Category: CBSE Class 12

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 5 Magnetism and Matter. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 5 Important Extra Questions Magnetism and Matter

Magnetism and Matter Important Extra Questions Very Short Answer Type

Question 1.
A small magnetic needle pivoted at the center is free to rotate In a magnetic meridian. At what place will the needle
be vertical?
Answer:
At the potes

Question 2.
What is the angle of dip at a place where the horizontal and vertical components of the earth’s magnetic field are equal?
Answer:
450

Question 3.
How does the intensity of a paramagnetic sample vary with temperature?
Answer:
it decreases with the increase in temperature.

Question 4.
What should be the orientation of a magnetic dipole in a uniform magnetic field so that its potential energy is maximum?
Answer:
It should be anti-parallel to the applied magnetic field.

Question 5.
What is the value of angle of dip at a place on the surface of the earth where the ratio of the vertical component to the horizontal component of the earth’s magnetic field is 1/\(\sqrt{3}\)?
Answer:
Using the expression tan δ = \(\frac{B_{V}}{B_{H}}=\frac{1}{\sqrt{3}}\)
Therefore, δ = 30°

Question 6.
Where on the surface of the earth is the angle of dip 90°? (CBSE Al 2011)
Answer:
Poles.

Question 7.
Where on the surface of the earth is the angle dip zero? (CBSE Al 2011)
Answer:
Magnetic equator

Question 8.
What are permanent magnets? Give one example. (CBSE Delhi 2013)
Answer:
It is an arrangement that has a permanent dipole moment, e.g. bar magnet.

Question 9.
At a place, the horizontal component of the earth’s magnetic field is B, and the angle of dip is 60°. What is the value of the horizontal component of the earth’s magnetic field at the equator? (CBSE Delhi 2017)
Answer:
Zero.

Question 10.
Is the steady electric current the only source of the magnetic field? Justify your answer. (CBSE Delhi 2013C)
Answer:
No, the magnetic field is also produced by alternating current.

Question 11.
Write one important property of a paramagnetic material. (CBSEAI2019)
OR
Do the diamagnetic substances have a resultant magnetic moment in an atom in the absence of an external magnetic field?
Answer:
It moves from the weaker to the stronger regions of the magnetic field.
OR
No

Question 12.
Steel is preferred for making permanent magnets, whereas soft iron is preferred for making electromagnets. Give one reason.
Answer:

1. The Retentivity of steel is more than that of soft iron.
2. Soft iron has a smaller value of coercivity than steel.

Question 13.
Define angle of inclination at a given place due to the earth’s magnetic field. (CBSE 2019C)
Answer:
The angle of inclination is the angle at which the earth’s magnetic field at a given place makes with the horizontal line in the magnetic meridian.

Question 14.
Relative permeability of a material p = 0.5. Identify the nature of the magnetic material and write its relation to magnetic susceptibility. (CBSE Delhi 2014C)
Answer:
Paramagnetic.
Susceptibility is small and greater than one.

Magnetism and Matter Important Extra Questions Short Answer Type

Question 1.
(a) Define the term magnetic susceptibility and write its relation in terms of relative magnetic permeability.
Answer:
It refers to the ease with which a substance can be magnetized. It is defined as the ratio of the intensity of magnetization to the magnetizing field. The required relation is µ_r = 1 + χ_m

(b) Two magnetic materials A and B have relative magnetic permeabilities of 0. 96 and 500. Identify the magnetic materials A and B. (CBSE Al, Delhi 2018C)
Answer:
A: Paramagnetic,
B: Ferromagnetic

Question 2.
A magnetic needle free to rotate in a vertical position orients itself with its axis vertical at a certain place on the earth. What are the values of
(a) the angle of dip and
(b) the horizontal component of the earth’s magnetic field at this place? Where will this place be on the earth?
Answer:
The angle of dip is 90° and the horizontal component of the earth’s magnetic field is zero. This place is the magnetic pole of the earth.

Question 3.
Out of the two magnetic materials ‘A’ has relative permeability slightly greater than unity while ‘B’ has less than unity. Identify the nature of the materials ‘A’ and ‘B’. Will their susceptibilities be positive or negative? (CBSE Delhi 2014)
Answer:

• ‘A’ is paramagnetic and ‘B’ is diamagnetic.
• ‘A’ will have positive susceptibility while
• ‘B’ will have negative susceptibility.

Question 4.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its northern tip down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place. (CBSE Delhi 2011)
Answer:
Given δ = 30°, B_H = 0.4 G, B = ?
Using the expression
B_H = Bcos δ we have
B = \(\frac{B_{H}}{\cos \delta}=\frac{0.4}{\cos 30^{\circ}}=\frac{0.4}{\sqrt{3} / 2}=\frac{0.8}{\sqrt{3}}\) G

Question 5.
The susceptibility of a magnetic material is -0.085. Identify the type of magnetic material. A specimen of this material is kept in a non-uniform magnetic field. Draw the modified field pattern.
Answer:
The material is a diamagnetic material as diamagnetic materials have negative susceptibility. The modified field pattern is as shown below.

Question 6.
A uniform magnetic field gets modified as shown below when two specimens X and Y are placed in it.

(a) Identify the two specimens X and Y.
Answer:
X is a diamagnetic substance and Y is a paramagnetic substance.

(b) State the reason for the behavior of the field lines in X and Y.
Answer:
This is because the permeability of a diamagnetic substance is less than one and that of a paramagnetic substance is greater than one.

Question 7.
Three identical specimens of magnetic materials nickel, antimony, and aluminum are kept in a non-uniform magnetic field. Draw the modification in the field lines in each case. Justify your answer.
Answer:
Nickel is ferromagnetic, antimony is diamagnetic and aluminium is paramagnetic. Therefore, they will show the behavior as shown in the following figures.

Question 8.
Define neutral point. Draw lines of force when two identical magnets are placed at a finite distance apart with their N-poles facing each other. Locate the neutral points.
Answer:
It is a point near a magnet where the magnetic field of the earth is completely balanced by the magnetic field of the magnet. The figure is as shown below.

The cross indicates the neutral point.

Question 9.
The susceptibility of a magnetic material is -2.6 x 10^-5. Identify the type of magnetic material and state its two properties. (CBSE Delhi 2012)
Answer:
Diamagnetic:

1. Very small and negative susceptibility
2. Permeability is less than one.

Magnetism and Matter Important Extra Questions Long Answer Type

Question 1.
Write the expression for the magnetic dipole moment for a closed current loop. Give its SI unit. Derive an expression for the torque experienced by a magnetic dipole in a uniform magnetic field.
Answer:
The required expression is m = nIA.
It is measured in A m².

Consider a uniform magnetic field of strength B. Let a magnetic dipole be suspended in it such that its axis makes an angle 6 with the field as shown in the figure below. If ‘m’ is the strength of each pole, the two poles experience two equal and opposite force ‘B’ each. These forces constitute a couple that tends to rotate the dipole. Suppose the couple exerts a torque of magnitude τ.

Then

τ = either force × arm of the couple
= mB × AN = mB × 2 L sin θ
or
Since m × 2L is the magnetic dipole moment of the magnet.
Therefore τ = MB sin θ in vector form

we have \(\vec{τ}\) = \(\vec{M}\) × \(\vec{B}\)

Question 2.
(a) State Gauss’s law for magnetism. Explain Its significance.
Answer:
(a) Gauss’s Law for magnetism states that “The total flux of the magnetic field, through any closed surface, is aLways
zero, i.e. ∮\(\vec{B}\).\(\vec{dL}\) = 0

This law implies that magnetic monopoles do not exist” or magnetic field lines form closed loops.

(b) Write the four Important properties of the magnetic field lines due to a bar magnet. (CBSE Delhi 2019)
Answer:
Four properties of magnetic field lines are as follows:

1. Magnetic field lines always form continuous closed loops.
2. The tangent to the magnetic field line at a given point represents the direction of the net magnetic field at that point.
3. The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field.
4. Magnetic field lines do not intersect.

Question 3.
A short bar magnet placed with its axis at 30° with an external field of 800 G experiences a torque of 0.016 Nm.
(a) What is the magnetic moment of the magnet?
Answer:
(a) From equation τ = mB sin θ,
For θ = 30°, sin θ = 0.5

Thus, 0.016 = m × 800 × 10^-4 × 0.5 or m = 160 × 2/800 = 0.40 Am²

(b) What is the work done in moving it from its most stable to a most unstable position?
Answer:
From equation U= -mB cos θ, the most stable position is for θ = 0° and the most unstable position is for θ = 180°.

Work done is given by
W = U_m (θ = 180°) – Um for (θ = 0°)
= 2 mB = 2 × 0.40 × 800 × 10^-4
= 0.064 J

(c) The bar magnet is replaced by a solenoid of cross-sectional area 2 × 10^-4 m² and 1000 turns but of the same magnetic moment. Determine the current flowing through the solenoid. (NCERT Delhi 2005C)
Answer:
From part (i), m = 0.40 Am² Using equation m = NIA.
0.40 = 1000 × l × 2 × 10^-4
l = 2 A

Question 4.
(a) What happens if a bar magnet is cut into two pieces: (i) transverse to its length and (ii) along its length?
Answer:
(a) In either case one gets two magnets each with a north pole and a south pole.

(b) A magnetized needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to torque. Why?
Answer:
The needle experiences no net force because the field is uniform.

The iron nail experiences a non¬uniform field due to the bar magnet. There is an induced magnetic moment in the nail; therefore, it experiences both force and torque.

The net force is attractive because the induced south pole (say) in the nail is closer to the north pole of the magnet than the induced north pole.

Question 5.
Name the elements of the earth’s magnetic field at a place. Explain their meaning.
Answer:
The elements required to completely specify the earth’s magnetic field are called magnetic elements. These elements are:
(a) Declination (θ)
(b) Dip or inclination (δ)
(c) Horizontal component of the earth’s magnetic field (B_H).

(a) Declination (θ): Declination at a place is defined as the acute angle between the magnetic meridian and the geographic meridian.
In the figure above ∠BAB’ = θ is called the angle of declination. There are several periodic variations in the value of declination for a given location.

(b) Dip (δ): Dip at a place is defined as the angle between the direction of the total intensity of the earth’s magnetic field and a horizontal line in the magnetic meridian at that place. It is denoted by 8.

Suppose a freely suspended magnet aligns itself along the direction AB as shown in the figure, then this gives the direction of the earth’s total magnetic field at that place. Then the angle ∠BAO = δ is called the angle of dip or simply dip at that point.

(c) Horizontal component of the earth’s magnetic field (BH): It is the component of the total intensity of the earth’s magnetic field along a horizontal line in the magnetic meridian. It is denoted by BH or H.

Question 6.
Name the three types of magnetic materials which behave differently when placed in a non-uniform magnetic field. Give two properties for each of them.
Answer:
The three types of magnetic materials are
(a) Diamagnetic
(b) Paramagnetic and
(c) Ferromagnetic

Diamagnetic	Paramagnetic	Ferromagnetic
These are the substances that are feebly repelled by a magnet, e.g., Bi, Zn, Cu, Ag, Au, diamond, C, NaCI, H20, Hg, N2, H2, etc. Exhibited by solids, liquids, and gases.	These are the substances that are feebly attracted by a magnet, e.g., AL, Na, Pt, Mn, CuCL2, O2, etc. Exhibited by solids, liquids, and gases.	These are the substances that are strongly attracted by a magnet, e.g., Fe, Ni, Co, Fe304, etc. Exhibited by solids only, that too crystalline.
When a diamagnetic substance is placed in a magnetizing field, the Lines of force prefer not to pass through the substance.	When a paramagnetic substance is placed in a magnetizing field, the Lines of force prefer to pass through the substance.	When a ferromagnetic substance is placed in a magnetizing field, the Lines of force prefer to pass through the substance.

Question 7.
The susceptibility of a magnetic material is 0-9853. Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field. (CBSEAI, Delhi 2018)
Answer:
We know that the susceptibility of a paramagnetic substance is positive but small. Hence, the given magnetic material is paramagnetic in nature.

When a specimen of a paramagnetic substance is placed in a magnetizing field, the magnetic field lines prefer to pass through the specimen rather than through air. Thus, magnetic induction 6 inside the sample is more than the magnetic induction B0 outside the sample.

Question 8.
(a) Draw the magnetic field lines due to a circular loop of area \(\vec{A}\) carrying current l. Show that it acts as a bar magnet of magnetic moment \(\vec{m}\) = I \(\vec{A}\).
Answer:
(a) The magnetic field lines are as shown.

Magnetic field due to circular loop on its axis at far off points B = \(\frac{\mu_{0}}{4 \pi} \frac{2 l A}{x^{3}}\)

Magnetic field due to a bar magnet at its axial point is B = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{x^{3}}\)
Comparing the above two we have m = IA

(b) Derive the expression for the magnetic field due to a solenoid of length ‘21’, radius ‘a’ having ‘n’ number of turns per unit length and carrying a steady current T at a point on the axial line, distant ‘r’ from the center of the solenoid. How does this expression compare with the axial magnetic field due to a bar magnet of the magnetic moment ‘m’? (CBSEAI 2015)
Answer:
(b) By the-Biot-Savarts law, the magnetic field at point P on the axis of the solenoid due to a small element of length dx at a distance ‘x’ from the center of the solenoid is

dB = \(\frac{\mu_{0} n d x / a^{2}}{2\left[(r-x)^{2}+a^{2}\right]^{3 / 2}}\)

Hence total magnetic field is given by
B = ∫dB = \(\frac{\mu_{0} / a^{2} n^{+L}}{2} \int_{-L}^{d x} \frac{d x}{\left[(r-x)^{2}+a^{2}\right]^{3 / 2}}\)

For r >> a (R >> L)
We have
B = ∫dB = \(\frac{\mu_{0} / a^{2} n^{+L}}{2 r^{3}} \int_{-L}^{+L} d x=\frac{\mu_{0} \ln }{2} \frac{2 L a^{2}}{r^{3}}\)

Magnetic moment of a solenoid
m = (n × 2L) I (πa²)

Therefore,
B = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}\) as that of a bar magnet.

Numerical Problems :

• Torque on a magnetic dipole τ = MB sin θ
• Potential energy of a magnetic dipole U = – MB cos θ
• \(B_{H}^{2}+B_{v}^{2}=B^{2}\)
• tan θ = \(\frac{B_{V}}{B_{H}}\)
• H = B cos δ
• V = B sin δ

Question 1.
A bar magnet of the magnetic moment 6 J T¹ is aligned at 60° with a uniform external magnetic field of 0 – 44 T. Calculate
(a) the work is done in turning the magnet to align its magnetic moment (i) normal to the magnetic field and (ii) opposite to the magnetic field and
Answer:

(b) the torque on the magnet in the final orientation in case (ii). (CBSEAI, Delhi 2018)
Answer:
τ = M × B sin θ₂
= 6 × 0.44 × sin 180° = 0

Question 2.
A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 x 10^-2 J. What is the magnitude of the magnetic moment of the magnet?
Answer:
Given θ = 30°, B = 0.25 T, τ = 4.5 × 10^-2 J,M = ?
Using the expression τ = MB sin θ we have
M = \(\frac{\tau}{B \sin \theta}=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}\) = 0.36 J T^-1

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 4 Chemical Kinetics. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 4 Important Extra Questions Chemical Kinetics

Chemical Kinetics Important Extra Questions Very Short Answer Type

Question 1.
Define rate of a reaction. (CBSE Delhi 2010)
Answer:
The rate of a reaction is the change of concentration in any one of the reactants or products per unit time.

