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In this page, we are providing Force and Laws of Motion Class 9 Extra Questions and Answers Science Chapter 9 pdf download. NCERT Extra Questions for Class 9 Science Chapter 9 Force and Laws of Motion with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 9 Extra Questions and Answers Force and Laws of Motion

Extra Questions for Class 9 Science Chapter 9 Force and Laws of Motion with Answers Solutions

Force and Laws of Motion Class 9 Extra Questions Very Short Answer Type

Force And Laws Of Motion Class 9 Numericals Extra Questions Question 1.
Define the following terms

1. Force
2. Balanced forces
3. Unbalanced forces
4. Inertia
5. Momentum
6. Conservation of momentum.

Answer:

1. Force: It is an entity which when applied on a body changes or tends to change its state of motion and shape.
2. Balanced forces: When a number of forces acting simultaneously on a body do not bring about any change in the state of rest or of uniform motion along a straight line, then the forces acting on the body are said to be balanced.
3. Unbalanced forces: When a number of forces acting simultaneously on a body bring about a change in its state of rest or of uniform motion along a straight line, then these forces acting on the body are said to be unbalanced forces.
4. Inertia: Inertia is the natural tendency of an object to resist a change in its state of motion or rest.
5. Momentum: Momentum of a body is the product of its mass and velocity.
6. Conservation of momentum: In an isolated system, the total momentum remains conserved.

Force And Laws Of Motion Class 9 Extra Questions Question 2.
Name the physical quantity that measures inertia.
Answer:
Mass of body measures its inertia.

Force And Laws Of Motion Extra Questions Numericals Question 3.
There are three solids made up of aluminum, steel, and wood, of the same shape and same volume. Which of them would have the highest inertia?
Answer:
Steel ball would have the highest inertia as steel is denser than the other two.

Force And Laws Of Motion Class 9 Extra Numericals Question 4.
Name the factors on which the momentum of a body depends.
Answer:

• Mass and
• The velocity of the body.

Class 9 Force And Laws Of Motion Extra Questions Question 5.
Define one newton of force.
Answer:
Force is said to be 1 newton if it produces an acceleration of 1 rn/s² in a body of 1 kg.

Force And Laws Of Motion Extra Questions Question 6.
An object is thrown vertically upwards. What is its momentum at the highest point?
Answer:
Since the velocity at the highest point is zero, so the momentum of the object is zero at the highest point.

Extra Numericals Of Force And Laws Of Motion Class 9 Question 7.
State action and reaction when a bullet is fired from the gun.
Answer:
Action: Force exerted by a spring on the bullet.
Reaction: Force exerted on the gun.

Class 9 Science Chapter 9 Extra Questions And Answers Question 8.
An athlete runs some distance before taking a jump. Why?
Answer:
To gain momentum so that he may jump higher.

Class 9 Physics Chapter 2 Extra Questions Question 9.
Mention any two effects of force.
Answer:

• It changes the state of rest or motion of a body.
• It changes the shape of the body.

Class 9 Force And Laws Of Motion Extra Numericals Question 10.
What happens to the momentum of a body whose velocity is halved?
Answer:
The momentum of the body becomes half.

Class 9 Science Ch 9 Extra Questions Question 11.
Give the magnitude and the direction of the net force acting on the cork of mass 10 g floating on water.
Answer:
Zero.

Class 9 Science Chapter 9 Extra Numericals Question 12.
A car of mass 1000 kg is moving with velocity 5 m/s. Calculate the momentum of the car.
Answer:
Given, mass, m = 1000 kg
Velocity, u = 5 m/s
Momentum, P = mυ
P= 1000 x 5 = 5000kg m/s
P= 5000 kg m/s

Class 9 Physics Force And Laws Of Motion Extra Questions Question 13.
A meteorite burns in the atmosphere before it reaches the Earth’s surface. What happens to the linear momentum?
Answer:
Meteorite burns due to heat produced by frictional force. The linear momentum of the meteorite decreases as a frictional force acts on it.

Class 9th Science Chapter 9 Extra Questions Question 14.
Find the acceleration produced by a force of 2000 N acting on a car of mass 800 kg.
Answer:
Given,
Mass, m =800kg
Force, F = 2000 N
Using F = ma
a = \(\frac{F}{m}\)
a = \(\frac{2000}{800}\) = 2.5 m/s²

Force and Laws of Motion Class 9 Extra Questions Short Answer Type 1

Question 1.
Write down SI unit of (i) force (ii) momentum.
Answer:

• newton (N)
• kg rn/s.

Question 2.
When a person hits a stone, his foot is injured. Why?
Answer:
When a person hits a stone, the stone exerts an equal force on his foot. Due to this force, his foot gets injured.

Question 3.
Why no force is required to move an object with a constant velocity?
Answer;
We know, F = ma
When, velocity is constant, then acceleration, a = 0. Hence, F = 0.

Question 4.
Why is it easier to catch a table tennis ball than a cricket ball, even both are moving with the same velocity?
Answer:
It is easier to catch a table tennis ball because the table tennis ball has less mass (inertia).

Question 5.
Write down the expression for the recoil velocity of the gun.
Answer:
Recoil velocity of gun is given by,
\(\mathrm{V}_{\mathrm{G}}=\frac{m_{b} v_{b}}{\mathrm{M}_{\mathrm{G}}}\)
m_b = mass of bullet
m_G = mass of gun
υ_b = velocity of bullet
V_G = recoil velocity of gun.

Force and Laws of Motion Class 9 Extra Questions Numericals

Question 6.
A car of mass 1000 kg moving with a velocity of 54 km/h hits a wall and comes to rest in 5 seconds. Find the force exerted by the car on the wall.
Answer:
Given, Mass = 1000 kg
Time = 5 s
Initial velocity, u = 54 km/h = 15 m/s
Final velocity, υ = 0 m/s
Time, t = 5s
using, F = m a
F = m\(\left(\frac{v-u}{t}\right)\) = 1000 \(\left(\frac{0-15}{5}\right)\) = -3000N

Question 7.
A body of mass 100g is at rest on a smooth surface. A force of 0.1-newton act on it for 5 seconds. Calculate the distance traveled by the body.
Answer:
Given, Mass of body = 100 g = 0.1 kg
Force, F = 0.1N
Time, t = 5s
Initial velocity, u = 0 m/s
Using formula, F = ma
⇒ a = \(\frac{F}{m}=\frac{0.1}{0.1}\) = 1 m/s²

Using formula, s = ut + at²
s = ut + \(\frac {1}{2}\)at²
⇒ s = 0 + \(\frac {1}{2}\) x 1 x (5)²
∴ s = 12.5 m

Question 8.
A bullet of mass 200 g is fired from a gun of mass 10 kg with a velocity of 100 m/s. Calculate the velocity of recoil.
Answer:
Given, Mass of bullet, m = 200 g = 0.2 kg
Massofgun, M = 10 kg
Velocity of bullet, V_b = 100 m/.s
Recoil velocity of gun, V_G = \(\frac{m v_{b}}{\mathrm{M}}\)
V_G = – \(\frac{0.2 \times 100}{10}\) = 2m/s
Recoil velocity of gun, V_G = 2 m/s

Question 9.
Two spheres of masses 20 g and 40 g moving in a straight line in the same direction with velocities of 3 mIs and 2 m/s respectively. They collide with each other and after the collision, the sphere of mass 20 g moves with a velocity of 2.5 miles. Find the velocity of the second ball after confusion.
Answer:
Given, m₁ = 20 g = 20 x 10^-3 kg
m₁ = 40g = 40g x 10^-3kg
u₁ = 3 m/s
u₂ = 2 m/s
υ₁ = 2.5 m/s
Applying conservation of linear momentum,
m₁u₁ + m₂u₂ = m₁υ₁ + m₂υ₂
20 x 10^-3 x 3 + 40 x 10^-3 x 2 = 20 x 10^-3 x 2.5 + 40 x 10^-3 x υ²
υ₂ = 2.25m/s.

Force and Laws of Motion Class 9 Extra Questions Short Answer Type 2

Question 1.
Newton’s first, second, and third law of motion.
Answer:
Newton’s first law of motion: An object remains in a state of rest or of uniform motion along a straight line unless compelled to change that state by an applied force.

Newton’s second law of motion: The second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.

Newton’s third law of motion: According to the third law of motion to every action, there is an equal opposite reaction and they act on two different bodies.

Question 2.
What is momentum? Write its SI unit. Interpret force in terms of momentum. Represent the following graphically.
(a) momentum versus velocity when mass is fixed.
(b) momentum versus mass when velocity is constant.
Answer:
Momentum gives an idea about the quantity of motion contained in a body.
The momentum of (P) of an object is defined as the product of its mass (m) and velocity (υ).
P = mυ
SI unit of momentum is kg ms^-1
Graphs
(a) If m = constant
P ∝ υ

(b) If v = constant
P ∝ m

Question 3.
“Action and reaction are equal and opposite but even they do not cancel each other.” Explain, why?
Answer:
Two equal and opposite forces can cancel each other if they act on the same body. But action and reaction do not act on the same body. They act on two different bodies. Hence cannot cancel each other.

Question 4.
Why are the wheels of vehicles provided with mudguards?
Answer:
The rotating wheels of a vehicle throw out mud sticking to it tangentially due to the inertia of direction. The mudguards stop this mud to fall on another vehicle just moving behind the vehicle.

Question 5.
A car weighing 1600 kg moving with a velocity of 30 mIs retards uniformly coming to rest in 20 seconds. Calculate the
1. Initial and final momentum of the car.
2. Rate of change of linear momentum of the car.
3. Acceleration of the car.
Answer:
Given, mass, m = 1600 kg
Initial velocity, u = 30 m/s
Final velocity, υ = 0 m/s
Time, t = 20s
1. Initial momentum, P_i = mu = 1600 x 30
P_i = 4800 kg m/s
Final momentum, P_f = mυ = 1600 x 0
P_f = 0 kg m/s

2. Rate of change of linear momentum
= \(\frac{P_{f}-P_{i}}{t}\) = \(\frac{0-4800}{20}\) = – 240N
a = \(\frac{v-u}{t}\)

3.  Acceleration,
\(\frac{0 – 30}{20}\)
a = – 1.5 m/s².

Question 6.
Using the second law of motion, derive the relation between force and acceleration. A bullet of 10 g strikes a sand-bag at a speed of 10³ ms² and gets embedded after traveling 5 cm. Calculate
1. the resistive force exerted by the sand on the bullet.
2. the time is taken by the bullet to come to rest.
Answer:
According to the second law of motion,
The rate of change of momentum is directly proportional to the force applied.
F ∝ \(\frac{m(v-u)}{t}\)
F = \(\frac{km(v-u)}{t}\)
∴ F = kma
Take k = 1 in SI system
∴ F = ma

1. Given, mass of bullet, m = 10g = 10 x 10^-3kg
Initial velocity of the bullet, u = 10³m/s
Distance travelled by the bullet, s = 5 cm = 5 x 10^-2 m
Final velocity of the bullet = 0
Using the formula, υ^-2 – u^-2 = 2as
⇒ 0 – (10³)² = 2 x a x 5 x 10^-2

f = m x a = 10 x 10^-3 x (10⁷) = – 10⁵ N

2. From the formula,
υ = u + at
⇒  t = \(\frac {υ – u}{a}\) = \(\frac{0-10^{3}}{-10^{7}}\) = 10^-4 s = 0.0001 s

Question 7.
An object is placed on a rough surface. The external force of 20 N is acting on the body and 10 N frictional force is acting on the body.
Find
1. The net force on an object
2. Acceleration of an object if the mass of an object is 2 kg
3. The velocity of the object after 2 seconds if the object is started from rest.
Answer:
Given,

∴ υ = 10 m/s

Force and Laws of Motion Class 9 Extra Questions Long Answer Type

Question 1.
Explain why?
(a) A cricket player lowers his hands while catching the ball.
(b) The vehicles are fitted with shockers.
(c) A karate player breaks the pile of tiles or bricks with a single blow.
(d) In a high jump athletic event, the athletes are allowed to fall either on a sand bed or cushioned bed.
(e) In a moving car, the drivers and other passengers are advised to wear seat belts.
(f) China and glassware are packed with soft materials.
(g) Athletes are advised to come to a stop slowly after finishing a fast race.
Answer:
(a) if a player does not lower his hands while catching the ball, the time to stop the ball is very small. So a large force has to be applied to reduce the velocity of the ball to zero or to change the momentum of the ball. When a player lowers his hands, the time to stop the ball is increased and hence less force has to be applied to cause the same change in the momentum of the ball. Therefore, the hands of the player are not injured.

(b) The vehicles are fitted with shockers (i.e., springs)
The floor of a vehicle is cónnected to the lower part of the vehicle by springs or shockers. When the vehicle moves over a rough road, the force due to jerks is transmitted to the floor of the vehicle through the shockers. The shockers increase the time of transmission of the force of jerk to reach the floor of the vehicle. Hence less force is experienced by the passengers in the vehicles.

(c) A karate player can break a pile of tiles with a single blow of his hand because he strikes the pile of tiles with his hand very fast, during which the entire linear momentum of the fast-moving hand is reduced to zero in a very short interval of time. This exerts a very large force on the pile of tiles which is sufficient to break them, by a single blow of his hand.

(d) In a high jump athletic event., the athletes are allowed to fall either on a sand bed or cushioned bed: This is done to increase the time of athletes fall to stop after making the high jump, which decreases the rate of change of linear momentum and decreases the impact.

(e) In a moving car, the drivers and other passengers are advised to wear seat belts: When brakes are applied sudden1y, the passengers of the car fall forward due to the inertia of motion. The seat belt worn by passengers of the car prevents them from falling forward suddenly. This enables the entire linear momentum of the passengers to reduce to zero over a long interval of time, hence it prevents injury.

(f) China and glassware are packed with soft material: China and glassware are wrapped in paper before packing to avoid breakage while transporting. During transportation, there may be collisions due to ta jerks of the packed wares. Soft material like paper slows down their rate of change of linear momentum. As a result, the impact is reduced and items are not broken.

(g) Athletes are advised to come to stop slowly after finishing a fast race: By doing so, he decreases the rate of change of linear momentum by increasing the time interval and hence, reducing the impact, which reduces injury.

Question 2.
Explain
(a) How do we swim?
(b) Why does a gun recoil?
(c) It is difficult to walk on sand or ice.
(d) The motion of rocket.
(e) Why does a fireman struggle to hold a hose-pipe?
(f) Rowing of a boat.
(g) When a man jumps out from a boat, the boat moves backward.
(h) Walking of a person.
Answer:
(a) While swimming, a swimmer pushes the water backward with his hands (i.e., he applies force in the backward direction, which is known as action.) The reaction offered by the water to the swimmer pushes him forward.

(b) Recoiling of a gun: When a bullet is fired from a gun, it exerts a forward force on the bullet and the bullet exerts an equal (in magnitude) and opposite (in direction) force on the gun. Due to the high mass of the gun, it moves a little distance backward and gives a backward jerk to the shoulder of the gunman.

(c) It is difficult to walk on sand or ice: When our feet press the sandy ground in the backward direction, the sand gets pushed away and as a result, we get only a small reaction (forward) from the sandy ground making it difficult to walk.

(d) Rocket propulsion Before firing the rocket, the total linear momentum of the system is zero because the rocket is in the state of rest. When it is fired, chemical fuels inside the rocket are burnt and the hot gases are passed through a nozzle with great speed. According to the law of conservation of linear momentum, the total linear momentum after firing must be equal to zero. As the hot gases gain linear momentum to the rear on leaving the rocket, the rocket acquires equal linear momentum in the upward i.e., opposite direction.

(c) A fireman has to make a great effort to hold a hose-pipe to throw a stream of water on the fire to extinguish it. This is because the stream of water rushing through the hose-pipe in the forward direction with a large speed exerts a large force on the hose-pipe in the backward direction which is known as the reaction force. This reaction force tends to move the hose-pipe in the backward direction. Therefore, a fireman struggles to hold the hose-pipe strongly to keep it at rest.

(f) Rowing of a boat: The boatman during the rowing of a boat pushes the water backward with oars (this is an action of the boatman). According to the third law of motion, water exerts an equal (in magnitude) and opposite (in direction) push on the boat which moves it forward (this is a reaction by water). Thus, harder the boatman pushes back the water with oars (i.e., greater is the action), greater is the reaction force exerted by water, and faster the boat moves in the forward direction.

(g) When a man jumps out from a boat, the boat moves backward: When the passengers start jumping out of a rowing boat, they push the boat backward with their feet. The boat exerts an equal (in magnitude) and opposite (in direction) force on passengers in the forward direction which enables them to move forward.

(h) Walking of a person: When a person walks on the ground, he pushes the ground with his foot in the backward direction by pressing the ground. This push is known as the action, According to Newton’s third law of motion, an equal and opposite reaction acts on the foot of the person by the ground. This reaction (force) of the ground on the person pushes him forward.

Force and Laws of Motion Class 9 Extra Questions HOTS

Question 1.
Two balls of the same size but of different materials, rubber and iron are kept on the smooth floor of a moving train. The brakes are applied suddenly to stop the train. Will the balls start rolling? If so, in which direction? Will they move with the same speed? Give reasons for your answer
Answer:
Yes, both the balls will start rolling in the direction opposite to the motion of the train. The speed of two balls will be different as the inertia of the two balls are different.

Question 2.
Two identical bullets are fired on by a light rifle and another by a heavy rifle with the same force. Which rifle will hurt the shoulders more and why?
Answer:
The light rifles will hurt more as the recoil velocity of the light rifle will be greater.

Question 3.
Velocity versus time graph of a ball of mass 50 g rolling on a concrete floor is shown in the figure. Calculate the acceleration and frictional force of the floor on the ball.

Answer:
Given, Mass of ball, m = 50 g = 50 x 10^-3 kg
Acceleration can be calculated by υ – t graph,

= \(\frac{80 \mathrm{m} / \mathrm{s}}{8 \mathrm{s}}\) = – 10 m/s²
a = – 10 m/s²
Friction force, F = ma
F = 50 x 10^-3 x 10
F = 0.5 N

Question 4.
A truck of mass M is moved under a force F. if the truck is then loaded with an object equal to the mass of the truck and the driving force is halved, then how does the acceleration change?
Answer:
Given, Initially
M₁ = M and F₁ = F
Finally
M₂ = 2M and F₂=\(\frac {F}{2}\)
Hence, a₁ = \(\frac{F_{1}}{M_{1}}\)
a₁ = \(\frac{F}{M}\)

and a₂ = \(\frac{F_{2}}{M_{2}}\)
a₂ = \(\frac{\mathrm{F}}{2 \times 2 \mathrm{M}}\) = \(\frac{F}{4M}\)
\(\frac{a_{1}}{a_{2}}=\frac{F / M}{F / 4 M}=4\)
a₂ = \(\frac{a_{1}}{4}\)
Acceleration will become one fourth.

Question 5.
Derive the unit of force using the second law of motion. A force of 5 N produces an acceleration of 8 ms^-2 on a mass m₁ and an acceleration of 24 ms^-2 on a mass m₂. What acceleration would the same force provide if both the masses are tied together?
Answer:
Given, Force, F = 5N
a₁ = 8 m/s²
and a₁ = 24 m/s²
From formula
m₁ = \(\frac{F}{a_{1}}\) = \(\frac{5}{8}\) kg
and m₂ = \(\frac{F}{a_{2}}\) = \(\frac{5}{24}\) kg
When both the masses tied together
a =\(\frac{\mathrm{F}}{m_{1}+m_{2}}\)
a = \(\frac{5}{\frac{5}{8}+\frac{5}{24}}\) = 6m/s²
a = 6m/s²

Force and Laws of Motion Class 9 Extra Questions Value Based (VBQs)

Question 1.
During a cricket match, a new player Ayush injured his hand while catching the ball. His friend Rudra advised him to catch the ball by lowering his hands backward. When Ayush got another chance to catch the ball, he successfully caught the ball without injuring his hands.
Answer the following questions:
(a) A cricket player lowers his hands while catching the ball. Explain why.
(b) Write down the values shown by Rudra.
Answer:
(a) If a player does not lower his hands while catching the ball, the time to stop the ball will be very small. So a large force has to applied to reduce the velocity of the ball to zero or to change the momentum of the ball. When a player lowers his hands, the time to stop the ball is increased and hence less force has to be applied to cause the same change in momentum of the ball. Therefore, the hands of the player are not injured.

(b) Rudra is a good person as he helped his friend Ayush. He is a knowledgeable person.

Question 2.
During servicing his bike Aman advised the mechanic to oil the shockers for its proper functioning.
Answer the following questions:
(a) The vehicles are fitted with shockers. Explain why.
(b) Write down the values shown by Aman.
Answer:
(a) The floor of a vehicle is connected to the lower part of the vehicle by springs or shockers. When a vehicle moves over a rough road, the force due to jerks is transmitted to the floor of the vehicle through the shockers. The shockers increase the time of transmission of the force of the jerk to reach the floor of the vehicle. Hence less force is experienced by the passengers in the vehicles.
(b) Aman is an intelligent and careful person.

Question 3.
Ranjan advised his son Ayush to wear a seat belt while driving the car.
Answer the following questions
(a) While driving the car, the drivers and other passengers are advised to wear seat belts. Explain why.
(b) Write down the values shown by Ranjan.
Answer:
(a) When brakes are applied suddenly, the passengers of the car fall forward due to the inertia of motion. The seat belt worn by passengers of the car prevents them from falling forward suddenly. This enables the entire linear momentum of the passengers to reduce to zero over a long interval of time, hence it prevents injury.
(b) Ranjan is a noble, intelligent, and careful person.

In this page, we are providing Structure of the Atom Class 9 Extra Questions and Answers Science Chapter 4 pdf download. NCERT Extra Questions for Class 9 Science Chapter 4 Structure of the Atom with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 4 Extra Questions and Answers Structure of the Atom

Extra Questions for Class 9 Science Chapter 4 Structure of the Atom with Answers Solutions

Structure of the Atom Class 9 Extra Questions Very Short Answer Type

Structure Of Atom Class 9 Extra Questions Question 1.
Is it possible for the atom of an element to have one electron, one proton and no neutron? If so, name the element. [NCERT Exemplar]
Answer:
Yes, it is true for hydrogen atom which is represented as \(_{ 1 }^{ 1 }{ H }\)

Class 9 Science Chapter 4 Extra Questions Question 2.
Will ³⁵Cl and ³⁵Cl have different valencies? Justify your answer. [NCERT Exemplar]
Answer:
No, ³⁵Cl and ³⁵Cl are isotopes of an element.

Class 9 Structure Of Atom Extra Questions Question 3.
Why did Rutherford select a gold foil in his a-ray scattering experiment? [NCERT Exemplar]
Answer:
Rutherford selected a gold foil in his a-ray scattering experiment because gold has high malleability.

Structure Of Atom Class 9 Important Questions With Answers Pdf Question 4.
One electron is present in the outermost shell of the atom of an element X. What would be the nature and value of charge on the ion formed if this electron is removed from the outermost shell? [NCERT Exemplar]
Answer:
+ 1.

Extra Questions On Structure Of Atom Class 9 Question 5.
Write any two observations which support the fact that atoms are divisible. [NCERT Exemplar]
Answer:
Discovery of electrons and protons.

Structure Of Atom Extra Questions Question 6.
Write down the electron distribution of chlorine atom. How many electrons are there in a L-shell? (Atomic number of chlorine is 17). [NCERT Exemplar]
Answer:
The electron distribution of chlorine atom is 2, 8, 7 and the L-shell has 8 electrons.

Class 9 Science Chapter 4 Extra Questions And Answers Question 7.
In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed? [NCERT Exemplar]
Answer:
– 2.

Extra Questions Of Structure Of Atom Class 9 Question 8.
Helium atom has 2 electrons in its valence shell but its valency is not 2, Explain.
Answer:
Helium atom has 2 electrons in its outermost shell and its duplet is complete. Hence the valency is zero.

Structure Of Atom Extra Questions Class 9 Question 9.
An element X has a mass number 4 and atomic number 2. Write the valency of this element.
Answer:
Valency is zero, as K-shell is completely filled.

Class 9 Science Ch 4 Extra Questions Question 10.
Who identified the sub-atomic particle electron?
Answer:
J.J. Thomson.

Class 9 Chapter 4 Science Extra Questions Question 11.
Name the rays discovered by E. Goldstein.
Answer:
Canal rays.

Structure Of An Atom Class 9 Extra Questions Question 12.
Give two examples from everyday life where we use cathode ray tubes.
Answer:
Television picture tube and Fluorescent light tubes.

Structure Of Atom Class 9 Extra Questions And Answers Question 13.
What are the actual values of charge and mass of the electron?
Answer:
Charge = 1.60 x 10^-19 C (one unit negative charge)
Mass = 9.11 x 10^-31 kg (1/1840th of the mass of hydrogen atom).

Chapter 4 Science Class 9 Extra Questions Question 14.
Compare the radius of the nucleus with that of the atom.
Answer:
Radius of the atom is of the order of 10^-10 m while that of the nucleus is of the order of 10^-15 m. Thus, nucleus is 10^-15/10^-10 i.e., 1/100,000^th of the size of the atom.

Ch 4 Science Class 9 Extra Questions Question 15.
Why are Bohr’s orbits called stationary states?
Answer:
According to Bohr, the orbits in which the electrons revolve have fixed energies. Hence, they are called stationary states (stationary means fixed).

Question 16.
Who discovered neutron?
Answer:
Chadwick.

Question 17.
What are nucleons? What is their number called?
Answer:
Protons and neutrons present in the nucleus are collectively called nucleons. Their total number is called ‘mass number’.

Question 18.
What do you mean by ‘Atomic number’?
Answer:
Atomic number of an element is the number of protons present in the nucleus of the atom of that element.

Question 19.
Who discovered the nucleus of the atom?
Answer:
Rutherford.

Question 20.
What is the charge on alpha particle?
Answer:
+ 2.

Question 21.
The mass number of an element is 18. It contains 7 electrons. What is the number of protons and neutrons in it?
Answer:
Number of protons = 7 Number of neutrons = 11 ,

Question 22.
What is the maximum number of electrons that can be present in M-shell?
Answer:
M-shell means 3rd shell for which n = 3. Hence, maximum number of electrons that can be present in M-shell = 2n² = 2 x 3² = 18.

Question 23.
Identify the isotopes out of A, B, C and D? ³³A₁₇, ⁴⁰B₂₀, ³⁷C₁₇, ³⁸D₁₉.
Answer:
³³A₁₄ and ³⁷C₁₇? are isotopes.

Question 24.
Represent the three isotopes of hydrogen and give their names.
Answer:
\(_{ 1 }^{ 1 }{ H }\) (Protium), \(_{ 2 }^{ 1 }{ H }\) (Deuterium), \(_{ 3 }^{ 1 }{ H }\) (Tritium).

Question 25.
How many electrons are present in the species He²⁺ ion? Suggest another name for it.
Answer:
He²⁺ ion has no electrons. It is also called alpha (α) particle.

Question 26.
What is the reason for the identical chemical properties of all the isotopes of an element?
Answer:
Isotopes have identical electronic configuration.

Question 27.
Give one similarity and one difference between a pair of isotopes.
Answer:
They have same number of protons but different number of neutrons in their nuclei.

Question 28.
Give the number of protons and neutrons in an atom of uranium (U-235) used in a nuclear reactor.
Answer:
In nuclear reactor U-235 is used. Atomic number of Uranium is 92.
∴ Number of protons = Atomic number = 92
Number of neutrons = Mass number – Atomic number = 235 – 92 = 143.

Question 29.
Out of proton and neutron, which is heavier?
Answer:
Neutron is slightly heavier (1.675 x 10^-27 kg) than proton (1.67 x 10^-27 kg).

Question 30.
If K and L shells of an atom are completely filled what will be its name?
Answer:
The atom will belong to the noble gas element neon (Ne).

Question 31.
Do isobars have also identical chemical characteristic like isotopes?
Answer:
No, these are not identical because the isobars have different atomic number as well as different electronic configuration.

Question 32.
What is the number of electrons in the valence shell of chlorine (Z = 17)?
Answer:
The electronic distribution of the element is:
K(2), L(8), M(7). This means that the valence shell of chlorine has 7 electrons.

Question 33.
Which radioisotope is used for the treatment of cancer?
Answer:
Radioisotope Co-60 is used for the treatment of cancer.

Question 34.
Out of C-12 and C-14 isotopes of carbon, which is of radioactive nature?
Answer:
C-14 isotope is of radioactive nature.

Question 35.
The electron configuration of an element is: 2(K), 8(L), 5(M). Predict its valency.
Answer:
The valency of the element is 3. It is calculated as: 8 – 5 = 3.

