of registered office and the alteration made in the memorandum of association within 30 days from the date of filing the documents of change.

3. Alteration of Object Clause: In order to alter its object clause, a company must pass a special resolution and obtain the permission of the Company Law Board.

The Company Law Board must satisfy the following conditions before granting permission:
(a) The objections, if any, of the Registrar of Companies have been obtained.
(b) Sufficient notice has been given to every creditor and other persons whose interest may be affected by the proposed alteration; and
(c) With respect to every creditor either his consent has been obtained or his debt has been discharged.

An alteration in the object clause is permissible if it is necessary to enable the company
(a) To carry on its business more economically or efficiently;
(b) To enlarge the area of its operations;
(c) To carry on some other business which can be profitably combined with the existing business;
(d) To amalgamate with any other company or association;
(e) To attain its main purpose by new or improved means;
(f) To restrict or abandon any of the objects specified in the Memorandum; and.
(g) To sell or dispose of the whole or part of the company’s property.

A special resolution must be passed at a general meeting of the company for alteration of the object clause. The resolution must state reasons or purposes for doing so.

4. Alteration of Liability Clause: If a company wants to make an alteration which imposes any additional liability on its members, it must pass a unanimous resolution in a meeting of its members. The liability of the members automatically becomes unlimited if their number is reduced below seven in case of a public company and below two in a private company. The liability of members may be increased if the members agree in writing consent may be given either before or after the alteration.

5. Alteration of Capital Clause: A company must pass a special resolution and obtain the approval of the Company Law Board. In case the capital is to be reduced, the permission offline court is also required. The court may direct that the company must write the words ‘reduced’ after its name and all the official documents of the company will bear this word. Where the alteration involves the return of capital or reduction in uncalled capital, the consent of creditors must be obtained.

A company may reduce its capital in the following ways:
(a) By extinguishing or reducing the liability of members for uncalled capital;
(b) By paying off some part of capital;
(c) By writing off or cancelling any paid-up capital which is lost; and
(d) By any other method approved by the court.

A company may alter its capital to:
(a) Increase the amount of share capital;
(b) Consolidate the divide its capital into shares of higher denominations;
(c) Subdivide the shares into those of smaller denominations;
(d) Cancel the unissued capital;
(e) Convert fully paid shares into stock or vice versa; and
(f) Reduce the amount of share capital.

6. Alteration of Subscription Clause: In this clause, the subscribers declare that they desire to be bound to assist information of a company and no subscriber can withdraw his name on any ground after registration of the company.

Question 2.
Explain the various contents of Article of Association and its alteration.
Articles of Association: The Articles of Association of a company contain the rules relating to the management of its internal affairs. The articles define the duties, the rights and the powers of the Board of Directors as between themselves and the company at large, and the mode and form in which the business of the company is to be carried on, and the mode and form in which changes in the internal regulations of the company may be made.

Articles of association contain the rules angulations for regulating the internal affairs of the company. The subscribers df the memorandum may include in it all such regulations as they deem fit. But everything included in it must be subject to the Companies Act and Memorandum of Company.

A public company limited by shares may opt for the adoption of Table A (i.e. the model set of 99 articles given in Schedule I of Companies Act). The other types of companies are required to file their articles of association along with the Memorandum at the time of registration. The Articles of Association should be printed, divided into paragraphs and numbered consecutively, and signed by each signatory to the memorandum in the presence of at least one attesting witness.

Contents of Articles of Association: The Articles of Association usually contain the provisions relating to the following matters:

1. The amount of share capital and different types of shares.
2. Rights of each class of shareholders.
3. Lien on shares, forfeiture of shares for non-payment of calls and transfer and transmission of shares.
4. Procedures tor conduct of meetings, voting, quorum, poll and proxy.
5. Appointment, removal and remuneration of directors and their powers and duties.
6. Procedure regarding alteration of share capital.
7. Matters relating to the distribution of dividend.
8. Matters relating to the keeping of statutory books.
9. Procedure regarding the winding up of the company.
11. Redemption of preference shares.
12. Rights of members.
13. Borrowing powers.
14. Reserves and capitalization of reserves.

Alteration of Articles: A company may change its Articles of Association by passing a special resolution subject to the following conditions.

1. The alteration must not be contrary to the Memorandum and the Companies Act. The alteration should not allow the company to do something which is forbidden by the Memorandum.
2. It must be bona fide and in the interest of the company as a whole.
3. It must not result in a breach of contract with outsiders.
4. The alteration must not purport sanction anything illegal or against public policy.
5. It must not have the effect of increasing the liability of members.
6. It should not amount of fraud by the majority on the monetary. Any alteration which constitutes the discrimination of interest of minority is a fraud on them.
7. If the alteration involves converting a public company into a private company, the approval of the Central Government must be obtained.

Question 3.
What is Prospectus? Mention its importance and various important contents.
Prospectus: The purpose of issuing a prospectus is to acquaint the investors with the proposed company and induce them to invest in its shares. The law with a view to protecting the interest of investors regulates the issue and the contents of the prospectus.