Question 2.
Define order of a reaction. (CBSE Delhi 2010, CBSE 2011)
Answer:
The sum of the powers of the concentration terms of the reactants in the rate law expression is called the order of that chemical reaction. For example, if rate law expression is
rate = K[A]^p[B]^q[C]^r
Order of reaction = p + q + r

Question 3.
For a chemical reaction R → P, the variation in the concentration (R) vs. time (t) plot is given as

(i) Predict the order of the reaction.
(ii) What is the slope of the curve? (CBSE 2014)
Answer:
(i) Zero order reaction
(ii) Slope = – k

Question 4.
Define ‘activation energy’ of a reaction. (CBSE Delhi 2011)
Answer:
Activation energy is the minimum energy that the reactants must possess for the conversion into the products during their collisions.
It is equal to difference between the threshold energy needed for the reaction and the average kinetic energy of the molecules.

Question 5.
Identify the order of a reaction if the units of its rate constant are:
(i) L^-1 mol s^-1
(ii) L mol^-1 s^-1 (CBSE All India 2011)
Answer:
(i) Zero order
(ii) Second order

Question 6.
For a reaction: A + B → P, the rate law is given by, r = k [A]^1/2 [B]² .
What is the order of reaction? (CBSE All India 2013)
Answer:
Order = \(\frac{1}{2}\) + 2 = 2\(\frac{1}{2}\) or 2.5.

Question 7.
In the Arrhenius equation, what does the factor e^-E₀/RT correspond to?
(CBSE Sample Paper 2017-18)
Answer:
e^-E₀/RT corresponds to the fraction of molecules having kinetic energy greater than E_a.

Chemical Kinetics Important Extra Questions Short Answer Type

Question 1.
For a reaction,
C₂H₄(g) + H₂(g) → C₂H₆(g),
rate = 5.5 × 10^-14 [C₂H₄].
(a) Write the unit of rate constant.
(b) Calculate its half-life (t_1/2). (CBSE 2019)
Answer:
(a) Hydrogenation of ethene is an example of first order reaction so the unit of rate constant is s^-1 or time^-1.

(b) t_1/2 = \(\frac{0.693}{k}=\frac{0.693}{5.5 \times 10^{-14}}\)
= 1.26 × 10¹³s
(or any other unit of time)

Question 2.
For a reaction

The proposed mechanism is as given below:
(1) H₂O₂ + I^– → H₂O + IO^– (slow)
(2) H₂O₂ + IO^– → H₂O + I^– + O₂ (fast)
(i) Write rate law for the reaction.
(ii) Write the overall order of reaction.
(iii) Out of steps (1) and (2), which one is rate-determining step? (CBSE Delhi 2019)
Answer:
(i) Rate = k[H₂O₂][I^–]
(ii) 2
(iii) Step 1 is rate determining.

Question 3.
A reaction is of first order in reactant A and of second order in reactant B. How is the rate of this reaction affected when
(i) the concentration of B alone is increased to three times?
(ii) the concentrations of A as well as B are doubled? (CBSE Delhi 2010)
Answer:
The rate law may be expressed as:
Rate = k[A] [B]²
(i) Rate becomes nine times
(ii) Rate becomes eight times.

Question 4.
What do you understand by the rate law and rate constant of a reaction? Identify the order of reaction if the units of its rate constant are:
(i) L^-1 mol s^-1
(ii) L mol^-1 s^-1 (CBSE 2011)
Answer:
Rate law is the representation of rate of reaction in terms of concentration of the reactants.
Rate constant is the rate of reaction when the concentration of reactants is unity. Order of reaction from rate constants
(i) Zero order
(ii) Second order

Question 5.
A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half? What is the unit of rate constant for such a reaction? (CBSE 2011)
Answer:
For a second order reaction,
rate = K[A]²
If the concentration of A is reduced to half, rate will become 1 /4 of the original value.
Units of k.

Question 6.
Distinguish between rate expression and rate constant of a reaction. (CBSE 2011)
Answer:
Rate expression is a mathematical expression which denotes the rate of a reaction in terms of molar concentration of reactants.

Rate constant is the rate of the reaction when the concentration of reactants is unity. It is proportionality constant in the rate law and is independent of initial concentrations of the reactants.

Question 7.
What do you understand by the ‘order of reaction’? Identify the reaction order from each of the following units of reaction rate constant:
(i) L^-1 mol s^-1
(ii) L mol^-1 s^-1 (CBSE Delhi 2012)
Answer:
The sum of the powers of the concentrations of reactants in the rate law expression is called order of a reaction. For example, if rate law expression for a reaction is:
rate = K[A]^a[B]^a[C]c
then order = a + b + c
(i) Zero order
(ii) Second order

Question 8.
(i) For a reaction A + B → P, the rate law is given by,
r = k[A]^1/2 [B]²
What is the order of this reaction?
(ii) A first order reaction is found to have a rate constant k = 5.5 × 10^-14 s^-1. Find the half life of the reaction. (CBSE 2013)
Answer:
(i) Order of reaction 1
= \(\frac{1}{2}\) + 2 = 2.5

(ii) t_1/2 = \(\frac{0.693}{k}=\frac{0.693}{5.5 \times 10^{-14}}\)
= 1.26 × 10¹³s

Question 9.
Write two differences between ‘order of reaction’ and ‘molecularity of reaction. (CBSE Delhi 2014)
Answer:

Order	Molecularity
1. Order is the sum of the powers of the concentration terms in the rate law expression.	1. Molecularity is the number of reacting species undergoing simultaneous collisions in the elementary or simple reaction.
2. Order of a reaction is determined experimentally and need not be a whole number, i.e. it can have fractional values also.	2. Molecularity is a theoretical concept and has whole number values only, i.e. 1, 2, 3, etc.

Question 10.
The rate constant for a reaction of zero order in A is 0.0030 mol L^-1s^-1. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M? (CBSE 2010)
Answer:
For a zero order reaction,
k = 0.0030 mol L^-1s^-1
[A]₀ = 0.10 M, [A] = 0.075M

Answer:

Question 11.
The rate of a reaction depends upon the temperature and is quantitatively expressed as k = \(A e^{-E_{a} / R T}\)
(i) If a graph Is plotted between log k and 1/T, write the expression for the slope of the reaction.
(ii) If at under different conditions, E_a1 and E_a2 are the activation energy of two reactions. If E_a1 = 40 J / mol and E_a2 = 80 J / mol which of the two has a larger value of the rate constant? (CBSE Sample Paper 2019)
Answer:
(ii) Arrhenius proposed a quantitative relationship between rate constant
and temperature as k = \(A e^{-E_{a} / R T}\) Taking logarithm,
In k = In A – \(\frac{E_{a}}{R} \times \frac{1}{T}\)
Converting to common logarithm (In X = 2.303 log X), we get
2.303 log k = 2.303 log A – \(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}\).
Dividing each side by 2.303, we get
log k = log A – \(\frac{E_{a}}{2.303 R T}\)
So, it is clear that increasing the temperature or decreasing the activation energy will result in an increase in the rate of reaction.

(i) Slope = – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\)

Plot of log k versus 1/T to calculate activation energy.

(ii) k₁ > k₂

Question 12.
The rate constant for a first order reaction is 60 s^-1. How much time will it take to reduce the concentration of the reactant to 1/10th of its initial value?
Answer:
For a first order reaction,

Question 13.
Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
log k = log A – \(\frac{E_{a}}{2.303 R}\left(\frac{1}{T}\right)\)
where E_a is the activation energy. When a graph is plotted for log k vs. \(\frac{1}{T}\), a straight line with a slope of – 4250 K is obtained. Calculate ‘E_a’ for the reaction (R = 8.314 JK^-1mol^-1). (CBSE Delhi 2013)
Answer:
Slope = –\(\frac{E_{a}}{2.303 R}\) = -4250K
E_a = – 2.303 × R × Slope
= – 2.303 × 8.314 JK^-1 mol^-1 × (-4250 K)
= 81375.3 J mol^-1 or 81.375 kJ mol^-1.

Question 14.
For the reaction
C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆
write:
(i) Rate of reaction expression,
(ii) Rate law equation,
(iii) Molecularity,
(iv) Order of reaction (CBSE Sample paper 2011)
Answer:
(i)

(ii) Rate law equation:
Rate = k[C₁₂H₂₂O₁₁]
(iii) Molecularity = 2
(iv) Order = 1

Question 15.
Consider the decomposition of hydrogen peroxide in alkaline medium which is catalysed by iodide ions.

This reaction takes place in two steps as given below:
Step -1 H₂O₂ + I^– → H₂O + IO^– (slow)
Step – II H₂O₂ + IO^– → H₂O + I^– (fast)
(i) Write the rate law expression and determine the order of reaction w.r.t. H₂O₂.
(ii) What is the molecularity of each individual step? (C8SE Sample paper 2011)
Answer:
(i) Rate = k[H₂O₂] [I^–]
Order w.r.t. H₂O₂ = 1
(ii) Molecularity: Step I = 2, Step II = 2

Question 16.
For a reaction: A + B → P, the rate law is given as: Rate = k[A][B]²
(i) How is the rate of reaction affected when the concentration of B is doubled?
(ii) What is the order of reaction if A is present in large excess? (CBSE All India 2015)
Answer:
(i) Rate becomes four times
(ii) Order = 2

Question 17.
What is the effect of adding a catalyst on
(i) Activation energy (E_a) and
(ii) Gibbs free energy (∆G) of a reaction? (CBSE All India 2017)
Answer:
(i) Activation energy decreases
(ii) No effect on ∆G because Gibbs free energies of products and reactants remain the same in the presence of catalyst.

Question 18.
For a reaction:

Rate = k
(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k. (CBSE Sample paper 2018)
Answer:

Rate = k
(i) Order of reaction: Zero order Molecularity = 2 (bi molecular)
(ii) Unit of k = mol L^-1 s^-1 or atm s^-1

Question 19.
Explain the following terms: (CBSE 2014)
(i) Rate constant (k).
(ii) Half life period of a reaction (t_1/2).
Answer:
(i) Rate constant is the rate of the reaction
when the molar concentration of each of the reactants is unity. For example, for a reaction,
A + B → AB
Rate of reaction is
Rate = k[A][B]
where k is the rate constant. When
[A] = 1 and [B] = 1
Rate of reaction = k × 1 × 1 = k
It is also catted specific reaction rate and is independent of the initial concentration of the reactants.

(ii) Half life period of a reaction is the time during which the concentration of a reactant Is reduced to half of its initial concentration. In other words, it is the time during which half of the reaction Is completed. It is denoted as t_1/2.

Chemical Kinetics Important Extra Questions Long Answer Type

Question 1.
Show that time required for 99% completion is twice the time required for the completion of 90% reaction. (CBSE Delhi 2013)
Answer:
For the first order reaction,
t = \(\frac{2.303}{k} \log \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]}\)
Let initial concentration, [A₀] = a
For 99% completion of reaction, [A]

For 90% completion of reaction,

Dividing equation (i) by (ii)
\(\frac{t_{(99 \%)}}{t_{(90 \%)}}\) = 2
or t_(99%) = 2 × t_(90%)

Question 2.
The decomposition of NH₃ on platinum surface is zero order reaction. If rate constant (k) Is 4 × 10^-3 M s^-1, how long will it take to reduce the initial concentration of NH₃ from 0.1 M to 0.064 M? (CBSE Delhi 2019)
Answer:
k = 4 × 10^-3M s^-1, [A]₀ = 0.1 M, [A] = 0.064 M

Question 3.
The thermal decomposition of HCO₂H is a first order reaction with a rate constant of 2.4 × 10^-3 s^-1 at a certain temperature. Calculate how long will it take for three fourth of initial quantity of HCO₂H to decompose. (log 4 = 0.6021) (CBSE 2011)
Answer:
We know that
t = \(\frac{2.303}{k} \log \frac{[A]_{0}}{[A]}\)
If initial conc. of A is a, then
[A] = a – \(\frac{3}{4}\)a = \(\frac{1}{4}\)a
k = 2.4 × 10^-3 s^-1

= \(\frac{2.303}{2.4 \times 10^{-3}}\) × 0.6021 = 578 s

Question 4.
For the reaction:
2N0(g) + Cl₂(g) → 2NOCl(g)
the following data were collected. All the measurements were taken at 263 K:

Experiment No.	Intial [NO] (M)	Intial [Cl2] (M)	Intial rate of Cl2 (M/min)
1	0.15	0.15	0.60
2	0.15	0.30	1.20
3	0.30	0.15	2.40
4	0.25	0.25	?

(i) Write the expression for rate law.
(ii) Calculate the value of rate constant and specify its units.
(iii) What is the initial rate of disappearance of Cl₂ In exp. 4? (CBSE 2012)
Answer:
(i) Rate = k[NO]²[Cl₂]
(ii) According to experiment 1,
[NO] = 0.15 M; [Cl₂] = 0.15 M
Rate = 0.6 M/min.
Rate = k[NO]²[Cl₂]
0.6 M/min = k(0.15M)² (0.15 M)
k = 177.8 M^-2 min^-1
or
= 177.8 mol^-2 L² min^-1
Therefore, units of k are M^-2 min^-1 or mol^-2 L² min^-1.

(iii) In experiment 4, [NO] = 0.25 M; [Cl₂] = 0.25 M and k = 177.8 M^-2 min^-1
Rate = k[NO]² [Cl₂]
= 177.8 M^-2 min^-1 × (0.25 M)² × 0.25 M
= 2.78 M min^-1.