Question 36.
Will \(_{ 12 }^{ 6 }{ C }\) and \(_{ 14 }^{ 6 }{ C }\) have different valencies?
Answer:
Both the species are the isotopes of the same element i.e., the carbon. Since their atomic numbers Eire same, their electronic configurations as well as valencies will also be the same.

Question 37.
Do the elements \(_{ 3 }^{ 1 }{ X }\) and \(_{ 3 }^{ 2 }{ Y }\) represent pair of isotopes?
Answer:
No, they do not because the two elements differ in their atomic numbers. Isotopes have the same atomic numbers.

Structure of the Atom Class 9 Extra Questions Short Answer Type 1

Question 1.
In response to a question, a student stated that in an atom, the number of protons is greater than the number of neutrons, which in turn is greater than the number of electrons. Do you agree with the statement? Justify your answer. [NCERT Exemplar]
Answer:
No, the statement is incorrect. In an atom the number of protons and electrons is always equal.

Question 2.
Calculate the number of neutrons present in the nucleus of an element X which is represented as \(_{ 31 }^{ 15 }{ X }\). [NCERT Exemplar]
Answer:
Mass number = No. of protons + No. of neutrons = 31
∴ Number of neutrons = 31 – number of protons
= 31 – 15
= 16

Question 3.
The atomic number of calcium and argon are 20 and 18 respectively, but the mass number of both these elements is 40. What is the name given to such a pair of elements? [NCERT Exemplar]
Answer:
∴ They are isobars because isobars have different atomic numbers but same mass number.

Question 4.
Why do Helium, Neon and Argon have a zero valency?
Answer:
Helium has two electrons in its only energy shell, while Argon and Neon have 8 electrons in their valence shells. As these have maximum number of electrons in their valence shells, they do not have any tendency to combine with other elements. Hence, they have a valency equal to zero.

Question 5.
What are isobars? Give one example.
Answer:
Isobars are the atoms of different elements which have different atomic numbers but same mass number.
Example is \(_{ 40 }^{ 18 }{ Ar }\), \(_{ 40 }^{ 20 }{ Ca }\) .

Question 6.
What is the number of valence electron in:
(i) Sodium ion (Na⁺)
(ii) Oxide ion (O^2-)?
Answer:
(i) Sodium ion (Na⁺)
No. of electrons: (11 – 1); Electronic configuration = 2, 8.
∴ Na⁺ ion has 8 valence electrons.

(ii) Oxide ion (O^2-)
No. of electrons: (8 + 2) = 10; Electronic configuration = 2, 8.
∴ O^2- ion has 8 valence electrons.

Question 7.
The number of electrons in the outermost ‘L’ shell of an atom is 5.
(a) Write its electronic configuration.
(b) What is its valency and why?
Answer:
(a) 2, 5
(b) 3, because it needs three more electrons to complete its octet.

Question 8.
Write two difference between isobars and isotopes.
Answer:
Isobars

• Same mass number but different atomic numbers.
• Different number of protons.

Isotopes:

• Same atomic number but different mass numbers.
• Same number of protons.

Question 9.
Mention the postulates Neils Bohr put forth to overcome the objections raised against Rutherford’s atomic model.
Answer:
Bohr put forward the following postulates in order to overcome the objections raised against Rutherford s atomic method: .

Only certain special orbits known as discrete orbits of electrons are allowed inside the atom.
While revolving in discrete orbits the electrons do not radiate energy. These orbits or shells are called energy levels.

Question 10.
Given that natural sample of iron has isotopes \(_{ 54 }^{ 28 }{ Fe }\), \(_{ 58 }^{ 28 }{ Fe }\) and \(_{ 57 }^{ 28 }{ Fe }\) in the ratio of 5%, 90% and 5% respectively. What will be the average atomic mass of iron (Fe)?
Answer:
Average atomic mass
= \(\frac { 54 × 5 + 56 × 90 + 57 × 5 }{ 100 }\)
= \(\frac { 270+5040+285 }{ 100 }\)
= 55.95 u.

Question 11.
(a) What are canal rays? Who discovered them?
What is the charge and mass of canal ray?
(b) How are the canal rays different from electron in terms of charge and mass?
Answer:
(a) New radiations in a gas discharge tube which are positively charged. E. Goldstein discovered them. Charge on canal rays is positive and mass is one unit.

(b) Electrons are negatively charged and their mass is approximately 1/2000 of that of canal rays.

Question 12.
The average atomic mass of a sample of an element y is 35.5 u. What are the percentages of isotopes \(_{ 37 }^{ 17 }{ y }\) and \(_{ 35 }^{ 17 }{ y }\), in the sample?
Answer:
Let the % composition of \(_{ 37 }^{ 17 }{ y }\) be x.
Then the % composition of \(_{ 37 }^{ 17 }{ y }\) will be 100 – x
Now,

Question 13.
(i) Write the postulates of Bohr model of atom.
(ii) Draw a sketch of Bohr model of an atom with atomic number 15.
Answer:
(i) Only certain special orbits-discrete orbits of electrons are allowed.

(ii) While revolving in discrete orbits, the electrons do not radiate energy.

Question 14.
List the main differences between an atom and an ion.
Answer:
Atom:

• It is electrically natural.
• The valence shell of atom may or may not have 8 electrons.
• The atoms are less stable.
• It may or may not exist freely in solution.

Ion:

• It has positive or negative charge.
• The valence shell of an ion has 8 electrons.
• The ions are more stable.
• It can exist freely in solution.

Question 15.
An ion has one neutron more than the number of protons but has one electron less than the number of protons. Which two ions are possible which satisfy this condition? Explain.
Answer:
(i) Na⁺ ion, because it has atomic number = 11 and mass number = 23. Thus, no. of protons = 11, no. of neutrons = 23 – 11 = 12 and number of electrons = 11 – 1 = 10.

(ii) K⁺ ion, because it has atomic number = 19 and mass number = 39. Thus, no. of protons = 19, No. of neutrons = 39 – 19 = 20 and no. of electrons = 19 – 1 = 18.

Question 16.
Give the main points of Dalton’s atomic theory.
Answer:

• All matter is made up of a large number of extremely small indivisible particles called atoms.
• Atoms of the different elements are different.
• Atom is the smallest unit of matter which takes part in a chemical reaction.
• Atoms can neither be created nor be destroyed.

Question 17.
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign proper symbol of the species.
Answer:
Number of electrons present = 18
Number of neutrons present = 16
Atomic number of the element = 16
The element with atomic number 16 is sulphur (S). Since it has 18 electrons and not 16, it is therefore, an anion (S^2- ion).

Question 18.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol. [NCERT Exemplar]
Answer:
We know
Mass number = No. of protons + No. of neutrons = 81
i.e., p + n = 81
Let number of protons = x

Structure of the Atom Class 9 Extra Questions Short Answer Type 2

Question 1.
Show diagrammatically the electron distributions in a sodium atom and a sodium ion and also given their atomic number. [NCERT Exemplar]
Answer:

Since the atomic number of sodium atom is 11, it has 11 electrons. A positively charged sodium ion (Na⁺) is formed by the removal of one electron from a sodium atom. So, a sodium ion has 11 – 1 = 10 electrons. Thus, electronic distribution of sodium ion will be 2, 8. The atomic number of an element is equal to the number of protons in its atom. Since, sodium atom and sodium ion contain the same number of protons, therefore, the atomic number of both is 11.

Question 2.
The ratio of the radii of hydrogen atom and its nucleus is ~ 10⁵. Assuming the atom and the nucleus to be spherical. [NCERT Exemplar]
(i) What will be the ratio of their sizes?
(ii) If atom is represented by planet Earth ‘R_e’ = 6.4 x 10⁵ m, estimate the size of the nucleus.
Answer:
(i) Volume of the sphere = \(\frac { 4 }{ 3 }\) πr³
Let R be the radius of the atom and r be that of the nucleus.

(ii) If the atom is represented by the planet earth (R_e = 6.4 x 10⁶ m) then the radius of the nucleus
would be r_n = \(\frac{\mathrm{R}_{e}}{10^{5}}\)
r_n = \(\frac{6.4 \times 10^{6} \mathrm{m}}{10^{5}}\)
= 6.4 x 10 m = 64 m.

Question 3.
The number of protons, neutrons and electrons in particles from A to E are given below:

(i) Which one is a cation?
(ii) Which one is an anion?
(iii) Which represent pair of isotopes?
Answer:
(i) B is a monovalent cation (B⁺)
(ii) E is a monovalent anion (E^–)
(iii) A and D represent pair of isotopes.

Question 4.
Explain Bohr Bury rules for distribution of electrons into different shells. Write the distribution of electrons in sodium atom (Z = 11).
Answer:
Bohr Bury Rules:

• The maximum no. of electrons present in a shell is given by the formula 2n² (where n is shell number).
• The maximum number of electrons that can be accommodated in the outermost orbit is 8.
• Electrons are not accommodated in a given shell, unless the inner shells are filled.
• Electronic configuration of Na = 2, 8, 1.

Question 5.
Atom A has a mass number 238 and atomic number 92 and atom B has mass number 235 and atomic number 92.
(i) How many protons, atoms A and B have?
(ii) How many neutrons, atoms A and B have?
(iii) Are atoms A and B isotopes of the same element? How?
Answer:
(i) Protons in atoms A and B = 92
(ii) Neutrons in atoms A and B = 146 and 143
(iii) Yes, atom A and B are isotopes of the element since they have the same atomic number or same electronic configuration.

Question 6.
State three features of the nuclear model of an atom put forward by Rutherford.
Answer:

1. There is a positively charged centre in an atom called the nucleus. Nearly, all the mass of an atom resides in the nucleus.
2. The electrons revolve around the nucleus in well-defined orbits.
3. The size of the nucleus is very small as compared to the size of the atom.

Question 7.
The nucleus of an atom is found to have a total mass of nearly 20.088 x 10^-27 kg and a total charge of 9.612 x 10^-19 coulombs. Calculate the atomic number and mass number of the atom. Name the element.
Answer:
Mass of a nucleon (proton or neutron)
= 1.674 x 10^-27 kg
Total mass of nucleus = 20.088 x 10^-27 kg
∴ No. of nucleons (protons + neutrons) present in the nucleus
= \(\frac{20.088 \times 10^{-27}}{1.674 \times 10^{-27}}\) = 12
∴ Mass number = 12u.
Charge on one proton = 1.602 x 10^-19 coulombs; Total charge on the nuclear = 9.612 x 10^-19
∴ No.of protons in the nucleus
= \(\frac{9.612 \times 10^{-19}}{1.602 \times 10^{-19}}\) = 6
Atomic number = 6. Thus, the element is carbon.

Question 8.
Give reasons why?
(а) Atom is electrically neutral.
(b) Atom as a whole is an empty space.
(c) Rutherford model of atom could not provide stability to the nucleus.
Answer:
(a) An atom is electrically neutral because the number of protons and number of electrons in it are equal.

(b) According to Rutherford’s experiment, the size of the nucleus is very small as compared to the size of an atom, therefore, atom as a whole is an empty space.

(c) According to Rutherford, the protons are present inside the nucleus and electrons revolve around the nucleus. According to the elecromagnetic theory, a charged particle moving in a circular path continuously loses energy in the form of electromagnetic radiations and finally falls into the nucleus.

Question 9.
Write the complete symbol for
(i) the nucleus with atomic number 56 and mass number 138.
(ii) the nucleus with atomic number 26 and mass number 55.
(iii) the nucleus with atomic number 4 and mass number 9.
Answer:
(i) The element with atomic number 56 is Ba. Its symbol is \(_{ 138 }^{ 56 }{ Ba }\)

(ii) The element with atomic number 26 is Fe. Its symbol is \(_{ 56 }^{ 26 }{ Fe }\)

(iii) The element with atomic number 4 is Be. Its symbol for \(_{ 9 }^{ 4 }{ Be }\)

Question 10.
How many protons, electrons and neutrons are there in the following nuclei?
(i) \(_{ 17 }^{ 8 }{ O }\)
(ii) \(_{ 25 }^{ 12 }{ Mg }\)
(iii) \(_{ 80 }^{ 35 }{ Br }\)
Answer:
(i) \(_{ 35 }^{ 79 }{ Br }\)
Atomic number, Z = 8
Mass number, A = 17
No. of protons = No. of electrons = Z = 8
No. of neutrons + No. of protons = A
No. of neutrons + 8 = 17
or No. of neutrons = 17 – 8 = 9

(ii) \(_{ 25 }^{ 12 }{ Mg }\)
Atomic number, Z = 12
Mass number, A = 25
No. of protons = No. of electrons = Z = 12
No. of neutrons = A – No. of protons = 25 – 12 = 13.

(iii) \(_{ 80 }^{ 35 }{ Br }\)
Atomic number, Z = 35
Mass number, A = 80
No. of protons = No. of electrons = Z = 35
No. of neutrons = A – No. of protons = 80 – 35 = 45.

Question 11.
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol for the ion.
Answer:
Since the iron carries 3 units of positive charge, it will have 3 electrons less than the number of protons. Let number of electrons = x
No. of protons = x + 3
No. of neutrons = x + \(\frac { x × 30.4 }{ 100 }\)
= x + 0.304x
= 1.304x
Now, No. of protons + No. of neutrons = 56
x + 3 + 1.304 x = 56
2.304 x = 53
x = \(\frac { 53 }{ 2.304 }\) = 23
No. of electrons = 23
No. of protons = 23 + 3 = 26
Symbol = \(_{ 56 }^{ 26 }{ Fe }\)³⁺.

Question 12.
The nuclear radius is of the order of 10^-13 cm while atomic radius is of the order 10^-8 cm. Assuming the nucleus and the atom to be spherical, what fraction of the atomic volume is occupied by the nucleus?
Answer:
The volume of a sphere = 4 πR^3/3 where R is the radius of the sphere.
∴ Volume of the nucleus = 4 πr^3/3 = 4π (10^-13)^3/3 cm³
Similarly,
Volume of the atom = 4 πR^3/3 = 4π (10^-8)^3/3 cm³
∴ Fraction of the volume of atom occupied by the nucleus
= \(\frac{4 \pi\left(10^{-13}\right)^{3} / 3 \mathrm{cm}^{3}}{4 \pi\left(10^{-8}\right)^{3} / 3 \mathrm{cm}^{3}}\)
= 10^-15.

Question 13.
Calculate the number of electrons, protons and neutrons in the following species:
(i) Phosphorus atom
(ii) Phosphide ion (P^3-)
(iii) Magnesium ion (Mg²⁺)
Mass numbers: P = 31; Mg = 24
Atomic numbers: P = 15, Mg = 12.
Answer:
(i) Phosphorus atom
Number of electrons = Atomic number = 15
Number of protons = Atomic number = 15
Number of neutrons = Mass number – Atomic number
= 31 – 15 = 16.

(ii) Phosphide ion (P^3-)
Phosphide ion (P^3-)
= Phosphorus atom + 3 electrons
P^3- = P + 3e^–
Thus, phosphide ion has same number of protons and neutrons as phosphorus atom but has three electrons more.
Number of electrons = 15 + 3 = 18
Number of protons = 15
Number of neutrons = 31 – 15 = 16

(iii) Magnesium ion (Mg²⁺)
Mg²⁺ ion is formed by loss of two electrons by Mg atom. Therefore, it has two electrons less than number of electrons in Mg atom.
Mg²⁺ = Mg – 2e^–
Number of electrons = 12 – 2 = 10
Number of protons = 12
Number of neutrons = (24 – 12) = 12

Structure of the Atom Class 9 Extra Questions Long Answer Type

Question 1.
(a) Enlist the conclusions drawn by Rutherford from his a-ray scattering experiment.
(b) In what way is the Rutherford’s atomic model different from that of Thomson’s atomic model?
Answer:
(a) Rutherford concluded from the a-particle scattering experiment that:

(i) Most of the space inside the atom is empty because the a-particles passed through the gold foil without getting deflected.

(ii) Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space.

(iii) A very small fraction of a-particles were deflected by 180°, indicating that all the positive charges and mass of the gold atom were concentrated in a very small volume within the atom.

From the data he also calculated that the radius of the nucleus is about 105 times less than the radius of the atom.

(b) Rutherford proposed a model in which electrons revolve around the nucleus in well-defined orbits. There is a positively charged center in an atom called the nucleus. He also proposed that the size of the nucleus is very small as compared to the size of the atom and nearly all the mass of an atom is centred in the nucleus.

Whereas, Thomson proposed the model of an atom to be similar to a Christmas pudding. The electrons are studded like currants in a positively charged sphere like Christmas pudding and the mass of the atom was supposed to be uniformly distributed.

Question 2.
(a) What were the drawbacks of Rutherford’s model of an atom?
(b) What are the postulates of Bohr’s model of an atom?
Answer:
(a) The orbital revolution of the electron is not excepted to be stable. Any particle in a circular orbit would undergo an acceleration and the charged particles would radiate energy. Thus, the revolving electron would lose energy and finally fall into the nucleus. If this were so, the atom should be highly unstable and hence matter would not exist in the form that we know.

(b) The postulates put forth by Neils Bohr’s about the model of an atom:
(i) Only certain special orbits known as discrete orbits of electrons, are allowed inside the atom.

(ii) While revolving in discrete orbits the electrons do not radiate energy.
These orbits are called energy levels. Energy levels in Em atom are shown by circles.
These orbits are represented by the letters K, L, M, N, …………….. or the numbers, n = 1, 2, 3, 4, ………………

Question 3.
An atom of an element has three electrons in the third shell which is the outermost shell. Write
(a) the electronic configuration
(b) the atomic number
(c) number of protons
(d) valency
(e) the name of the element
(f) its nature whether metal or non-metal.
Answer:
The third shell is M-shell. If the atom of the element has three electrons in the third shell, this means that K and L shells are already filled.
(a) Electronic configuration: 2, 8, 3
(b) Atomic number = No. of electrons = 13
(c) Number of protons = No. of electrons = 13
(d) Valency of the element = 3
(e) The element with Z = 13 is aluminium (Al)
(f) It is a metal.

Question 4.
Answer the following in one line or two:
(a) What is the maximum number of electrons that can be accommodated in the outermost energy shell in an atom?
(b) On the basis of Thomson’s model of an atom, explain how an atom is neutral as a whole.
(c) How many neutrons are present in hydrogen atom?
(d) Do isobars belong to the same element?
(e) An element has five electrons in the M shell which is the outermost shell. Write its electronic configuration.
Answer:
(a) The outermost energy shell in an atom can have a maximum of eight electrons.

(b) According to Thomson’s model of an atom, all the protons in an atom are present in the positively charged sphere while negatively charged electrons are studded in this sphere. Since the electrons and protons are equal in number each carrying one unit charge, the atom as a whole is electrically neutral.

(c) Hydrogen atom has no neutron.

(d) No, isolars belong to different elements since they differ in their atomic numbers.

(e) The electronic configuration of the element is K(2), L(8), M(5).

Question 5.
Explain why did Rutherford select a gold foil in his alpha particle scattering experiments? What observations in a-scattering experiment led Rutherford to make the following observations:
(i) Most of the space in an atom is empty.
(ii) Nucleus is positively charged.
Mention any two drawbacks of Rutherford’s model.
Answer:
Rutherford selected a gold foil because he wanted a very thin layer as possible. This gold foil was about 1000 atoms thick.

(i) As most of the alpha particles passed through the foil undeviated, it means that they did not come across any obstruction in their part. Thus, most of the space in an atom was thought to be empty.

(ii) Very few particles deviated by small angles from their path which suggested that nucleus is positively charged.

The revolution of the electron in any circular orbit is not expected to be stable. Any particle in a circular orbit would undergo acceleration. During acceleration, charged particles would radiate energy. Thus, the revolving electrons would lose energy and finally fall into the nucleus. If this were so, the atom should be highly unstable and matter would not exist in the form that we know. We know that atoms are quite stable.

Question 6.
(a) Why are anode rays called canal rays?
(6) Mention two postulates of J.J. Thomson’s model.
(c) Compare the properties of protons and electrons.
Answer:
(a) The anode rays produced at the anode of the discharge tube are called canal rays because they pass through the holes of the cathode.

(b)

• Atom consists of positively charged sphere and electrons are embedded in it.
• The negative and positive charges are equal in magnitude.

(c) Protons:

• Positively charged.
• Mass of 1 proton is equal to mass of H atom.

Electrons:

• Negatively charged.
• Mass of electron is 1/1840 times that of a proton.

Question 7.
(a) What are isobars?

(b) Atomic number of an element Y is 17.
(i) Write its electronic configuration.
(ii) What is the number of valence electron in Y?
(iii) How many electrons are needed to complete the octet of Y?
(iv) Is it a metal or non-metal?

(c) The valency of Na is 1 and not 7. Give reason.
Answer:
(a) Atoms of different elements with different atomic numbers which have same mass number are called isobars.

(b) (i) 2,8, 7
(ii) 7
(iii) 1
(iv) Non-metal

(c) Electronic configuration of Na is 2, 8,1. It can lose one electron to attain the electronic configuration of neon. Therefore, the valency of Na is 1. The valency of Na may be 7 when it gains 7 electrons in its valence shell, but gain ring 7 electrons is difficult. Therefore, the valency of Na is not 7.

Question 8.
An atom of an element has 2 electrons in the M-shell i.e., third shell. What will be the atomic number of this element? Name this element. Find the valency of this element. Also, find the number of neutrons in the atom of this element.
Answer:
(i) Atomic number is 12

(ii) Element is magnesium.
(iii) Valency is 2 (⁺²)
(iv) No. of neutrons = Atomic mass – No. of protons = 24 – 12 = 12.

Structure of the Atom Class 9 Extra Questions HOTS

Question 1.
What information do you get from the figure given below about the atomic number, mass number and valency of atoms X, Y and Z? Give your answer in a tabular form. [NCERT Exemplar]

Answer:

Question 2.
In the Gold foil experiment of Geiger and Marsdein, that paved the way for Rutherford’s model of an atom, ~ 1.00% of the a-particles were found to deflect at angles > 50°. If one mole of α – particles were bombarded on the gold foil, compute the number of α – particles that would deflect at angles less than 50°. [NCERT Exemplar]
Answer:
% of α – particles deflected more than 50° = 1% of α – particles
% of α – particles deflected less than 50° = 100 – 1 = 99%
Number of α – particles bombarded = 1 mole = 6.022 x 10²³ particles
Number of particles that deflected at an angle less than 50°
= \(\frac { 99 }{ 100 }\) x 6.022 x 10²³
= \(\frac { 596.178 }{ 100 }\) x 10²³
= 5.96 x 10²³.

Question 3.
An ion Y^3- contains 18 electrons and 16 neutrons. Calculate the atomic number and mass number of the element Y. Name the element Y.
Answer:
Number of electrons in Y^3- ion = 18
Since negative charge is formed by gain of electrons by the neutral atom and the number of electrons gained is equal to the number of units of negative charge on the ion.
∴ Number of electrons in the neutral atom = 18-3 =15.
Now, for a neutral atom, Atomic number = Number of protons = Number of electrons.
∴ Atomic number of element Y = 15
Mass number of the element = Number of protons + Number of neutrons = 15 + 16 = 31
The given element Y with atomic number 15 is phosphorus.

Question 4.
The following data represent the distribution of electrons, protons and neutrons in atoms of four elements A, B, C, D.

Answer the following questions:
(a) Give the electronic distribution of element B.
(b) The valency of element A.
(c) The atomic number of element B.
(d) The mass number of element D.
Answer:
(а) Electronic distribution of elements B : 2, 8, 6.
(b) Valency of A is 1(-1).
(c) Atomic number of element B is 16.
(d) Mass number of element D is 39.

Question 5.
Give a reason to explain why:
(a) isotopes of an element show identical chemical properties?
(b) the atomic masses of elements are in fractions?
(c) atoms combine with other atoms?
Answer:
(a) Since all the isotopes of an element have identical electronic configuration containing the same number of valence electrons. Therefore all the isotopes of an element show identical chemical properties.

(b) The fractional atomic masses of elements are due to their isotopes having different masses.

(c) The atoms combine with other atoms to achieve the electronic configuration of the nearest noble gas and thus, become more stable.

Question 6.
(a) Describe the main features of Bohr’s model of an atom. Draw a neat and labelled diagram of energy levels.
(b) Which of the following pairs are isotopes and which are isobars?
(i) \(_{ 58 }^{ 26 }{ A }\), \(_{ 58 }^{ 28 }{ B }\)
(ii) \(_{ 79 }^{ 35 }{ X }\), \(_{ 80 }^{ 35 }{ Y }\)
Give reasons for your choice.
(c) Elements A and B have atomic numbers 18 and 16 respectively. Which of these two would be more reactive and why?
Answer:
(a) Features of Bohr’s Models:
(i) Electrons revolve in certain permitted orbits which are associated with fixed amount of energy. So they are called energy levels (K, L, M, N) or sub-shells.
(ii) As long as electron revolves in the same energy level, they do not lose energy.
(iii) The energy of orbit closest to the nucleus is lowest and farthest away is highest.
(iv) When we supply energy to an electron, it goes to a higher energy level and when it comes back to a lower level it radiates energy.

(b) \(_{ 79 }^{ 35 }{ X }\), \(_{ 80 }^{ 35 }{ Y }\) are isotopes because their atomic no. is same but mass no. is different. \(_{ 58 }^{ 26 }{ A }\), \(_{ 58 }^{ 28 }{ B }\) are iso¬bars, as mass no. is same but atomic no. is different.

(c) B would be more reactive as its electronic configuration is 2, 8, 6 and it requires two more electrons to complete its octet.

Question 7.
Choose the noble gases from the elements shown in the table below:

Answer:

A, B and D have completely filled valence shells, i.e., either 2 or 8 electrons in their valence shells.
∴ A, B and D are noble gases.

Question 8.
The mass number and electronic configuration of five elements A, B, C, D and E are as follows:

(а) Name the elements which has 22 neutrons in nucleus.
(b) What is atomic number of C?
(c) Name the elements which will form most stable ionic bond.
(d) Give the formation of the compound between B and D.
(e) Name the elements which will not take part in chemical combination.
Answer:
(a) E
(b) 11
(c) B and C
(d) DB₄
(e) E

Question 9.
Write the electronic configurations of the following elements and write the number of valence electrons present in it.
(a) \(_{ 14 }^{ 7 }{ N }\)
(b) \(_{ 28 }^{ 14 }{ Si }\)
(c) \(_{ 40 }^{ 20 }{ Ca }\)
(d) \(_{ 40 }^{ 18 }{ Ar }\)
(e) \(_{ 9 }^{ 4 }{ Be }\)
Answer:

Question 10.
Complete the following table.

Answer:
For F atom
Atomic no. = 9 (Give)
No. of electrons = 9
No. of protons = 9
No. of neutrons = 10 (Given)
Mass no. = Z + n = 9 + 10 = 19

For Mg²⁺ ion
No. of protons = 12 (Given)
Atomic no. = 12
No. of electrons = 12 – 2 = 10
Mass no. = 24 (Give)
No. of neutrons = A – Z = 24 – 12 = 12

For S atom
No. of protons = 16 (Given)
Atomic no. = 16
No. of protons = 16
Mass no. = 32 (Given)
No. of neutrons = A – Z = 32 – 16 = 16

For P^3- ion
No. of electrons = 18 (Given)
No. of protons = 18 – 3 = 15
Atomic no. = 15
No. of neutrons = 16 (Given)
Mass no. = Z + n = 15 + 16 = 31

The complete table is as follows:

Note: The value of positive charge on cation shows that the number of electrons in it are less than the number of protons by the same value. Similarly, the value of negative charge on anion shows that the number of electrons in it are more than the number of protons by the same value.

In this page, we are providing Diversity in Living Organisms Class 9 Extra Questions and Answers Science Chapter 7 pdf download. NCERT Extra Questions for Class 9 Science Chapter 7 Diversity in Living Organisms with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 7 Extra Questions and Answers Diversity in Living Organisms

Extra Questions for Class 9 Science Chapter 7 Diversity in Living Organisms with Answers Solutions

Diversity in Living Organisms Class 9 Extra Questions Very Short Answer Type

Class 9 Science Chapter 7 Extra Question Answer Question 1.
Who proposed the two kingdom classification?
Answer:
Carolus Linnaeus.

Diversity In Living Organisms Class 9 Extra Questions And Answers Question 2.
What is biodiversity?
Answer:
The variety of life forms including plants, animals and microscopic organisms which inhabit this earth constitute biodiversity.

Diversity In Living Organisms Class 9 Questions Question 3.
What is classification?
Answer:
The grouping of organisms on the basis of their similarities and differences is called classification.

Diversity In Living Organisms Class 9 Extra Questions Question 4.
What is the taxonomy?
Answer:
The branch of science which deals with the classification of organisms is called taxonomy.