A prospectus means “any prospectus, notice, circular, advertisement or other document inviting deposits from the public or inviting offers from the public for the subscription or purchase of any shares or debentures of a company.” The term “Prospectus”, therefore, includes any document which invites deposits from the public or invites offers from the public to purchase shares or debentures of a company.

The essential elements of a prospectus are as follows:

1. There must be an invitation to the public.
2. The invitation must be made “by or on behalf of the company”.
3. The invitation must be “to subscribe or purchase its shares or debentures or such other instruments”.
4. Every prospectus issued must contain the matters specified by the provisions of the Companies Act.

Importance of Prospectus: Investors make up their minds about investment in a company primarily on the basis of the information contained in the prospectus. Therefore, there must not be a misstatement in the prospectus and all significant information must be duly disclosed.

The prospectus of a company serves the following purposes:

1. It reflects the future policies and programmes of the company.
2. It serves as an invitation to the public to subscribe to the shares and debentures of the company.
3. It provides a legal document of the terms and conditions on which shares and debentures have been issued.
4. It identifies the persons who can be held responsible for any untrue statements made in it.

Contents of a Prospectus: Prospectus is the only document through which the prospective investors can evaluate the soundness of the company.

It must contain at least the following broad particulars:

1. Company’s name and address of its registered office, nature of business of the company.
2. The main objects of the company and other objects.
3. The number and classes of shares, if any, and the nature and extent of the interest of the holders in the property and profits of the company. Rights attached to the shareholders are also mentioned.
4. The details about the redeemable preference shares intended to be issued, if any, i.e., the date and mode of redemption etc, details regarding debentures also if any.
5. Qualification shares of directors, if any.
6. Any provision in the articles as to the remuneration of the directors, managing director or otherwise.
7. The names, addresses, and occupations of the directors, managing director or manager.
8. The “minimum subscription” that is, the minimum amount which, in the opinion of directors must be raised by the issue of shares.
9. The time of the opening and closing of the subscription list.
10. To amount payable on application and allotment of each class of share.
11. Rights, privileges and restrictions attached to each class of shares.
12. A reasonable time and place at which copies of audited balance sheets and profit and loss accounts of the company may be inspected.
13. Names and addresses of auditors, bankers, brokers and solicitors.
14. Names and addresses of underwriters and commission payable to underwriters.

Question 4.
What is a misleading statement in a prospectus and give its consequences?
Mis-statement or untrue statement in the prospectus: If a prospectus contains misstatements or omits material facts, it will give rise to many consequences. A misstatement or an untrue statement is one which is misleading in the form and context in which it has been included in the prospectus. A prospectus shall also be deemed to have an untrue statement if the omission of any matter from it is calculated to mislead those who act on the faith of the prospectus.

Consequences of Misleading Prospectus: If a prospectus contains misstatements, the aggrieved party will have remedies against the company, directors, promoters and legal experts. A prospectus must be prepared very carefully. In an effort to influence the investors, a company may make exaggerated statements about its operations and future prospects. But a prospectus must contain full and honest disclosure of all material facts. To avoid preparing a misleading prospectus, it should be ensured that material facts are not concealed and no false details are given.

The persons responsible for issuing a misleading prospectus face civil and criminal liability.
1. Civil Liability: If the prospectus contains false information or incomplete details due to the negligence of the directors, it amounts to misrepresentation. The persons who invest money in the company on the basis of such misrepresentation have the right to avoid the contract and claim refund of their money. They can also claim damages from the company and from those who have authorised the issue of a misleading prospectus. An aggrieved subscriber of shares or debentures may sue directors, promoters etc. for payment of loss or damages caused to him by the misrepresentation in the prospectus issued. Subscriber must prove that he has suffered loss by reason of the omission of a matter from the prospectus.

A person accused of misrepresentation in a prospectus can avoid liability if he proves that:
(a) He had withdrawn his consent before the issue of the prospectus and that it was issued without his consent or authority;
(b) The prospectus was issued without his knowledge or consent and he had given reasonable notice to this effect after he came to know about it;
(c) After the issue of the prospectus but before allotment, on becoming aware of the misrepresentation, he had withdrawn his consent and given public notice to that effect;
(d) The statement was based on the statement of an expert who had given his consent to it;
(e) He genuinely believed the statement to be true; or
(f) The statement was based on an official document.

The directors etc., however, will not be liable for the compensation of loss, if it is proved that they honestly believed the statement to be true.

2. Criminal Liability: If the directors of a company deliberately, conceal information, they are punishable with a fine extending to Rs. 5,000 or imprisonment up to two years or both. If the public is fraudulently induced to invest money, the penalty may extend to a fine of Rs. 10,000 or imprisonment of five years or both.

A person accused of deliberate or fraudulent misrepresentation may avoid his ability by proving that:
(a) The statement was immaterial; or
(b) He had reasonable grounds to believe and did believe that the statement was true.

An expert who has given his consent to issue the prospectus containing an untrue statement made by him shall be liable to every subscriber who takes shares on the faith of his statement. The aggrieved shareholder is entitled to claim damages under the general law i.e. Contract Act. An expert shall not be liable if he proves that he was competent to make the statement and that he had reasonable ground to believe that statement was true at the time of allotment of securities. Expert includes an engineer, a valuer, an accountant and any other person whose profession gives authority to a statement made by him.