Question 5.
The following data were obtained during the first order thermal decomposition of
SO₂Cl₂ at a constant volume:
SO₂Cl₂(g) → SO₂(g) + Cl₂(g)

Experiment	Time/s-1	Total Pressure/atm
1	0	0.4
1	100	0.7

Calculate the rate constant.
(Given: log 4 = 0.6021, log 2 = 0.3010) (CBSE Delhi 2014)
Answer:
SO₂Cl₂ → SO₂ + Cl₂
At t = 0 s     0.4 atm            0            0
At t = 100 s (0.4 – x) atm   x atm   x atm
P_t = 0.4 – x + x + x = 0.4 + x
0.7 = 0.4 + x      or    x = 0.3
k = \(\frac{2.303}{t}\) log \(\frac{p_{0}}{2 p_{0}-p_{t}}\)
t = 100 s
∴ k = \(\frac{2.303}{100}\) log \(\frac{0.4}{2 \times 0.4-0.7}\)
= \(\frac{2.303}{100}\) log \(\frac{0.4}{0.1}\)
∴ k = \(\frac{2.303}{100}\) × 0.6021 = 1.38 × 10^-2 s^-1

Question 6.
The rate constant for the first order decomposition of H₂O₂ is given by the following equation:
log k = 14.2 – \(\frac{1.0 \times 10^{4} \mathrm{~K}}{\mathrm{~T}}\)
Calculate E_a for this reaction and rate constant k if Its half-life period be 200 minutes. (Given: R = 8.314 JK^-1 mol^-1) (CBSE Delhi 2016)
Answer:
According to Arrhenius equation,
log k = log A – \(\frac{E_{a}}{2.303 \mathrm{RT}}\) …….. (i)
The given equation is
log k = 14.2 – \(\frac{1.0 \times 10^{4} \mathrm{~K}}{\mathrm{~T}}\) ………. (ii)
Comparing eq. (1) and (ii)
\(\frac{E_{a}}{2.303 \mathrm{RT}}\) = 1.0 × 10⁴
∴ E_a = 1.0 × 10⁴ × 2.303 × 8.314
= 19.147 × 10⁴ J mol^-1
Now t_1/2 = \(\frac{0.693}{k}\)
or k = \(\frac{0.693}{t_{1 / 2}}\)
= \(\frac{0.693}{200 \mathrm{~min}}\) = 3.465 × 10^-3 min ^-1

Question 7.
The rate of a reaction increases four times when the temperature changes from 300 K to 320 K. Calculate the activation energy of the reaction assuming that it does not change with temperature (R = 8.3 14 JK^-1 mol^-1). (CBSE All India 2010)
Answer:
T₁ = 300 K, T₂ = 320 K, R = 8.314 JK^-1 mol^-1

Question 8.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours? (CBSE Sample Paper 2011)
Answer:
k = \(\frac{0.693}{t_{1 / 2}}\) = \(\frac{0.693}{3.0}\) = 0.231 hr^-1
Let initial cone, of sucrose = 1 M
Cone, after 8 hr. = (1 – x) where x is the amount of sucrose decomposed

Conc. of sucrose left after 8 hr = 1 – 0.842
= 0.158 M

Question 9.
The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E_a) of the reaction assuming that it does not change with temperature.
[R = 8.314 J K^-1 mol^-1, log 4 = 0.6021] (CBSE All India 2013)
Answer:

Question 10.
The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at constant volume.
SO₂Cl₂(g) → SO₂(g) + Cl₂(g)

Experiment	Times/s	Total Pressure/atm
1	0	0.5
2	100	0.6

Calculate the rate of reaction when total pressure is 0.65 atm. (CBSE Sample Paper 2011)
Answer:
For the decomposition of SO₂Cl₂
SO₂Cl₂(g) → SO₂(g) + Cl₂(g)
At t = 0          p₀ atm              0        0
At time, t      (p₀ – x) atm       x        x
p_t = p₀ – x + x + x = p₀ + x
or x = p_t – p₀
Now at time t, p(SOCl₂) = p₀ – x
= P₀ – (P_t – P₀) = 2 p₀ – p_t
∴ k = \(\frac{2.303}{t}\) log \(\frac{p_{0}}{2 p_{0}-p_{t}}\)
when t = 100 s
k = \(\frac{2.303}{100}\) log \(\frac{0.5}{2 \times 0.5-0.6}\)
= \(\frac{2.303}{100}\) log 1.25 = 2.2316 × 10^-3 s^-1
When p_t = 0.65 atm, i.e. p₀ + p = 0.65 atm
p = 0.65 – p₀ = 0.65 – 0.50 = 0.15 atm
∴ Pressure of SO₂Cl₂ at time t = 0.50 – 0.15
= 0.35 atm
Rate = 2.2316 ×10^-3 × 0.35
= 7.8 × 10^-5 atm s^-1

Question 11.
A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed. {CBSE All India 2017)
Answer:
For the first order reaction:

Question 12.
A first order reaction takes 23.1 minutes for 50% completion. Calculate the time required for 75% completion of this reaction. (log 2 = 0.301, log3 0.4771, log4 = 0.6021) (CBSE All India 2015)
Answer:
For the first order reaction:

Question 13.
The decomposition of phosphine, PH₃, proceeds according to the following equation:
4PH₃(g) → P₄(g) + 6H₂(g)
It is found that the reaction follows the following rate equation:
Rate = k[PH₃].
The half-life of PH₃ is 37.9 sat 120 C.
(i) How much time is required for 3/4th of PH₃ to decompose?
(ii) What fraction of the original sample of PH₃ remains behind after 1 minute? (C8SE All India 2010)
Answer:

Question 14.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with t_1/2 = 3 hrs. Calculate the fraction of sucrose which remains after 8 hrs. (CBSE Sample Paper 2011)
Answer:

Question 15.
Starting from 10 g of a radioactive element, 0.25 g was left after 5 years. Calculate
(i) Rate constant for the decay of the radioactive element.
Answer:

(ii) The amount left after one year.
Answer:
Amount left after 1 year

(iii) The time required for half of the element to decay.
Answer:
Time for decay of half of the element
t_1/2 = \(\frac{0.693}{k}=\frac{0.693}{0.7379}\) = 0.9392 year

(iv) Average life of the element. (CBSE All India 2011)
Answer:
t = \(\frac{1}{k}=\frac{1}{0.7379}\) = 1.3552 year

Question 16.
In general, It is observed that the rate of a chemical reaction becomes double with every 1o°. rise in temperature. If this generalization holds for a reaction In the temperature range 298 to 308 K, what would be the value of activation energy for this reaction? (R = 8.3 14 JK^-1 mol^-1) (CBSE Delhi 2000)
Answer:
Arrhenius equation is:

Question 17.
A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction.
(Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.3 14 Jk^-1 mol^-1) (CBSE All India 2018)
Answer:
Let us first calculate k₁ and k₁ at temperatures 300 K and 320 K. We know that

Question 18.
The rate constant of a first order reaction increases from 2 × 10^-2 to 6 × 10^-2 when the temperature changes from 300 K to 320 K. Calculate the energy of activation. (Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) (CBSE 2019 C)
Answer:

Question 19.
For the reaction A + B → products, the following initial rates were obtained at various given initial concentrations:

[A] mol / L	[B] mol / L	Intial rate M/S
1. 0.1	0.1	0.05
2. 0.2	0.1	0.10
3. 0.1	0.2	0.05

Determine the half-life period.
OR
A first order reaction is 50% complete In 50 minutes at 300 K and the same reaction is again 50% complete In 25 minutes at 350 K. Calculate activation energy of the reaction. (CBSE Sample Paper 2019)
Answer:
Rate = k [A]^x [B]^y
0.05 = k [0.1]^x [0.1]^y …… (i)
0.10 = k [0.2]^x [0.1]^y ……. (ii)
0.05 = k [0.1]^x [0.2]^y …….. (iii)
Dividing equation (ii) by (i)
\(\frac{0.10}{0.05}\) = (2)^x
x = 1
Dividing equation (iii) by (i)
\(\frac{0.05}{0.05}\) = (2)^y
y = 0
Rate = k [A]¹ [B]⁰

Question 20.
For a certain chemical reaction variation in concentration [A] vs. time (s) plot as shown in given figure:

(i) Predict the order of the given reaction?
(ii) What does the slope of the line and intercept indicate?
(iii) What is the unit of rate constant k? (CBSE Sample paper 2018-19)
Answer:
(i) Zero order reaction
(ii) Slope represents k; Intercept represents [R]₀
(iii) mol L^-1 s^-1

Question 21.
The following data were obtained during the first order thermal decomposition of N₂O₅(g) at a constant volume:
2N₂O₅(g) → 2N₂O₄(g) + O₂(g)

S.NO	Time (sec.)	Total pressure (atm)
1.	0	0.5
2.	100	0.512

Calculate the rate constant.
OR
Two reactions of the same order have equal pre-exponential factors but their activation energies differ by 24.9 kJ mol^-1. Calculate the ratio between the rate constants of these reactions at 27°C. (Gas constant R 8.314 J K^-1 mol^-1) (CBSE 2018)
Answer:
2N₂O₅(g) → 2N₂O₄(g) + O₂(g)
At t = 0       0.5 atm       0 atm      0 atm
At time t   0.5-2x atm   2x atm    x atm
P_t = P_N2O5 + P_N2O5 + P_O2
= (0.5 – 2x) + 2x + x = 0.5 + x
x = p_t – 0.5
P_N2O5 = 0.5 – 2x
= 0.5 – 2(p_t – 0.5)
= 1.5 – 2p_t
At t = 100s; p_t = 0.512 atm
P_N2O5 = 1.5 – 2 × 0.512 = 0.47 atm

OR
The Arrhenius equation: k = Ae^-Ea/RT
Taking log on both sides:
log k = log A – \(\frac{E_{a}}{2.303 R T}\)
For reaction (i) log k₁ = log A – \(\frac{E_{o}(1)}{2.303 R T}\)
For reaction (ii) log k₂ = logA – \(\frac{E_{o}(2)}{2.303 R T}\)
Subtracting (i) from (ii)

Question 22.
(a) Define order of reaction. How does order of a reaction differ from molecularity for a complex reaction?
Answer:
(a) Order of reaction is the sum of the power to which the concentration terms are raised in the rate law equation. Order of a reaction is applicable for the complex reaction but molecularity has no meaning for the complex reaction. 0.693

(b) A first order reaction is 50% complete in 25 minutes. Calculate the time for 80% completion of the reaction.
Answer:

(a) The decomposition of a hydrocarbon has value of rate constant as 2.5 × 10⁴ s^-1 at 27 °C. At what temperature would rate constant be 7.5 × 10⁴ s^-1 if energy of activation is 19.147 × 10³ J mol^-1?
Answer:

0.1431 T₂ = T₂ – 300
0.1431 T₂ – T₂ = 300
0.8569T₂ = – 300
T₂ = \(\frac{300}{0.8569}\) = 350 k

(b) Write a condition under which a bimolecular reaction is kinetically first order. Give an example of such a reaction.
(Given: log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990) (CBSE AI 2019)
Answer:
A bimolecular reaction is kinetically of first order when one of the reactant is in excess. For example, hydrolysis of methyl acetate.
CH₃COOCH₃ + H₂O (excess) → CH₃COOH + CH₃OH

Question 23.
Nitrogen pentoxide decomposes according to equation:
2N₂O₅ (g) →4NO₂(g) + O₂(g)
The first order reaction was allowed to proceed at 40° C and the data below were collected:

[N2O5] (M)	Time (min)
0.400	0.00
0.289	20.0
0.209	40.0
0.151	60.0
0.109	80.0

(i) Calculate the rate constant. Include units with your answer.
Answer:
Since it is a first order reaction,
k = \(\frac{2.303}{t}\) log \(\frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]}\)
k can be calculated as:
at t = 20 min


The average value of rate constant,
k = 0.01624 min^-1 or = 1.624 × 10^-2 min^-1

(ii) What will be the concentration of N₂O₅ after 100 minutes?
Answer:
Concentration of [N₂O₅] at t = 100 min

(iii) Calculate the initial rate of reaction. (CBSE 2011)
Answer:
Initial rate of reaction,
rate = k[N₂O₅]
= 0.01624 × 0.40
= 6.496 × 10^-3

Question 24.
(i) Explain the following terms:
(a) Rate of a reaction
Answer:
The rate of reaction is the change of concentration in any one of the reactants or products per unit time. For a reaction, A → B rate may be expressed as
= \(\frac{-d[A]}{d t}\) or \(\frac{d[B]}{d t}\)

(b) Activation energy of a reaction
Answer:
Activation energy is the minimum energy which the reacting molecules should acquire so that they react to give the products. It is also defined as the energy required to form the products.

(ii) The decomposition of phosphine, PH₃, proceeds according to the following equation:
4PH₃(g) → P₄ (g) + 6H₂(g)
It Is found that the reaction follows the following rate equation:
Rate = k[PH₃]
The half life of PH₃ is 37.9 s at 120°C.
(a) How much time is required for 3/4th of PH₃ to decompose?
Answer:

(b) What fraction of the original sample of PH₃ remains behind after 1 minute? (CBSE 2010)
Answer:

Question 25.
Explain the following terms:
(a) Order of a reaction
Answer:
The sum of the powers of the concentration of reactants in the rate law expression is called the order of the reaction.
For example, if rate law expression for a reaction is:
Rate = k[A]^a[B]^b[C]^c
then order of reaction is
order = a + b + c

(b) Molecularity of a reaction
Answer:
Molecularity is the number of molecules taking part in a reaction. For example, for the reaction:
aA + bB → Product
Molecularity is a + b

(ii) The rate of a reaction Increases four times when the temperature changes from 300 K to 320 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature (R = 8.314 JK^-1 mol^-1). (CBSE 2010)
Answer:
T₁ = 300 k, T₂ = 320 k,
R = 8.314 Jk^-1 mol^-1

Question 26.
(i) A reaction is second order in A and first order in B.
(a) Write the differential rate equation.
(b) How is the rate affected on Increasing the concentration of A three times?
(C) How is the rate affected when the concentrations of both A and B are doubled?
Answer:
(a) Rate = k[A]²[B]
(b) 9 times
(C) 8 times

(ii) A first order reaction takes 40 minutes for 30% decomposition. Calculate t_1/2 for this reaction. (Given log 1.428 = 0.1548)
Answer:
Let initial conc. = a
Conc. after 40 min

OR

(i) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer:
For a first order reaction,
t = \(\frac{2.303}{k}\) log \(\frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]}\)
Let intial concentration, [A]₀ = a
For 99% completion of reaction,

Dividing equation (i) by (ii)
\(\frac{t_{(99 \%)}}{t_{(90 \%)}}\) = 2 or t_(99%) = 2 × t_(90%)

(ii) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
log k = log A – \(\frac{E_{a}}{2.303 R}\left(\frac{1}{T}\right)\)
where E_a is the activation energy. When a graph is plotted for log k vs. \(\frac{1}{\mathrm{~T}}\), a straight line with a slope
of – 4250 K is obtained. Calculate ‘E_a’ for the reaction (R = 8.3 14 JK^-1 mol^-1). (CBSE Delhi 2013)
Answer:
Slope = – \(\frac{E_{a}}{2.303 R}\) = – 4250 k
∴ E_a = 2.303 × 8.314 Jk^-1 mol^-1 × 4250 k
= 81.375 kJ mol^-1

Question 27.
(i) What is rate of reaction? Write two factors that affect the rate of reaction.
Answer:
Rate of reaction is defined as the change in concentration of reactants or products per unit time. Factors that affect rate of a reaction
(a) Concentration of reactant
(b) Temperature

(ii) The rate constant of a first order reaction increases from 4 × 10^-2 to 8 × 10^-2 when the temperature changes from 27 °C to 37°C. Calculate the energy of activation (E_a).
(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
Answer:
log \(\frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right]\)
k₁ = 4 × 10^-2,
k₂ = 8 × 10^-2,
T₁ = 273 + 27 = 300 K
T₂ = 273 + 37 = 310 K,
R = 8.314 JK^-1 mol^-1

= 53598 J mol^-1
= 53.598 KJ mol^-1

OR

(i) For a reaction A + B → P, the rate is given by.
Rate = k [A] [B]²
(a) How is the rate of reaction affected if the concentration of B is doubled?
(b) What is the overall order of reaction if A is present in large excess?
Answer:
Rate = K[A][B]²
(a) becomes four times
(b) Second order

(ii) A first order reaction takes 23.1 minutes for 50% completion. Calculate the time required for 75% completion of this reaction.
(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021) (CBSE 2015)
Answer:
For a first order reaction,

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 1 The Solid State. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 1 Important Extra Questions The Solid State

The Solid State Important Extra Questions Very Short Answer Type

Question 1.
A metal crystallises in a body centred cubic structure. If ‘a’ is the edge length of its unit cell, ‘r’ is the radius of the sphere. What is the relationship between ‘r’ and ‘a’? (CBSE Sample Paper 2017-18)
Answer:
r = \(\frac{\sqrt{3}}{4}\)a

Question 2.
If the radius of the octahedral void is ‘r’ and radius of the atoms in close packing is ‘R’. What is the relation between ‘r’ and ‘R’? (CBSE Sample Paper 2017-18)
Answer:
r = 0.414 R

Question 3.
How many octahedral voids are there in 1 mole of a compound having cubic closed packed structure? (CBSE Sample Paper 2007)
Answer:
4 mole

Question 4.
Write a feature which will distinguish a metallic solid from an ionic solid.
Answer:
Metallic solids are good conductors of heat and electricity whereas ionic solids are insulators in solid state but conductors in molten state and in aqueous solution.