Diversity In Living Organisms Extra Questions Question 5.
Who proposed the five kingdom system of classification?
Answer:
R.H. Whittaker.

Diversity In Living Organisms Class 9 Questions And Answers Question 6.
Name the fundamental unit of classification.
Or
Which is the lowermost category in the hierarchy of classification?
Answer:
Species.

Diversity In Living Organisms Class 9 Extra Questions And Answers Pdf Question 7.
Give one point of difference between gymnosperms and angiosperms.
Answer:
The seeds of gymnosperms are naked whereas the seeds of angiosperms are enclosed within a fruit.

Diversity In Living Organisms Class 9 Important Questions Question 8.
Name the largest phylum of kingdom Animalia.
Answer:
Arthropoda.

Class 9 Science Ch 7 Extra Questions Question 9.
Name the class to which the sea horse belongs.
Answer:
Sea horse belongs to class Pisces.

Questions On Diversity In Living Organisms Class 9 Question 10.
Which group of plants is referred to as vascular cryptogams?
Answer:
Pteridophytes.

Class 9 Diversity In Living Organisms Extra Questions Question 11.
Which phylum in animial kingdom consists of pseudocoelomate organisms?
Answer:
Nematoda

Class 9 Science Chapter 7 Extra Questions Answers Question 12.
Which group of organisms are called as the ‘Amphibians of plant kingdom*?
Answer:
Bryophytes

Diversity In Living Organisms Class 9 Fill In The Blanks Question 13.
Name the two classes of angiosperms with one example of each.
Answer:
The two classes are –

• Monocots: Wheat
• Dicots: Pea

Diversity In Living Organisms Important Questions Question 14.
Name the division of cryptogams to which algae belong.
Answer:
Thallophyta

Class 9 Diversity In Living Organisms Questions Question 15.
Which is the highest unit of classification?
Answer:
Kingdom

Question 16.
What is the characteristic feature due to which echinoderms are named?
Answer:
Spiny skin (Echino- spiny; derm-skin)

Question 17.
Name the group which comprises of bacteria and blue green algae.
Answer:
Monera

Question 18.
Name an organism which is called saprophyte. Why is it called so?
Answer:
Yeast. It is called so as it feeds on dead and decaying matter to obtain its nutrition.

Question 19.
Identify the kingdom in which the organisms do not have a well-defined nucleus and are not able to show multicellular designs.
Answer:
Monera

Question 20.
Give reason, why blue green algae are classified along with bacteria and placed in the kingdom Monera.
Answer:
As the blue green algae are unicellular prokaryotes like bacteria.

Question 21.
Mention any one characteristic feature of saprophytes.
Answer:
The saprophytes feed on dead and decaying organic matter.

Question 22.
Write one point of difference between monocotyledonous and dicotyledonous plants.
Answer:
Monocots are plants which bear single cotyledon in their seeds. Dicots are the plants which bear two cotyledons in their seeds.

Question 23.
Poriferans have hole or pores all over the body that lead to a system that helps in circulating water to bring in food and oxygen. Name the system.
Answer:
The system is called as the canal system.

Question 24.
State the phylum to which liver fluke and Planaria belong.
Answer:
Phylum Platyhelminthes

Question 25.
Write the type of body cavity and symmetry possessed by nematodes.
Answer:
Body cavity is pseudocoelom and symmetry is bilateral symmetry.

Question 26.
Name one mammal that lays eggs.
Answer:
Platypus

Question 27.
Name the substance which Coelomic cavity of arthropods is filled with. What type of symmetry do they have?
Answer:
Coelomic cavity of arthropods is filled with blood. They show bilateral symmetry.

Question 28.
Name the phylum to which centipede and prawn belong.
Answer:
Phylum Arthropoda

Question 29.
Echinoderms are marine animals. What is their skeleton made up of?
Answer:
Their skeleton is made up of calcium carbonate.

Question 30.
Name one reptile with a four chambered heart.
Answer:
Crocodile

Question 31.
Shyam knew the correct scientific name of mango but did not follow the convention while writing it and wrote it as Mangifera Indica. Rewrite the scientific name as per the convention.
Answer:
Mangifera indica

Question 32.
Rewrite the scientific names correctly:
(a) Panthera tigris
(b) Periplaneta Americana
Answer:
(a) Panthera tigris
(b) Periplaneta americana

Diversity in Living Organisms Class 9 Extra Questions Short Answer Type 1

Question 1.
How can we say that classification of organisms is closely related to their evolution?
Answer:
We can say that classification of organisms is closely related to their evolution because the simple organisms have a primitive body design as they appeared earlier whereas the complex organisms have more advanced body designs as they are more recent. This shows that during the course of evolution more complex body designs were formed from simpler ones.

Question 2.
What is the difference between algae and fungi?
Answer:
Algae:

• Have chlorophyll.
• Autotrophic mode of nutrition.
• Cell wall made of cellulose.
• Food stored in the form of starch.
• Examples: Chlamydomonas, Spirogyra, Ulo-thrix

Fungi:

• Lack chlorophyll.
• Heterotrophic mode of nutrition.
• Cell wall made of chitin.
• Food stored in the form of glycogen.
• Examples: Rhizopus, Agaricus, Yeast

Question 3.
Pick the odd one out and justify your choice by giving reasons:
(a) Moss, Fern, Pinus, Spirogyra
(b) Sea cucumber, Octopus, Feather star, Star fish
Answer:
(a) The odd one out in this case is Pinus as it is a phanerogams having covered reproductive parts whereas the other three are cryptogams which bear hidden reproductive organs.

(b) The odd one out in this case is Octopus as it belongs to phylum Mollusca while others are the members of phylum Echinodermata.

Question 4.
Why were fungi and bacteria considered as plants even though they do not have chlorophyll?
Answer:
Fungi and bacteria were considered as plants as they have cell wall which is a characteristic feature of the plants. So, some earlier classification systems included them under plants.

Question 5.
Why do bryophytes and pteridophytes grow in moist and shady places?
Answer:
Bryophytes and pteridophytes grow in moist and shady places as they need water for their reproduction. Male gametes are carried towards the female gamete by water in order to bring about fertilisation in them.

Question 6.
Which divisions of the plant kingdom are called cryptogams? Why are they called so?
Answer:
Thallophyta, Bryophyla and Pteridophyta are considered as cryptogams. They are called so because they bear hidden and inconspicuous reproductive orgAnswer:

Question 7.
Characteristics of some organisms are given. Identify their group and give one example of each.
(а) Single celled, eukaryotic and photosynthetic
(b) The body is divided into segments, may be unisexual or hermaphrodite
Answer:
(a) Protista: Euglena
(b) Annelida: Earthworm

Question 8.
How do saprophytes get their food? Give one example of saprophyte.
Answer:
Saprophytes derive their nutrition from the dead and decaying materials. For example, Agaricus (Mushroom), Rhizopus (Bread mould), Yeast etc.

Question 9.
What are phanerogams? How are they classified?
Answer:
The phanerogams are the plants which produce seeds and have a well differentiated body with true roots, stem and leaves. They are advanced members of kingdom Plantae. It includes gymnosperms and angiosperms.

Question 10.
What are gymnosperms? Give two examples.
Answer:
The plants which bear naked seeds which are not enclosed in fruit are called gymnosperms. For example, Cycas, Pinus, etc.

Question 11.
Write two peculiar characters of sponges.
Answer:
The two peculiar features of sponges are:
(i) They have pores called ostia all over their body and a single large opening at the top, called osculum which helps to develop a canal system for water movement.

(ii) They have a skeleton made up of calcareous or siliceous spicules or spongin fibres which gives strength and support.

Question 12.
Classify the following in their respective phylum/class: Jellyfish, earthworm, cockroach, rat
Answer:

• Jellyfish: Phylum Coelenterata
• Earthworm: Phylum Annelida
• Cockroach: Phylum Arthropoda
• Rat: Phylum Vertebrata, Class: Mammalia

Question 13.
What are the two peculiar features of phylum Echinodermata?
Answer:
The phylum Echinodermata has organisms which have

• spiny skin
• water vascular system

Question 14.
What are the three germ layers present in the organisms? What are the two groups of organisms on the basis of germ layers?
Answer:
The three germ layers are ectoderm, mesoderm and endoderm. The groups are diploblastic and triploblastic.

Question 15.
How are vertebrates different from the other chordates?
Answer:
The notochord is present at any stage of their life cycle in the case of chordates. In the vertebrates the notochord gets replaced by the vertebral column.

Question 16.
How are pteridophytes are different form Phanerogams? Give one example for each group.
Answer:
Pteridophytes have hidden, inconspicuous reproductive organs example ferns.
Phanerogams have well differentiated reproductive organs which are not hidden, example rose, apple.

Question 17.
(a) Given below are few plant species. Identify the divisions to which they belong and write the major characteristic of each division.
(i) Spirogyra
(ii) Deodar
(iii) Moss
(b) What is the mode of nutrition for all of them?
Answer:
(a) (i) Thallophyta: Plant body is not well differentiated
(ii) Gymnosperms: have naked seeds
(iii) Bryophyta: have rhizoids for absorption of water, have stem-like and leaf-like structures.

(b) All of them are autotrophic organisms.

Question 18.
What do you understand by the term ‘naked embryo’? Name any two divisions in kingdom Plantae that have naked embryo. Give one example of each division.
Answer:
Naked embryo is the term which refers to an embryo which is not borne inside the seed. The pteridophytes and gymnosperms bear naked embryo. For example, Ferns and horsetail are pteridophytes. Pinus and Cycas are gymnosperms.

Question 19.
Write the difference between Gymnosperms and Angiosperms giving example of each type.
Or
What are gymnosperms? Give two characteristics.
Answer:
Gymnosperm:

• Bear naked seeds.
• Are woody, evergreen, perennials.
• Examples: Pinus and Cycas

Angiosperm:

• Bear seeds enclosed in fruit.
• Can be woody, non-woody annual, biennial or perennials.
• Examples: Mango, Neem

Question 20.
Thallophyta, bryophyta and pteridophyta are classified as cryptogams whereas gymnosperms and angiosperms are classified as phanerogams, why?
Answer:
Due to the presence of hidden and inconspicuous reproductive organs, thallophyta, bryophyta and pteridophyta are called as cryptogams. Gymnosperms and angiosperms are phanerogams as they have well developed and distinct reproductive organs, flowers, fruits and seeds.

Question 21.
State reasons for the following:
(а) Platyhelminthes are called so.
(b) Birds have hollow bones.
Answer:
(a) Platyhelminthes are called so because they have a dorsoventrally flattened body.
(b) Presence of hollow bones is an adaptation in birds which helps them to keep low body weight which is helpful in flight.

Question 22.
Identify the phylum of animals by the given characteristics and give an example of each.
(a) The coelomic cavity is blood-filled and the animals have jointed legs.
(b) The animals are called as flatworms and are either free living or parasitic.
Answer:
(a) Phylum arthropoda: eg., cockroach, butterfly, spider, etc.
(b) Phylum Platyhelminthes: Planaria, liver fluke, tapeworm, etc.

Question 23.
Write one point of difference between the following:
(а) Bilateral symmetry and radial symmetry
(b) Annelids and Nematodes
Answer:
(a) Bilateral symmetry – Body can be divided into two exact halves from one plane only.
Radial symmetry – Body can be divided into equal halves from any plane.

(b) Annelids – Have true coelom
Nematodes – Have pseudocoelom.

Question 24.
Give two examples of each:
(а) Egg laying mammals
(b) Organisms with open circulatory system.
Answer:
(а) Duck-billed platypus and Echidna are the egg laying mammals.
(b) Cockroach and Octopus have open circulatory system.

Question 25.
Select the odd one out with respect to classification. Also give reason for your choice: prawn, scorpion, octopus, butterfly.
Answer:
Octopus is the odd one out as it belongs to phylum Mollusca whereas others belong to phylum Arthropoda.

Question 26.
Given below are the two groups of organisms belonging to kingdom Animalia. Write the names of the phylum to which they belong.
(a) Octopus, Pila, Unio
(b) Centipede, prawn, scorpion
Answer:
(а) Octopus, Pila, Unio belong to Phylum Mollusca.
(b) Centipede, prawn, scorpion belong to phylum Arthropoda.

Question 27.
What is binomial nomenclature? Who introduced it?
Answer:
The system of scientific naming of the organism which consists of a generic name and a specific epithet is called as binomial nomenclature. It was introduced by Carolus Linnaeus.

Question 28.
Give two differences between bony fish and cartilaginous fish. Give one example of each.
Answer:
Bony Fish:

• Endoskeleton is made up of bones.
• Operculum covers the gill slits.
• Terminal mouth Example: Rohu, sea horse

Cartilaginous Fish:

• Endoskeleton is made up of cartilage.
• Operculum absent.
• Ventral mouth Example: Shark, saw fish

Question 29.
State any two characteristics of Mammalia. Name two egg laying mammals.
Answer:
Two characteristics of Mammalia are:

• Presence of mammary glands, four chambered heart and are warm-blooded
• Skin has hairs, sweat glands and oil glands Egg laying mammals are platypus and Echidna.

Question 30.
Write appropriate terms for the following.
(a) Animals which are able to maintain a certain body temperature over a wide range of temperature in the environment.
(b) Animals which have pseudocoelom.
Answer:
Kingdom Mammalia: Warm-blooded animals
Phylum Nematoda: Its members have pseudocoelom.

Diversity in Living Organisms Class 9 Extra Questions Short Answer Type 2

Question 1.
Explain the three basic features for grouping all organisms into five major kingdoms.
Answer:
The three basic features for grouping the organisms into five kingdoms are
(i) Cell structure: On the basis of this the two groups are prokaryotes and the eukaryotes which are distinguished on the basis of absence or presence of well defined nuclear membrane.

(ii) Thallus organisation: The organisms are grouped as unicellular or multicellular organisms on the basis of their being composed of a single cell or of many cells respectively.

(iii) Mode of nutrition: The organisms are grouped as autotrophs or heterotrophs on the basis of their ability to synthesise their own food or being dependent on other organisms for their food.

Question 2.
What are the steps in building a hierarchy of classification?
Answer:
The characteristics chosen for developing a hierarchy of classification should start with the characteristic which forms the broadest division and then the next characteristic chosen should be dependent on the previous characteristic and division, besides having its own new characteristic features.

This process should be continued for the next levels in order to build a hierarchy in classification. For example, the classification of organisms into two broad categories prokaryotes and eukaryotes forms the basis of further characteristics on which their classification is based.

Question 3.
Differentiate between Bryophyta and Pteridophyta. Give example of each group.
Or
Write four main features of pteridophyta and give two examples.
Answer:
Bryophyta:

• They are called the ‘amphibians of the plant kingdom’.
• They lack vascular tissues.
• Body is not well-differentiated into true root, stem or leaves.
• The dominant phase or the main plant body is gametophyte (haploid).
• Sporophyte depends upon gametophyte for its support and nutrition.
• Spores are formed in capsule of sporophyte.
• Examples: Liverworts, Mosses

Pteridophyta:

• They are the first land plants.
• They have vascular tissues xylem and phloem.
• Body is well-differentiated into true roots, stem and leaves.
• The dominant phase or the main plant body is sporophyte (diploid).
• Sporophyte and gametophyte are independent structures in them.
• Spores are produced inside the sporangia borne on leaves or cones.
• Examples: Ferns, Horsetail, Marsilea

Question 4.
Name three groups which are placed under Cryptogamae. State and explain two characteristics which are exhibited by each category of these plant bodies.
Answer:
The three groups placed under Cryptogamae are: Thallophyta, Bryophyta and Pteridophyta.

The two characteristic features of cryptogams are:

• They have inconspicuous and hidden reproductive organs.
• They produce naked embryos called spores.

Question 5.
Define the following: (a) Radial symmetry (b) Bilateral symmetry
Answer:
When any plane passing through the central axis can divide the organism into identical halves, it is called radial symmetry, example coelenterates and the adults of Echinoderms.

When only one plane can divide the body of the organism into identical right and left halves, the symmetry is called as bilateral symmetry, example annelids, arthropods and humans.

Question 6.
Give main features of phylum Chordata.
Answer:
The main features of phylum Chordata are presence of:

• Notochord
• Dorsal Central nervous system
• Pharynx perforated by gill slits
• Ventral heart
• A post-anal tail

Question 7.
(a) Write any two features that are present in all chordates.
(b) Write one difference between pseudocoelom and true coelom.
Answer:
(a) The features present in all chordates are:

• Dorsal central nervous system and dorsal nerve cord.
• Triploblastic, coelomate with organ system level of organisation.

(b) If the body cavity of an organism is not lined by mesoderm and the space is filled with vacuolated cells, then the body cavity called pseudocoelom.

Question 8.
Define classification. Give any two of its significance.
Answer:
Grouping of organisms on the basis of their similarities and differences is called classification. Classification is important as it helps to make the study of vast variety of organisms easier and also helps us to understand the inter-relationships which occur among the organisms.

Question 9.
Classify the following plants into different plant divisions. Spirogyra, Fern, Funaria, Pinus, Apple tree, Mustard plant
Answer:
Spirogyra: Thallophyta, Fern: Pteridophyta, Funaria: Bryophyta, Pinus: Gymnosperm, Apple tree: Angiosperm, Mustard plant: Angiosperm

Question 10.
To which group do the following organisms belong and give one reason for each.
(a) Cyanobacteria
(b) Euglena
(c) Ulothrix
Answer:
(a) Cyanobacteria: Kingdom Monera; does not have a well-defined nucleus and lacks membrane bound cell organelles, prokaryotes.

(b) Euglena: Kingdom Protista; are aquatic, unicellular eukaryotes with well-defined nucleus and membrane bound cell organelles.

(c) Ulothrix: Kingdom Plantae, Thallophyta; Have a thalloid body, photosynthetic and eukaryotic.

Question 11.
Why are bryophytes called as ‘Amphibians of plant kingdom9?
Answer:
Bryophytes are called as ‘amphibians of plant kingdom’ because:

• They live in moist, damp places in order to get water from soil either directly or with the help of their rhizoids.
• They require water for the transfer of their gametes and are dependent on water for sexual reproduction.

Question 12.
(а) What is coelom?
(b) Presence of ‘coelom’ in an animal is considered advantageous. Why?
Answer:
(a) Coelom is a body cavity lined by mesodermal cells and lies between the body wall and alimentary canal (gut) of the organism.

(b) The coelom is considered advantageous as it helps to accommodate the various organs of the body in a proper way and gives a greater flexibility to the body of the organism.

Question 13.
(а) Write one characteristic each of amphibia and aves.
(b) Write the name of the class to which following belong:
(i) Sea-horse
(ii) King cobra
Answer:
(a) Amphibia: Can live both on land and in water, slimy to touch, cold blooded with three chambered heart.
Aves: Have forelimbs modified into wings for flight. Body covered with feathers, are warm-blooded with four-chambered heart

(b) Sea-horse: Class Pisces;
King cobra: Class Reptilia

Question 14.
Give the three characteristic features of Class Mammalia.
Answer:
The characteristic features of Class Mammalia are:

• Presence of Mammary glands.
• The skin has hairs.
• Their body has sweat glands and oil glands.
• They have a four-chambered heart, are warm-blooded.

Question 15.
Distinguish between monocots and dicots.
Answer:
Monocots:

• Possess a single cotyledon.
• Have fibrous roots.
• Have parallel venation. Examples: Wheat, Maize

Dicots:

• Possess two cotyledons.
• Have tap root.
• Have reticulate venation. Examples: Pea, gram

Question 16.
(a) What is symbiotic relationship?
(b) Name a symbiotic life form that grows on the bark of tree as large coloured patches.
Answer:
(a) The mutually dependent relationship between two organisms where both are benefitted is called symbiotic relationship.

(b) Lichen is a symbiotic life form that grows on the bark of trees as large coloured patches. It is symbiotic association between fungi and algae.

Question 17.
State the bases of classifying plants and animals into different categories.
Answer:
The bases for classifying animals and plants into different categories are:

• Mode of nutrition – Plants are autotrophic whereas animals heterotrophic.
• Presence or absences of cell wall – Plant cells have cell wall whereas animal cells do not have a cell wall.

Question 18.
(a) Differentiate between Fungi and Plantae.
(b) Mention the basis of classification among plants to different levels.
Answer:
Fungi:

• They are non-chlorophyllous and heterotrophic.
• Their cell wall is made of chitin.
• They are mostly saprophytes or parasites.

Plantae:

• They have chlorophyll and are autotrophic.
• Their cell wall is made of cellulose.
• They are autotrophs.

(b) The basis of classification of plants to different levels is –

• Presence or absence of well-differentiated body tissues
• Presence or absence of a well developed transport system for water, minerals and organic substances.
• Seed bearing ability and presence of seeds enclosed in fruits or not.

Question 19.
Name the group which is called as Amphibian of the plant kingdom. Cite an example of this group, also mention one important feature of the same group.
Answer:
Bryophytes are called amphibians of the plant kingdom, example Marchantia, Funaria.

Characteristic feature: Plant body is thalloid, not well-differentiated and has root-like, stem-like and leaf-like structures.

Question 20.
Identify the following:
(a) Amphibians of the plant kingdom.
(b) Plants with hidden reproductive organs
(c) Mutually benefitted relationship between two organisms.
Answer:
(a) Bryophytes
(b) Phanerogams comprising of Algae, Bryophytes and Pteridophytes.
(c) Symbiotic relationship as shown by algae and fungi in Lichen.

Question 21.
Describe the following:
(a) Lichens
(b) Cryptogams
(c) Phanerogams
Answer:
(a) Lichen is a symbiotic relationship between an algae and fungi.
(b) Cryptogams have hidden or inconspicuous reproductive organs.
(c) Phanerogams have well differentiated reproductive tissues which result in formation of seed.

Question 22.
(а) Which group of plants is known as ‘flowering plants’?
(b) On the basis of seed how is a maize plant different from a pea plant?
Answer:
(a) Angiosperms are called the flowering plants.
(b) Maize is a monocot plant as it bears one cotyledon in its seed whereas pea is a dicot as it has two cotyledons in its seed.

Question 23.
Name the largest group of animals. Write the salient features of this group. Give two examples.
Answer:
Phylum Arthropoda is the largest group of animals. They have jointed legs, bilateral symmetry and a body cavity filled with fluid. For example, Cockroach, spider, butterfly, etc.

Question 24.
What is binomial nomenclature? Who gave it? Write its advantage.
Answer:
The system of scientific naming of organisms which consists of a generic name and a specific epithet is called binomial nomenclature. It was given by Carolus Linnaeus. Its advantage is that it helps in the systematic study and understanding of the living organisms.

Question 25.
List the conventions used for writing a scientific name. What is the importance of scientific name?
Answer:
Convention for writing the scientific names:

• The name of the genus begins with a capital letter.
• The name of the species begins with a small letter.
• When printed, the scientific name is given in italics.
• When written by hand, the genus name and the species name have to be underlined separately.

Common names cannot be used in the same way by the scientist world over and can often result in confusion. To avoid this, a system of scientific names has been proposed.

Question 26.
Write true (T) or false (F)
(а) Whittaker proposed five kingdom classification.
(b) Monera is divided into Archaebacteria and Eubacteria.
(c) Starting from Class, Species comes before the Genus.
(d) Anabaena belongs to the kingdom Monera.
(e) Blue-green algae belongs to the kingdom Protista.
(f) All prokaryotes are classified under Monera.
Answer:
(a) True
(b) True
(c) False
(d) True
(e) False
(f) True

Question 27.
Fill in the blanks:
(a) Fungi shows _______ mode of nutrition.
(b) Cell wall of fungi is made up of _______
(c) Association between blue-green algae and fungi is called as _______
(d) Chemical nature of chitin is _______
(e) _______ has smallest number of organisms with maximum number of similar characters.
(f) Plants without well differentiated stem, root and leaf are kept in _______
(g) _______ are called the amphibians of the plant kingdom.
Answer:
(a) saprophytic
(b) chitin
(c) lichens
(d) Carbohydrate
(e) Species
(f) Thallophyta
(g) Bryophytes

Question 28.
You are provided with the seeds of gram, wheat, rice, pumpkin, maize and pea. Classify them whether they are monocot or dicot.
Answer:
Gram – dicot
Wheat – monocot
Rice – monocot
Pumpkin – dicot
Maize – monocot
Pea -dicot

Question 29.
Match items of column (A) with items of column (B).

Answer:
(a) (ii)
(b) (i)
(c) (iv)
(d) (iii)
(e) (vi)
(f) (u)
(g) (vii)

Question 30.
Match items of column (A) with items of column (B).

Answer:
(a) (iii)
(b) (ii)
(c) (vi)
(d) (i)
(e) (v)
(f) (iv)

Question 31.
Classify the following organisms based on the absence/presence of true coelom (i.e., acoelomate, pseudocoelomate and coelomate) Spongilla, Sea anemone, Planaria, Liver fluke, Wuchereria, Ascaris, Nereis, Earthworm, Scorpion, Birds, Fishes, Horse.
Answer:

• Spongilla: Acoelomate
• Sea anemone: Acoelomate
• Planaria: Acoelomate
• Liver fluke: Acoelomate
• Wuchereria: Pseudocoelomate
• Ascaris: Pseudocoelomate
• Nereis: Coelomate
• Scorpion: Coelomate
• Earthworm: Coelomate
• Birds, Fishes and Horse: Coelomate

Question 32.
Endoskeleton of fishes are made up of cartilage and bone; classify the following fishes as cartilagenous or bony:
Torpedo, Sting ray, Dog fish, Rohu, Angler fish, Exocoetus.
Answer:

• Torpedo: cartilagenous
• Sting ray: cartilagenous
• Dog fish: cartilagenous
• Rohu: bony
• Angler fish: cartilagenous
• Exocoetus: bony

Question 33.
Classify the following based on number of chambers in their heart.
Rohu, Scoliodon, Frog, Salamander, Flying lizard, King Cobra, Crocodile, Ostrich, Pigeon, Bat, Whale
Answer:

• Rohu, Scoliodon: 2 chambered,
• Frog, Salamander, Flying lizard, King Cobra: 3 chambered,
• Crocodile, Ostrich, Pigeon, Bat, Whale: 4 chambered

Question 34.
Classify Rohu, Scolidon, Flying lizard, King Cobra, Frog, Salamander, Ostrich, Pigeon, Bat, Crocodile and Whale into the cold-blooded/warm-blooded animals.
Answer:

• Cold-blooded: Rohu, Scolidon, Flying lizard, King Cobra, Frog, Salamander Crocodile
• Warm-blooded: Ostrich, Pigeon, Bat, Whale

Question 35.
Name two egg laying mammals.
Answer:

• Platypus
• Echidna.

Question 36.
Fill in the blanks:
(а) Five kingdom classification of living organisms is given by _______
(b) Basic smallest unit of classification is _______
(c) Prokaryotes are grouped in Kingdom _______
(d) Paramecium is a Protista because it is an _______
(e) Fungi do not contain _______
(f) A fungus _______ can be seen without microscope.
(g) Common fungi used in preparing bread is _______
(h) Algae and fungi form symbiotic association called _______
Answer:
(a) Robert Whittaker
(b) species
(c) Monera
(d) eukaryotic unicellular organism
(e) chlorophyll
(f) mushrooms
(g) yeast
(h) lichens

Question 37.
Give True (T) and False (F).
(а) Gymnosperms differ from Angiosperms in having covered seed.
(b) Non flowering plants are called Cryptogamae.
(c) Bryophytes have conducting tissue.
(d) Funaria is a moss.
(e) Compound leaves are found in many ferns.
(f) Seeds contain embryo.
Answer:
(a) False
(b) True
(c) False
(d) True
(e) True
(f) True

Question 38.
Give examples for the following.
(a) Bilateral, dorsiventral symmetry is found in _______
(b) Worms causing the disease elephantiasis is _______
(c) Open circulatory system is found in _______ where coelomic cavity is filled with blood.
(d) _______ are known to have pseudocoelom.
Answer:
(a) Liver Fluke
(b) Filarial worms
(c) Arthropods
(d) Nematodes

Question 39.
Label a, b, c and d, given in the figure below. Give the function of (b).
Answer:

(a) Dorsal fin
(b) Caudal fin
(c) Pelvic fin
(d) Pectoral fin
Function of Caudal fins: Caudal fins help in streamlined movement in water.

Question 40.
Fill in the boxes given in figure with appropriate characteristics/plant groups.

(a) Thallophyta
(b) Without specialized vascular tissue
(c) Pteridophyta
(d) Phanerogams
(e) Bear naked seeds
(f) Angiosperm
(g) Have seeds Thallophyta with two cotyledons
(h) Monocots

Diversity in Living Organisms Class 9 Extra Questions Long Answer Type

Question 1.
Explain the various categories of taxonomical hierarchy.
Answer:
The categories in taxonomical hierarchy are

• Kingdom: The highest category of classification
• Phylum (for animals) / Division (for plants): Group of related classes
• Class: Group of related orders.
• Order: Group of related families.
• Family: Group of related genus.
• Genus: Group of related species.
• Species: Group of organisms which can interbreed among themselves to produce a fertile offspring.