Question 5.
Write a distinguishing feature of metallic solids.
Answer:
Metallic solids are good conductors of heat and electricity.

Question 6.
KF has ccp structure. Calculate the radius of the unit cell if the side of the cube or edge length is 400 pm. How many F ions and octahedral voids are there in the unit cell? (CBSE Sample Paper 2011)
Answer:
For ccp lattice,
r = \(\frac{a}{2 \sqrt{2}}=\frac{400}{2 \times 1.414}\) = 141.4 pm
There are four (4) F ions and four (4) octahedral voids per unit cell.

Question 7.
Which point defect in crystals lowers the density of a crystal?
Answer:
Schottky defect

Question 8.
What type of interactions hold the molecules together in a polar molecular solid?
Answer:
Dipole-dipole forces

Question 9.
Which stoichiometric defect in crystals increases the density of a solid? (CBSE Delhi 2012)
Answer:
Interstitial defect

Question 10.
What type of semiconductor is obtained when silicon is doped with arsenic? (C.B.S.E. 2010)
Answer:
n-type semiconductor

Question 11.
Define forbidden zone of an insulator. (C.B.S.E. Delhi 2008)
Answer:
The large energy gap between the filled valence band and the empty conduction band in an insulator is called forbidden zone.

Question 12.
How do metallic and ionic substances differ in conducting electricity? (C.B.S.E. 2009)
Answer:
Metallic solids conduct electricity in the solid state whereas ionic substances conduct electricity in molten state or in solution, or metallic substances conduct electricity through electrons while ionic substances conduct electricity through ions.

Question 13.
What is the coordination number of each type of ions in a rock salt type crystal structure? (C.B.S.E. Delhi 2008)
Answer:
C.N. of Na⁺ = 6, C.N. of Cl^– = 6.

Question 14.
What is meant by ‘doping’ in a semiconductor? (C.B.S.E. Delhi 2012)
Answer:
The process of introduction of small amounts of impurities in the lattice of the crystal is called doping.

Question 15.
How may the conductivity of an intrinsic semiconductor be increased? (C.B.S.E. Delhi 2012)
Answer:
The conductivity of an intrinsic semiconductor may be increased by adding an appropriate amount of suitable impurity. This process is called doping.

Question 16.
How many atoms constitute one unit cell of a face-centred cubic crystal? (C.B.S.E. Delhi 2013)
Answer:
Four

Question 17.
What type of magnetism is shown by a substance if magnetic moments of domains are arranged in the same direction? (C.B.S.E. Delhi 2016)
Answer:
Ferromagnetism

Question 18.
Out of NaCI and AgCI, which one shows Frenkel defect and why? (C.B.S.E. Delhi 2019)
Answer:
AgCI shows Frenkel defect. This is because of small size of Ag⁺ ions; these can fit into interstitial sites of Cl^– ions.

Question 19.
What type of stoichiometric defect is shown by ZnS and why? (C.B.S.E. Delhi 2019)
Answer:
Frenkel defect. This is caused due to the large difference in size of ions.

Question 20.
Why conductivity of silicon increases on doping with phosphorus? (CBSE Delhi 2019)
Answer:
When silicon is doped with phosphorus, an extra electron is introduced after forming four covalent bonds. This extra electron gets delocalised and serves to conduct electricity. Hence, conductivity increases.

The Solid State Important Extra Questions Short Answer Type

Question 1.
Calculate the number of unit cells in 8.1 g of aluminium if it crystallises in a face-centred cubic (fcc) structure. (Atomic mass of Al = 27 g mol^-1).
Answer:
Moles of aluminium = \(\frac{8.1}{27}\) mol
No. of atoms of Al in 8.1 g = \(\frac{8.1}{27}\) × 6.022 × 10²³
No. of atoms in fee unit cell = 4
No. of unit cells = \(\frac{8.1}{27} \times \frac{6.022 \times 10^{23}}{4}\)
= 4.5 × 10²²

Question 2.
Tungsten crystallises in body centred cubic unit cell. If edge of the unit cell is 316.5 pm, what is the radius of the tungsten atom? (CBSE Sample Paper 2017-18)
Answer:
If a is the edge length of bcc unit cell, then radius of an atom,
r = \(\frac{\sqrt{3}}{4}\)a
Here, a = 316.5 pm
∴ r = \(\frac{\sqrt{3}}{4}\) × 316.5 pm = 137.04 pm

Question 3.
Silver metal crystallises with a face centred cubic lattice. The length of the unit cell is found to be 4.077 × 10^-8 cm. Calculate atomic radius and density of silver. (Atomic mass of Ag = 108u, N_A = 6.02 × 10²³). (C.B.S.E. Sample Paper 2012)
Answer:
Edge length of unit cell, a = 4.077 × 10^-8 cm For fcc lattice, radius of an atom is related to edge length, a as:

Question 4.
Analysis shows that FeO has a non- stoichiometric composition with molecular formula Fe_0.950. Give reason.
Answer:
It shows metal deficiency defect. In FeO, some Fe²⁺ ions are replaced by Fe³⁺ ions. Three Fe²⁺ ions are replaced by two Fe³⁺ ions to maintain electrical neutrality.

Question 5.
Following is the schematic alignment of magnetic moments:

Identify the type of magnetism. What happens when these substances are heated? (CBSE Sample Paper 2017-18)
Answer:
Ferrimagnetism. These substances lose ferrimagnetism on heating and become paramagnetic.

Question 6.
Identify the type of defect shown in the following figure:
What type of substances show this defect? (CBSE Sample Paper 2017-18)

Answer:
Schottky defect.
This type of defect is shown by ionic compounds in which
(i) the ions have high coordination number and
(ii) ions (cations and anions) are of almost similar sizes.

Question 7.
Classify each of the following as being either a p-type or n-type semiconductor:
(i) Ge doped with In.
(ii) Si doped with B. (CBSE Sample Paper 2011)
Answer:
(i) Ge belongs to group 14 and In belongs to group 13.
Therefore, an electron deficient hole is created and it is a p-type semiconductor.

(ii) B belongs to group 13 and Si belongs to group 14. Therefore, an electron deficient hole is created and it is a p-type semiconductor.

Question 8.
Give reason:
(a) Why is Frenkel defect found in AgCI?
(b) What is the difference between phosphorus doped and gallium doped silicon semiconductors? (CBSE Sample Paper 2011)
Answer:
(a) Due to small size of Ag⁺ ion, it can fit into interstitial sites.
(b) Phosphorus doped silicon is n-type semiconductor while gallium doped silicon is p-type semiconductor.

Question 9.
Why does LiCl acquire pink colour when heated in Li vapours? (CBSE Sample Paper 2011)
Answer:
On heating LiCl in Li vapours, the excess of Li atoms deposit on the surface of the crystal. The CT ions diffuse to the surface of the crystal and combine with Li atoms to form LiCl. The electrons produced by ionisation of Li atoms diffuse into the crystal and get trapped at anion vacancies called F-centres. These absorb energy from visible light and radiate pink colour.

Question 10.
Write a feature which will distinguish a metallic solid from an ionic solid. (C.B.S.E. Delhi 2010, 2012)
Answer:
Metallic solids are good conductors of heat and electricity whereas ionic solids are insulators in solid state but conductors in molten state and in aqueous solution.

Question 11.
Crystalline solids are anisotropic in nature. What does this statement mean? (C.B.S.E. Delhi 2011)
Answer:
This means that crystalline solids have different physical properties such as electrical resistance or refractive index in different directions. This is because of different arrangement of particles in different directions.

Question 12.
What are n-type semiconductors? (C.6.S.E. Delhi 2012)
Answer:
These are the semiconductors in which the current is carried by the electrons in the normal way. For example, germanium doped with impurity containing five valence electrons (e.g., P).

Question 13.
What is the formula of a compound in which the element Y forms hep lattice and atoms of X occupy 2/3rd of tetrahedral voids? (C.B.S.E. 2015)
Answer:
Atoms Y adopt hep arrangement and there are two tetrahedral sites per atom of Y. Since 2/3rd tetrahedral sites are occupied by atoms of X, then for each atom of Y, the number of X atoms will be = 2 × \(\frac{2}{3}=\frac{4}{3}\)
Formula of compound – X_4/3Y or X₄Y₃

Question 14.
Define the following terms in relation to crystalline solids:
(i) Unit cell
(ii) Coordination number
Give one example in each case. (C.B.S.E. 2011)
Answer:
(i) Unit cell is the smallest portion of a crystal lattice which when repeated over and over again in different directions results in the entire lattice. For example, simple cubic unit cell.

(ii) The number of nearest neighbours in a packing is called coordination number. For example, in a body centred cubic structure, the coordination number is eight.

Question 15.
Account for the following:
(i) Schottky defects lower the density of related solids.
(ii) Conductivity of silicon increases on doping it with phosphorus. (C.B.S.E. 2013)
Answer:
(i) In Schottky defect, there are holes due to missing cations and anions. Due to the presence of holes in solid, the density decreases.

(ii) Pure silicon has a network lattice in which all the four valence electrons are bonded to four other atoms. Therefore, it is an insulator. However, when silicon is doped with phosphorus having five valence electrons, the impurity leads to excess of electrons after forming four covalent bonds like silicon. The extra electrons serve to conductivity and therefore, the conductivity of silicon doped with phosphorus increases.

Question 16.
(i) What type of non-stoichiometric point defect is responsible for the pink colour of LiCl?
Answer:
Metal excess defect due to anion vacancies filled by free electrons i.e. F-centres.

(ii) What type of stoichiometric defect is shown by NaCI?
Answer:
Schottky defect.

OR

How will you distinguish between the following pairs of terms:
(i) Tetrahedral and octahedral voids
Answer:
Tetrahedral void is surrounded by 4 constituent particles (atoms, molecules or ions). Octahedral void is surrounded by 6 constituent particles (atoms, molecules or ions).

(ii) Crystal lattice and unit cell? (C.B.S.E. 2014)
Answer:
A regular three dimensional arrangement of points in space is called crystal lattice.
The smallest repeating pattern in crystal lattice which when repeated in three dimensional space gives the entire lattice is called the unit cell.

Question 17.
The well known mineral fluorite is chemically calcium fluoride. It is known that in one unit cell of this mineral there are 4 Ca²⁺ ions and 8 F^– ions and that Ca²⁺ ions are arranged in a fcc lattice. The fluoride ions fill all the tetrahedral holes in the face centred cubic lattice of Ca²⁺ ions. The edge of the unit cell is 5.46 × 10^-8 cm in length. The density of the solid is 3.18 g cm^-3. Use this information to calculate Avogadro’s number. (Molar mass of CaF₂ = 78.08 g mol”1) (C.B.S.E. Delhi 2010)
Answer:
Density, d = \(\frac{Z \times M}{a^{3} \times N_{A}}\)
d = 3.18 g cm^-3, M = 78.08 g mol^-1
a = 5.46 × 10^-8 cm
Since the lattice is fcc, Z = 4
Substituting the values,

Question 18.
The density of copper metal is 8.95 g cm^-3. If the radius of copper atom be 127.8 pm, is the copper unit cell simple cubic, body centred cubic or face centred cubic? (C.B.S.E. 2010, C.B.S.E. Delhi 2010) ZxM
Answer:
Density, d = \(\frac{Z \times M}{a^{3} \times N_{A}}\)
Assuming fcc lattice for copper
a = 2√2 r
= 2 × 1.414 × 127.8 × 10^-10 cm
∴ a³ = (2 × 1.414 × 127.8 × 10^-10 cm)³
= 4.721 × 10^-23 cm³
∴ d = \(\frac{4 \times 63.54}{4.721 \times 10^{-23} \times 6.02 \times 10^{23}}\)
= 8.94 g cm^-3
Since the density is same as observed, the lattice is fcc lattice.

Question 19.
An element with density 10 g cm^-3 forms a cubic unit cell with edge length of 3 × 10^-8 cm. What is the nature of the cubic unit cell if the atomic mass of the element is 81 g mol^-1? (C.B.S.E. 2015)
Answer:
Density, d = \(\frac{z \times M}{a^{3} \times N_{A}}\)
d = 10 g cm^-3, M = 81 g mol^-1
a³ = (3 × 10^-8 cm)³
= 27 × 10^-24 cm³
N_A = 6.022 × 10²³
Therefore, 10 = \(\frac{Z \times 81}{\left(27 \times 10^{-24}\right) \times 6.022 \times 10^{23}}\)
Z = \(\frac{10 \times 27 \times 10^{-24} \times 6.022 \times 10^{23}}{81}\)
= 2.007
Nature of cubic unit cell = bcc

Question 20.
Aluminium crystallises in a fcc structure. Atomic radius of the metal is 125 pm. What is the length of the side of unit cell of the metal? (C.B.S.E. Delhi 2019C)
Answer:
For fcc,
r = \(\frac{a}{2 \sqrt{2}}\)
a = 2r × √2 = 2 × 125 pm × 1.414 = 353.5 pm

Question 21.
Answer the following:
(a) What is the formula of a compound in which element Y forms ccp lattice and atom X occupy \(\frac{1}{3}\)rd of tetrahedral voids?
(b) What type of non-stoichiometric point defect leads to colour in alkali metal halides? (C.B.S.E. Delhi 2019C)
Answer:
(a) X₂Y₃
(b) Metal excess defect due to anionic vacancies / F-centres.

Question 22.
(a) Atoms of element B form hep lattice and those of the element A occupy \(\frac{2}{3}\)rd of octahedral voids. What is the formula of the compound formed by the elements A and B?
(b) What type of stoichiometric defect is shown by ZnS and why? (C.B.S.E. Delhi 2019C)
Answer:
(a) A₂B₃
(b) Frenkel defect, due to small size of Zn²⁺ ion.

Question 23.
What happens when AgCl is doped with CdCl₂? What is the name of this defect?
OR
What type of defect is shown by NaCl in
(a) stoichiometric defects, and
(b) non-stoichiometric defects? (C.B.S.E. Delhi 2019C)
Answer:
When CdCl₂ is added to AgCl, it introduces impurity defect. The addition of one Cd²⁺ ion will replace two Ag⁺ ions to maintain electrical neutrality. One of the position of Ag⁺ ions will be occupied by Cd²⁺ ion and other will be left as a hole. Thus, cationic vacancies are produced. This is called impurity defect.
OR
(a) In stoichiometric defects: Schottky defect.
(b) In non-stoichiometric defects: Metal excess due to anionic vacancies.

The Solid State Important Extra Questions Long Answer Type

Question 1.
Copper crystallises with face centred cubic unit cell. If the radius of copper atom is 127.8 pm, calculate the density of copper metal. (Atomic mass of Cu = 63.55 u and Avogadro’s number, N_A = 6.02 × 10²³ mol^-1). (CBSE 2012)
Answer:
For a fcc unit cell, edge length (a) is related to radius of atom as:
r = \(\frac{a}{2 \sqrt{2}}\)
or a = 2√2.r = 2 × 1.414 × 127.8
= 361.42 pm
or = 361.42 × 10^-10 cm
Since the lattice is fcc, the number of copper atoms per unit cell, Z = 4.
Density, d = \(\frac{Z \times M}{a^{3} \times N_{A}}\)
M = 63.55 u, N_A = 6.02 × 10²³
∴ d = \(\frac{4 \times 63.55}{\left(361.42 \times 10^{-10}\right)^{3} \times\left(6.02 \times 10^{23}\right)}\)
= 8.94 g cm^-3

Question 2.
An element crystallises in a fcc lattice with cell edge of 250 pm. Calculate its density if 300 g of this element contain 2 × 10²⁴ atoms.
Answer:
Length of edge, a = 250 pm = 250 × 10^-12 m
= 250 × 10^-10 cm
Volume of unit cell = (250 × 10^-10 cm)³
= 15.625 × 10^-24 cm³
Mass of unit cell = No. of atoms in unit cell × Mass of each atom
Since the element has fcc arrangement, the number of atoms per unit cell, Z = 4
Mass of an atom = \(\frac{300}{2 \times 10^{24}}\) g
∴ Mass of unit cell = \(\frac{300}{2 \times 10^{24}}\) x 4
= 6.0 × 10^-22 g

Question 3.
An element with molar mass 27 g mol^-1 forms a cubic unit cell with edge length 4.05 × 10^-8 cm. If its density is 2.7 g cm^-3, what is the nature of the unit cell?
Answer:
Density of unit cell, d = \(\frac{\mathrm{Z} \times \mathrm{M}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}\)
d = 2.7 g cm^-3, a = 4.05 × 10^-8 cm,
M = 27 g mol^-1, NA = 6.022 × 10²³

= 4.0
Since Z = 4, the unit cell is face centred cubic (fcc) unit cell.

Question 4.
An element with density 11.2 g cm^-3 forms a fee lattice with edge length of 4 × 10^-8 cm. Calculate the atomic mass of the element.
Answer:
Edge length of the unit cell
a = 4 × 10^-8 cm
Density = 11.2 g cm^-3
No. of atoms per unit cell in fcc lattice, Z = 4
Density, d = \(\frac{z \times M}{a^{3} \times N_{A}}\)
11.2 g cm^-3 = \(\frac{4 \times M}{\left(4 \times 10^{-8} \mathrm{~cm}\right)^{3} \times\left(6.02 \times 10^{23} \mathrm{~mol}^{-1}\right)}\)
or M = \(\frac{\left(11.2 \mathrm{~g} \mathrm{~cm}^{-3}\right) \times\left(4 \times 10^{-8} \mathrm{~cm}\right)^{3} \times\left(6.02 \times 10^{23} \mathrm{~mol}^{-1}\right)}{4}\)
= 107.9 g mol^-1
Atomic mass of element = 107.9 u.

Question 5.
The density of lead is 11.35 g cm^-3 and the metal crystallises with fcc unit cell. Estimate the radius of lead atom. (At. Mass of lead = 207 g mol^-1 and N_A = 6.02 × 10²³ mol^-1) (CBSE Delhi 2011)
Answer:
Let length of edge = a cm
Density = 11.35 g cm^-3
No. of atoms per unit cell in fcc Lattice = 4
Atomic mass, M = 207 g mol^-1
Density, d = \(\frac{Z \times M}{a^{3} \times N_{A}}\)

= 121.14 × 10^-24 cm³
Edge length, a = (121.14)^1/3 × 10^-8
= 4.948 × 10^-8 cm
or = 4.948 × 10^-10 m
or = 494.8 × 10^-12 m = 494.8 pm
Now, radius in fcc = \(\frac{a}{2 \sqrt{2}}\)
= \(\frac{494.8 \mathrm{pm}}{2 \times 1.414}\) = 174.96 pm

Question 6.
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain. (C.B.S.E. 2011)
Answer:
Consider a unit cell of edge a cm
Volume of unit cell = a³ cm³

Mass of unit cell = No. of atoms in a unit cell × Mass of each atom = Z × m …… (ii)
Mass of an atom present in a unit cell

Knowing density (d), edge length (a), number of atoms per unit cell (Z) and Avogadro’s number (6.02 × 10²³), atomic mass can be calculated.

Question 7.
ZnO turns yellow on heating. Why? (C.B.S.E. 2016)
Answer:
When ZnO is heated, it loses oxygen as:

The Zn²⁺ ions are entrapped in the interstitial sites and electrons are entrapped in the neighbouring interstitial sites to maintain electrical neutrality. This results in metal excess defect. Due to the presence of free electrons in the interstitial sites the colour is yellow.

Question 8.
(a) What type of semiconductor is obtained when silicon is doped with boron?
Answer:
(a) When silicon is doped with boron having three valence electrons, the bonds formed create electron deficient sites called holes. Under the influence of applied electric field, one electron from neighbouring atom moves to fill the hole but creates another hole at its own place. Therefore, the electrical conductance is due to movement of positive holes. Therefore, this type of semi-conductor is called p-type semi-conductor.

(b) What type of magnetism is shown in the following alignment of magnetic moments?
↑    ↑    ↑    ↑   ↑   ↑    ↑
Answer:
Ferromagnetic.

(c) What type of point defect is produced when AgCI is doped with CdCl₂? (C.B.S.E. Delhi 2013)
Answer:
CdCl₂ on adding to AgCl introduces impurity defect. The addition of one Cd²⁺ ion will replace two Ag⁺ ions to maintain electrical neutrality. One of the positions of Ag⁺ will be occupied by Cd²⁺ ion and the other will be left as a hole. Thus, a hole is created similar to Schottky defect.

Question 9.
Account for the following:
(i) Schottky defects lower the density of related solids.
Answer:
In Schottky defect, there are holes due to missing cations and anions. Due to the presence of holes in solid, the density decreases.

(ii) Conductivity of silicon increases on doping it with phosphorus. (C.B.S.E. 2013)
Answer:
Pure silicon has a network lattice in which all the four valence electrons are bonded to four other atoms. Therefore, it is an insulator. However, when silicon is doped with phosphorus (s²p³) having five valence electrons, the impurity leads to excess of electrons after forming four covalent bonds like silicon. The extra electrons serve to conductivity and therefore, the conductivity of silicon doped with phosphorus increases.

Question 10.
Examine the given defective crystal

Answer the following questions:
(i) What type of stoichiometric defect is shown by the crystal?
Answer:
Schottky defect

(ii) How is the density of the crystal affected by this defect?
Answer:
Density of the crystal decreases.

(iii) What type of ionic substances show such defect? (C.B.S.E. Delhi 2014)
Answer:
Crystals having
(a) high coordination number and
(b) ions (cations and anions) of almost similar sizes. For example: KCl, KBr.

Question 11.
An element crystallises in a fcc lattice with edge length of 400 pm. The density of the element is 7 g cm^-3. How many atoms are present in 280 g of the element? (C.B.S.E. 2016)
Answer:
Edge length, a = 400 pm
= 400 × 10^-10 cm
Volume of unit cell = (400 × 10^-10)³
= 64 × 10^-24 cm^-3
Mass of element = 280 g
Density = 7 g cm^-3
Volume of 280 g of the element = \(\frac{280 \mathrm{~g}}{7 \mathrm{~g} \mathrm{~cm}^{-3}}\)
= 40 cm³
No. of unit cells in this volume
= \(\frac{40 \mathrm{~cm}^{3}}{64 \times 10^{-24} \mathrm{~cm}^{3}}\)
= 6.25 × 10²³ unit cells
Since the structure is fcc, number of atoms in a unit cell = 4
∴ No. of atoms in 280 g of element
= 4 × 6.25 × 10²³
= 2.5 × 10²⁴ atoms

Question 12.
An element crystallises in a fcc lattice with cell edge of 250 pm. Calculate the density of 300 g of this element containing 2 × 1024 atoms. (C.B.S.E. Delhi 2016)
Answer:
Length of edge, a = 250 pm = 250 × 10^-10 cm
Volume of unit cell = (250 × 10^-10)³
= 15.625 × 10^-24 cm³
Mass of unit cell = No. of atoms in unit cell × mass of each atom
Since the element has fcc arrangement, number of atoms per unit cell, Z = 4
Mass of an atom = \(\frac{300}{2 \times 10^{24}}\) = 1.50 × 10^-22 g

Question 13.
An element crystallises in fcc lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm^-3. Calculate the number of atoms in 108 g of the element. (C.B.S.E. Delhi 2019)
Answer:
Edge length of the unit cell = 300 pm
= 300 × 10^-10 cm
Volume of unit cell = (300 × 10^-10) ³
= 27 × 10^-24 cm³
Mass of element = 108 g
Volume of 108 g element = \(\frac{108 \mathrm{~g}}{10.8 \mathrm{~g} \mathrm{~cm}^{-3}}\)
= 10 cm³
No. of unit cells = \(\frac{10}{27 \times 10^{-24}}\)
= 3.7 × 10²³ unit cells
Since the structure is fcc, number of atoms per unit cell = 4
Number of atom is 108 g = 3.7 × 10²³ × 4
= 1.48 × 10²⁴.

Question 14.
(a) An element crystallises in bcc lattice with a cell edge of 3 × 10^-8 cm. The density of the element is 6.89 g cm^-3. Calculate the molar mass of the element. (N_A = 6.022 × 10²³ mol^-1)
Answer:
Density of element, d = 6.89 gcm^-3
Cell edge, a = 3 × 10^-8 cm
Z = 2 (bcc)

(b) What type of semiconductor is obtained when
(i) Ge is doped with In?
(ii) Si is doped with P? (C.B.S.E. 2019 Al)
Answer:
(i) p-type
(ii) n-type

Question 15.
(a) A compound is formed by two elements M and N. The element N forms ccp and M atoms occupy 1/3rd of tetrahedral voids. What is the formula of the compound?
(b) Which of the following lattices has the highest packing efficiency
(i) simple cubic
(ii) body centred cubic and
(iii) hexagonal close packed lattice?
(c) An element with molar mass 2.7 × 10^-2 kg mol^-1 forms a cubic unit cell with edge length 405 pm. If the density is 2.7 × 10³ kg m^-3. What is the nature of the cubic unit cell?
Answer:
(a) Since N forms ccp arrangement, it will have 4 atoms in a unit cell.
Number of N atoms in unit cell = 4
For each atom, there are two tetrahedral voids so that there are 8 tetrahedral voids per unit cell.
No. of M atoms = \(\frac{1}{3}\) × 8 = \(\frac{8}{3}\)
Formula = M_8/3N₄
or = M₂ N₃.

(b) The packing efficiencies are:
simple cubic = 52.4%
body centred cubic = 68%
hexagonal close packed = 74%
∴ Hexagonal close packed lattice has highest packing efficiency.

(c) Edge length = 405 pm = 405 × 10^-12 m
Density of the cell = 2.7 × 10^-3 kg m^-3
Molar mass = 2.7 × 10^-2 kg mol^-1

Since the unit cell contains 4 atoms, it is cubic close packed structure, ccp.

Question 16.
How will you distinguish between the following pair of terms:
(i) Hexagonal close packing and cubic close packing
(ii) Crystal lattice and unit cell.
(iii) Tetrahedral void and octahedral void.
Answer:
(i)

Hexagonal close packing	cubic close packing
(i) In hcp, the spheres of the third layer are exactly aligned with those of the first layer. This arrangement is represented as AB AB …………… type.	(i) In ccp, the spheres of the third layer are not aligned with those of the first layer or second layer. The layers of fourth layer are aligned with those of the first layer. This pattern is represented as ABC ABC ………
(ii) In hcp, the tetrahedral voids of the second layer may be covered by the spheres of the third layer.	(ii) In ccp, the third layer may be placed above the second layer in a manner such that its spheres cover the octahedral voids.

(ii) The three dimensional arrangement of constituent particles of a substance (atoms, ions or molecules) is called crystal lattice.
The smallest repeating pattern in a crystal lattice which when repeated in three dimensions gives the crystal is called unit cell.

(iii) A void surrounded by four spheres is called a tetrahedral void while a void surrounded by six spheres is called an octahedral void.

Question 17.
(i) How many lattice points are there in one unit cell of each of the following lattice?
(a) face centred cubic
(b) face centred tetragonal
(c) body centred.
Answer:
(i) (a) In face centred cubic arrangement, number of lattice points are:
8 (at corners) + 6 (at face centres)
Lattice points per unit celt
= 8 × \(\frac{1}{8}\) + 6 × \(\frac{1}{2}\) = 4

(b) In face centred tetragonal, number of lattice points are:
= 8 (at corners) + 6 (at face centres) Lattice points per unit cell 1 1
= 8 × \(\frac{1}{8}\) + 6 × \(\frac{1}{2}\) = 4

(c) In body centred cubic arrangement number of lattice points are:
= 8 (at corners) + 1 (at body center) Lattice points per unit cell
= 8 × \(\frac{1}{8}\) + 1 = 2

(ii) Explain: (a) The basis of similarities and differences between metallic and ionic crystals.
(b) Ionic solids are hard and brittle.
Answer:
(a) Basis of similarities
1. Both ionic and metallic crystals have electrostatic forces of attraction. In ionic crystals, these are between the oppositely charged ions while in metals, these are among the valence electrons and the kernels. That is why both ionic and metallic crystals have high melting points.
2. In both cases, the bond is non- directional.