Question 2.
(a) Draw labelled diagrams of three protozoa.
(b) Euglena is a dual organism. Why?
Answer:
(a)

(b) Euglena is a dual organism as:

• It does not have a cell wall and behaves as a heterotrophic organism in the absence of sunlight.
• It contains chloroplast and can perform photosynthesis and behaves as autotroph in the presence of sunlight.

Question 3.
Mention the class to which they belong and give one characteristic feature of each. Frog, fish, lizard, pigeon, bat.
Answer:
(i) Fish: Class Pisces; Aquatic, respiration with gills, moist scales present on the body, two-chambered heart, cold-blooded

(ii) Frog: Class Amphibia; Can live both on land and in water, cold-blooded, three-chambered heart, slimy to touch

(iii) Lizard: Class Reptilia; Cold-blooded, three-chambered heart, hard covering present on eggs to prevent from desiccation

(iv) Pigeon: Class Aves; Forelimbs modified into wings for flight and have feathers, have beak, warm-blooded, four-chambered heart

(v) Bat: Class Mammalia; has mammary glands, warm-blooded, four-chambered heart.

Question 4.
(i) Draw a neat labelled diagram of Hydra.
(ii) Label mesoglea and gastro-vascular cavity.
(iii) Name the group of animals it belongs to.
(iv) Name one species of this group that lives in colonies.
Answer:
(i) and (ii)-See figure

(iii) Phylum Coelenterata
(iv) Corals live in colonies.

Question 5.
Write one difference for each of the following pairs:
(i) Thallophyta and Bryophyta
(ii) Nematoda and Annelida
(iii) Amphibians and Reptiles
Answer:
(i)

• Thallophyta Bryophyta – Plant body is not well differentiated into root, stem and leaves. It is a thalloid example – Spirogyra.
• Bryophyta – Plant body slightly more differentiated than Thallophyta with root-like, stem-like and leaf-like structures, called as Amphibians of the plant kingdom.

(ii)

• Nematoda – Called as roundworms, bilaterally symmetrical having pseudocoelom, example Ascaris
• Annelida – Have a metamerically segmented body and a true coelom, example Earthworm

(iii) Amphibians

• Can live both in land as well as water.
• Lack scales on body and have slimy skin.
• Lay eggs in water.
• Eggs are devoid of tough covering.

Reptiles:

• Most of them are terrestrial animals.
• Have dry scales on body.
• Lay their eggs mostly on land.
• Eggs have a tough covering to protect from drying.

Question 6.
(a) In which two ways are amphibians different from fishes?
(b) Identify the phylum of organisms having the characteristics:
(i) Pore bearing animals and radial symmetry
(ii) Body spiny and radial symmetry
(c) Why do gymnosperms not require water for fertilisation?
Answer:
(a) Amphibians can live both on land as well as water, have three-chambered heart and have lungs for respiration in adult stage.
Fishes are aquatic, have two-chambered heart and have gills for respiration.

(b) (i) Phylum Porifera
(ii) Phylum Echinodermata

(c) Gymnosperms do not require water for fertilisation as they bear pollen grains which are carried away by agents like wind to cause pollination.

Question 7.
Give four features of phylum Coelenterata. Give two examples.
Answer:
The four distinguishing features of phylum Coelenterata are:

• They have a radially symmetrical body.
• The body is made of two layers of cells, so called diploblastic.
• Has tentacles which surround their mouth. Tentacles bear cnidoblasts, stinging cells.
• They have a characteristic cavity called as coelenteron.
• They are solitary like Hydra and can be colonial like Obelia. example Hydra, Obelia, Jelly fish.

Question 8.
List the convention that is followed while writing the scientific names. Give scientific name of Mango and Tiger.
Answer:
The conventions followed while writing the scientific names are:

• The name of the genus begins with a capital letter.
• The name of the species begins with a small letter.
• When printed, the scientific name is given in italics.
• When written by hand, the genus name and the species name have to be underlined separately.

Scientific name of Mango is Mangifera indica and of Tiger is Panthera tigris.

Question 9.
Enlist the features of organisms placed in Protista. Give two examples.
Answer:
The members of the kingdom Protista have the following features:

• All of them are unicellular and eukaryotic.
• Their mode of nutrition is either autotrophic or heterotrophic.
• They are usually aquatic but some of them are parasitic.
• Have cilia, flagella or pseudopodia for movement. example Amoeba, Plasmodium

Question 10.
Give the general characteristics of fungi. Give two examples.
Answer:
The general characteristics of fungi are:

• They are heterotrophic organisms which can occur as saprophytes, parasites or symbionts.
• Their cell wall is made up of chitin and lack chlorophyll.
• Food is stored in the form of glycogen in them instead of starch as stored in plants.
• They are eukaryotes as they have membrane bound nucleus.
• For example: Yeast, Rhizopus, Agaricus, Penicillium, etc.

Question 11.
What are the five kingdoms of Whittaker? Give the most important characteristic feature of each kingdom.
Answer:
The five kingdoms proposed by R. H. Whittaker comprises of – Monera, Protista, Fungi, Plantae and Animalia.

• Kingdom Monera: Unicellular, prokaryotes
• Kingdom Protista: Unicellular eukaryotes
• Kingdom Fungi: Non-photosynthetic, chlorophyll absent, heterotrophic nutrition
• Kingdom Plantae: Chlorophyll bearing, photosynthetic autotrophs
• Kingdom Animalia: Heterotrophic nutrition, cell wall absent

Question 12.
(a) To which group do algae belong? Write one characteristic of the division. Give two examples.
(b) Name the group:
(i) Which includes unicellular eukaryotic organisms
(ii) In which mode of nutrition is saprophytic.
(iii) In which seeds are not enclosed in fruits.
(c) Classify flowering plants on the basis of number of cotyledons present in the seed.
Answer:
(a) Algae are members of division Thallophyta. Characteristic – Aquatic, chlorophyll bearing autotrophs.

(b) (i) Protista
(ii) Fungi
(iii) Gymnosperms

(c) The flowering plants are classified as monocots and dicots on the basis of presence of one cotyledon or two cotyledon respectively.

Question 13.
List three groups of plants and tell which plants are referred to as vascular plants? Also mention out of these which group is further classified on the basis of number of cotyledon? State its two characteristics.
Answer:
The pteridophytes, gymnosperms and angiosperms are the vascular plants. Out of these three, angiosperms are classified on the basis of number of cotyledon as – monocots having one cotyledon and dicots having two cotyledons.

Characteristic features of angiosperms are:

• Their seeds are enclosed within fruit.
• Embryo bears the cotyledons also called seed leaves which provide nutrition to the young plant/seedling on the germination of the seed.

Question 14.
(a) On what basis does the embryo of cryptogam differ from that of phanerogams?
(b) Describe the feature that divides the angiosperms into two groups.
(c) State the two sub-groups of angiosperms.
Answer:
(a) Cryptogams bear naked embryos whereas the phanerogams have embryos which are enclosed in seed.
(b) The feature that divides the angiosperms into two groups is the number of cotyledons present in their embryo.
(c) The two sub-groups of angiosperms are monocots (bear single cotyledon) and dicots (have two cotyledons).

Question 15.
Draw a neat diagram of Spirogyra and label the following parts:
(a) Outermost layer of the cell
(b) Organelle that performs the function of photosynthesis
(c) Jelly-like substance in the cell where all organelles are suspended.
(d) Dark coloured and dot-like structure generally present in the centre of the cell.
Answer:

(а) Outermost layer of the cell – Cell wall
(b) Organelle that performs the function of photosynthesis – Chloroplast
(c) Jelly-like substance in the cell where all organelles are suspended – Cytoplasm
(d) Dark coloured and dot-like structure generally present in the centre of the cell – Nucleus

Question 16.
Thallophyta, bryophyta and pteridophyta are called as ‘Cryptogams’. Gymnosperms and Angiosperms are called as ‘phanerogams’. Discuss why. Draw one example of Gymnosperm.
Answer:
Thallophyta, bryophyta and pteridophyta are called as ‘Cryptogams’ because they have hidden or inconspicuous reproductive organs. Spores are formed in them instead of seeds. Gymnosperms and Angiosperms are called as ‘phanerogams’ as they have well differentiated reproductive tissue/organs. Seed harbours the embryo and provides it nourishment too.

Question 17.
Define the terms and give one example of each
(a) Bilateral symmetry
(b) Coelom
(c) Triploblastic
Answer:
(a) If the organism can be divided exactly into two halves from one median plane only, the symmetry is called bilateral symmetry, example liver fluke.

(b) The internal body cavity present between visceral organs and body wall in which well developed organs can be accommodated is called as coelom, example butterfly.

(c) The organisms who have three embryonic layers are called as triploblastic organisms example star fish.

Question 18.
You are given leech, Nereis, Scolopendra, prawn and scorpion; and all have segmented body organisation. Will you classify them in one group? If no, give the important characters based on which you will separate these organisms into different groups.
Answer:
No, all the organisms given in the question do not belong to one group.

Leech and Nereis belong to phylum Annelida because they have metamerically segmented body i.e., body is divided into many segments internally by septa. Body segments are lined up one after the other from head to tail.

The characteristic identifying feature of Scolopendra, prawn and scorpion are the jointed legs and open circulating system due to which they are placed in phylum Arthropoda.

Question 19.
Which organism is more complex and evolved among Bacteria, Mushroom and Mango tree? Give reasons.
Answer:
Mango tree is more complex and evolved because, it is eukaryotic, autotrophic, terrestrial sporophyte with covered seed. Bacteria is unicellular prokaryote and fungi are the heterotrophic, simple thallophyte with no tissue systems.

Question 20.
Differentiate between flying lizard and bird. Draw the diagram.
Answer:
Flying lizard belongs to the group reptiles and is characterised as cold-blooded, body covered with scales and have three-chambered heart, while birds belong to group aves and are characterised as warm¬blooded, having feather covered body, forelimbs modified as wings and having four-chambered heart.

Question 21.
List out some common features in cat, rat and bat.
Answer:
Bat, rat and cat belong to the class Mammalia and have following common features:
(а) All have notochord at some stage of their life cycle.
(b) All are warm-blooded:
(c) All have four-chambered heart.
(d) All have skin covered with hair and with sweat and oil glands.

Question 22.
Why do we keep both snake and turtle in the same class?
Answer:
Because both are –

• Cold-blooded
• Have scales
• Breathe through lungs
• Have three-chambered heart
• They lay eggs with tough covering.

Diversity in Living Organisms Class 9 Extra Questions HOTS

Question 1.
Which group among the Amphibia and Pisces is more advanced and why?
Answer:
The amphibians are more advanced than the Pisces as

• Amphibians can survive both on land as well as in water whereas the members of Pisces can survive only in water.
• Amphibians have a three-chambered heart compared to the two-chambered heart present in the Pisces.
• Amphibians have the ability to respire through lungs but fishes respire through gills.

Question 2.
The protozoans have been included in Protista and not in kingdom Animalia. Give reason.
Answer:
Protozoans are unicellular, eukaryotes so they have been kept in the Kingdom Protista comprising of only the unicellular, eukaryotic organisms. Kingdom Animalia consists of multicellular eukaryotes.

Question 3.
Three types of animal specimen were collected by Rajeev and labelled as A, B and C. The specimen A had slimy skin, respired through lungs, B had dry scales with eggs having tough covering and the specimen C had moist scales with terminal mouth. Identify the class to which the specimens belong.
Answer:
A: Amphibia
B: Reptilia
C: Pisces

Question 4.
While playing near a pond Anmol experienced a pain on his feet. He saw a black coloured organism with metamerically segmented body was clinging to his foot and trying to suck blood from his foot. On the basis of this identify the organism and the phylum to which it may belong to. Name two other members of the phylum to which this organism belongs.
Answer:

• Organism: Leech
• Phylum: Annelida
• Other members: Nereis, Earthworm

Question 5.
Shobha went for a school trip and was shown some organisms which were told by their teachers as the members of the second largest phylum of kingdom Animalia. The organisms had a shell on their body and moved around with their muscular foot. Name the phylum to which the mentioned organism will belong. Give one more characteristic feature and two examples of such organisms.
Answer:
Phylum : Mollusca
Feature : Reduced Coelomic cavity
Examples : Snails and mussels

Question 6.
Three types of plant specimen were observed by Renu and labelled as A, B and C. The specimen A had green colour with undifferentiated thalloid body, B had well differentiated body which formed spores and had a vascular system and the specimen C had cones which had seeds but no fruits were formed in them. Identify the class to which the specimens belong.
Answer:
A: Thallophyta
B: Pteridophyta
C: Gymnosperms

Diversity in Living Organisms Class 9 Extra Questions Value Based (VBQs)

Question 1.
Reena read an article in newspaper regarding the problems caused by the chemical fertilisers. She surfed through the internet and came to know that certain blue-green algae are a good source of fertilisers. She advised the farmers of the area to use them and helped them to stop using chemical fertilisers.
(a) Which group comprises of the blue-green algae?
(b) How do these blue-green algae increase soil fertility?
(c) What are the characteristic features of the group to which they belong?
(d) What values are shown by Reena by her work?
Answer:
(a) Kingdom Monera
(b) They fix atmospheric nitrogen and increase soil fertility
(c) They are unicellular prokaryotes which do not have a well defined nucleus.
(d) Values shown by Reena are concern for environment, helpfulness, eco-friendliness and scientific attitude.

Question 2.
During their trip to a hill station, Shiksha observed some thalloid plants having little differentiation of body which were growing on the moist and damp surface in the form of dense mats. She asked her teacher about them and their role. Her teacher told her which group of organism they were and said that these dense mats were helpful in preventing soil erosion and need water for fertilisation.
(a) What is the probable group of organisms which her teacher would have told her?
(b) What is the peculiar feature of this group of organisms?
(c) Give two examples of members of the group.
(d) What are the values shown by Shiksha?
Answer:
(a) Bryophytes
(b) Can live both on land and in water
(c) Marchantia, Funaria
(d) The values shown by Shiksha are keen observation and scientific attitude.

In this page, we are providing Tissues Class 9 Extra Questions and Answers Science Chapter 6 pdf download. NCERT Extra Questions for Class 9 Science Chapter 6 Tissues with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 6 Extra Questions and Answers Tissues

Extra Questions for Class 9 Science Chapter 6 Tissues with Answers Solutions

Tissues Class 9 Extra Questions Very Short Answer Type

Class 9 Science Chapter 6 Tissue Extra Question Answer Question 1.
Name any two types of simple permanent plant tissues.
Answer:
The simple permanent tissues of plants are: Parenchyma, collenchyma and sclerenchyma.

Tissues Class 9 Extra Questions With Answers Question 2.
What are blood platelets?
Answer:
Blood platelets are the cell fragments present in the plasma of blood which help in the clotting of blood.

Tissue Class 9 Extra Questions And Answers Question 3.
Name the connective tissue that is found between skin and muscles.
Answer:
Areolar tissue.

Tissues Extra Questions And Answers Question 4.
Name the tissue present in the brain.
Answer:
Nervous tissue which comprises of its basic unit called neurons.

Class 9 Tissue Extra Questions Question 5.
In which of the simple plant tissue, deposition of lignin is found?
Answer:
Sclerenchyma.

Tissues Class 9 Extra Questions Question 6.
Name the basic packing tissue of plant.
Answer:
Parenchyma.

Tissue Extra Questions Class 9 Question 7.
Name the tissue which is present in the veins of leaves.
Answer:
Sclerenchyma.

Class 9 Science Chapter 6 Extra Questions And Answers Question 8.
Why is cork impervious to gases and water?
Answer:
Due to presence of a chemical substance called suberin.

Class 9 Tissues Extra Questions Question 9.
What is the function of phloem?
Answer:
Phloem helps in the transport of food from leaves to the various parts of the plant.

Extra Questions On Tissues Class 9 Question 10.
Which body cell provides resistance against infections?
Answer:
White blood cells (WBC) provide resistance against infections.

Extra Questions Of Tissue Class 9 Question 11.
Which biochemicals compose the solid matrix of cartilage?
Answer:
Proteins and sugars make up the solid matrix of cartilage.

Tissue Class 9 Extra Questions Question 12.
Name the connective tissue which helps in the repair of tissues.
Answer:
Areolar connective tissue helps in the repair of tissue.

Class 9 Science Ch 6 Extra Questions Question 13.
Which connective tissue is specialised for fat storage and acts as heat insulator?
Answer:
Adipose tissue helps in storage of fats and acts as heat insulator.

Tissue Chapter Class 9 Extra Questions Question 14.
Which muscle has spindle-shaped cells?
Answer:
Smooth muscle cells have spindle shaped cells.

Ch 6 Science Class 9 Extra Questions Question 15.
Which meristem is present at growing tips of stems and roots?
Answer:
Apical meristem is present at the growing tips of the stem and roots.

Tissues Class 9 Extra Questions Short Answer Type 1

Question 1.
List any four salient features of meristematic tissue.
Answer:
The salient features of meristematic tissue are:

• This tissue consists of cells which continuously divide to produce new cells.
• The cells of this tissue lack vacuoles.
• The cells of this tissue have dense cytoplasm.
• The cells of this tissue have thin cellulosic cell walls and prominent nuclei.

Question 2.
Write the four elements of xylem.
Answer:
The four elements of xylem are tracheids, vessels, xylem parenchyma and xylem fibres.

Question 3.
How is ligament different from tendons?
Or
Differentiate between tendon and ligament.
Answer:
Ligament is a connective tissue which joins bone to bone and is elastic in nature.

Tendons join bone to muscles and are less elastic as compared to the ligaments.

Question 4.
Write a short note on the different types of meristematic tissue with their location and functions in the plants.
Answer:
The meristematic tissues are classified as apical, lateral and intercalary meristematic tissue based on the region where they are present.

Apical meristem – It is present at the growing tips of stem and roots and results in increase in the length of the stem and the root. Lateral meristem (cambium): It is present on the lateral sides of stem and roots. It helps to increase the girth of the stem or root.

Intercalary meristem – It is present at the base of the leaves or internodes. It helps in the longitudinal growth of plants.

Question 5.
Show the diagrammatic representation of location of lateral meristem and intercalary meristem in plant body.
Answer:

Question 6.
Differentiate between chlorenchyma and aerenchyma.
Or
Write the difference between aerenchyma and chlorenchyma.
Answer:
Chlorenchyma:

• It is a type of parenchyma which contains chlorophyll.
• It helps to perform photosynthesis.
• It is present in green parts of plants like the leaves.

Aerenchyma:

• This type of parenchyma that has large air cavities in it.
• It helps to provide buoyancy to the plants.
• It is present in the aquatic plants, example in their floating leaves.

Question 7.
Write the functions of collenchyma in plants.
Answer:
Collenchyma allows easy bending in various parts of a plant (leaf, stem) without breaking. It also provides mechanical support to plants like in the leaf stalks below the epidermis.

Question 8.
What are the roles of epidermis in plants?
Answer:
The functions of epidermis are:

• Epidermis is usually made up of a single layer of cells and gives protection.
• The epidermis may be thicker in some plants living in dry habitats or often secrete a waxy, water- resistant layer on their outer surface called cutin (chemical substance with waterproof quality) to prevent water loss.
• The epidermis of leaves have small pores called as stomata which help in gaseous exchange and transpiration.
• The epidermal cells of roots bear root hairs that greatly increase the total absorptive surface area of the roots for absorption of water.

Question 9.
Answer the following:
(i) How is the epidermis of the plants living in very dry habitats adapted?
(ii) Write functions of guard cells of stomata in the leaf.
Answer:
(i) The epidermis of plants living in dry habitats may be thicker or often secrete a waxy, water-resistant layer on their outer surface called cutin (chemical substance with waterproof quality) to prevent water loss.

(ii) The guard cells of stomata in the leaf help in gaseous exchange and transpiration.

Question 10.
What is the function of areolar tissues?
Answer:
The functions of areolar tissues are:

• It fills the space inside the organs
• It supports internal organs.
• It helps in repair of tissues.

Question 11.
Determine the location of the following tissues:

1. Unstriated muscle fibres
2. Cuboidal epithelium
3. Adipose tissue
4. Striated muscle fibres

Answer:

1. Unstriated muscle fibres: Present in iris of the eye, ureters, blood vessels, alimentary canal and bronchi of lungs.
2. Cuboidal epithelium: Present in lining of kidney tubules and ducts of salivary glands.
3. Adipose tissue: It is found below the skin and between internal orgAnswer:
4. Striated muscle fibres: It is present in muscles of our limbs

Tissues Class 9 Extra Questions Short Answer Type 2

Question 1.
Explain how the bark of a tree is formed. How does it act as a protective tissue?
Answer:
In the older stem, a strip of secondary meristem replaces the epidermis. The secondary meristem cuts off cells towards outside to form a several-layer thick tissue; This is called the cork or the bark of the tree.

Cells of cork or bark are dead, compactly arranged without intercellular spaces and have a chemical called suberin in their walls that makes them impervious to gases and water. In this way it acts as a protective tissue.

Question 2.
Draw a diagrammatic labelled sketch of stem tip to show location of meristematic tissue. Mention the functions of different types of meristematic tissue.
Answer:

The types of meristematic tissue are:
(i) Apical meristem: It is present at the growing tips of stems and roots and results in increase in the length of the stem and the root.

(ii) Lateral meristem (cambium): It is present on the lateral sides of stems and roots. It helps to increase the girth of the stem or root.

(iii) Intercalary meristem: It is present at the base of the leaves or internodes. It helps in the longitudinal growth of plants.

Question 3.
What are the two main components of blood? Why is blood considered a type of connective tissue?
Answer:
Blood is a special connective tissue consisting of a fluid matrix, plasma, and formed elements. The formed elements are red blood cells (RBCs), white blood cells (WBCs) and blood platelets. Blood is considered as a type of connective tissue as they have the same origin as other types of connective tissue and helps to connect the different parts of the body to facilitate exchange of various components like nutrients and gases.

Question 4.
Give one function of each of the following.
(i) Stomata
(ii) Root nodules
(iii) Cardiac muscle fibres
Answer:
(i) Stomata: Help in exchange of gases in the plants.

(ii) Root nodules: In leguminous plants, the root nodules harbour nitrogen fixing bacteria which convert atmospheric nitrogen into nitrates.

(iii) Cardiac muscle fibres: They help in rhythmic contraction and relaxation of the heart.

Question 5.
Differentiate between bone and cartilage.
Or
Differentiate between bone and cartilage with respect to structure, function and location.
Answer:
Bone:

• Bones have a hard and non-pliable ground substance.
• Its matrix is rich in calcium salts and collagen fibres.
• It is the main tissue that provides structural frame to the body.
• The bone cells (Osteocytes) are present in the spaces called lacunae.
• Bones are present in the limbs and form main skeletal framework of the body.

Cartilage:

• Cartilage is pliable, flexible and resist compression.
• Its matrix is rich in protein called chondrin and sugars.
• It is present in bones of the vertebral column, limbs and hands in adults.
• Cells of this tissue (chondrocytes) are enclosed in small cavities within the matrix secreted by them.
• Cartilage is present in the tip of nose, outer ear joints, between adjacent bones of the vertebral column.

Question 6.
Explain the basic criteria for classification of permanent tissue in plants.
Answer:
The permanent tissues are classified on the basis of the following criteria:

1. Simple (made of one type of cell) or complex (made of more than one type of cells)
2. Cell wall: Thin or thick
3. Type of cell: living or dead
4. Type of function the tissue performs: epidermis is protective, parenchyma is packing or supportive tissue and sclerenchyma makes up conducting tissue.

Question 7.
Identify the given two slides A and B as a parenchyma or sclerenchyma. Sclerenchyma can be identified by which characteristic?

Answer:
Slide A is parenchyma and Slide B is sclerenchyma.
Sclerenchyma can be identified by the type of cells which are long and narrow as the walls are thickened due to presence of lignin.

Question 8.
(i) Identify the given figures.

(ii) Give any two major differences between the structures identified.
(iii) Describe the role performed by these two in the plant body.
Answer:
(i) Structure (A) is a tracheid and structure (B) is a vessel.

(ii) Tracheid:

• Tracheids are elongated or tube-like cells with thick and lignified walls and tapering ends.
• They are in the form of single cells.

Vessel:

• Vessel is a long cylindrical tube-like structure made up of many cells called vessel members.
• They are composed of a number of cells fused together.

(iii) Tracheids and vessels help in vertical transport of water and minerals in the plants. They also help to provide mechanical strength to the plants.

Question 9.
Draw a well labelled diagram of cardiac muscle found in the human body. Write two differences between striated and smooth muscles.
Answer:

Question 10.
Draw a labelled diagram of unstriated muscle tissue and mention its occurrence, features and functions.
Answer:

(i) The cells are long and spindle-shaped.
(ii) They do not have striations.
(iii) Involuntary in nature as they are not under control of our will.
(iv) The cells of smooth muscles are uninucleate.
(v) Smooth muscle fibres are present in iris of the eye, ureters, blood vessels, alimentary canal and bronchi of lungs.

Question 11.
Name the kinds of muscles found in your limbs and lungs. How do they differ from each other structurally and functionally?
Answer:
Striated muscle fibres are found in limbs whereas smooth muscle fibres are present in lungs. The differences in their structure are:
(i) Striated muscle fibres have alternate light and dark bands which are not present in the smooth muscle fibres.

(ii) Striated muscle fibres are cylindrical and multinucleate whereas the smooth muscle fibres are spindle-shaped and uninucleate.

(iii) Striated muscles are voluntary in nature (under control of our will) whereas the smooth muscle fibres are involuntary in nature (not under control of our will).

Question 12.
What are neurons? Where are they found in the body? What function do they perform in the body of an organism?
Answer:
The cells of nervous tissue are called nerve cells or neurons. Neurons are the structural and functional unit of the nervous system. They are found in the brain, spinal cord and nerves.

Their functions are:

• They are highly specialised for transmitting the stimulus from one place to another within the body on being stimulated.
• They help to coordinate the various functions of the body.

Question 13.
Animals of colder regions and fishes of cold water have thicker layer of subcutaneous fat. Describe why?
Answer:
The thick layer of subcutaneous fat acts as insulator and prevents the heat of the body to escape out. The layer of fat acts as a subcutaneous insulation of body for thermoregulation.

Question 14.
Match the column (A) with the column (B).

Answer:
(a) (v)
(b) (iv)
(c) (iii)
(d) (i)
(e) (ii)
(f) (vi)

Question 15.
Match the column (A) with the column (B).

Answer:
(a) (i)
(b) (c)
(d) (iii)
(e) (iv)

Question 16.
If a potted plant is covered with a glass jar, water vapours appear on the wall of glass jar. Explain why.
Answer:
The water is lost by the plant in the form of water vapour due to the process of transpiration. These water vapours appear on the wall of the glass jar.

Question 17.
Name the different components of xylem and draw a living component.
Answer:
Xylem consists of four elements which are:
(a) tracheids
(b) vessels
(c) xylem parenchyma
(d) xylem fibres
The only living component of xylem is xylem parenchyma whose basic structure is shown below:

Question 18.
Draw and identify different elements of phloem.
Answer:
Phloem has four elements called sieve tubes, companion cells, phloem fibres and the phloem parenchyma.

Question 19.
Write true (T) or false (F).
(a) Epithelial tissue is protective tissue in animal body.
(b) The lining of blood vessels, lung alveoli and kidney tubules are all made up of epithelial tissue.
(c) Epithelial cells have a lot of intercellular spaces.
(d) Epithelial layer is permeable layer.
(e) Epithelial layer does not allow regulation of materials between body and external environment.
Answer:
(a) True
(b) True
(c) False
(d) True
(e) False

Question 20.
Differentiate between voluntary and involuntary muscles. Give one example of each type.
Answer:
Voluntary muscles are present in our limbs as skeletal muscles and can be moved by our conscious will whenever we want. Involuntary muscles cannot function on their own. They cannot be controlled by our will or desire. The cardiac muscle and the smooth muscles are involuntary in nature.

Question 21.
Differentiate the following activities on the basis of voluntary (V) or involuntary (IV) muscles.
(a) Jumping of frog
(b) Pumping of the heart
(c) Writing with hand
(d) Movement of chocolate in your intestine
Answer:
(a) (V)
(b) (IV)
(c) (V)
(d) (IV)

Question 22.
Fill in the blanks.
(a) Lining of blood vessels is made up of _______
(b) Lining of small intestine is made up of _______
(c) Lining of kidney tubules is made up of _______
(d) Epithelial cells with cilia are found in _______ of our body.
Answer:
(a) Squamous epithelium
(b) Columnar epithelium
(c) Cuboidal epithelium
(d) Respiratory tract

Question 23.
Water hyacinth floats on water surface. Explain.
Answer:
The parenchyma present in the swollen petiole of water hyacinth is called aerenchyma which has large cavities to provide buoyancy and help them float on the water surface.