Basis of differences
1. In ionic crystals, the ions are not free to move and therefore, they do not conduct electricity in the solid state. They conduct electricity in the molten state or in their aqueous solution. However, in metals, the valence electrons are free to move and hence they conduct electricity in the solid state.
2. Ionic bond in ionic crystals is strong due to electrostatic forces of attraction. However, metallic bond may be weak or strong depending upon the number of valence electrons and the size of the kernels.

(b) Ionic crystals are hard because there are strong electrostatic forces of attraction among the oppositely charged ions. They are brittle because the ionic bond is non-directional.

Question 18.
(a) If the radius of the octahedral void is r and radius of the atoms in close packing is R. Derive relation between r and R.
Answer:
A sphere of radius r filling in an octahedral void of spheres of radius R is shown in figure.

If the length of the unit cell is a cm, then
In right angled ∆ABC, AB = BC = a cm
The diagonal AC is:

Thus, for an atom to occupy an octahedral void, its radius must be 0.414 times the radius of the sphere.

(b) Copper crystallises into a fcc lattice with edge length 3.61 × 10^-8 cm. Show that the calculated density is in agreement with its measured value of 8.29 g cm^-3.
Answer:
Density, d = \(\frac{Z \times M}{a^{3} \times N_{A}}\)
For fcc lattice, Z = 4
Atomic mass, M of copper = 63.5 g/mol^-1
a = 3.61 × 10^-8 cm
∴ p = \(\frac{4 \times 63.5}{\left(3.61 \times 10^{-8}\right)^{3} \times 6.022 \times 10^{23}}\)
= 8.96 g cm^-3
This value is close to measured value.

Question 19.
What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
Answer:
The substances whose conductance lies between that of conductors (metals) and insulators are called semiconductors. They have conductivity values ranging from 10^-6 to 10⁴Ω^-1m^-1. Two main types of semiconductors are n-type and p-type.

(i) n-type semiconductors: These are the semiconductors in which the current is carried by the electrons in the normal way. For example, silicon and germanium belong to group 14 of the periodic table and have four valence electrons each. In their crystals, each atom forms four covalent bonds with its neighbours.

When it is doped with an element of group 15 such as P or As, which contains five valence electrons, they occupy some of the lattice sites in silicon or germanium crystal. Four of the five valence electrons are used in the formation of four covalent bonds with the four neighbouring Si atoms and the fifth electron is extra and becomes delocalised.

This delocalised electron will be able to conduct electricity and such type of conduction is called n-type conduction. Hence silicon or germanium doped with extra electron of impurity are called n-type semiconductors.

(ii) p-type semiconductors: These are the semiconductors in which current is carried by the movement of positive holes. For example, when Ge or Si are doped with electron deficient atoms such as Ga or In (of Group 13) containing three valence electrons, the atoms of Ga or In will replace Ge atoms.

Each In or Ga atom will use its three electrons for forming three covalent bonds with neighbouring Ge atoms and the place for fourth bond will remain missing and is called electron vacancy or hole. Such holes can move through the crystals like a positive charge giving rise to conductivity. This type of conduction is called p-type conduction. Hence, Si or Ge doped with electron deficient atoms as impurity are called p-type semiconductors.

Question 20.
Explain the following with suitable examples:
(a) Ferromagnetism
(b) Paramagnetism
(c) Ferrimagnetism
(d) Anti-ferromagnetism
(e) 12-16 and 13-15 group compounds.
Answer:
(a) Ferromagnetism: When there is spontaneous alignment of magnetic moments of domains in the same direction, we get ferromagnetism. These have strong magnetic effects and ordering of domains persists even when magnetic field is removed and – ferromagnetic substance becomes permanent magnet.

(b) Paramagnetism: The substances which have permanent magnetic dipoles and are attracted by the magnetic field are called paramagnetic substances. This property of attraction by the magnetic field is known as paramagnetism. The paramagnetic substance are atoms, molecules or ions having unpaired electrons. For example, O₂, Cu²⁺, Fe²⁺, etc. They lose their magnetism in the absence of magnetic field.

(c) Ferrimagnetism: When the magnetic moments of the domains are aligned in parallel or anti-parallel directions in unequal numbers resulting in net magnetic moment, we get ferrimagnetism.

(d) Anti-ferromagnetism: It arises when the alignment of magnetic moments is in a compensatory way so that the resultant magnetic moment is zero (e.g., MnO).

(e) The solid binary compounds prepared by combining elements of group 12 and 16 are called 12-16 compounds. For example CdS, ZnS, etc. The compounds prepared by combining elements of group 13 and 15 are called 13-15 compounds. For example, AlP, GaAs, etc. These compounds are used as semiconductors.

Question 21.
Give the points of differences between Schottky defect and Frenkel defect ?
OR
Explain:
(i) Why is Frenkel defect not found in pure alkali metal halides ?
(ii) Zinc oxide is white but it turns yellow on heating.
(iii) CaCl₂ will introduce Schottky defect when added to AgCl crystal.
Answer:
The important differences between Schottky and Frenkel defects are given ahead:

Schottky defect	Frenkel defect
1. It is produced be cause of missing atoms or ions from their normal sites.	1. It is produced when some atoms or ions leave their normal sites and occupy interstitial sites.
2. The presence of Schottky defect lowers the density of the crystal.	2. It does not affect the density of the crystal.
3. It is generally exhibited by ionic solids having high coordination number and in which cations and anions are of equal size. e.g., NaCl, CsCl.	3. It is generally exhibited by ionic solids having low coordination number and in which anions are larger in size than cation e.g., ZnS, AgCl.

OR
(i) Frenkel defect is not found in pure alkali metal halides because the ions cannot get into the interstitial sites due to their larger size.

(ii) When ZnO is heated it loses oxygen as:

The Zn²⁺ ions are entrapped in the interstitial sites and electrons are entrapped in the neighbouring interstitial sites to maintain electrical neutrality. This results in metal excess defect. Due to the presence of free electrons in the interstitial sites the colour is yellow.

(iii) CaCl₂ on adding to AgCl introduces impurity defect. The addition of one Ca²⁺ ion will replace two Ag⁺ ions to maintain electrical conductivity. One of the position of Ag⁺ will be occupied by Ca²⁺ ion and other will be left as a hole. Thus, a hole is created similar to Schottky defect.

Question 22.
What do you understand by a space lattice and a unit cell?
Answer:
(a) Space lattice is a regular arrangement of the constituent particles (atoms, ions or molecules) of a crystalline solid in three dimentional space.

The positions which are occupied by atoms, ions or molecules in the crystal lattice are called lattice points or lattice sites. A two dimensional and a three dimensional space lattice is shown in Fig. 1 (a) and 1 (b) respectively.


Fig. 1. (a) Two dimensional and
(b) Three dimensional space lattice.

A unit cell is the smallest repeating unit in space lattice which when repeated over and over again results in the crystal of the given substance. The unit cell gives the shape of the entire crystal. The crystal may be considered to consist of an infinite number of unit cells. The unit cell in the above crystal lattice [Fig. 1 (b)] is shown in The complete crystal lattice can be obtained by extending the unit cell in all the three directions.

Question 23.
What are tetrahedral and octahedral holes (voids) in close packed stacks of spheres? Explain.
Answer:
Tetrahedral and octahedral holes. The close packed arrangement of spheres has two types of holes or voids:
Tetrahedral holes: Tetrahedral hole Is produced when one sphere rests upon three other touching spheres. The space which is left between these touching spheres is called tetrahedral hole or tetrahedral site. The tetrahedral hole is shown in Fig. (a). It may be noted that in close packed arrangement (ccp or hcp), there are twice as many tetrahedral holes as there are spheres.

Octahedral hole.: This type of hole is formed at the centre of six spheres which form a regular octahedral. This is shown in Fig. (b). From the figure, it is clear that the octahedral hole is formed by two sets of equilateral triangles which point in opposite directions. It may be noted that there are same number of octahedral holes as the number of spheres.

Fig. (a) Tetrahedral hole (b) Octahedral hole.

Question 24.
(a) Why does table salt, NaCl, some times appear yellow in colour?
(b) Why is FeO(s) not formed in stoichiometric composition?
(c) Why does white ZnO (s) becomes yellow upon heating?
Answer:
(a) The yellow colour of sodium chloride crystals is due to metal excess defect. In this defect, the unpaired electrons get trapped in anion vacancies. These sites are called F-centres. The yellow colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.

(b) In the crystals of FeO, some of the Fe²⁺ cations are replaced by Fe³⁺ ions. To balance the charge, three Fe²⁺ ions are replaced by two Fe³⁺ ions to make up for the loss of positive charge. As a result, there would be less amount of metal as compared to stoichiometric proportion.

(c) When ZnO is heated it loses oxygen as:

The Zn²⁺ ions are entrapped in the interstitial sites and electrons are entrapped in the neighbouring interstitial sites to maintain electrical neutrality. This results in metal excess defect and F-centres are created. Due to the presence of electrons in the interstitial voids, the colour is yellow.

Question 25.
(a) Explain why does conductivity of germanium crystals increase on doping with gallium.
(b) In a compound, nitrogen atoms (N) make cubic close packed lattice and metal atoms (M) occupy one-third of the tetrahedral voids present. Determine the formula of the compound formed by M and N?
(c) Under which situations can an amorphous substance change to crystalline form?
Answer:
(a) On doping germanium with gallium, some of the positions of lattice of germanium are occupied by gallium. Gallium atom has only three valence electrons. Therefore, fourth valency of nearby germanium atom is not satisfied and this site remains vacant. This place is deficient of electrons and is called electron hole or electron vacancy.

Electron from neighbouring atom moves to fill the gap, thereby creating a hole in its original position. Under the influence of electric field, electrons move towards positively charged plates through these holes and conduct electricity. The holes appear to move towards negatively charged plates. The movement of electrons (or electron holes) results in increase in conductivity of germanium.

(b) N atoms make up ccp arrangement and there are two tetrahedral sites per atom of N.
No. of N atoms per unit cell = 4
No. of tetrahedral sites = 4 × 2 = 8
No. of sites occupied by M = \(\frac{1}{3}\) × 8 = \(\frac{8}{3}\)
∴ No. of M atoms per unit cell = \(\frac{8}{3}\)
Formula M_8/3N₄ or M₈N₁₂ or M₂N₃

(c) On heating, amorphous solids become crystalline at some temperature. For example, some glass objects from ancient civilisations are found to become milky in appearance because of some crystallisation.

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 4 Moving Charges and Magnetism. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 4 Important Extra Questions Moving Charges and Magnetism

Moving Charges and Magnetism Important Extra Questions Very Short Answer Type

Question 1.
Under what condition is the force acting on a charge moving through a uniform magnetic field minimum?
Answer:
When the charge moves parallel to the direction of the magnetic field.

Question 2.
What Is the nature of the magnetic field in a moving coil galvanometer?
Answer:
Radial magnetic field.

Question 3.
State two properties of the material of the wire used for suspension of the coil in a moving coil galvanometer.
Answer:

1. High tensile strength.
2. SmalL vaLue of torque per unit twist.

Question 4.
Write one condition under which an electric charge does not experience a force in a magnetic field.
Answer:
When it moves parallel to the direction of the magnetic field.

Question 5.
Mention the two characteristic properties of the material suitable for making the core of a transformer. (CBSE AI 2012)
Answer:

1. Low retentivity
2. High permeability

Question 6.
Write the expression, in a vector form, for the Lorentz magnetic force due to a charge moving with velocity \(\vec{V}\) in a magnetic field \(\vec{B}\). What is the direction of the magnetic force? (CBSE Delhi 2014)
Answer:
The expresion is \(\vec{F}\) = q(\(\vec{V}\) × \(\vec{B}\)). The force is perpendicular to both the velcoity and the magentic field vector.

Question 7.
Write the condition under which an electron will move undeflected in the presence of crossed electric and magnetic fields. (CBSE Al 2014C)
Answer:
An electron moves perpendicular to both fields.

Question 8.
What can be the cause of the helical motion of a charged particle? (CBSE Al 2016)
Answer:
The charge enters the magnetic field at any angle except 0°, 180°, and 90°.

Question 9.
Write the underlying principle of a moving coil galvanometer. (CBSE Delhi 2016)
Answer:
A current-carrying loop placed in a magnetic field experiences a torque.

Question 10.
A proton and an electron traveling along parallel paths enter a region of the uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with a higher frequency? (CBSEAI and Delhi 2018)
Answer:
The frequency of revolution is given by
v = \(\frac{B q}{2 \pi m}\) ⇒ v ∝ \(\frac{1}{m}\) .

As for m_e < m_p
therefore v_e >v_p

Question 11.
Two protons of equal kinetic energies enter a region of the uniform magnetic field. The first proton enters normal to the field direction while the second enters at 30° to the field direction. Name the trajectories followed by them. (CBSEAI and Delhi 2018C)
Answer:
Normal: circular
At an angle of 30°, it will follow a helical path.

Question 12.
Consider the circuit shown here where APB and AQB are semi-circles. What will be the magnetic field at the center C of the circular loop?

Answer:
Zero, because magnetic fields due to APB and AQB are equal in magnitudes but opposite in directions.

Question 13.
Which one of the following will have a minimum frequency of revolution,
when projected with the same velocity v perpendicular to the magnetic field B: (i) α – particle and (ii) β – particle.
Answer:
Frequency of revolution v = \(\frac{q B}{2 \pi m}\) and \(\frac{q}{m}\) of α – particle is less, hence α – particle will have minimum frequency of revolution.

Question 14.
An ammeter and a milli-ammeter are converted from the same galvanometer. Out of the two, which current-measuring instrument has higher resistance?
Answer:
A milli-ammeter has higher resistance.

Question 15.
Equal currents I and I are flowing through two infinitely long parallel wires. What will be the magnetic field at a point mid-way when the currents are flowing in the same direction?
Answer:
Zero, because fields due to two wires will be equal but opposite.

Question 16.
The figure shows a circular loop carrying current l. Show the direction of the magnetic field with the help of lines of

Answer:
The magnetic field lines are as shown

Question 17.
An electron is moving with velocity v along the axis of a long straight solenoid carrying current I. What will be the force acting on the electron due to the magnetic field of the solenoid?
Answer:
Zero, as a force on a charged particle moving in a magnetic field, is F= Bqv sin θ
Here both v and B are along the axis of – solenoid, so θ = 0° between them. Hence F= qvB sin θ = 0.