Question 24.
Which structure protects the plant body against the invasion of parasites?
Answer:
The epidermis of plants has thick cuticle and waxy substances to prevent the invasion of parasites.

Question 25.
Fill in the blanks.
(а) Cork cells possesses _______on their walls that makes it impervious to gases and water.
(b) _______ have tubular cells with perforated walls and are living in nature.
(c) Bone possesses a hard matrix composed of and _______ and _______
Answer:
(a) suberin
(b) sieve tubes
(c) calcium and phosphorus

Question 26.
Why is epidermis important for the plants?
Answer:
The outermost layer of cells covering an organism is called epidermis. It is usually made up of a single layer of cells and gives protection.

The epidermis may be thicker in some plants living in dry habitats or often secrete a waxy, water- resistant layer on their outer surface called cutin (chemical substance with waterproof quality) to prevent water loss.

The stomata present on the epidermis of leaves helps in gaseous exchange and the loss of water vapour by transpiration.

The epidermal cells of roots bear root hairs that greatly increase the total absorptive surface area of the roots for absorption of water.

Question 27.
Fill in the blanks.
(a) _______ are forms of complex tissue.
(b) _______ have guard cells.
(c) cells of cork contain a chemical called _______
(d) Husk of coconut is made of _______ tissue.
(e) _______ gives flexibility in plants.
(f) _______ and _______ are both conducting tissues.
(g) Xylem transports and _______ and _______ from soil.
(h) Phloem transport from _______ and _______ to other parts of the plant.
Answer:
(a) Xylem and phloem
(b) Stomata
(c) suberin
(d) sclerenchyma
(e) Collenchyma
(g) water; minerals
(h) Food; leaf

Tissues Class 9 Extra Questions Long Answer Type

Question 1.
Differentiate between
(i) Xylem and phloem
(ii) Vessel and sieve tube
(iii) Tracheid and vessel
Answer:
(i) Xylem and phloem –
Xylem:

• Xylem consists of tracheids, vessels, xylem parenchyma and xylem fibres.
• All the cells of xylem except the xylem parenchyma are dead.
• Xylem helps to transport water and minerals.
• The transport is unidirectional through xylem.

Phloem:

• Phloem has four elements called sieve tubes, companion cells, phloem fibres and the phloem parenchyma.
• All cells of phloem are living except the phloem fibres.
• Phloem transports food from leaves to other parts of the plant.
• The transport is bidirectional through the phloem.

(ii) Vessel and sieve tube –
Vessel:

• They are tubular structures having a hollow lumen and composed of dead cells.
• Vessel helps to conduct water and minerals in plants.
• The walls of vessels are lignified.
• They also provide mechanical strength to the plants.
• Their end walls are completely dissolved.

Sieve Tube:

• They are tubular structures having vacuolated cytoplasm and composed of living cells.
• They help to transport food from leaves to other parts of the plant.
• Their walls are not lignified.
• They do not provide mechanical strength to the plants.
• Their end walls have perforations in form of sieve plate.

(iii) Tracheid and vessel _
Tracheid:

• Tracheids are elongated or tube-like cells with thick and lignified walls and tapering ends.
• They are in the form of single cells.
• The inner layers of the cell walls are more thickened.
• They have narrow lumen.
• They have pointed ends.

Vessel:

• Vessel is a long cylindrical tube-like structure made up of many cells called vessel members.
• They are composed of a number of cells fused together.
• Their walls are less thickened.
• They have wide lumen.
• They have blunt ends.

Question 2.
Differentiate between striated, unstriated and cardiac muscle fibres.
Answer:

Question 3.
(i) What is nervous tissue?
(ii) Draw a well labelled diagram of neuron. (Label any 4 parts)
Answer:
Nervous tissue is a tissue made of neurons. It is divided into two parts: the central nervous system (CNS) consisting of the brain and spinal cord; and the peripheral nervous system (PNS) which regulates and controls the various functions and activities of the body.

Question 4.
Write the differences between animal tissue and plant tissue.
Answer:
Plant Tissue:

• The tissue is well differentiated into meristematic tissue and permanent tissue.
• The tissue can grow throughout life due to activity of meristematic tissue.
• They are autotrophic in nature.
• The tissue has more amount of dead tissue which provides mechanical strength to the plants.
• The tissue organisation is comparatively simple.

Animal Tissue:

• The tissue is not much differentiated like the plant tissue.
• The tissue does not show growth throughout life.
• They are heterotrophic in nature.
• The tissue has more amount of living tissue than dead tissue.
• The tissue is complex as it is organised into organs and organ systems.

Question 5.
Write a note on the protective tissue in plants. (Give appropriate diagram also)
Answer:
The protective tissues in plants are epidermis and the cork.
(i) Epidermis: The outermost layer of cells covering an organism is called epidermis. It is usually made up of a single layer of cells and gives protection.

The epidermis may be thicker in some plants living in dry habitats or often secrete a waxy, water- resistant layer on their outer surface called cutin to prevent water loss.

The epidermis of leaves have small pores called stomata which are enclosed by two kidney-shaped cells called guard cells. Stomata help in gaseous exchange and transpiration.

The epidermal cells of roots bear root hairs that greatly increase the total absorptive surface area of the roots for absorption of water.

(ii) Cork: A strip of secondary meristem replaces the epidermis of the older stem and cuts off cells towards compactly arranged without intercellular spaces and have a chemical called suberin in their walls that makes them impervious to gases and water.

Question 6.
Explain the significance of the following:
(i) Hair-like structures on epidermal cells.
(ii) Epidermis has thick waxy coating of cutin in desert plants.
(iii) Small pores in epidermis of leaf.
(iv) Numerous layers of epidermis in cactus.
(v) Presence of a chemical suberin in cork cells.
Answer:
(i) To increase the total absorptive surface area for absorption of water.
(ii) To prevent water loss by transpiration and protection from pathogens.
(iii) To help in gaseous exchange and transpiration.
(iv) To prevent water loss by transpiration.
(v) To make tissue impervious to gases and water.

Question 7.
Differentiate between sclerenchyma and parenchyma tissues. Draw well labelled diagram.
Answer:
Parenchyma:

• Cells are thin walled and thickened with cellulose.
• It is made up of living cells.
• Cells are usually loosely pac ked with large intercellular spaces.
• Helps to store nutrients and water in stem and roots.
• It is called chlorenchyma if it contains chlorophyll and performs photos ynthesis. The parenchyma of
• aquatic plants have large cavities to provide buoyancy to the plants to help them float, it is then called aerenchyma.

Sclerenchyma:

• Cells are thick and thickened with lignin.
• This tissue is made up of dead cells.
• There are no intercellular spaces between the cells.
• Provides strength to the various parts of the plant.
• The cells are long and narrow, make the plant hard and stiff. This tissue provides strength to the plants and is present in stems, around vascular bundles, in the veins of leaves and in the hard covering of seeds and nuts.

Question 8.
Describe the structure and function of different types of epithelial tissues. Draw diagram of each type of epithelial tissue.
Answer:
Epithelial tissues are the covering or protective tissues and cover most organs and cavities in the animal body. These cells are tightly packed, form a continuous sheet and are almost without any intercellular spaces between them. E.g., skin, the lining of the mouth, the lining of blood vessels, lung alveoli and kidney tubules are all made of epithelial tissue.

All epithelium is usually separated from the underlying tissue by an extracellular fibrous basement membrane. The types of epithelium on the basis of their structure and functions are:

(a) Squamous epithelium: Consists of flattened cells. Present in oesophagus and lining of the mouth. Skin epithelial cells are arranged in many layers to prevent wear and tear and are called stratified squamous epithelium.

(b) Columnar epithelium: Has tall or ‘pillar-like’ cells. It forms the inner lining of the intestine.

(c) Cuboidal epithelium: Has cube-shaped cells. It forms the lining of kidney tubules and ducts of salivary glands, where it provides mechanical support.

(d) Ciliated epithelium: Have cilia on the outer surfaces of epithelial cells. The cilia can move and their movement pushes the mucus in the respiratory tract forward to clear it.

(e) Glandular epithelium: Has gland cells which secrete substances at the epithelial surface.

Question 9.
Give reasons for
(a) Meristematic cells have a prominent nucleus and dense cytoplasm but they lack vacuole.
(b) Intercellular spaces are absent in sclerenchymatous tissues.
(c) We get a crunchy and granular feeling, when we chew pear fruit.
(d) Branches of a tree move and bend freely in high wind velocity.
(e) It is difficult to pull out the husk of a coconut tree.
Answer:
(a) Because the meristematic cells are actively dividing cells and there is no need of storage.
(b) Because they have a thick deposition of lignin in them.
(c) Due to the presence of stone cells (sclerenchyma) in the pear fruit.
(d) Due to the presence of collenchyma which provides flexibility to the various parts of the plant.
(e) Due to the sclerenchyma present in the husk of the coconut.

Question 10.
List the characteristics of cork. How are they formed? Mention their role.
Answer:
The characteristics of cork are:

• Cells of cork are dead at maturity.
• These cells are compactly arranged.
• Cells do not possess intercellular spaces.
• Cells possess a chemical substance suberin in their walls.
• They are several layers thick.

A strip of secondary meristem replaces the epidermis of the older stem and cuts off the outside cells to form a several-layer thick cork or the bark of the tree. Cells of cork are dead, compactly arranged without intercellular spaces and have a chemical called suberin in their walls that makes them protective in function and impervious to gases and water.

Question 11.
(a) Differentiate between meristematic and permanent tissues in plants.
(b) Define the process of differentiation.
(c) Name any two simple and two complex permanent tissues in plants.
Answer:
(a) Meristematic tissue:

• This tissue consists of cells which continuously divide to produce new cells.
• They are located at specific regions of the plant, i.e., apical, lateral and intercalary.
• The cells of this tissue are very active, lack vacuoles, have dense cytoplasm, thin cell walls and prominent nuclei.

Permanent tissue:

• Consists of cells which have taken up a specific role and lost the ability to divide.
• They are distributed throughout the plant body.
• They are vacuolated, vary in shape and size. Their cell wall may be thick.

(b) The process of taking up a permanent shape, size, and a function by the cells is called differentiation.

(c) Simple: Parenchyma/collenchyma/sclerenchyma Complex: Phloem/xylem

Tissues Class 9 Extra Questions HOTS

Question 1.
The walls of the sclerenchymatous cells are thickened and have narrow lumen. Which substance thickens it and what is its role?
Answer:
The walls of the sclerenchymatous cells are thickened due to presence of lignin. It helps in providing mechanical strength to the various parts of the plant.

Question 2.
Which type of muscle fibres will contract to remove your hands instantly when you touch a hot object?
Answer:
Striated muscle fibres will contract to remove our hands instantly when we touch a hot object.

Question 3.
Which tissue helps the leaves of lotus plant to float on water? Why?
Answer:
Aerenchyma helps the leaves of lotus plant to float on water. Aerenchyma has large cavities to provide buoyancy to the parts of aquatic plants.

Question 4.
A tissue present in plants helps in storing food and sideways conduction of water. Identify the type of tissue.
Answer:
The tissue is xylem parenchyma.

Question 5.
Which tissue enables the heart to pump blood to various parts of the body? Why?
Answer:
The cardiac muscles help the heart to pump blood to various parts of the body as they show rhythmic contraction and relaxation throughout their life.

Question 6.
What will be the consequence of
(i) removal of blood platelets from blood?
(ii) removal of cutin from the layer of epidermis?
Answer:
(i) Removal of blood platelets from blood will inhibit clotting of blood if an injury occurs and the person may bleed to death.

(ii) Removal of cutin would increase the amount of water loss from the leaves of the plants.

Question 7.
Some actions of our body are under our control but many of them are not under our control. Why is it so?
Answer:
The actions of our body are controlled by our muscles. The voluntary actions are under the control of our will and are caused by the activity of striated muscles, eg., movement of our limbs. The involuntary . actions are not under the control of our will and are performed by the smooth muscles, e.g., the activity of bronchi of lungs. Even the activity of cardiac muscles which helps in the rhythmic contraction and relaxation of heart are involuntary in nature.

Question 8.
Which kind of meristem can help grasses to regenerate parts removed by the grazing herbivores?
Answer:
Intercalary meristem can help grasses to regenerate parts removed by the grazing herbivores.

Question 9.
Name the tissue which replaces the epidermal tissue in older stem and is rich in suberin. What is the function of suberin?
Answer:
Cork is the tissue which replaces the epidermal tissue in older stem and is rich in suberin. Suberin present in the walls of cork cells makes them impervious to gases and water.

Question 10.
The process of transpiration does not occur properly when the leaves are covered by a layer of oily substance. Why? Which other functions will get affected due to this covering?
Answer:
The layer of oily substance will close the stomata present in leaves and this would decrease the rate of transpiration. The rate of exchange of gases decreases and consequently the rate of photosynthesis would also decrease.

Tissues Class 9 Extra Questions Value Based (VBQs)

Question 1.
Raman got injured while playing football. His injured leg started bleeding and his friends immediately rushed to take him to the doctor to give him first aid. The blood flowing from the wound stopped after some time and the doctor applied antiseptic on the wound.
(i) Why did the blood stop flowing after some time from the wound?
(ii) What kind of tissue is blood? Why?
(iii) What values are shown by Raman’s friends?
Answer:
(i) The blood stopped flowing after some time from the wound as the blood platelets present in blood helped in clotting of blood.

(ii) Blood is a connective tissue. Blood is considered as a type of connective tissue as they have the same origin as other types of
connective tissue and helps to connect the different parts of the body.

(iii) The values shown by Raman’s friends are presence of mind, helpful and a caring nature.

Question 2.
During a sports event, Shivani suffered a sprain due to which she was not able to run. Her teacher gave her support and told her that it was due to a ligament tear. She also called the doctor to give treatment to Shivani.
(i) What is a ligament? What kind of tissue is it?
(ii) Which type of fibrous tissue has great strength, limited flexibility and is similar to ligament?
(iii) What values are shown by Shivani’s teacher?
Answer:
(i) Ligament is the connective tissue which connects two bones. It is a kind of connective tissue.

(ii) Tendon is a fibrous tissue that has great strength, limited flexibility and is similar to ligament.

(iii) The values shown by her teacher are knowledge, scientific approach and a caring nature.

Question 3.
Rishi brought an aquatic plant which was floating on the surface to the science laboratory of water. He cut a section of the leaf of the plant and saw a tissue with lot of air cavities in it. He went to his teacher and discussed about the role of the air cavities in the leaves of the aquatic plant.
(i) Which type of tissue present in plants has air cavities?
(ii) What is the role of large air cavities in the leaves of such plants?
(iii) What values are shown by Rishi?
Answer:
(i) Aerenchyma is the tissue present in the plants and has large air cavities.

(ii) The large air cavities in the leaves of such plants help in providing buoyancy to the leaves to help them float on water.

(iii) Rishi shows a scientific attitude, inquisitive nature and empirical approach.

Here we are providing Probability Class 9 Extra Questions Maths Chapter 15 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Probability with Answers Solutions

Extra Questions for Class 9 Maths Chapter 15 Probability with Solutions Answers

Probability Class 9 Extra Questions Very Short Answer Type

Probability Class 9 Extra Questions Question 1.
The blood groups of some students of Class IX were surveyed and recorded as below :

If a student is chosen at random, find the probability that he/she has blood group A or AB.
Solution:
Here,
total number of students = 19 + 6 + 13 + 12 = 50
Number of students has blood group A or AB = 19 + 13 = 32
Required probability = \(\frac{38}{50}\) = \(\frac{16}{25}\)

Class 9 Probability Extra Questions Question 2.
A group of 80 students of Class X are selected and asked for their choice of subject to be
taken in Class XI, which is recorded as below :

If a student is chosen at random, find the probability that he/she is a student of either commerce or humanities stream.
Solution:
Here, total number of students = 80
Total number of students of Commerce or Humanities stream = 33
Required probability = \(\frac{33}{80}\)

Probability Extra Questions Class 9 Question 3.
A box contains 50 bolts and 150 nuts. On checking the box, it was found that half of the bolts and half of the nuts are rusted. If one item is chosen at random, find the probability that it is rusted.
Solution:
Total number of nuts and bolts in the box = 150 + 50
= 200
Number of nuts and bolts rusted = \(\frac{1}{2}\) × 200 = 100
P(a rusted nut or bolt) = \(\frac{100}{200}\) = \(\frac{1}{2}\)

Class 9 Maths Chapter 15 Extra Questions Question 4.
A dice is rolled number of times and its outcomes are recorded as below:

Find the probability of getting an odd number.
Solution:
Total number of outcomes = 250
Total number of outcomes of getting odd numbers = 35 + 50 + 53 = 138
∴ P(getting an odd number) = \(\frac{138}{250}\) = \(\frac{69}{125}\)

Extra Questions Of Probability Class 9 Question 5.
The probability of guessing the correct answer to a certain question is \(\frac{x}{2}\) If probability of
not guessing the correct answer is \(\frac{2}{3}\), then find x.
Solution:
Here, probability of guessing the correct answer = \(\frac{x}{2}\)
And probability of not guessing the correct answer = \(\frac{x}{2}\)
Now, \(\frac{x}{2}\) + \(\frac{2}{3}\) = 1
⇒ 3x + 4 = 6
⇒ 3x = 2
⇒ x = \(\frac{2}{3}\)

Extra Questions On Probability Class 9 Question 6.
A bag contains x white, y red and z blue balls. A ball is drawn at the random, then what is the probability of drawing a blue ball.
Solution:
Number of blue balls = Z
Total balls = x + y + Z
∴ P(a blue ball) = \(\frac{z}{x+y+z}\)

Probability Class 9 Extra Questions Short Answer Type 1

Probability Questions For Class 9 Question 1.
750 families with 3 children were selected randomly and the following data recorded:

If a family member is chosen at random, compute the probability that it has :
(i) no boy child
(ii) no girl child
Solution:
(i) P(no boy child) = \(\frac{100}{750}\) = \(\frac{2}{15}\)
and P (no girl child) = \(\frac{120}{750}\) = \(\frac{4}{25}\)

Probability Questions Class 9 Question 2.
If the probability of winning a race of an athlete is \(\frac{1}{6}\) less than the twice the probability of losing the race. Find the probability of winning the race.
Solution:
Let probability of winning the race be p
∴ Probability of losing the race = 1 – p
According to the statement of question, we have
p = 2 (1 – p) – \(\frac{1}{6}\)
⇒ 6p = 12 – 12p – 1
⇒ 18p = 11
⇒ p = \(\frac{11}{18}\) .
Hence, probability of winning the race is \(\frac{11}{18}\).

Questions On Probability Class 9 Question 3.
Three coins are tossed simultaneously 150 times with the following frequencies of different outcomes :

Compute the probability of getting :
(i) At least 2 tails
(ii) Exactly one tail
Solution:
Here, total number of chances = 150
(i) Total number of chances having at least 2 tails = 32 + 63 = 95
∴ Required probability = \(\frac{95}{150}\) = \(\frac{19}{30}\)
(ii) Total number of chances having exactly one tail = 30
∴ Required probability = \(\frac{30}{150}\) = \(\frac{1}{5}\)

Probability Class 9 Extra Questions Short Answer Type 2

Probability Class 9 Extra Questions Pdf Question 1.
The table shows the marks obtained by a student in unit tests out of 50 :

Find the probability that the student get 70% or more in the next unit test. Also, the probability that student get less than 70%.
Solution:
Here, the marks are out of 50, so we first find its percentage (i.e., out of 100)

Total number of outcomes = 5
Probability of getting 70% or more marks = \(\frac{3}{5}\)
Probability of getting less than 70% = \(\frac{2}{5}\)

Class 9 Maths Probability Extra Questions Question 2.
Books are packed in piles each containing 20 books. Thirty five piles were examined for defective books and the results are given in the following table :

One pile was selected at random. What is the probability that it has :
(i) no defective books ?
(ii) more than 0 but less than 4 defective books ?
(iii) more than 4 defective books ?
Solution:
Total number of books = 700
(i) P(no defective books) = \(\frac{400}{700}\) = \(\frac{4}{7}\)
(ii) P(more than 0 but less than 4 defective books) = \(\frac{269}{700}\)
13 (iii) P(more than 4 defective books) = \(\frac{13}{700}\)

Probability Class 9 Extra Questions Short Answer Type 1 and 2

Ch 15 Maths Class 9 Extra Questions Question 1.
Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and number of times ‘1 Tail’ appeared is double the number of times ‘No Tail’ appeared. Find the probability of getting ‘Two tails’.
Solution:
Total number of outcomes = 360
Let the number of times ‘No Tail’ appeared be x
Then, number of times 2 Tails’ appeared = 3x
Number of times ‘1 Tail’ appeared = 2x
Now, x + 2x + 3x = 360
⇒ 6x = 360
⇒ x = 60
P(of getting two tails) = \(\frac{3 \times 60}{360}\) = \(\frac{1}{2}\)

Probability Class 9 Important Questions Question 2.
A die was rolled 100 times and the number of times, 6 came up was noted. If the experimental probability calculated from this information is \(\frac{2}{5}\), then how many times 6 came up ? Justify your answer.
Solution:
Here, total number of trials = 100
Let x be the number of times occuring 6.
We know, probability of an event
= \(\frac { Frequency\quad of\quad the\quad event\quad occuring }{ Total\quad number\quad of\quad trails }\)
⇒ \(\frac{x}{100}\) = \(\frac{2}{5}\) [∵ Probability is given)
⇒ x = 40

Probability Class 9 Extra Questions Long Answer Type

Probability Class 9 Extra Questions With Answers Question 1.
Three coins are tossed simultaneously 250 times. The distribution of various outcomes is listed below :
(i) Three tails : 30,
(ii) Two tails : 70,
(iii) One tail : 90,
(iv) No tail : 60
Find the respective probability of each event and check that the sum of all probabilities is 1.
Solution:
Here, the total number of chances = 250
Total number of three tails = 30

Probability Extra Questions For Class 9 Question 2.
A travel company has 100 drivers for driving buses to various tourist destination. Given
below is a table showing the resting time of the drivers after covering a certain distance (in km).

What is the probability that the driver was chosen at random :
(a) takes a halt after covering 80 km?
(b) takes a halt after covering 115 km?
(c) takes a halt after covering 155 km?
(d) takes a halt after crossing 200 km?
Solution:
Total number of drivers = 100
(a) P (takes a halt after covering 80 km) = \(\frac{13}{100}\)
(b) P (takes a halt after covering 115 km) = \(\frac{60}{100}\) = \(\frac{3}{5}\)
(c) P (takes a halt after covering 155 km) = \(\frac{90}{100}\) = \(\frac{9}{10}\)
(d) P (takes a halt after crossing 200 km) = \(\frac{10}{100}\) = \(\frac{1}{10}\)

Probability Class 9 Extra Questions With Solutions Question 3.
A company selected 2300 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed in the table below :

If a family is chosen at random, find the probability that the family is :
(i) earning ₹7000 – ₹13000 per month and owning exactly 1 vehicle.
(ii) owning not more than one vehicle. (iii) earning more than ₹13000 and owning 2 or more than 2 vehicles. (iv) owning no vehicle
Solution:
Here, we have a total number of families = 2300
(i) Number of families earning ₹7000 to ₹13000 per month and owning exactly 1 vehicle = 295 + 525 = 820

Class 9 Maths Ch 15 Extra Questions Question 4.
A survey of 2000 people of different age groups was conducted to find out their preference
in watching different types of movies :
Type I + Family Type II → Comedy and Family
Type III → Romantic, Comedy, and Family 242.
Type IV → Action, Romantic, Comedy and Family

Find the probability that a person chosen at random is :
(a) in 18-29 years of age and likes type II movies
(b) above 50 years of age and likes all types of movies
(c) in 30-50 years and likes type I movies. :
Solution:
(a) Let E₁ be the event, between the age group (18 – 29) years and liking type II movies
Favourable outcomes to event E₁ = 160
∴ P(E₁) = \(\frac{160}{2000}\) = \(\frac{160}{2000}\)
(b) Let E₂ be the event, of age group above 50 years and like all types of movies
Favourable outcomes to event E₂ = 9
∴ P(E₂) = \(\frac{9}{2000}\)
(c) Let E₃ be the event, between age group (30 – 50) years and liking type I movies
Favourable outcomes to event E₃ = 505
∴ P(E₃) = \(\frac{505}{2000}\) = \(\frac{101}{400}\)

Probability Class 9 Extra Questions HOTS

Question 1.
In a kitchen, there are 108 utensils, consisting of bowls, plates, and glasses. The ratio of bowls, plates the glasses is 4:2:3. A utensil is picked at random. Find the probability that :
(i) it is a plate.
(ii) it is not a bowl.
Solution:
Total utensils in the kitchen = 108
Let number of bowls be 4x, number of plates be 2x and number of glasses be 3x
∴ 4x + 2x + 3x = 108
9x = 108
x = \(\frac{108}{9}\) = 12
Thus, number of bowls = 4 × 12 = 48
Number of plates = 2 × 12 = 24
Number of glasses = 3 × 12 =
(i)P (a plate) = \(\frac{24}{108}\) = \(\frac{2}{9}\)
(ii) P (not a bowl) = \(\frac{24+36}{108}\) = \(\frac{60}{108}\) = \(\frac{5}{9}\)

Question 2.
A bag contains 20 balls out of which x are white.
(a) If one ball is drawn at random, find the probability that it is white.
(b) If 10 more white balls are put in the bag, the probability of drawing a white ball will be double that in part (a), find x.
Solution:
Here, total no. of balls = 20
No. of white balls = x
∴ P(white ball) = \(\frac{x}{20}\)
Now, 10 more white balls are added
∴ Total no. of balls = 20 + 10 = 30
Total no. of white balls = x + 10
According to statement of question, we have
\(\frac{x+10}{30}\) = 2 × \(\frac{x}{20}\)
⇒ \(\frac{x+10}{3}\) = x
⇒ x + 10 = 3x
⇒ 2x = 10
⇒ x = 5

Question 3.
Here is an extract from a mortality table.

(i) Based on this information, what is the probability of a person aged 60′ of dying within a year ?
(ii) What is the probability that a person aged 61′ will live for 4 years ?
Solution:
(i) Clearly 16090 persons aged 60, (16090 – 11490), i.e., 4600 died before reaching their 61st birthday.
Therefore, P (a person aged 60 die within a year) = \(\frac{4600}{16090}\) = \(\frac{460}{1609}\)
(ii) Number of persons aged 61 years = 11490
Number of persons surviving for 4 years = 2320
P (a person aged 61 will live for 4 years) = \(\frac{2320}{11490}\) = \(\frac{232}{1149}\)

Probability Class 9 Extra Questions Value Based (VBQs)

Question 1.
An insurance company selected 2000 drivers at random (i.e., without any preference of one driver over another) in a particular city to find a relationship between age and accidents. The data obtained are given in the following table :

Find the probability of the following events for a driver-selected at random from the city :
(i) being 18 – 29 years of age and having exactly 3 accidents in one year.
(ii) being 30 – 50 years of age and having one or more accidents in a year.
(iii) having no accident in one year.
(iv) Which value would you like to remember from this data?
Solution:
Here, total number of drivers = 2000
(i) The number of drivers in the age group 18 – 29 having exactly 3 accidents = 70
So, P (driver in age group 18 – 29 having exactly 3 accidents in one year) = \(\frac{70}{2000}\) = 0.035
(ii). The number of drivers in the age group 30 – 50 and having one or more than one accidents in one year = 118 + 65 + 20 + 21 = 224
P (driver in age group 30 – 50 having one or more accidents in one year) = \(\frac{224}{2000}\) = 0.112
(iii) The number of drivers having no accident in one year = 395 + 520 + 390 = 1305
So, P (driver having no accident) = \(\frac{1305}{2000}\) = 0.6525
(iv) Most people in India died or injured due to accidents as compared to any other country. So, we should obey the traffic rules as life is very precious.

Question 2.
100 plants were sown in six different colonies A, B, C, D, E, and F. After 31 days, the number of plants survived as follows :

What is the probability of :
(i) more than 80 plants survived in a colony?
(ii) less than 82 plants survived in a colony?
(iii) which values are depicted from the above data?
Solution:
Here, we have total number of colonies = 6
(i) Number of colonies in which more than 80 plants survived = 4 (i.e., B, C, E and F)
∴ P(more than 80 plants survived in a colony) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
(ii) Number of colonies in which less than 82 plants survived = 2 (i.e., A and D)
∴ P (less than 82 plants survived in a colony) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
(iii) In order to keep environment safe, clean and green, we should grow more and more plants.