Moving Charges and Magnetism Important Extra Questions Short Answer Type

Question 1.
A charged particle having a charge q is moving with a speed of v along the X-axis. It enters a region of space where the electric field is \(\overrightarrow{\boldsymbol{E}}(\boldsymbol{E} \hat{\boldsymbol{j}})\) and a magnetic field \(\vec{B}\) are both present. The particle, on emerging from the region, is observed to be moving, along the X-axis only. Obtain an expression for the magnitude of \(\vec{B}\) in terms of v and E. Give the direction of \(\vec{B}\).
Answer:
Since the particle continues to move along the X-axis, therefore, the magnetic force acting on it should be completely balanced by the electric force. Since the electric force acts along the Y-axis, therefore, the magnetic force must be along the Z-axis.
Thus is equilibrium q E = B q v or v = E/B

Question 2.
A stream of electrons traveling with speed v m s^-1 at right angles to a uniform magnetic field ‘B’ is reflected in a circular path of radius ‘r’ . Prove that \(\frac{e}{m}=\frac{v}{r B}\)
Answer:
Let a stream of electrons be traveling with speed v at right angles to a uniform magnetic field B then force due to magnetic field provides the required centripetal force which deflects the electron beam along a circular path of radius ‘r’ such that
Bev = \(\frac{m v^{2}}{r}\)
or
\(\frac{e}{m}=\frac{v}{r B}\)
where e = electronic charge and m = mass of the electron.

Question 3.
Which one of the two, an ammeter or a milliammeter, has a higher resistance and why?
Answer:
The shunt resistance connected to convert a galvanometer into an ammeter or a milliammeter is given by the expression S = \(\frac{I_{g} G}{I-l_{g}}\) where S is shunt resistance, G galvanometer resistance, l total current through G and S, and l_g galvanometer current. In the case of milliammeter, l is small.

Therefore S_milliammeter > S_ammeter . Hence the resistance of a milliammeter is greater than that of an ammeter.

Question 4.
A straight wire of length L carrying a current l stays suspended horizontally in mid-air in a region where there is a uniform magnetic field \(\vec{B}\). The linear mass density of the wire is l. Obtain the magnitude and direction of the magnetic field.
Answer:
The magnetic force acting on the straight wire balances the weight of the wire.
Therefore, in equilibrium we have Mg = BIL, here M = L l, therefore we have L l g = BlL or B = l/ l g
This field acts vertically upwards.

Question 5.
In the figure below, the straight wire AB Is fixed while the loop Is free to move under the influence of the electric currents flowing in them. In which direction does the loop begin to move? Give a reason for your answer.

Answer:
The loop moves towards the straight wire AB. In the loop in the side nearer to the wire AB current l₂ is in the same direction as l₁ and hence attractive force acts. However, on the side farther away from the wire AB current l₂ is in the opposite direction and the force is repulsive. But as the magnitude of attractive force is greater than the repulsive force, the net force is attractive in nature and hence, the loop moves towards the wire AB.

Question 6.
A coil of ‘N’ turns and radius ‘R’ carries a current ‘l’. It is unwound and rewound to make a square coil of side ‘a’ having the same number of turns (N). Keeping the current ‘l’ same, find the ratio of the magnetic moments of the square coil and the circular coil. (CBSE Delhi 2013C)
Answer:
The magnetic moment of a current loop is given by the relation M = nlA
For the circular loop M_c = NlπR² …(1)

Now when the coil is unwound and rewound to make a square coil, then
2 πR = 4a or a = πR/2

Hence magnetic moment of the square coil is
M_s = Nl a² = Nl (πR/2)² = Nl π²R²/4 …(2)

From (1) and (2) we have
\(\frac{M_{S}}{M_{C}}=\frac{N l \pi^{2} R^{2} / 4}{N l \pi R^{2}}=\frac{\pi}{4}\)

Question 7.
Write the expression for Lorentz magnetic force on a particle of charge ‘q’ moving with velocity v in a magnetic field B. Show that no work is done by this force on the charged particle. (CBSE Al 2011)
Answer:
The expression is \(\vec{F}\) = q(\(\vec{v}\) × \(\vec{B}\)). This force always acts perpendicular to the direction of motion of the charged particle. Therefore the angle between \(\vec{F}\) and \(\vec{r}\) is 90°. Hence work done is W = \(\vec{F}\). r = Fr cos 90° = 0

Question 8.
(a) State Biot-Savart law in vector form expressing the magnetic field due to an element \(\vec{dl}\) carrying current l at a distance \(\vec{r}\) from the element.
Answer:
It states that for a small current element dl the magnetic field at a distance r is given by
\(\overrightarrow{d B}=\frac{\mu_{0}}{4 \pi} \frac{I \overline{(d L} \times \hat{r})}{r^{2}}\).

(b) Write the expression for the magnitude of the magnetic field at the center of a circular loop of radius r carrying a steady current l. Draw the field lines due to the current loop. (CBSE Al 2014C)
Answer:
The magnetic field at the center of a circular loop is given by
B = \(\frac{\mu_{0} l}{2 r}\)

The field lines are as shown.

Question 9.
Draw the magnetic field lines due to current passing through a long solenoid. Use Ampere’s circuital law, to obtain the expression for the magnetic field due to the current l in a long solenoid having n number of turns per unit length. (CBSE Delhi 2014C)
Answer:
The diagram is as shown.

Consider a rectangular path abed of length L as shown in the figure below. Let us apply Ampere’s circuital law to this rectangular path so that we have



But by Ampere’s circuital law we have
\(\oint \vec{B} \cdot \vec{d} L\) = μ₀Nl = μ0(nL)l …(2)

From equations (1) and (2) we have
BL = μ₀nLl
or
B = μ₀ n l

But n = N/L, therefore we have
B = μ₀ \(\frac{N l}{L}\)

This gives the value of a magnetic field inside a solenoid.

Question 10.
(a) Why do we use a shunt to convert a galvanometer into an ammeter?
Answer:
Since an ammeter is an instrument used to measure the current in the circuit, so it has to be connected in series in the circuit to measure the whole current. Hence its resistance must below. A low shunt resistance makes it suitable for measuring current.

(b) A galvanometer of resistance 15 Ω shows a full-scale deflection on the meter scale for a current of 6 mA. Calculate the value of the shunt resistance required to convert the galvanometer into an ammeter of range 0-6 A. (CBSE 2019C)
Answer:
Given G = 15 Ω, l_g = 6 mA = 6 × 10^-3 A, l = 6 A, S = ?

Question 11.
An electron beam projected along + X-axis experiences a force due to a magnetic field along the + Y-axis. What is the direction of the magnetic field?
Answer:
The direction of the magnetic field is along  Z-axis. This is because the direction of motion, the magnetic field, and the force are perpendicular to one other as an electron carries a negative charge.

Question 12.
A current is set up in a long copper pipe. Is there a magnetic field
(i) inside,
Answer:
There is no magnetic field inside the pipe.

(ii) outside the pipe?
Answer:
There is a magnetic field outside the pipe.

Question 13.
Which one of the following will experience a maximum force, when projected with the same velocity V perpendicular to the magnetic field (i) alpha particle and (ii) beta particle?
Answer:
The force experienced by a charged particle is given by the expression F = B q v. Since an alpha particle has a moving charge than the beta particle, therefore the alpha particle will experience more force.

Question 14.
An electron and a proton moving parallel to each other in the same direction with equal momenta enter into a uniform magnetic field which is at right angles to their velocities. Trace their trajectories in the magnetic field.
Answer:
Because both electron and proton have the same charge and momentum, therefore they will describe circles of equal radii as shown.

Question 15.
Is the steady electric current the only source of the magnetic field? Justify your answer. (CBSE Delhi 2013C)
Answer:
No, the magnetic field is also produced by alternating current.

Question 16.
A deuteron and an alpha particle having the same momentum is in turn allowed to pass through a magnetic field B, acting normal to the direction of motion of the particles. Calculate the ratio of the radii of the circular paths described by them. (CBSE Delhi 2019)
Answer:
P_d = P_α
Now r = \(\frac{mv}{Bq}\) ⇒ r ∝ \(\frac{1}{q}\)

Therefore, \(\frac{r_{d}}{r_{\alpha}}=\frac{q_{\alpha}}{q_{d}}=\frac{2 e}{e}\) = 2

Question 17.
Two wires of equal length are bent in the form of two loops. One of the loops is square-shaped and the other is circular. These are suspended in a uniform magnetic field and the same current is passed through them. Which loop will experience a greater torque? Give reasons.
Answer:
Torque experienced by a current-carrying loop placed in a uniform magnetic field is given by the expression τ = BlnA. In other words, torque is directly proportional to the area of the loop. Since a circular wire has more area than a square wire for the same dimension, therefore the circular wire experiences more torque than the square wire.

Question 18.
Which one of the following will have a minimum frequency of revolution, when projected with the same velocity v perpendicular to the magnetic field B: (i) alpha particle and (ii) beta particle?
Answer:
The frequency of revolution of a charged particle in a magnetic field is given by Bq
v = \(\frac{Bq}{2πm}\).

The ratio of q/m for an alpha particle is less than that for a beta particle; therefore, the alpha particle will have a minimum frequency of revolution.

Question 19.
Using the concept of force between two infinitely long parallel current-carrying conductors, define one ampere of current. (CBSE AI 2014)
Answer:
One ampere is that much current which when flowing through each of the two infinitely long straight conductors held 1 meter apart in space, produce a force of F= 2 × 10^-7 N per meter of their length.

Question 20.
(a) Write the expression for the force \(\vec{F}\) , acting on a charged particle of charge ‘q’, moving with a velocity v in the presence of both electric field \(\vec{E}\) and magnetic field \(\vec{B}\) . Obtain the condition under which the particle moves undeflected through the fields.
Answer:
(a) The required expression is \(\vec{F}=q \vec{E}+q(\vec{v} \times \vec{B})\)

The particle will move undeflected if the force acting on it due to the electric field balances the force acting on it due to the magnetic field. Thus qE = Bqv or v=E/B

(b) A rectangular loop of size l × b carrying a steady current I is placed in a uniform magnetic field \(\vec{B}\) . Prove that the torque τ acting on the loop is given by \(\vec{τ}\) = \(\vec{m}\) × \(\vec{B}\) , where m is the magnetic moment of the loop. (CBSE AI 2012, Delhi 2013)
Answer:
The figure below shows a rectangular loop of wire with length ‘a’ and breadth ‘b’. A line perpendicular to the plane of the loop (i.e. a normal to the plane) makes an angle Φ with the direction of the magnetic field B, and the loop carries a current / as shown. Let the forces acting on the various sides of the loop be \(\vec{F}_{1}\), \(\vec{F}_{2}\), \(\vec{F}_{3}\) and \(\vec{F}_{4}\) as shown.

It follows from the expression for the force experienced by a conductor in a magnetic field that force on arm AB is
\(\vec{F}_{1}=I(\overrightarrow{\mathrm{AB}} \times \vec{B})\) …(1)
(in the plane of the paper and is directed upwards as shown).

The force on arm CD is given by \(\vec{F}_{2}=l(\overrightarrow{\mathrm{CD}} \times \vec{B})\) …(2)
(in the plane of the paper and is directed downwards as shown.)

Since these two forces are equal and opposite and have the same line of action therefore they cancel out each other’s effect and their resultant effect on the coil is zero.

Now the force on arm BC is
\(\vec{F}_{3}=l(\overrightarrow{\mathrm{BC}} \times \vec{B})\) …(3)
(acts perpendicular to the plane of the paper and is directed outwards as shown) Finally the force on arm DA is
\(\vec{F}_{4}=I(\overrightarrow{\mathrm{DA}} \times \vec{B})\) …(4)
(acts perpendicular to the plane of the paper and is directed inwards as shown in figure).

Both forces F₃ and F₄ make an angle of 90° with the direction of the magnetic field. Therefore, in magnitude, these forces are given by
F₃ = F₄ = laB sin90° = laB …(5)

The lines of action of both these forces are perpendicular to the plane of the paper. The two forces F3 and F4 lie along different lines and each give rise to a torque about the X-axis. The two torques produce a resultant torque in + X direction. The arm of the couple (perpendicular distance between the lines of action of the two forces) from the figure below is given by
An arm of couple = b Sin Φ …(6)

Therefore, by the definition of torque we have
Torque = either force × arm of couple Using equations (5) and (6) we have
Torque = l B a × b Sin Φ
But a b = A, area of the coil, therefore τ = l B A Sin Φ

Question 21.
Show that a force that does no work must be a velocity-dependent force. (NCERT Exemplar)
Answer:
We know that work is the dot product of force and displacement, therefore
dW = \(\vec{F} \cdot d \vec{l}\) = 0
Or
dW = \(\vec{F}\) .\(\vec{v}\)dt = 0
Or
\(\vec{F} \cdot \vec{v}\) = 0

Thus F must be velocity dependent which implies that the angle between F and v is 90°. If v changes direction then the direction of F should also change so that the above condition is satisfied.

Question 22.
Five long wires A, B, C, D, and E, each carrying current l is arranged to form edges of a pentagonal prism as shown in the figure. Each carries current out of the plane of the paper.
(a) What will be magnetic induction at a point on axis 0? Axis is at a distance R from each wire.
(b) What will be the field if the current in one of the wires (say A) is switched off?
(c) What if the current in one of the wires (say) A is reversed? (NCERT Exemplar)

Answer:
(a) Zero Mn I
(b) \(\frac{\mu_{0}}{2 \pi} \frac{1}{R}\) perpendicular to AO towards left
(c) \(\frac{\mu_{0}}{\pi} \frac{1}{R}\) perpendicular to AO towards left

Moving Charges and Magnetism Important Extra Questions Long Answer Type

Question 1.
(a) A particle of charge ‘q’ and mass ‘m’, moving with velocity \(\vec{v}\) is subjected to a uniform magnetic field \(\vec{B}\) perpendicular to its velocity. Show that the particle describes a circular path. Obtain an expression for the radius of the circular path of the particle.
Answer:
Let a charged particle of charge q and mass m be moving with velocity \(\vec{v}\) right angle to the field (i.e. in the plane of the paper), then magnetic force \(\vec{F}\) acting on the charge q will be
\(\vec{F}=q(\vec{v} \times \vec{B})\)
or
F = qvB sin 90°
or
F = qvB … (1)

As this forces fact at a right angle to the velocity V of the charged particle, the slot is unable to change the velocity but can make the charged particle move In a circular path.

If r is the radius of the circle, then the centripetal force required by the charged particle will be
\(F_{c}=\frac{m v^{2}}{r}\)

This centripetal force Is provided by the magnetic force acting on the charged particle.

(b) Explain, how its path will be affected if the velocity \(\vec{v}\) makes an angle (θ ≠ 90°) with the direction of the magnetic field. (CBSE 2019C)
Answer:
If (θ ≠ 90°), the velocity \(\vec{v}\) of the moving charge can be resolved into two components v cos θ, in the direction of the magnetic field and make it v sin θ, in the direction perpendicular to the magnetic field. The charged particle under the combined effect of the two components of velocities wilt cover linear as well as a circular path, i.e. helical path whose axis is parallel to the magnetic field.