In this page, we are providing Is Matter Around Us Pure Class 9 Extra Questions and Answers Science Chapter 2 pdf download. NCERT Extra Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 2 Extra Questions and Answers Is Matter Around Us Pure

Extra Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure with Answers Solutions

Is Matter Around Us Pure Class 9 Extra Questions Very Short Answer Type

Class 9 Science Chapter 2 Extra Questions And Answers Question 1.
What is the reason for the difference in properties of solutions, colloids and suspensions?
Answer:
Due to different particle size.

Is Matter Around Us Pure Extra Questions Question 2.
Is rain water or distilled water a pure substance?
Answer:
Yes, because it contains particles (molecules) of only water.

Is Matter Around Us Pure Class 9 Extra Questions Question 3.
Can we separate a mixture of alcohol and water by a separating funnel?
Answer:
No, the two liquids are miscible.

Ch 2 Science Class 9 Extra Questions Question 4.
Give two examples of metalloids.
Answer:
Silicon and germanium.

Class 9 Chemistry Chapter 2 Extra Questions Question 5.
Give an example of a solution in which solid is a solute as well as the solvent.
Answer:
Alloys are solid in solid solutions. For example, brass contains about 30% zinc and 70% copper. Here, zinc is the solute while copper is the solvent.

Class 9 Science Ch 2 Extra Questions Question 6.
What type of liquid mixture will kerosesne oil and water form? How will you separate it?
Answer:
Immiscible. We can separate this mixture by using a separating funnel.

Class 9 Is Matter Around Us Pure Extra Questions Question 7.
Define the term heterogeneous.
Answer:
Heterogeneous means that the substance does not have the same properties or characteristics throughout its bulk.

Chapter 2 Science Class 9 Extra Questions Question 8.
Write the constituent element of potassium hydroxide and sodium chloride.
Answer:
The constituent element of potassium hydroxide is K, H and O and sodium chloride is Na and Cl.

Ncert Class 9 Science Chapter 2 Extra Questions Question 9.
On the basis of composition, how is matter classified?
Answer:

1. Pure substance
2. Mixture.

Extra Questions Of Chapter 2 Science Class 9 Question 10.
What are different categories of pure substances?
Answer:

1. Elements
2. Compounds.

Class 9 Chapter 2 Science Extra Questions Question 11.
What are the different kinds of mixture?
Answer:

1. Homogeneous mixture
2. Heterogeneous mixture.

Is Matter Around Us Pure Class 9 Important Questions With Answers Question 12.
What are the constituents of brass?
Answer:
Brass is an alloy and is a mixture of zinc (30%) and copper (70%).

Class 9 Science Chapter 2 Extra Questions Question 13.
How are elements further classified?
Answer:
Metals, non-metals, metalloids.

Extra Questions Of Is Matter Around Us Pure Question 14.
A solution of water and alcohol contains 30 g of water and 60 g of alcohol. What is the concentration of solution?
Answer:
\(\frac { 30 }{ 30 + 60 }\) x 100 = \(\frac { 30 }{ 90 }\) x 100 = 33.3%.

Class 9th Science Chapter 2 Extra Questions Question 15.
What are aqueous solutions?
Answer:
Solutions in which water is the solvent are called aqueous solutions, e.g., sugar solution, in which sugar is dissolved in water.

Question 16.
What is an unsaturated solution?
Answer:
A solution in which some more solute can be dissolved at any fixed temperature is called an unsaturated solution.

Question 17.
How many gram of water is needed to make 8% mass by mass percentage of sodium carbonate solution if 4 g of sodium carbonate is a variable to make a solution?
Answer:
8% means 8 g in 100 g of solution. So if 4 g Na₂CO₃ is present, it means solution must be 50 g.

Question 18.
What are the conditions required to convert air into liquid air?
Answer:
200 atmospheric pressure and – 200°C.

Question 19.
Define the term inter-conversion of matter.
Answer:
The phenomenon of change of one state of matter into another and back to the original state is called inter-conversion of matter.

Question 20.
Why do fish go in deep water during day light?
Answer:
During day time, the shallow water is warmed and hence it has less dissolved oxygen. Therefore fish tend to go in deep water during day time.

Question 21.
Out of colloid, solution and a suspension which one can be separated by filtration.
Answer:
Suspension.

Question 22.
Out of colloid, solution and a suspension which has the smallest particle?
Answer:
Solution.

Is Matter Around Us Pure Class 9 Extra Questions Short Answer Type 1

Question 1.
Suggest separation technique(s) one would need to employ to separate the following mixtures. [NCERT Exemplar]
(a) Mercury and water
(b) Potassium chloride and ammonium chloride
(c) Common salt, water apd sand
(d) Kerosene oil, water and salt.
Answer:
(a) Separation by using separating funnel
(b) Sublimation
(c) Filtration followed by evaporation
Or
Centrifugation followed by evaporation/distillation
(d) Separation by using separating funnel to separate kerosene oil followed by distillation.

Question 2.
A salt can be recovered from its solution by evaporation. Suggest some other technique for the same? [NCERT Exemplar]
Answer:
Crystallisation.

Question 3.
The ‘sea water’ can be classified as a homogeneous as well as heterogeneous mixture. Comment.
Answer:

• Homogeneous – mixture of salts and water only.
• Heterogeneous – contains salts, water, mud, decayed plant, etc.

Question 4.
Fill in the following blanks:
(i) Milk is a ……………… solution but vinegar is a solution.
(ii) Milk is a colloidal solution in which ……………… is the dispersed phase and ……………… is the dispersion medium.
Answer:
(i) colloidal, true
(ii) fat, water.

Question 5.
(a) Classify Brass and Diamond as element and mixture.
(b) How is a chemical change different from a physical change?
Answer:
(a) Brass is homogeneous mixture also called alloy. The constituents are Cu and Zn. Diamond is an element. It is an allotropic form of carbon.

(b) In a chemical change, a new substance is formed as a result of chemical reaction. No new substance is formed in a physical change.

Question 6.
Identify the dispersed phase and dispersion medium in the following examples of colloids:
(a) Fog
(b) Cheese
(c) Coloured gemstone.
Answer:
(a) Fog: Liquid (water drops) acts as dispersed phase and gas (air) as the dispersion medium.
(b) Cheese: Solid (fat) acts as the dispersed phase and water (liquid) as the dispersion medium.
(c) Coloured gemstone: Solids act the dispersed phase as well as the dispersion medium.

Question 7.
Explain, why particles of a colloidal solution do not settle down when left undisturbed, while in the case of a suspension they do. [NCERT Exemplar]
Answer:
Particle size in a suspension is larger than those in a colloidal solution. Also molecular interaction in a suspension is not strong enough to keep the particles suspended and hence they settle down.

Question 8.
Smoke and fog both are aerosols. In what way are they different? [NCERT Exemplar]
Answer:
Both fog and smoke have gas as the dispersion medium. The only difference is that the dispersed phase in fog is liquid and in smoke it is a solid.

Question 9.
How will you bring about the following separation?
(i) Fine mud particles floating in water.
(ii) Carbon particles present in smoke.
Answer:
(i) By coagulation using alum and then filtering.
(ii) By passing smoke through electric plates maintained at a high potential difference. The colloidal particles of carbon get neutralised and fall down while air escapes out.

Question 10.
Classify the following as Physical or chemical properties. [NCERT Exemplar]
(a) The composition of a sample of steel is 98% iron, 1.5% carbon and 0.5% other elements.
(b) Zinc dissolves in hydrochloric acid with the evolution of hydrogen gas.
(c) Metallic sodium is soft enough to be cut with a knife.
(d) Most metal oxides form alkalies on interacting with water.
Answer:
Physical properties: (a) and (c)
Chemical properties: (b) and (d)

Question 11.
Suggest a suitable separation technique for the following: [NCERT Exemplar]
(a) Mercury and water
(b) Coloured components from blue ink
(c) Ammonium chloride and potassium chloride
(d) Mixture of alcohol and water.
Answer:
(a) The separation can be done by the use of a separating funnel. Mercury forms the lower layer (heavier) and water the upper layer (lighter).
(b) The separation can be done with the help of chromatography.
(c) Process of sublimation can be used. Ammonium chloride collects as the sublimate while potassium chloride remains in the dish.
(d) Process of fractional distillation can be used. Alcohol (ethyl alcohol) with lower boiling point (78°C or 351 K) gets distilled leaving behind water with higher boiling point (100°C or 373 K) in the distillation flask.

Question 12.
Name the type of colloids in each of the following giving an example of each.

	Dispersed Phase	Dispersing Medium
A	Liquid	Gas
B	Liquid	Liquid
C	Liquid	Solids

Answer:

A	Aerosol	(Fog)
B	Emulsion	(Milk)
C	Gel	(Jelly)

Question 13.
You are given two samples of water labelled as ‘A’ and ‘B’ Sample ‘A’ boils at 100°C and sample ‘B’ boils at 102°C. Which sample of water will not freeze at 0°C? Comment. [NCERT Exemplar]
Answer:
Sample ‘B’ will not freeze at 0°C because it is not pure water. At 1 atm, the boiling point of pure water is 100°C and the freezing point of pure water is 0°C.

Question 14.
Identify colloids from the following: Copper sulphate solution, milk, smoke, muddy water, butter, sugar solution, face cream, lemonade.
Answer:
Colloids: milk, smoke, muddy water, butter, face cream, lemonade.

Question 15.
What are the favourable qualities given to gold when it is alloyed with copper or silver for the purpose of making ornaments? [NCERT Exemplar]
Answer:
Pure gold is very soft as compared to gold alloyed with silver or copper. Thus for providing strength to gold, it is alloyed.

Question 16.
12 mL of dettol is added to a beaker containing 500 mL of water and stirred. State four observations that you make.
Answer:
When dettol is added to a beaker containing of water, the following observations are made.

• An emulsion is formed which is of colloidal nature.
• The colour of emulsion is milky.
• It gives characteristic smell of dettol.
• The solution can pass through a filter paper.

Is Matter Around Us Pure Class 9 Extra Questions Short Answer Type 2

Question 1.
During an experiment the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10 g of sugar in water to make 100 g of the solution. [NCERT Exemplar]
(a) Are the two solutions of the same concentration?
(b) Compare the mass % of the two solutions.
Answer:
(a) No.
\(Mass\%=\frac{\text { Mass of solute }}{\text { Mass of solute }+\text { Mass of solvent }} \times 100\)

(b) Solution made by Ramesh
Mass % = \(\frac { 10 }{ 10 + 100 }\) = \(\frac { 10 }{ 110 }\) x 100 = 9.09%
solution made by sarika
Mass % = \(\frac { 10 }{ 100 }\) x 100 = 10%
The solution prepared by Sarixa has a higher mass % than that prepared by Ramesh.

Question 2.
State which of the following solutions exhibit Tyndall effect:
Starch solution, Sodium chloride solution, Tincture of iodine, Air.
Answer:
(i) Tyndall effect is shown both by starch solution and air which are heterogeneous mixtures and have the capacity to scatter a beam of light as it passes through them.

(ii) Sodium chloride solution and tincture of iodine (iodine crystals dissolved in ethyl alcohol) are both homogeneous in nature and do not exhibit any Tyndall effect.

Question 3.
While diluting a solution of salt in water, a student by mistake added acetone (boiling point 56°C). What technique can be employed to get back the acetone? Justify your choice. [NCERT Exemplar]
Answer:
Distillation, since acetone is more volatile it will separate out first.

Question 4.
(a) Why is crystallisation technique better than evaporation?
(b) Write any two physical properties of each of metals and non-metals.
(c) Name the technique used to separate butter from curd.
Answer:
(a) Both these techniques are used to separate solid substances from their solutions. But crystallisation is considered better because during evaporation certain solids may decompose or some of them like sugar get charred when the solution is evaporated completely to dryness. As a result of crystallisation, even the shapes of the crystals do not change.

(b) (i) Metals have a shiny surface known as lustre.
(ii) Metals are malleable and ductile.
(iii) Non-metals are mostly poor conductors of electricity.
(iv) Non-metals are generally soft.

(c) Butter can be separated from curd by the process of centrifugation. This is usually done by churning which is very common as well as convenient.

Question 5.
Name the process associated with the following:
(a) Dry ice is kept at room temperature and at one atmospheric pressure.
(b) A drop of ink placed on the surface of water contained in a glass spreads throughout the water.
(c) A potassium permanganate crystal is in a beaker and water is poured into the beaker with stirring.
(d) A acetone bottle is left open and the bottle becomes empty.
(e) Milk is churned to separate cream from it.
(f) Settling of sand when a mixture of a sand and water is left undisturbed for some time.
Answer:
(a) Sublimation
(b) Diffusion
(c) Dissolution/diffusion
(d) Evaporation, diffusion
(e) Centrifugation
(f) Sedimentation

Question 6.
On dissolving chalk powder in water, a suspension is obtained. Give any four reasons to support the fact that the mixture so obtained is a suspension only.
Answer:
It is supported by the following reasons:

• White particles of chalk powder can be seen with the naked eyes.
• The particles can be separated by ordinary filter paper.
• Upon shaking, a white turbidity reappears in solution.
• Light cannot pass through the suspension which shows that it is of opaque nature.

Question 7.
Give an example for each of the following:
(a) Solid-liquid homogeneous mixture
(b) Gas-gas homogeneous mixture
(c) Liquid-liquid heterogeneous mixture.
Answer:
(a) Mixture of sodium chloride in water.
(b) Air. It is a homogeneous mixture of a number of gases.
(c) Emulsion of oil and water.

Question 8.
What would you observe when
(a) a saturated solution of potassium chloride prepared at 60°C is allowed to cool to room temperature?
(b) an aqueous sugar solution is heated to dryness?
(c) a mixture of iron filings and sulphur powder is heated strongly?
Answer:
(a) Solid potassium chloride will separate out.
(b) Initially the water will evaporate and then sugar will get charred.
(c) Iron sulphide will be formed.

Question 9.
(a) Arrange solids, liquids and gases in increasing order of the following properties of matter
(i) rigidity
(ii) diffusion
(iii) compressibility.
(b) Write one example from your daily life which is based on diffusion of gases.
Answer:
(a) (i) Rigidity: Gases < Liquids < Solids
(ii) Diffusion: Solids < Liquids < Gases
(iii) Compressibility: Solids < Liquids < Gases.

(b) Smell of aroma or perfume released in one corner of the room soon spreads in the whole room.

Question 10.
Is air a mixture or a compound? Give three reasons.
Answer:
Air is a mixture and not a compound as discussed below:
(i) The properties of a mixture are in between those of its constituents. The two major components of air are oxygen (20.9% by volume) and nitrogen (78.1% by volume). In oxygen, any fuel bums brightly but in nitrogen it gets extinguished. In contrast, in air the fuel bums slowly.

(ii) The components of a mixture can be separated by simple physical methods. For example, the components of air can be separated by fractional distillation of liquid air.

(iii) The composition of a mixture is variable. The composition of air is also variable. It has more oxygen in the country side than in big cities.

(iv) When air is obtained by mixing its constituent gases, no heat is either evolved or absorbed.

(v) Liquid air does not have a fixed boiling point.

Is Matter Around Us Pure Class 9 Extra Questions Long Answer Type

Question 1.
Classify each of the following, as a physical or a chemical change. Give reasons.
(a) Drying of a shirt in the Sun.
(b) Rising of hot air over a radiator.
(c) Burning of kerosene in a lantern.
(d) Change in the colour of black tea on adding lemon juice to it.
(e) Churning of milk cream to get butter.
Answer:
Physical changes: (a), (b), (e)
Chemical changes: (c), (d)

Question 2.
Iron filings and sulphur were mixed together and divided into two parts, ‘A’ and ‘B’. Part ‘A’ was heated strongly while Part ‘B’ was not heated. Dilute hydrochloric acid was added to both the parts and evolution of gas was seen in both the cases. How will you identify the gases evolved?
Answer:
Part A

Part B

When dilute HCl is added to it, only the iron filings in the mixture react and sulphur remains unreacted
Fe (s) + 2 HCl (aq) → FeCl₂ + H₂ gas
H₂S gas formed has a foul smell and on passing through lead acetate solution, it turns the solution black. Hydrogen gas burns with a pop sound.

Question 3.
Fill in the blanks:
(a) A colloid is a ……………….. mixture and its components can be separated by the technique known as ………………..
(b) Ice, water and water vapour look different and display different ……………….. properties but they are the same.
(c) A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for sometime. The upper layer in the separating funnel will be of ……………….. and the lower layer will be that of
(d) A mixture of two or more miscible liquids, for which the difference in the boiling points is less than 25 K can be separated by the process called ………………..
(e) When light is passed through water containing a few drops of milk, it shows a bluish tinge. This is due to the ……………….. of light by milk and the phenomenon is called ……………….. This indicates that milk is a ……………….. solution. [NCERT Exemplar]
Answer:
(a) heterogeneous, centrifugation
(b) physical, chemically
(c) water, chloroform (hint-density of water is less than that of chloroform)
(d) fractional distillation scattering, Tyndall effect, colloidal.

Question 4.
Give an example each for the mixture having the following characteristics. Suggest a suitable method to separate the components of these mixtures. [NCERT Exemplar]
(a) A volatile and a non-volatile component.
(b) Two volatile components with appreciable difference in boiling points.
(c) Two immiscible liquids.
(d) One of the components changes directly from solid to gaseous state.
(e) Two or more coloured constituents soluble in some solvent.
Answer:
(a) Evaporation or distillation
(b) Distillation
(c) By using a separating funnel
(d) Sublimation
(e) Chromatography

Question 5.
Which separation techniques you will apply for the separation of the following mixtures?
(a) Oil from water
(b) Camphor from sand
(c) Sodium chloride from its solution in water
(d) Cream from milk
(e) Metal pieces from engine oil of a car.
Answer:
(a) By the use of a separating funnel.
(b) With the help of sublimation technique
(c) By evaporation crystallisation technique.
(d) By the use of a centrifuge
(e) By the use of filtration technique

Is Matter Around Us Pure Class 9 Extra Questions HOTS

Question 1.
Which of the tubes shown below will be more effective as a condenser in the distillation apparatus? [NCERT Exemplar]

Answer:
The presence of beads in tube (a) would provide a larger surface area for cooling. So, this tube will be more effective as a condenser.

Question 2.
Non-metals are usually poor conductors of heat and electricity. They are non-lustrous, non- sonorous, non-malleable and are coloured.
(a) Name a lustrous non-metal.
(b) Name a non-metal which exists as a liquid at room temperature.
(c) The allotropic form of a non-metal is a good conductor of electricity. Name the allotrope.
(d) Name a non-metal which is known to form the largest number of compounds.
(e) Name a non-metal other than carbon which shows allotropy.
(f) Name a non-metal which is required for combustion.
Answer:
(a) Iodine
(b) Bromine
(c) Graphite
(d) Carbon
(e) Sulphur, phosphorus
(f) Oxygen.

Question 3.
The teacher instructed three students ‘A’, ‘B’ and ‘C’ respectively to prepare a 50% (mass by volume) solution of sodium hydroxide (NaOH). ‘A’ dissolved 50 g of NaOH in 100 mL of water, ‘B’ dissolved 50 g of NaOH in 100 g of water while ‘C’ dissolved 50 g of NaOH in water to make 100 mL of solution. Which one of them has made the desired solution and why?
Answer:
‘C’ has made the desired solution.

= \(\frac { 50 }{ 100 }\) x 100
= 50% mass by volume.

Question 4.
On heating calcium carbonate gets converted into calcium oxide and carbon dioxide.
(a) Is this a physical or a chemical change?
(b) Can you prepare one acidic and one basic solution by using the products formed in the above process? If so, write the chemical equation involved.
Answer:
(a) Chemical change

(b) Acidic and basic solutions can be prepared by dissolving the products of the above process in water
CaO + H₂O → Ca(OH)₂ (basic solution)
CO₂ + H₂O → H₂CO₃ (acidic solution)

Question 5.
Arun has prepared 0.01% (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solutions?
(a) 1.00 g of NaCl + 100 g water
(b) 0.11 g of NaCl + 100 g of water
(c) 0.01 g of NaCl + 99.99 g of water
(d) 0.10 g of NaCl + 99.90 g of water
Answer:
(c) 0.01 g of NaCl + 99.99 g of water

= \(\frac { 0.01 }{ 0.01 + 99.99 }\) x 100
= \(\frac { 0.01 }{ 100 }\) x 100
= 0.01 g

Question 6.
A child wanted to separate the mixture of dyes constituting a sample of ink. He marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in the figure. The filter paper was removed when the water moved near the top of the filter paper. [NCERT Exemplar]

1. What would you expect to see, if the ink contains three different coloured components?
2. Name the technique used by the child.
3. Suggest one more application of this technique.

Answer:

1. Three different bands will be observed.
2. Chromatography
3. To separate the pigments present in chlorophyll.

Question 7.
A group of students took an old shoe box and covered it with a black paper from all sides. They fixed a source of light (a torch) at one end of the box by making a hole in it and made another hole on the other side to view the light. They placed a milk sample contained in a beaker/tumbler in the box as shown in the figure. They were amazed to see that milk taken in the tumbler was illuminated. They tried the same activity by taking a salt solution but found that light simply passed through it?

1. Explain why the milk sample was illuminated. Name the phenomenon involved.
2. Same results were not observed with a salt solution. Explain.
3. Can you suggest two more solutions which would show the solution?

Answer:

1. Milk is a colloid and would show Tyndall effect.
2. Salt solution is a true solution and would not scatter light.
3. Detergent solution, sulphur solution.

Question 8.
Sudha tested the solubility of four salts X, Y, Z and T at different temperatures and collected the following data. (Solubility refers to the amount in grams dissolved in 100 g of water to give a saturated solution.)

Answer the following questions from the table:

1. Which salt has the highest and lowest solubility at 323 K?
2. A student prepared a saturated solution of X at 323 K and then added 25 g water to it. What mass of X must be added to again make the solution saturated?
3. The solubility of which salt is least affected by increase in temperature?
4. What mass of T would be required to make saturated solution in 200 g of water at 290 K?

Answer:
1. At 323 K, salt Y has the highest solubility in water while salt Z has the lowest solubility.

2. By definition of saturated solution,
100 g of water at 323 K contain salt = 40 g
125 g of water at 323 K contain salt = \(\frac { 40 g }{ 100 g }\) x (125 g) = 50 g
∴ Mass of salt to be added to make the solution again saturated = (50 – 40) = 10 g

3. The data show that the solubility of the salt Y is least affected with increase in temperature.

4. At 290 K, mass of T required to make a saturated solution in 200 g of water = \(\frac { 25 g }{ 100 g }\) x (200 g) = 50 g

Is Matter Around Us Pure Class 9 Extra Questions Value Based (VBQs)

Question 1.
Mallika’s mother was suffering from cold and cough. Mallika prepared tea for her mother. She boiled water in a pan, then she added tea leaves, sugar and milk to it. She filtered the tea in a cup and served it to her mother.
(a) Explain the values shown by Mallika.
(b) Identify solute, solvent, residue and filtrate in this activity.
Answer:
(a) Mallika used the knowledge of chemistry to provide relief to her mother. Actually she prepared an extract of tea leaves which is helpful in curing cold and cough and gives warmth to the body.

(b) Solute: Tea leaves and sugar.
Solvent: water and milk.
Filtrate: homogeneous mixture of water, milk, sugar and extract of tea leaves.

Question 2.
Amit was asked by his teacher to separate a liquid mixture of acetone and ethyl alcohol. He set up a distillation apparatus and tried to distil the mixture. To his surprise, both the liquids got distilled. Teacher told Amit to repeat the experiment by using a fractionating column in the distillation flask. Amit followed the advice of the teacher and he was able to separate the two liquids.

1. Why was Amit not successful in separating the liquid mixture earlier?
2. Why did teacher ask him to use the fractionating column?
3. Which liquid was distilled first?
4. As a student of chemistry, what value based information you have gathered?

Answer:
1. The difference in boiling point temperatures of acetone (56°C) and ethyl alcohol (78°C) is only 22°C.
Therefore, process of simple distillation fails in this case.

2. Fractionating column is quite effective in this case because it obstructs the distillation of ethyl alcohol which is high boiling and at the same time helps in the distillation of acetone which is low boiling.

3. Acetone was distilled first since it has comparatively low boiling point.

4. The process of simple distillation can be used only in case, the liquids present in the mixture differ in their boiling point by 25°C or more.

Here we are providing Statistics Class 9 Extra Questions Maths Chapter 14 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Statistics with Answers Solutions

Extra Questions for Class 9 Maths Chapter 14 Statistics with Solutions Answers

Statistics Class 9 Extra Questions Very Short Answer Type

Statistics Class 9 Extra Questions Question 1.
The points scored by a basketball team in a series of matches are follows:
17, 7, 10, 25, 5, 10, 18, 10 and 24. Find the range.
Solution:
Here, maximum points = 25 and
minimum points = 5
Range = Maximum value – Minimum value
= 25 – 5 = 20

Class 9 Statistics Extra Questions Question 2.
The points scored by a basketball team in a series of matches are as follows:
17, 2, 7, 27, 25, 5, 14, 18, 10. Find the median.
Solution:
Here, points scored in ascending order are 2, 5, 7, 10, 14, 17, 18, 25, 27, we have n = 9 terms

Statistics Class 9 Extra Questions Pdf Question 3.
The scores of an English test (out of 100) of 20 students are given below :
75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. .:: Find the median and mode of the data.
Solution:
The scores of an English test (out of 100) in ascending order are
44, 55, 59, 64, 67, 69, 73, 75, 75, 88, 88, 88, 88, 88, 90, 95, 95, 95, 98, 99
Here, n = 20

= Mean of 10^th and 11^th term Median
= Mean of 88 and 88 = 88
Mode = 88 [∵ 88 occured max. no. of times i.e., 5 times]

Statistics Extra Questions Class 9 Question 4.
Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean.
Solution:
Since mean of 20 observations is 17
∴ Sum of the 20 observations = 17 × 20 = 340
New sum of 20 observations = 340 – 40 + 12 = 312
New mean = \(\frac{312}{20}\) = 15.6

Extra Questions On Statistics Class 9 Question 5.
Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct
mean.
Solution:
Mean of 36 observations = 12
Total of 36 observations = 36 × 12 = 432
Correct sum of 36 observations = 432 – 74 + 47 = 405
Correct mean of 36 observations = \(\frac{405}{36}\) = 11.25

Class 9 Maths Chapter 14 Extra Questions Question 6.
The median of the data 26, 56, 32, 33, 60, 17, 34, 29, 45 is 33. If 26 is replaced by 62, then
find the new median.
Solution:
Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62

Hence, new median is 34.

Statistics Class 9 Extra Questions With Solutions Question 7.
There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers.
Solution:
Let x be the mean of 50 numbers.
i. Sum of 50 numbers = 50x
Since each number is subtracted from 53.
According to question, we have

Class 9 Maths Statistics Extra Questions Question 8.
To draw a histogram to represent the following frequency distribution :

Find the adjusted frequency for the class 25-45.
Solution:
Adjusted frequency of a class

Statistics Class 9 Extra Questions Short Answer Type 1

Extra Questions Of Statistics Class 9 Question 1.
For a particular year, following is the distribution of ages (in years) of primary school teachers in a district :

(i) Write the lower limit of first class interval.
(ii) Determine the class limits of the fourth class interval.
(iii) Find the class mark of the class 45 – 50.
(iv) Determine the class size.
Solution:
(i) First class interval is 15 – 20 and its lower limit is 15.
(ii) Fourth class interval is 30 – 35
Lower limit is 30 and upper limit is 35.
(iii) Class mark of the class 45 – 50 = \(\frac{45+50}{2}\) = \(\frac{95}{2}\) = 47.5
(iv) Class size = Upper limit of each class interval – Lower limit of each class interval
∴ Here, class size = 20 – 15 = 5

Statistics Class 9 Important Questions Question 2.
Find the mean of the following distribution :

Solution:

Class 9 Maths Ch 14 Extra Questions Question 3.
In figure, there is a histogram depicting daily wages of workers in a factory. Construct the frequency distribution table.

Solution:

Questions On Statistics Class 9 Question 4.
Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending
order. The median of the data is 24. Find the value of x.
Solution:
Here, the arranged data is 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43
Total number of observations = 10

But median of data is 24 (given)
⇒ \(\frac{3 x-12}{2}\) = 24
⇒ 3x – 12 = 48
⇒ 3x = 60
⇒ x = 20
∴ The value of x = 20

Statistics Class 9 Extra Questions Short Answer Type 2

Chapter 14 Maths Class 9 Extra Questions Question 1.
Draw a histogram for the given data :

Solution:
Let us represent class-intervals along x-axis and corresponding frequencies along y-axis on
a suitable scale, the required histogram is as under :

Question 2.
Given are the scores (out of 25) of 9 students in a Monday test :
14, 25, 17, 22, 20, 19, 10, 8 and 23
Find the mean score and median score of the data.
Solution:
Ascending order of scores is :
8, 10, 14, 17, 19, 20, 22, 23, 25

Question 3.
Draw a histogram of the weekly pocket expenses of 125 students of a school given below :

Solution:
Here, the class sizes are different, so calculate the adjusted frequency for each class by using the formula.
Minimum class size Frequency density or adjusted frequency for a class

Here, the minimum class size = 10 – 0 = 10

Let us represent weekly pocket money along x-axis and corresponding adjusted frequencies along y-axis on a suitable scale, the required histogram is as given below :

Question 4.
The weight in grams of 35 mangoes picked at random from a consignment are as follows:
131, 113, 82, 75, 204, 81, 84, 118, 104, 110, 80, 107, 111, 141, 136, 123, 90, 78, 90, 115, 110, 98, 106, 99, 107, 84, 76, 186, 82, 100, 109, 128, 115, 107, 115 From the grouped frequency table by dividing the variable range into interval of equal width of 20 grams, such that the mid-value of the first class interval is 70 g. Also, draw a histogram.
Solution:
It is given that the size of each class interval = 20 and the mid-value of the first class interval is 70.
Let the lower limit of the first class interval be a, then its upper limit = a + 20.
\(\frac{a+(a+20)}{2}\) = 70
⇒ a = 70 – 10 = 60
Thus, the first class interval is 60 – 80 and the other class-intervals are 80 – 100, 100 – 120, 120 – 140, 140 – 160, 160 – 180, 180 – 200 and 200 – 220.
So, the grouped frequency table is as under :


Let us represent weight (in g) along x-axis and corresponding frequencies along y-axis on a suitable scale, the required histogram is as under :

Statistics Class 9 Extra Questions Long Answer Type

Question 1.
Find the mean salary of 60 workers of a factory from the following table :

Solution:

Hence, mean salary of 60 workers is ₹5083.33.

Question 2.
In a school marks obtained by 80 students are given in the table. Draw a histogram. Also,
make frequency polygon.

Solution:
∴ Lower limit of first class interval is 305 – \(\frac{10}{2}\) = 300
Upper limit of first class interval is 305 + \(\frac{10}{2}\) = 310
Thus, first class interval is 300 – 310

Required histogram and frequency polygon is given on the graph paper.

Question 3.
The following two tables gives the distribution of students of two sections according to the marks obtained by them :

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Solution:
The class marks are as under :

Let us take class marks on X-axis and frequencies on Y-axis. To plot frequency polygon of Section-A, we plot the points (5, 3), (15, 9), (25,17), (35,12), (45, 9) and join these points by (15,19). line segments. To plot frequency polygon of Section-B, we plot the points (5, 5), (15, 19), (25, 15), (35, 10), (45, 1) on the same scale and join these points by dotted line segments.

From the above two polygons, clearly the performance of Section A is better.

Statistics Class 9 Extra Questions HOTS

Question 1.
The mean weight of 60 students of a class is 52.75 kg. If mean weight of 25 students of this class is 51 kg, find the mean weight of remaining 35 students of the class.
Solution:
Total weight of 60 students = 60 × 52.75 kg = 3165 kg
Total weight of 25 students = 25 × 51 kg = 1275 kg
∴ Total weight of 35 students = (3165 – 1275) kg = 1890 kg
∴ Mean weight of 35 students = \(\frac{1890}{35}\) = 54 kg

Question 2.
Find the missing frequencies in the following frequency distribution. If it is known that the mean of the distribution is 50.16 and the total number of items is 125.

Solution:
Since total number of items = 125
∴ 17 + f₁ + 32 + f₁ + 19 = 125
f₁ + f₂ = 125 – 17 – 32 – 19
f₁ + f₂ = 57 …..(i)
Now, mean of data = 50.16
We know that

Statistics Class 9 Extra Questions Value Based (VBQs)

Question 1.
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of the Indian society is given below :

(i) Represent the information above by a bar graph.
(ii) In the classroom, discuss what conclusions can be arrived at from the graph.
(iii) What steps should be taken to improve the situation ?
Solution:
(i) The required graph is given below :

In the graph, different sections of the society is taken on X-axis and number of girls per thousand boys is taken on the Y-axis.
[Scale : 1 cm = 10 girls.]
(ii) From the graph, the number of girls to the nearest ten per thousand boys are maximum in scheduled tribes whereas they are minimum in urban.
(iii) Pre-natal sex determination should strictly banned in urban.

Question 2.
Shimpi, a class IX student received cash award of 10000 (Ten thousand) in the singing competition. Her father advised her to make a budget plan for spending this amount. She
made the following plan :

Make a bar graph for the above data.
From above answer the following questions :
(i) Which mathematical concepts have been covered in this ?
(ii) How will you rate her budget plan ? In your opinion which head has been given
(a) more than deserved and (b) less than it deserved ?
(iii) Which values are depicted in her plan?
Solution:
The bar graph of given data is given below :

In the graph, head is taken on X-axis and amount is taken on Y-axis.
(i) Representation of data using bar graph.
(ii) Very good
(a) Picnic for family
(b) Tuition fee for needy child
(iii) Help the needy people and respect the elders.

Here we are providing Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Heron’s Formula with Answers Solutions

Extra Questions for Class 9 Maths Chapter 12 Heron’s Formula with Solutions Answers

Heron’s Formula Class 9 Extra Questions Very Short Answer Type

Heron’s Formula Class 9 Extra Questions Question 1.
Find the area of an equilateral triangle having side 6 cm.
Solutioin:
Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) × (side)² = \(\frac{\sqrt{3}}{4}\) × 6 × 6 = 9√3 cm²

Herons Formula Class 9 Extra Questions Question 2.
If the perimeter of an equilateral triangle is 90 m, then find its area.
Solutioin:

Herons Formula Extra Questions Question 3.
If every side of a triangle is doubled, then find the percent increase in area of triangle so formed.
Solutioin:
Let the sides of the given triangle be, a units, b units and c units.

Hence, percent increase = 300%

Class 9 Herons Formula Extra Questions Question 4.
If the length of a median of an equilateral triangle is x cm, then find its area.
Solutioin:

Let each equal sides of given equilateral triangle be 2y². We know that median is also perpendicular bisector.
∴ y² + x² = 4y²
⇒ x² = 3y²
⇒ x = √3y
or
⇒ y = \(\frac{x}{\sqrt{3}}\)
Now, area of given triangle = \(\frac{1}{2}\)
× 2y × X = y × x = \(\frac{x}{\sqrt{3}}\)× x = \(\frac{x^{2}}{\sqrt{3}}\)

Heron’s Formula Class 9 Extra Questions Short Answer Type 2

Class 9 Maths Chapter 12 Extra Questions Question 1.
Find the area of a triangle whose sides are 11 m, 60 m and 61 m.
Solutioin:
Let a = 11 m, b = 60 m and c = 61 m :

Chapter 12 Maths Class 9 Extra Questions Question 2.
Suman has a piece of land, which is in the shape of a rhombus. She wants her two sons to work on the land and produce different crops. She divides the land in two equal parts by drawing a diagonal. If its perimeter is 400 m and one of the diagonals is of length 120 m, how much area each of them will get for his crops ?
Solutioin:
Here, perimeter of the rhombus is 400 m.
∴ Side of the rhombus = \(\frac{400}{4}\) = 100 m
Let diagonal BD = 120 m and this diagonal divides the rhombus ABCD into two equal parts.

Hence, area of land allotted to two sons for their crops is 4800 m² each.

Extra Questions For Class 9 Maths Chapter 12 With Solution Question 3.
The perimeter of a triangular field is 144 m and its sides are in the ratio 3:4:5. Find the length of the perpendicular from the opposite vertex to the side whose length is 60 m.
Solutioin:
Let the sides of the triangle be 3x, 4x and 5x
∴ The perimeter of the triangular field = 144 m
⇒ 3x + 4x + 5x = 144
⇒ 12x = 144

Extra Questions On Herons Formula Class 9 Question 4.
Find the area of the triangle whose perimeter is 180 cm and two of its sides are of lengths 80 cm and 18 cm. Also, calculate the altitude of the triangle corresponding to the shortest side.
Solutioin:
Perimeter of given triangle = 180 cm
Two sides are 18 cm and 80 cm
∴ Third side = 180 – 18 – 80 = 82 cm

Hence, area of triangle is 720 cm² and altitude of the triangle corresponding to the shortest side is 80 cm.

Heron’s Formula Class 9 Extra Questions Long Answer Type

Extra Questions Of Herons Formula Class 9 Question 1.
Calculate the area of the shaded region.

Solutioin:

= 2 × 2 × 3 × 7 = 84 cm²
Area of shaded region = Area of ∆ABC – Area of ∆AOB
= 84 cm² – 30 cm² = 54 cm²

Class 9 Maths Herons Formula Extra Questions Question 2.
The sides of a triangular park are 8 m, 10 m and 6 m respectively. A small circular area of diameter 2 m is to be left out and the remaining area is to be used for growing roses. How much area is used for growing roses ? (use n = 3.14)
Solutioin:
The sides of the triangular park are 8 m, 10 m and 6 m.

Radius of the circle = \(\frac{2}{2}\) = 1 m
Area of the circle = πr² = 3.14 × 1 × 1 = 3.14 m²
∴ Area to be used for growing roses = Area of the park – area of the circle
= 24 – 3.14 = 20.86 m²

Heron’s Formula Class 9 Extra Questions HOTS

Herons Formula Class 9 Extra Questions With Solutions Pdf Question 1.
OPQR is a rhombus, whose three vertices P, Q and R lie on the circle with centre 0. If the radius of the circle is 12 cm, find the area of the rhombus.
Solutioin:
Since diagonals bisect each other at 90°.
∴ In right ∆QLR, (LR)² + (LQ)² = (QR)²

Class 9 Maths Ch 12 Extra Questions Question 2.
How much paper of each shade is needed to make a kite given in the figure, in which ABCD is a square with diagonal 60 cm?

Solutioin:
Since diagonals of a square are of equal length and bisect each other at right angles, therefore,
Area of ∆AOD = \(\frac{1}{2}\) × 30 × 30 = 450 cm²
Area of ∆AOD = Area of ∆DOC = Area of ∆BOC
= Area of ∆AOB = 450 cm²
[∵ ∆AOD = ∆AOB ≅ ∆BOC ≅ ∆COD, ∵ they² have equal area]
Now, area of ACEF (by Heron’s formula)
Here a = 20 cm, b = 20 cm and c = 30 cm

Now, area of orange shaded paper in kite
= Area of ∆AOD + Area of ∆CEF
= 450 cm² + 198.4 cm²
= 648.4 cm²
Area of blue shaded paper in kite
= Area of ∆AOB + Area of ∆COD
= 450 cm² + 450 cm² = 900 cm²
Area of black shaded paper in kite = Area of ∆BOC = 450 cm².

Heron’s Formula Class 9 Extra Questions Value Based (VBQs)

Heron’s Formula Class 9 Extra Questions With Solutions Question 1.
Sister Nivedita has trapezium shaped plot which she divided into three triangular portion for different purposes. I – for providing free education for orphan children, II – for providing dispensary for the needy villagers and III – for the library for villagers. Find the area of trapezium plot given in the figure. Which qualities of sister Nivedita are being depicted in question ?

Solutioin:
Here, ABCD is the trapezium with AB || DC.
Through C, Draw CF ⊥ AB
For – I: For area of ∆EBC

But, we know, Area of ∆EBC = \(\frac{1}{2}\)(base × height)
⇒ \(\frac{1}{2}\)1 × 30 × CF = 336
15 × CF = 336
⇒ CF = \(\frac{336}{15}\) = 22.4 m
Now, area of trapezium shaped plot = \(\frac{1}{2}\) (20 + 50)(22.4)
= 35 x 22.4 = 784 m²
Caring, kind, social, generous and visionary lady.

Ch 12 Maths Class 9 Extra Questions Question 2.
In an exhibition, an umbrella is made by stiching 10 triangular pieces of cloth with same message written on two triangular pieces. If each piece of cloth measures 60 cm, 60 cm and 20 cm, find how much cloth is required for each message.
Why we should respect women, educate women, save women and empower women ?

Solutioin:
Here, each triangular piece is an isosceles triangle with sides 60 cm, 60 cm and 20 cm.

Now, there are 2 triangular pieces with same message.
∴ Total area of cloth for each message = 2 × 591.61 = 1183.22 cm²
We should respect women, educate women, save women and empower women to serve humanity and give them equal opportunity so far they deprived.

Here we are providing Areas of Parallelograms and Triangles Class 9 Extra Questions Maths Chapter 9 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Areas of Parallelograms and Triangles with Answers Solutions

Extra Questions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles with Solutions Answers

Areas of Parallelograms and Triangles Class 9 Extra Questions Very Short Answer Type

Areas Of Parallelograms And Triangles Class 9 Extra Questions With Answers Question 1.
Two parallelograms are on equal bases and between the same parallels. Find the ratio of their areas.
Solution:
1:1 [∵ Two parallelograms on the equal bases and between the same parallels are equal in
area.]

Class 9 Maths Chapter 9 Extra Questions Question 2.
In ∆XYZ, XA is a median on side YZ. Find ratio of ar(∆XYA) : ar(∆XZA).

Solution:
Here, XA is the median on side YZ.
∴ YA = AZ
Draw XL ⊥ YZ

Parallelogram Questions For Class 9 Question 3.
ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (fig.). E and F are the mid-points of the non parallel sides. Find the ratio of ar(ABFE) and ar(EFCD).

Solution:

Area Of Parallelograms And Triangles Class 9 Extra Questions Question 4.
ABCD is a parallelogram and Q is any point on side AD. If ar(∆QBC) = 10 cm², find ar(∆QAB) + ar(∆QDC).

Solution:
Here, ∆QBC and parallelogram ABCD are on the same base BC and lie between the same parallels BC || AD.
∴ ar(||^gm ABCD) = 2 ar(∆QBC) ar(∆QAB) + ar(∆QDC) + ar(∆QBC) = 2 ar(∆QBC)
ar(∆QAB) + ar(∆QDC) = ar(∆QBC)
Hence, ar(∆QAB) + ar(∆QDC) = 10 cm² [∵ ar(∆QBC) = 10 cm² (given)]

Areas Of Parallelograms And Triangles Class 9 Questions With Answers Question 5.
WXYZ is a parallelogram with XP ⊥ WZ and ZQ ⊥ WX. If WX = 8 cm, XP = 8 cm and ZQ = 2 cm, find YX.

Solution:
ar(||^gm WXYZ) = ar(||^gm WXYZ)
WX × ZQ = WZ × XP
8 × 2 = WZ × 8
⇒ WZ = 2 cm
Now, YX = WZ = 2 cm [∵ opposite sides of parallelogram are equal]

Questions On Areas Of Parallelograms And Triangles Class 9 Question 6.
In figure, TR ⊥ PS, PQ || TR and PS || QR. If QR = 8 cm, PQ = 3 cm and SP = 12 cm, find ar(quad. PQRS).

Solution:
Here,
PS || QR [given]
∴ PQRS is a trapezium
Now, TR ⊥ PS and PQ || TR [given]
⇒ PQ ⊥ PS
∴ PQ = TR = 3 cm [given]
Now, ar(quad. PQRS) = \(\frac{1}{2}\) (PS + QR) × PQ = \(\frac{1}{2}\)(12 + 8) × 3 = 30 cm²

Area Of Parallelogram Class 9 Extra Questions Question 7.
In the given figure, ABCD is a parallelogram and L is the mid-point of DC. If ar(quad. ABCL) is 72 cm, then find ar(∆ADC).

Solution:
In ||^gm ABCD, AC is the diagonal

Areas Of Parallelograms And Triangles Questions With Answers Question 8.
In figure, TR ⊥ PS, PQ || TR and PS || QR. If QR = 8 cm, PQ = 3 cm and SP = 12 cm, find ar (PQRS).

Solution:
Here, PS || QR
∴ PQRS is a trapezium in which PQ = 3 cm, QR = 8 cm and SP = 12 cm
Now, TR I PS and PQ || TR
∴ PQRT is a rectangle [∵ PQ || TR, PT || QR and ∠PTR = 90°]
⇒ PQ = TR = 3 cm
Now, ar(PQRS) = \(\frac{1}{2}\)(PS + QR) × TR = \(\frac{1}{2}\)(12 + 8) × 3 = 30 cm².

Areas of Parallelograms and Triangles Class 9 Extra Questions Short Answer Type 1

Area Of Parallelogram Class 9 Questions Question 1.
ABCD is a parallelogram and O is the point of intersection of its diagonals. If ar(A AOD) = 4 cm\(²\) find area of parallelogram ABCD.

Solution:
Here, ABCD is a parallelogram in which its diagonals AC and BD intersect each other in O.
∴ O is the mid-point of AC as well as BD.
Now, in ∆ADB, AO is its median
∴ ar(∆ADB) = 2 ar(∆AOD)
[∵ median divides a triangle into two triangles of equal areas]
So, ar(∆ADB) = 2 × 4 = 8 cm²
Now, ∆ADB and ||^gm ABCD lie on the same base AB and lie between same parallels AB and CD
∴ ar(ABCD) = 2 ar(∆ADB).
= 2 × 8
= 16cm²

Areas Of Parallelograms And Triangles Question 2.
In the given figure of ∆XYZ, XA is a median and AB || YX. Show that YB is also a median.

Solution:
Here, in ∆XYZ, AB || YX and XA is a median.
∴ A is the mid-point of YZ. Now, AB is a line segment from mid-point of one side (YZ) and parallel to another side (AB || YX), therefore, it bisects the third side XZ.
⇒ B is the mid-point of XZ.
Hence, YB is also a median of ∆XYZ.

Question 3.
ABCD is a trapezium. Diagonals AC and BD intersect each other at O. Find the ratio ar (∆AOD) : ar (∆BOC).
Solution:

Here, ABCD is a trapezium in which diagonals AC and BD intersect each other at O. ∆ADC and ABCD are on the same base DC and between the same ‘parallels i.e., AB || DC.
∴ ar(∆ADC) = ar(∆BCD)
⇒ ar(∆AOD) + ar(∆ODC)
= ar(ABOC) + ar(AODC)
⇒ ar(∆AOD) = ar(∆BOC)
⇒ \(\frac { ar(∆AOD) }{ ar(∆BOC) } \) = 1

Question 4.
ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ (fig.). If AQ intersects DC at P, show that ar(∆BPC) = ar(∆DPQ).

Solution:
In ||^gm ABCD,
ar(∆APC) = ar(∆BCP) …(i)
[∵ triangles on the same base and between the same parallels have equal area]
Similarly, ar(∆ADQ) = ar(∆ADC) i …(ii)
Now, ar(∆ADQ) – ar(∆ADP) = ar(∆ADC) – ar(∆ADP)
ar(∆DPQ) = ar(∆ACP) … (iii)
From (i) and (iii), we have
ar(∆BCP) = ar(∆DPQ)
or ar(∆BPC) = ar(∆DPQ)

Question 5.
In the figure, PQRS is a parallelogram with PQ = 8 cm and ar(∆PXQ) = 32 cm². Find the altitude of gm PQRS and hence its area.

Solution:
Since parallelogram PQRS and APXQ are on the same base PQ and lie between the same
parallels PQ || SR
∴ Altitude of the ∆PXQ and ||^gm PQRS is same.
Now, \(\frac{1}{2}\)PQ × altitude = ar(∆PXQ)
⇒ \(\frac{1}{2}\) × 8 × altitude = 32
altitude = 8 cm
ar(||^gm PQRS) = 2 ar(∆PXQ)
= 2 × 32 = 64 cm²
Hence, the altitude of parallelogram PQRS is 8 cm and its area is 64 cm².

Question 6.
In ∆ABC. D and E are points on side BC such that CD = DE = EB. If ar(∆ABC) = 27 cm, find ar(∆ADE)

Solution:
Since in ∆AEC, CD = DE, AD is a median.
∴ ar(∆ACD) = ar(∆ADE)
[∵ median divides a triangle into two triangles of equal areas]
Now, in ∆ABD, DE = EB, AE is a median
ar(∆ADE) = ar(∆AEB)… (ii)
From (i), (ii), we obtain
ar(∆ACD) = ar(∆ADE) = ar(∆AEB)\(\frac{1}{3}\)ar(∆ABC)
∴ ar(∆ADE) = \(\frac{1}{3}\) × 27 = 9 cm²

Areas of Parallelograms and Triangles Class 9 Extra Questions Short Answer Type 2

Question 1.
For the given figure, check whether the following statement is true or false. Also justify your answer. PQRS is a trapezium with PQ || SR, PS || RU and ST || RQ, then ar(PURS) = ar(TQRS)

Solution:
Since ST || RQ and SR || TQ [given]
⇒ STQR is a ||^gm
Similarly, PS || UR and SR || PU [given]
⇒ PSRU is a ||^gm
Also, similarly, ||^gm STQR and ||^gm PSRU lie on same base SR and between same parallels PQ and SR.
∴ ar(||^gm STQR) = ar(||^gm PSRU)
Hence, the given statement is true.

Question 2.
In the given figure, WXYZ is a quadrilateral with a point P on side WX. If ZY || WX, show that :
(i) ar(∆ZPY) = ar(∆ZXY)
(ii) ar(∆WZY) = ar(∆ZPY)
(iii) ar(∆ZWX) = ar(∆XWY)

Solution:
∆ZPY and ∆ZXY lie on same base ZY and between same parallels ZY and WX
∴ ar(∆ZPY) = ar(∆ZXY)
Again, (∆WZY) and (∆ZPY) lie on same base ZY and between same parallels ZY and WX
∴ ar(∆WZY) = ar(∆ZPY)
Also, ∆zwX and ∆XWY lie on same base XW and between same parallels ZY and WX
∴ ar(∆ZWX) = ar(∆XWY)

Question 3.
In ∆ABC ; D, E and F are mid-points of sides BC, AC and AB respectively. A line through C drawn parallel to DE meets FE produced to G. Show that ar(∆FDE) = ar(∆EGC).

Solution:
Here, in ∆ABC; D, E and F are the mid-points of sides BC, AC and AB respectively. A line through C is drawn parallel to DE meets FE produced at G.
Since a line segment drawn through the mid-points of two sides, is parallel to third side and is half of it.
∴ DE || AB, EF || BC and FD || AC
⇒ AEDF, EFDC and EFDB are parallelograms
Also, a diagonal of a parallelogram divides it into two congruent triangles
∴ ∆AFE ≅ ∆DEF
∆DEF ≅ ∆FDB and
∆DEF ≅ ∆EDC
∴ ar(∆FDE) ≅ ar(∆EDC) …(i)
Again, in quad. EGCD, we have
CG || DE and DC || EG [given]
∴ EGCD is a parallelogram .
∴ ∆EDC = ∆CGE
⇒ ar(AEDC) = ar(ACGE) … (ii)
From (i) and (ii), we obtain
ar(∆FDE) = ar(∆EGC)

Question 4.
In ∆PQR, A and B are points on side QR such that they trisect QR. Prove that ar(∆PQB) = 2ar(∆PBR).
Solution:
Here, in ∆PQR, A and B are points on side QR
such that QA = AB = BR.

Through P, draw a line / parallel to QR
Now, APQA, APAB and APBR on the equal bases
and between the same parallels l || QR
⇒ ar(∆PQA) = ar(∆PAB) = ar(∆PBR) …. (i)
Now, ar(∆PQB) = ar(∆PQA) + ar(∆PAB)
= 2ar(∆PQA)[using (i)]
= 2ar(∆PBR) [using (i)]

Question 5.
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). Show that ar(||^gm ABCD) = ar(||^gm PBQR).

Solution:

Join AC and QP, also it is given that ∆Q || CP
∴ ∆ACQ and ∆APQ are on the same base ∆Q and lie between the same parallels ∆Q || CP.
∴ ar(∆ACQ) = ar(∆APQ)
or ar(∆ABC) + ar(∆ABQ) = ar(ABPQ) + ar(∆ABQ)
ar(∆ABC) = ar(ABPQ)
or \(\frac{1}{2}\)ar(||^gm ABCD) = \(\frac{1}{2}\)ar (||^gm PBQR)
or ar(||^gm ABCD) = ar(||^gm PBQR)

Areas of Parallelograms and Triangles Class 9 Extra Questions Long Answer Type

Question 1.
EFGH is a parallelogram and U and T are points on sides EH and GF respectively. If ar(∆EHT) = 16 cm, find ar(∆GUF).

Solution:
∴ ar(∆EHT) = \(\frac{1}{2}\) ar(||^gm EFGH) …..(i)
Similarly, ∆GUF and parallelogram EFGH are on the same base GF and lie between the same parallels GF and HE
∴ ar(∆GUF) = \(\frac{1}{2}\) ar(||^gm EFGH) …..(ii)
From (i) and (ii), we have
ar(∆GUF) = ar(∆EHT)
= 16 cm² [∵ ar(∆EHT) = 16 cmcm²] [given]

Question 2.
ABCD is a parallelogram and P is any point in its interior. Show that :
ar(∆APB) + ar(∆CPD) = ar(∆BPC) + ar(∆APD)

Solution:
Through P, draw a line LM || DA and EF || AB
Since ∆APB and ||^gm ABFE are on the same base AB and lie between the same parallels AB and EF.
∴ ar(∆APB) = \(\frac{1}{2}\) ar(||^gm ABFE) … (i)

Similarly, ACPD and parallelogram DCFE are on the same base DC and between the same parallels DC and EF.
∴ ar(∆CPD) = \(\frac{1}{2}\) ar(||^gm DCFE) … (ii)
Adding (i) and (ii), we have
ar(∆APB) + ar(∆CPD) = \(\frac{1}{2}\) ar (||^gm ABFE) + ar(||^gm DCFE)
= \(\frac{1}{2}\) ar(|lgm ABCD) … (iii)
Since ∆APD and parallelogram ADLM are on the same base AB and between the same parallels AD and ML
∴ ar(∆APD) = \(\frac{1}{2}\) ar(||^gm ADLM) …..(iv)
Similarly, ar(∆BPC) = \(\frac{1}{2}\) = arc||^gm BCLM) ….(v)
Adding (iv) and (u), we have
ar(∆APD) + ar (∆BPC) = \(\frac{1}{2}\) ar(||^gm ABCD) ….(vi)
From (iii) and (vi), we obtain
ar(∆APB) + ar(∆CPD) = ar(∆APD) + ar(ABPC)

Question 3.
In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP.

Solution:
In ∆PAD, ∠A = 90° and DA = PA = AB
⇒ ∠ADP = ∠APD = \(\frac{90^{\circ}}{2}\) = 45°
Similarly, in ∆QBC, ∠B = 90° and BQ = BC = AB
⇒ ∠BCQ = ∠BQC = \(\frac{90^{\circ}}{2}\) = 45°
In ∆PAD and ∆QBC, we have
PA = BQ [given]
∠A = ∠B [each = 90°]
AD = BC [sides of a square]
⇒ ∠PAD ≅ ∆QBC [by SAS congruence rule]
⇒ PD = QC [c.p.c.t.]
Now, in APDC and ∆QCD
DC = DC [common]
PD = QC [prove above]
∠PDC = ∠QCD [each = 90° + 45° = 135°]
⇒ ∆PDC = ∆QCD [by SAS congruence rule]
⇒ PC = QD or DQ = CP

Areas of Parallelograms and Triangles Class 9 Extra Questions HOTS

Question 1.
In the given figure, PQRS, SRNM and PQNM are parallelograms, Show that :
ar(∆PSM) = ar(∆QRN).

Solution:
Since PQRS is a parallelogram.
∴ PS = QR and PS || QR
Since SRNM is also a parallelogram.
∴ SM = RN and SM || RN
Also, PQNM is a parallelogram
∴ PM || QM and PM = QM
Now, in APSM and ∆QRN
PS = QR
SM = RN
PM = QN
∆PSM ≅ ∆QRN [by SSS congruence axiom]
∴ ar (∆PSM) = ar (∆QRN) [congruent triangles have same areas)

Areas of Parallelograms and Triangles Class 9 Extra Questions Value Based (VBQs)

Question 1.
Naveen was having a plot in the shape of a quadrilateral. He decided to donate some portion of it to construct a home for orphan girls. Further he decided to buy a land in lieu of his donated portion of his plot so as to form a triangle.
(i) Explain how this proposal will be implemented?
(ii) Which mathematical concept is used in it?
(iii) What values are depicted by Naveen?
Solution:

(i) Let ABCD be the plot and Naveen decided to donate some portion to construct a home for orphan girls from one corner say C of plot ABCD. Now, Naveen also purchases equal amount of land in lieu of land CDO, so that he may have triangular form of plot. BD is joined. Draw a line through C parallel to DB to meet AB produced in P.
Join DP to intersect BC at 0.
Now, ABCD and ABPD are on the same base and between same parallels CP || DB.
ar(∆BCD) = ar(∆BPD) ar(∆COD) + ar(∆DBO) = ar(∆BOP) + ar(∆DBO)
ar(ACOD) = ar(ABOP) ar(quad. ABCD)
= ar(quad. ABOD) + ar(∆COD)
= ar(quad. ABOD) + ar(∆BOP)
[∵ ar(ACOD) = ar(ABOP)] (proved above]
= ar(∆APD)
Hence, Naveen purchased the portion ABOP to meet his requirement.
(ii) Two triangles on the same base and between same parallels are equal in area.
(iii) We should help the orphan children.

Question 2.
A flood relief camp was organized by state government for the people affected by the natural calamity near a city. Many school students volunteered to participate in the relief work. In the camp, the food items and first aid centre kits were arranged for the flood victims. The piece of land used for this purpose is shown in the figure.
(a) If EFGH is a parallelogram with P and Q as mid-points of sides GH and EF respectively, then show that area used for first aid is half of the total area.
(b) What can you say about the student volunteers working for the relief work?

Solution:
(a) Here, EFGH is a ||^gm
∴ EF = GH and EF || GH

Hence, area used for first aid is half of the total area.

(b) Students working for the noble cause show compassion towards the affected people. They also realize their social responsibility to work for helping the ones in need.

Here we are providing Quadrilaterals Class 9 Extra Questions Maths Chapter 8 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Quadrilaterals with Answers Solutions

Extra Questions for Class 9 Maths Chapter 8 Quadrilaterals with Solutions Answers

Quadrilaterals Class 9 Extra Questions Very Short Answer Type

Quadrilateral Class 9 Extra Questions Question 1.
If one angle of a parallelogram is twice of its adjacent angle, find the angles of the parallelogram.
Solution:
Let the two adjacent angles be x and 2x.
In a parallelogram, sum of the adjacent angles are 180°
∴ x + 2x = 180°
⇒ 3x = 180°
⇒ x = 60°
Thus, the two adjacent angles are 120° and 60°. Hence, the angles of the parallelogram are 120°, 60°, 120° and 60°.

Class 9 Quadrilaterals Extra Questions Question 2.
If the diagonals of a quadrilateral bisect each other at right angles, then name the
quadrilateral.
Solution:
Rhombus.

Quadrilaterals Class 9 Extra Questions With Solutions Question 3.
Three angles of a quadrilateral are equal and the fourth angle is equal to 144o. Find each of the equal angles of the quadrilateral.
Solution:
Let each equal angle of given quadrilateral be x.
We know that, sum of all interior angles of a quadrilateral is 360°
∴ x + x + x + 144° = 360°
3x = 360° – 144°
3x = 216°
x = 72°
Hence, each equal angle of the quadrilateral is of 72o measures.

Extra Questions On Quadrilaterals Class 9 Question 4.
If ABCD is a parallelogram, then what is the measure of ∠A – ∠C ?
Solution:
∠A – ∠C = 0° (opposite angles of parallelogram are equal]

Quadrilaterals Class 9 Extra Questions Question 5.
PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram.
Solution:

Here, PQ = SR = 12 cm
Let PS = x and PS = QR
∴ x + 12 + x + 12 = Perimeter
2x + 24 = 40
2x = 16
x= 8
Hence, length of each side of the parallelogram is 12 cm, 8 cm, 12 cm and 8 cm.

Class 9 Maths Quadrilaterals Extra Questions Question 6.
Two consecutive angles of a parallelogram are (x + 60)° and (2x + 30)°. What special name can you give to this parallelogram?
Solution:
We know that consecutive interior angles of a parallelogram are supplementary.
∴ (x + 60° + (2x + 30)° = 180°
⇒ 3x° + 90° = 180°
⇒ 3x° = 90°
⇒ x° = 30°
Thus, two consecutive angles are (30 + 60)°, 12 x 30 + 30)”. i.e., 90° and 90°.
Hence, the special name of the given parallelogram is rectangle.

Class 9 Quadrilaterals Extra Questions Pdf Question 7.
ONKA is a square with ∠KON = 45°. Determine ∠KOA.
Solution:
Since ONKA is a square
∴ ∠AON = 90°
We know that diagonal of a square bisects its ∠s
⇒ ∠AOK = ∠KON = 45°
Hence, ∠KOA = 45°

Quadrilaterals Extra Questions Class 9 Question 8.
In quadrilateral PQRS, if ∠P = 60° and ∠Q : ∠R : ∠S = 2 : 3 : 7, then find the measure of ∠S.
Solution:
Let ∠Q = 2x, ∠R = 3x and ∠S = 7x
Now, ∠P + ∠Q + ∠R + ∠S = 360°
⇒ 60° + 2x + 3x + 7x = 360°
⇒ 12x = 300°
x = \(\frac{300^{\circ}}{12}\) = 25°
∠S = 7x = 7 x 25° = 175°

Quadrilaterals Class 9 Extra Questions Short Answer Type 1

Questions On Quadrilaterals For Class 9 Question 1.
ABCD is a parallelogram in which ∠ADC = 75° and side AB is produced to point E as shown in the figure. Find x + y.
Solution:

Here, ∠C and ∠D are adjacent angles of the parallelogram.
∴ ∠C + ∠D = 180°
⇒ x + 75° = 180°
⇒ x = 105°
Also, y = x = 105° [alt. int. angles]
Thus, x + y = 105° + 105° = 210°

Class 9 Maths Chapter 8 Extra Questions With Solutions Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:

Given: A parallelogram ABCD, in which AC = BD.
To Prove: ΔBCD is a rectangle.
Proof : In ΔABC and ΔBAD
AB = AB (common]
AC = BD (given]
BC = AD(opp. sides of a ||^gm]
⇒ ΔABC ≅ ΔBAD
[by SSS congruence axiom]
⇒ ∠ABC = ∠BAD (c.p.c.t.)
Also, ∠ABC + ∠BAD = 180° (co-interior angles)
∠ABC + ∠ABC = 180° [ ∵ ∠ABC = ∠BAD ]
2∠ABC = 180°
∠ABC = 1/2 x 180° = 90°
Hence, parallelogram ABCD is a rectangle.

Quadrilateral Extra Questions Class 9 Question 3.
In the figure, ABCD is a rhombus, whose diagonals meet at O. Find the values of x and y.

Solution:
Since diagonals of a rhombus bisect each other at right angle.
In ∴ ΔAOB, we have
∠OAB + ∠x + 90° = 180°
∠x = 180° – 90° – 35°
= 55°
Also,
∠DAO = ∠BAO = 35°
∠y + ∠DAO + ∠BAO + ∠x = 180°
⇒ ∠y + 35° + 35° + 55° = 180°
⇒ ∠y = 180° – 125o = 55°
Hence, the values of x amd y are x = 55°, y = 55°.

Extra Questions For Class 9 Maths Quadrilaterals With Solutions Question 4.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see fig.). Show that :
(i) ΔAPB = ΔCQD
(ii) AP = CQ
Solution:

Given : In ||gm ABCD, AP and CQ are perpendiculars from the
vertices A and C on the diagonal BD.
To Prove: (i) ΔAPB ≅ ΔCQD
(ii) AP = CQ
Proof : (i) In ΔAPB and ΔCQD
AB = DC (opp. sides of a ||gm ABCD]
∠APB = ∠DQC (each = 90°)
∠ABP = ∠CDQ (alt. int. ∠s]
⇒ ΔAPB ≅ ΔCQD[by AAS congruence axiom]
(ii) ⇒ AP = CQ [c.p.c.t.]

Quadrilaterals Class 9 Extra Questions Short Answer Type 2

Extra Questions Of Quadrilaterals Class 9 Question 1.
The diagonals of a quadrilateral ABCD are perpendicular to each other. Show that the quadrilateral formed by joining the mid-points of its sides is a rectangle.

Solution:
Given: A quadrilateral ABCD whose diagonals AC and BD are perpendicular to each other at O. P, Q, R and S are mid-points of side AB, BC, CD and DA respectively are joined are formed quadrilateral PQRS.
To Prove: PQRS is a rectangle.
Proof : In ∆ABC, P and Q are mid-points of AB and BC respectively.
∴ PQ || AC and PQ = \(\frac{1}{2}\) AC … (i) (mid-point theorem]
Further, in SACD, R and S are mid-points of CD and DA respectively.
SR || AC and SR = \(\frac{1}{2}\) AC … (ii) (mid-point theorem]
From (i) and (ii), we have PQ || SR and PQ = SR
Thus, one pair of opposite sides of quadrilateral PQRS are parallel and equal.
∴ PQRS is a parallelogram.
Since PQ|| AC PM || NO
In ∆ABD, P and S are mid-points of AB and AD respectively.
PS || BD (mid-point theorem]
⇒ PN || MO
∴ Opposite sides of quadrilateral PMON are parallel.
∴ PMON is a parallelogram.
∠ MPN = ∠ MON (opposite angles of ||gm are equal]
But ∠MON = 90° [given]
∴ ∠MPN = 90° ⇒ ∠QPS = 90°
Thus, PQRS is a parallelogram whose one angle is 90°
∴ PQRS is a rectangle.

Class 9 Maths Chapter 8 Extra Questions Question 2.
In the fig., D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. Prove that DEF is also an equilateral triangle.

Solution:
Since line segment joining the mid-points of two sides of a triangle is half of the third side.
Therefore, D and E are mid-points of BC and AC respectively.
⇒ DE = \(\frac{1}{2}\)AB …(i)
E and F are the mid-points of AC and AB respectively.
∴ EF = \(\frac{1}{2}\)BC … (ii)
F and D are the mid-points of AB and BC respectively.
∴ FD = \(\frac{1}{2}\) AC … (iii)
Now, SABC is an equilateral triangle.
⇒ AB = BC = CA
⇒ \(\frac{1}{2}\)AB = \(\frac{1}{2}\)BC = \(\frac{1}{2}\)CA
⇒ DE = EF = FD (using (i), (ii) and (iii)]
Hence, DEF is an equilateral triangle.

Quadrilateral Class 9 Questions Question 3.
In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY is a parallelogram. Show that ABCD is a parallelogram.

Solution:
Since BXDY is a parallelogram.
XO = YO
DO = BO
[∵ diagonals of a parallelogram bisect each other]
But AX = CY …. (iii) (given]
Adding (i) and (iii), we have
XO + AX = YO + CY
⇒ AO = CO …. (iv)
From (ii) and (iv), we have
AO = CO and DO = BO
Thus, ABCD is a parallelogram, because diagonals AC and BD bisect each other at O.

Class 9 Maths Ch 8 Extra Questions Question 4.
ABCD is a quadrilateral in which the bisectors of ∠A and ∠C meet DC produced at Y and BA produced at X respectively. Prove that
∠X +∠Y = \(\frac{1}{2}\)(∠A + ∠C)
Solution:

Here, ∠1 = ∠2 and ∠3 = ∠4
In ΔXBC, we have
∠X + ∠B + ∠4 = 180°
∠X + ∠B + \(\frac{1}{2}\)∠C = 180
In ΔADY, we have
∠2 + ∠D + ∠Y= 180°
\(\frac{1}{2}\) ∠A + ∠D + ∠Y = 180°
Adding (i) and (ii), we have
∠X + ∠Y + ∠B + ∠D + \(\frac{1}{2}\) ∠C + \(\frac{1}{2}\) ∠A = 360°
Also, in quadrilateral ABCD
∠A + ∠B + ∠C + ∠D = 360°
∠X + ∠Y + ∠B + ∠D + \(\frac{1}{2}\) ∠C + \(\frac{1}{2}\) ∠A = ∠A + ∠B + ∠C + ∠D
∠X + ∠Y = ∠A – \(\frac{1}{2}\) ∠A + \(\frac{1}{2}\) ∠C – \(\frac{1}{2}\) ∠C
∠X+ ∠Y = \(\frac{1}{2}\) (∠A + ∠C)

Quadrilaterals Class 9 Extra Questions Long Answer Type

Quadrilateral Questions For Class 9 Question 1.
In the figure, P, Q and R are the mid-points of the sides BC, AC and AB of ΔABC. If BQ and PR intersect at X and CR and PQ intersect at Y, then show that XY = \(\frac{1}{4}\) BC.

Solution:
Here, in ΔABC, R and Q are the mid-points of AB and AC respectively.
∴ By using mid-point theorem, we have
RQ || BC and RQ = \(\frac{1}{2}\) BC
∴ RQ = BP = PC [∵ P is the mid-point of BC]
∴ RQ || BP and RQ || PC
In quadrilateral BPQR
RQ || BP, RQ = BP (proved above]
∴ BPQR is a parallelogram. [∵ one pair of opp. sides is parallel as well as equal]
∴ X is the mid-point of PR. [∵ diagonals of a ||gm bisect each other]
Now, in quadrilateral PCQR
RQ || PC and RQ = PC [proved above)
∴ PCQR is a parallelogram [∵ one pair of opp. sides is parallel as well as equal]
∴ Y is the mid-point of PQ [∵ diagonals of a ||gm bisect each other]
In ΔPQR
∴ X and Y are mid-points of PR and PQ respectively.

Class 9 Quadrilaterals Important Questions Question 2.
In the given figure, AE = DE and BC || AD. Prove that the points A, B, C and D are concyclic. Also, prove that the diagonals of the quadrilateral ABCD are equal.

Solution:
Since AE = DE
∠D = ∠A …. (i) [∵ ∠s opp. to equal sides of a Δ]
Again, BC || AD
∠EBC = ∠A …. (ii) (corresponding ∠s]
From (i) and (ii), we have
∠D = ∠EBC …. (iii)
But ∠EBC + ∠ABC = 180° (a linear pair]
∠D + ∠ABC = 180° (using (iii)]
Now, a pair of opposite angles of quadrilateral ABCD is supplementary
Thus, ABCD is a cyclic quadrilateral i.e., A, B, C and D’are concyclic. In ΔABD and ΔDCA
∠ABD = ∠ACD [∠s in the same segment for cyclic quad. ABCD]
∠BAD = ∠CDA [using (i)]
AD = AD (common]
So, by using AAS congruence axiom, we have
ΔABD ≅ ΔDCA
Hence, BD = CA [c.p.c.t.]

Question 3.
In ΔABC, AB = 8 cm, BC = 9 cm and AC = 10 cm. X, Y and Z are mid-points of AO, BO and CO respectively as shown in the figure. Find the lengths of the sides of ΔXYZ.

Solution:
Here, in ΔABC, AB = 8 cm, BC = 9 cm, AC = 10 cm.
In ΔAOB, X and Y are the mid-points of AO and BO.
∴ By using mid-point theorem, we have
XY = \(\frac{1}{2}\) AB = \(\frac{1}{2}\) x 8 cm = 4 cm
Similarly, in Δ𝜏BOC, Y and Z are the mid-points of BO and CO.
∴ By using mid-point theorem, we have
YZ = \(\frac{1}{2}\) BC = \(\frac{1}{2}\) x 9cm = 4.5 cm
And, in Δ𝜏COA, Z and X are the mid-points of CO and AO.
∴ ZX = \(\frac{1}{2}\) AC = \(\frac{1}{2}\) x 10 cm = 5 cm
Hence, the lengths of the sides of ΔXYZ are XY = 4 cm, YZ = 4.5 cm and ZX = 5 cm.

Question 4.
PQRS is a square and ∠ABC = 90° as shown in the figure. If AP = BQ = CR, then prove that ∠BAC = 45°

Solution:
Since PQRS is a square.
∴ PQ = QR … (I) [∵ sides of a square are equal]
Also, BQ = CR … (ii) [given]
Subtracting (ii) from (i), we obtain
PQ – BQ = QR – CR
⇒ PB = QC … (iii)
In Δ𝜏APB and Δ𝜏BQC
AP = BQ[given
∠APB = ∠BQC = 90°](each angle of a square is 90°)
PB = QC (using (iii)]

So, by using SAS congruence axiom, we have
ΔAPB ≅ ΔBQC
∴ AB = BC [c.p.c.t.]
Now, in ΔABC
AB = BC [proved above]
∴ ∠ACB = ∠BAC = x° (say) [∠s opp. to equal sides]
Also, ∠B + ∠ACB + ∠BAC = 180°
⇒ 90° + x + x = 180°
⇒ 2x° = 90°
x° = 45°
Hence, ∠BAC = 45°

Question 5.
ABCD is a parallelogram. If the bisectors DP and CP of angles D and C meet at P on side AB, then show that P is the mid-point of side AB.
Solution:

Since DP and CP are angle bisectors of ∠D and ∠C respectively.
: ∠1 = ∠2 and ∠3 = ∠4
Now, AB || DC and CP is a transversal
∴ ∠5 = ∠1 [alt. int. ∠s]
But ∠1 = ∠2 [given]
∴ ∠5 = ∠2

Now, in ABCP, ∠5 = ∠2
⇒ BC = BP … (I) [sides opp. to equal ∠s of a A]
Again, AB || DC and DP is a transversal.
∴ ∠6= ∠3 (alt. int. Δs]
But ∠4 = ∠3 [given]
∴ ∠6 = ∠4
Now, in ΔADP, ∠6 = ∠4
⇒ DA = AP …. (ii) (sides opp. to equal ∠s of a A]
Also, BC = DA… (iii) (opp. sides of parallelogram)
From (i), (ii) and (iii), we have
BP = AP
Hence, P is the mid-point of side AB.

Question 6.
In the figure, ΔBCD is a trapezium in which AB || DC. E and F are the mid-points of AD and BC respectively. DF and AB are produced to meet at G. Also, AC and EF intersect at the point O. Show that :
(i) EO || AB
(ii) AO = CO

Solution:
Here, E and F are the mid-points of AD and BC respectively.
In ΔBFG and ΔCFD
BF = CF [given]
∠BFG = ∠CFD (vert. opp. ∠s]
∠BGF = ∠CDF (alt. int. ∠s, as AB || DC)
So, by using AAS congruence axiom, we have
ΔBFG ≅ ΔCFD
⇒ DF = FG [c.p.c.t.)
Now, in ΔAGD, E and F are the mid-points of AD and GD.
∴ By mid-point theorem, we have
EF || AG
or EO || AB
Also, in ΔADC, EO || DC
∴ EO is a line segment from mid-point of one side parallel to another side.
Thus, it bisects the third side.
Hence, AO = CO

Here we are providing Number Systems Class 9 Extra Questions Maths Chapter 1 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Number Systems with Answers Solutions

Extra Questions for Class 9 Maths Chapter 1 Number Systems with Solutions Answers

Number Systems Class 9 Extra Questions Very Short Answer Type

Number System Class 9 Extra Questions With Solutions Question 1.
Simplify: (√5 + √2)².
Solution:
Here, (√5 + √2² = (√5² + 2√5√2 + (√2)²
= 5 + 2√10 + 2 = 7 + 2√10

Question 2.
Find the value of √(3)^-2.
Solution:

Number System Class 9 Extra Questions Question 3.
Identify a rational number among the following numbers :
2 + √2, 2√2, 0 and π
Solution:
O is a rational number.

Class 9 Maths Chapter 1 Extra Questions Question 4.
Express 1.8181… in the form \(\frac{p}{q}\) where p and q are integers and q ≠ 0.
Solution:
Let x =1.8181… …(i)
100x = 181.8181… …(ii) [multiplying eqn. (i) by 100]
99x = 180 [subtracting (i) from (ii)]
x = \(\frac{180}{99}\)
Hence, 1.8181… = \(\frac{180}{99}\) = \(\frac{20}{11}\)

Class 9 Number System Extra Questions Question 5.
Simplify : √45 – 3√20 + 4√5
Solution:
√45 – 3√20 + 4√5 = 3√5 – 6√5 + 4√5 = √5.

Class 9 Maths Number System Extra Questions With Solutions Question 6.
Find the value of’

Solution:

Class 9 Maths Ch 1 Extra Questions Question 7.
Find the value of

Solution:

Number Systems Class 9 Extra Questions Short Answer Type 1

Extra Questions For Class 9 Maths Chapter 1 With Solution Pdf Question 1.
Evaluate : (√5 + √2² + (√8 – √5)²
Solution:
(√5 + √2)² + (√8 – √5² = 5 + 2 + 2√10 + 8 + 5 – 2√40
= 20 + 2√10 – 4√10 = 20 – 2√10

Extra Questions On Number System Class 9 Question 2.
Express \(23 . \overline{43}\) in \(\frac{p}{q}\) form, where p, q are integers and q ≠ 0.
Solution:
Let x = \(23 . \overline{43}\)
or x = 23.4343…         ….(i)
100x = 2343.4343…    …(ii) [Multiplying eqn. (i) by 100]
99x = 2320 [Subtracting (i) from (ii)
⇒ x = \(\frac{2320}{99}\)
Hence, \(23 . \overline{43}\) = \(\frac{2320}{99}\)

Number System Extra Questions Question 3.
Let ‘a’ be a non-zero rational number and ‘b’ be an irrational number. Is ‘ab’ necessarily an irrational ? Justify your answer with example.
Solution:
Yes, ‘ab’ is necessarily an irrational.
For example, let a = 2 (a rational number) and b = √2 (an irrational number)
If possible let ab = 2√2 is a rational number.
Now, \(\frac{ab}{a}\) = \(\frac{2 \sqrt{2}}{2}\) = √2 is a rational number.
[∵ The quotient of two non-zero rational number is a rational]
But this contradicts the fact that √2 is an irrational number.
Thus, our supposition is wrong.
Hence, ab is an irrational number.

Chapter 1 Maths Class 9 Extra Questions Question 4.
Let x and y be a rational and irrational numbers. Is x + y necessarily an irrational number? Give an example in support of your answer.
Solution:
Yes, x + y is necessarily an irrational number.
For example, let x = 3 (a rational number) and y = √5 (an irrational number)
If possible let x + y = 3 + √5 be a rational number.
Consider \(\frac{p}{q}\) = 3 + √5, where p, q ∈ Z and q ≠ 0.
Squaring both sides, we have

∵ \(\frac{p}{q}\) is a rational
⇒ √5 is a rational
But this contradicts the fact that √5 is an irrational number.
Thus, our supposition is wrong.
Hence, x + y is an irrational number.

Number Systems Class 9 Extra Questions Short Answer Type 2

Ch 1 Maths Class 9 Extra Questions Question 1.
Represent √3 on the number line.
Solution:


On the number line, take OA = 1 unit. Draw AB = 1 unit perpendicular to OA. Join OB.
Again, on OB, draw BC = 1 unit perpendicular to OB. Join OC.
By Pythagoras Theorem, we obtain OC = √3. Using
compasses, with centre O and radius OC, draw an arc, which intersects the number line at point
D. Thus, OD = √3 and D corresponds to √3.

Class 9 Chapter 1 Maths Extra Questions Question 2.
Represent √3.2 on the number line.
Solution:
First of all draw a line of length 3.2 units such that AB = 3.2 units. Now, from point B, mark a distance of 1 unit. Let this point be ‘C’. Let ‘O’ be the mid-point of the distance AC. Now, draw a semicircle with centre ‘O’ and radius OC. Let us draw a line perpendicular to AC passing through the point ‘B’ and intersecting the semicircle at point ‘D’.
∴ The distance BD = √3.2

Now, to represent √3.2 on the number line. Let us take the line BC as number line and point ‘B’ as zero, point ‘C’ as ‘1’ and so on. Draw an arc with centre B and radius BD, which intersects the number line at point ‘E’.
Then, the point ‘E’ represents √3.2.

Class 9 Ch 1 Maths Extra Questions Question 3.
Express 1.32 + 0.35 as a fraction in the simplest form.
Solution:
Let . x = 1.32 = 1.3222…..(i)

Multiplying eq. (i) by 10, we have
10x = 13.222…
Again, multiplying eq. (i) by 100, we have
100x = 132.222… …(iii)
Subtracting eq. (ii) from (iii), we have
100x – 10x = (132.222…) – (13.222…)
90x = 119
⇒ x = \(\frac{119}{90}\)
Again, y = 0.35 = 0.353535……
Multiply (iv) by 100, we have …(iv)
100y = 35.353535… (v)
Subtracting (iv) from (u), we have
100y – y = (35.353535…) – (0.353535…)
99y = 35
y = \(\frac{35}{99}\)

Extra Questions For Class 9 Maths Ch 1 Question 4.
Find the square root of 10 + √24 + √60 + √40.
Solution:

Question 5.
If x = 9 + 4√5, find the value of √x – \(\frac{1}{\sqrt{x}}\).
Solution:
Here,
x = 9 + 4√5
x = 5 + 4 + 2 x 2√5
x = (√5² + (2² + 2 x 2x √5).
x = (√5 + 2)²
√x = √5 + 2
Now, \(\frac{1}{\sqrt{x}}\) = \(\frac{1}{\sqrt{5}+2}\)

Question 6.
If x = \(\frac{1}{\sqrt{5}-2}\) , find the value of x³ – 3² – 5x + 3
Solution:

∴ x – 2 = √5
Squaring both sides, we have
x² – 4x + 4 = 5
x² – 4x – 1 = 0 …(i)
Now, x³ – 3² – 5x + 3 = (x² – 4x – 1) (x + 1) + 4
= 0 (x + 1) + 4 = 4 [using (i)]

Question 7.
Find ‘x’, if 2^x-7 × 5^x-4 = 1250.
Solution:
We have 2^x-7 × 5^x-4 = 1250
⇒ 2^x-7 × 5^x-4 = 2 5 × 5 × 5 × 5
⇒ 2^x-7 × 5^x-4 = 21 × 54
Equating the powers of 2 and 5 from both sides, we have
⇒ x – 7 = 1 and x – 4 = 4
⇒ x = 8 and x = 8
Hence, x = 8 is the required value.

Question 8.
Evaluate:

Solution:

Number Systems Class 9 Extra Questions Long Answer Type

`Question 1.
If x = \(\frac { \sqrt { p+q } +\sqrt { p-q } }{ \sqrt{p+q}-\sqrt{p-q} }\), then prove that q² – 2px + 9 = 0.
Solution:


Squaring both sides, we have
⇒ q²x² + p² – 2pqx = p² – q²
⇒ q²x² – 2pqx + q² = 0
⇒ q(q² – 2px + q) = 0
⇒ qx² – 2px + q = 0 (∵ q ≠ 0)

Question 2.
If a = \(\frac{1}{3-\sqrt{11}}\) and b = \(\frac{1}{a}\), then find a² – b²
Solution:

Question 3.
Simplify:

Solution:

Question 4.
Prove that:

Solution:

Question 5.
Find a and b, if

Solution:

Number Systems Class 9 Extra Questions HOTS

Question 1.
If x^a = y, y^b = z and z^c = x, then prove that abc = 1.
Solution:
We have x^a = y, y^b = z and z^c = x

Question 2.
Prove that:

Solution:
Taking L.H.S., we have

Question 3.
Show that:

Solution:

Number Systems Class 9 Extra Questions Value Based (VBQs)

Question 1.
Sudhir and Ashok participated in a long jump competition along a straight line marked as a number line. Both start the jumps one by one but in opposite directions. From ‘O’ Ashok jumps one unit towards the positive side while Sudhir jumps double in units as Ashok jumps, along negative side. After jumping 4 jumps each, at which point Ashok and Sudhir reached. What is the distance between their final positions ? Ashok argue that he is the winner since Sudhir is at negative side. Who do you think is winner and why? What is the value of the competition ?

Solution:
After jumping four jumps each, Ashok reached at 4 in positive direction and Sudhir reached at -8 i.e., in negative direction. Distance between their final positions is 12 units. Here, distance covered by Sudhir is 8 units and distance covered by Ashok is 4 units. Thus, Sudhir is the winner. Competition inculcate spirit of performance.

Question 2.
Manu went to his mathematics teacher and asked him “Sir, I want some chocolates to distribute among my classmates for my birthday but I have no money. Can you provide me some chocolates”. Teacher told Manu, I am giving you two numbers \(\frac{1}{3+2 \sqrt{2}}\) and \(\frac{1}{3-2 \sqrt{2}}\) and if you can find the value of sum of their squares, then I will provide you as many chocolates as the resulting value of sum of squares of given numbers. Find the number of chocolates. What value is depicted from this action?
Solution:

= (3 – 2√2)² + (3 + 2√2²
= 9 + 8 – 2 × 3 × (2√2) + 9 + 8 + 2 × 3 × 2√2 = 34.
Hence, resulting value of sum of squares of numbers = number of chocolates = 34. By doing this, teacher motivates the students to use their knowledge and apply it in day to day life with caring and kindness.