Question 2.
(a) Obtain the conditions under which an electron does not suffer any deflection while passing through a magnetic field.
Answer:
No deflection suffered by the electron if it moves parallel or anti-parallel to the magnetic field.

(b) Two protons P and Q moving with the same speed pass through the magnetic fields \(\vec{B}_{1}\) and \(\vec{B}_{2}\) respectively, at right angles to the field directions. If \(\left|\vec{B}_{2}\right|>\left|\vec{B}_{1}\right|\), which of the two protons will describe the circular path of smaller radius? Explain. (CBSEAI 2019)
Answer:
The radius of the circular path traveled by a charged particle in a magnetic field is given by
r = \(\frac{mv}{Bq}\)

Therefore, \(\frac{r_{1}}{r_{2}}=\frac{B_{2}}{B_{1}}\)

As \(\left|\vec{B}_{2}\right|>\left|\vec{B}_{1}\right|\) therefore, r₂ < r₁

Question 3.
A straight wire of length L is bent into a semi-circular loop. Use Biot-Savart law to deduce an expression for the magnetic field at its center due to the current l passing through it. (CBSE Delhi 2011C)
Answer:
Consider a straight wire of length L. Let it be bent into a semicircular arc of radius r as shown,

Then πr = L or r = L/π

Let a current I be passed through it. Divide the semi-circular loop into a large number of elements; consider one such element PQ of length dl. Then the small magnetic field cfB produced at the point O is

dB = \(\frac{\mu_{0}}{4 \pi} \frac{l d l \sin 90^{\circ}}{r^{2}}=\frac{\mu_{0}}{4 \pi} \frac{l d l}{r^{2}}\) outwards at point O.

Therefore total magnetic field at point O is
B = \(\int_{0}^{\pi r} \frac{\mu_{0}}{4 \pi} \frac{l d l}{r^{2}}=\frac{\mu_{0} l}{4 r}=\frac{\mu_{0} l \pi}{4 L}\)

Question 4.
A circular coil of N turns and radius R carries a current l. It is unwound and rewound to make another coil of radius R/2, current l remaining the same. Calculate the ratio of the magnetic moments of the new coil and the original coil. (CBSE AI 2012)
Answer:
The magnetic moment of a current-carrying coil is given by
M = n lA = n l πR²

When the coil is unwound and wound into another coil of radius R/2, the number of turns will double, i.e. n = 2
Therefore, n₁ = 1, n₂ = 2, R₁ = R, R₂ = R/2, hence

Therefore \(\frac{M_{2}}{M_{1}}=\frac{n_{2} / \pi R_{2}^{2}}{n_{1} / \pi R_{1}^{2}}=\frac{2}{4}=\frac{1}{2}\)

Question 5.
Two identical coils P and Q each of radius R are lying in perpendicular planes such that they have a common center. Find the magnitude and direction of the magnetic field at the common center when they carry currents equal to l and \(\sqrt{3}\) l respectively. (CBSE Al 2019)

Answer:
Magnetic field at the common centre due to coil P
\(B_{1}=\frac{\mu_{o} l}{2 R}\)

Magnetic field at the common centre due to coil Q
\(B_{2}=\frac{\mu_{0} \sqrt{3} l}{2 R}\)

The two fields are mutually perpendicular, therefore, the resultant field at the common centre is
B = \(\sqrt{B_{1}^{2}+B_{2}^{2}}=\sqrt{\left(\frac{\mu_{o} l}{2 R}\right)^{2}+\left(\frac{\mu_{0} \sqrt{3} 1}{2 R}\right)^{2}}\)
= \(\frac{\mu_{o} l}{R}\)

Let the resultant field make an angle θ with magnetic field B₂ as shown
tan θ = \(\frac{B_{1}}{B_{2}}=\frac{1}{\sqrt{3}}\) ⇒ θ = 30°

Question 6.
Find the condition under which the charged particles moving with different speeds in the presence of electric and magnetic field vectors can be used to select charged particles of a particular speed. (CBSE A! 2017)
Answer:
Consider crossed electric and magnetic fields. Let the particle enter perpendicular to both these fields. Due to this, the electric and magnetic forces are in opposite directions. Suppose, we adjust the value of E and B such that the magnitudes of the two forces are equal. Then, the total force on the charge is zero and the charge will move in the fields undeflected.

This happens when, or qE = qvB or v = E/B

This condition can be used to select charged particles of a particular velocity out of a beam containing charges moving at different speeds (irrespective of their charge and mass). The crossed E and B fields, therefore, serve as a velocity selector. Only particles with speed E/B pass undeflected through the region of crossed fields.

Question 7.
(a) Define the SI unit of current in terms of the force between two parallel current-carrying conductors.
Answer:
One ampere is that much current which when flowing through each of the two infinitely long straight conductors held 1 meter apart in space, produces a force of F = 2 × 10^-7 N per meter of their length.

(b) Two long straight parallel conductors carrying steady currents l_a and l_b along the same direction are separated by a distance d. How does one explain the force of attraction between them? If a third conductor carrying a current l in the opposite direction is placed just in the middle of these conductors, find the resultant force acting on the third conductor. (CBSE AI, Delhi 2018C)
Answer:
The magnetic field produced by one current applies a force on the other current-carrying conductor. By Fleming’s left-hand rule the forces acting on each due to other is directed towards each other. This shows attraction.

Force on C conductor due to conductor A
F_ca = \(\frac{\mu_{0} l_{a} l_{c}}{\pi d}\) repulsive

Force on C conductor due to conductor B
F_cb = \(\frac{\mu_{0} l_{b} l_{c}}{\pi d}\)

\(F_{\mathrm{ac}}-F_{\mathrm{bc}}=\frac{\mu_{0} l_{\mathrm{c}}}{\pi d}\left(l_{a}-l_{b}\right)\)

Question 8.
ExplaIn how will you convert a galvanometer into a voltmeter to read a maximum potential of ‘V’ volt. Can one use a voltmeter to measure the emf of a cell? Justify your answer.
Answer:
Suppose a galvanometer having resistance G is to be converted into a voltmeter, which can measure the potential difference from O to V volt. Let a high resistance R be joined in senes with the galvanometer for this purpose. Its value is so chosen that when the galvanometer with the resistance is connected between two points having a potential difference of V volt, the gaLvanometer gives full-scale deflection. It is clear from the figure below that
V = l_g (R + G)
or
R = \(\frac{v}{l_{g}}\) – G

On connecting the above high resistance in series with a galvanometer, the galvanometer is converted into a voltmeter of range V volt.

A voltmeter cannot measure the emf of the cell as it draws current from the cell while measuring the potential differences.

Question 9.
Explain how will you convert a galvanometer into an ammeter to read a maximum current of ‘l’ ampere. An ammeter is always connected in series with a circuit. Why? (CBSE AI 2019)
Answer:
Suppose a galvanometer of resistance G is to be converted into an ammeter having range 0 to l ampere. Let lg be the current,

which gives full-scale deflection in the galvanometer. Suppose S is the appropriate shunt required for this purpose, i.e. when shunt S Is used, current ‘g passes through the galvanometer and the remaining (l – l_g) passes through the shunt as shown in the figure below. Since the shunt and the galvanometer are connected In parallel therefore the potential differences across both wilL be the same. Hence
l_g G = (l – l_g) S ….(1)

S = \(\frac{l_{s} G}{l-l_{s}}\) ….(2)

An ammeter is used to measure current; therefore, it is connected in senes so that the entire current passes through it. Moreover, an ammeter is a low resistance device.

Question 10.
A steady current (l₁) flows through a long straight wire. Another wlrè carrying steady current (l₂) In the same direction is kept close and parallel to the first wire. Show with the help of a diagram how the magnetic field due to the current l₁ exerts a magnetic force on the second wire. Write the expression for this force. (CBSE AI 2011)
Answer:
The diagram is as shown below.

Magnetic field B1 is produced by the wire carrying current l₁ on a wire carrying current l₂. Thus the second current-carrying wire is placed in the magnetic field produced by the first as a result it experiences a force is given by F = \(\frac{\mu_{0} l_{1} l_{2}}{2 \pi r}\) per unit length.

Question 11.
(a) State Ampere’s circuital law expressing it in the integral form.
(b) Two long coaxial insulated solenoids S₁ and S₂ of equal lengths are wound one over the other as shown in the figure. A steady current ‘l’ flows through the inner solenoid S1 to the other end B, which is connected to the outer solenoid S2 through which the same current ‘l’ flows in the opposite direction so as to come out at end A. If n₁ and n₂ are the numbers of turns per unit length, find the magnitude and direction of the net magnetic field at a point
(i) inside on the axis and
(ii) outside the combined system. (CBSE Delhi 2014)

Answer:
(a) The line integral of B around any closed path equals µ_ol, where l is the total steady current passing through any surface bounded by the closed path.” Mathematically
\(\oint \vec{B} \cdot \overrightarrow{d L}=\mu_{0} I\)

(b) (i) The magnetic field due to a current-carrying solenoid:
B = µ_on l

where n = number of turns per unit length
l = current through the solenoid

Now, the magnetic field due to solenoid S₁ will be in the upward direction and the magnetic field due to S₂ will be in the downward direction (by right-hand screw rule).

In the upward direction

(ii) The magnetic field is zero outside a solenoid.

Question 12.
Three long straight parallel wires are kept as shown in the Figure. The wire (3) carries a current l

(i) The direction of flow of current l in wire (3) is such that the net force, on a wire (1), due to the other two wires, is zero.
(ii) By reversing the direction of l, the net force on the wire (2), due to the other two wires, becomes zero. What will be the direction of current l, in the two cases? Also obtain the relation between the magnitudes of currents l₁, l_2, and l. (CBSE Delhi 2016C)
Answer:
Case 1: The direction of flow of current in the wire (3) will be opposite to the direction of flow of current in the wire (1), i.e. downwards.
Also \(\frac{\mu_{0} l_{1} l}{2 \pi(2 a)}=\frac{\mu_{0} l_{1} l_{2}}{2 \pi a}\)
or
l = 2l₂

Case 2: The direction of flow of current in wire (3) will be same as the direction of flow of current in wire (1), i.e. upwards. For zero force on wire (2) we have
\(\frac{\mu_{0} l_{1} l_{2}}{2 \pi a}=\frac{\mu_{0} l l_{2}}{2 \pi a}\)
or
l = l₁
Therefore l = l₁ = 2 l₂

Question 13.
Show mathematically that the cyclotron frequency does not depend upon the speed of the particle.
Answer:
Due to the presence of the perpendicular magnetic field the particle moves in a circle of radius r given by
r = \(\frac{m v}{q B}\) ….(1)

The path of the particles in the dees is a semicircle and the time the particle spends in each dees is length of the semi circular path velocity
t = \(\frac{\text { length of the semi circular path }}{\text { velocity }}\)
= \(\frac{\pi r}{v}=\frac{\pi m}{B q}\) …(2)

using equation (1)
The above time is independent of the radius of the path and the velocity of the charged particle. Now the time period of the cyclotron is twice the time spent by the particle in each dee. Thus

T = 2t = \(\frac{2 \pi m}{B q}\) …(3)

Hence cyclotron frequency or the magnetic resonance frequency is given by
v = \(\frac{1}{T}=\frac{B q}{2 \pi m}\) …(4)

which is independent of the speed of the particle.

Question 14.
State the principle of a cyclotron. Show that the time period of revolution of particles in a cyclotron is independent of their speeds. Why is this property necessary for the operation of a cyclotron? (CBSE Al 2016)
Answer:
Principle of cyclotron: It is based on the principle that the positive ions can be accelerated to high energies with a comparatively smaller alternating potential difference by making them cross the electric field again and again, by making use of a strong magnetic field.

The necessary centripetal force required by charged particle to revolve in a circular path in magnetic field is provided by force due to magnetic field, i.e.
\(\frac{m v^{2}}{r}\) = Bqv
or
v = \(\frac{\text { Bqr }}{m}\)

So, the frequency of revolution is then given by
T = \(\frac{2 \pi r}{v}=2 \pi r \times \frac{m}{B q r}=\frac{2 \pi m}{B q}\)

It is clear from the expression that T is Independent of speed.

If this condition is not met the charged particle will very soon go out of step with the applied electricity and will not be accelerated.

Question 15.
(a) Write the expression for the magnetic force acting on a charged particle moving with velocity v in the presence of magnetic field B.
Answer:
\(\vec{F}\) = q\((\vec{v} \times \vec{B})\)

(b) A neutron, an electron and an alpha particle moving with equal velocities enter a uniform magnetic field going into the plane of the paper as shown. Trace their paths in the field and justify your answer. (CBSE Delhi 2016)

Answer:
The path area is shown.

The radius of the circular path traveled by each particle is given by the expression
r = \(\frac{m v}{B q}\) since B and v is the same therefore

we have r ∝ \(\frac{m}{q}\). Since neutron does not have a charge therefore it passes straight without deflection. The ratio m/q for an alpha particle is greater for an alpha particle therefore its path will be less curved.

Also by Fleming’s left-hand rule the alpha particle and the electron will experience a force in the direction as shown.

Question 16.
A long solenoid of length ‘L’ having N turns carries a current l. Deduce the expression for the magnetic field in the interior of the solenoid. (CBSE AI 2011C)
Answer:
Consider a rectangular path abed of length L as shown in the figure below. Let us apply Ampere’s circuital law to this rectangular path so that we have



But by Ampere’s circuital law we have
\(\oint \vec{B} \cdot \vec{d} L\) = μ₀Nl = μ0(nL)l …(2)

From equations (1) and (2) we have
BL = μ₀nLl
or
B = μ₀ n l

But n = N/L, therefore we have
B = μ₀ \(\frac{N l}{L}\)

This gives the value of a magnetic field inside a solenoid.

Question 17.
Using Biot-Savart’s law, derive an expression for the magnetic field intensity at the center of a current-carrying circular coil.
Answer:
Consider a circular loop of radius r carrying a current l and having a center at O as shown in the figure below.

Consider a small current element dL on the loop. Then by Biot-Savart’s law the magnitude of the magnetic field at the center of the loop due to the current element we have
dB = \(\frac{\mu_{0}}{4 \pi} \frac{l d L \sin \theta}{r^{2}}\) …(1)

In this case, the angle between the current element dL and the radius vector is 90° therefore equation (1) can be written as
dB = \(\frac{\mu_{0}}{4 \pi} \frac{l d L}{r^{2}}\) ….(2)

The circular loop can be considered to be consisting of such small elements placed side by side, and then the magnetic intensities of these elements will be in the same direction. Thus, the net intensity of B at the center of the loop is given by

Question 18.
A charged particle q is moving in the presence of a magnetic field B which is inclined to an angle 300 with the direction of the motion of the particle. Draw the trajectory followed by the particle In the presence of the field and explain how the particle describes this path. (CBSE Delhi 2019)
Answer:
Two components of velocity vector V are responsible for the helical motion. Force on the charged particle due to the component normal to the magnetic field acts perpendicular to the velocity and the magnetic field and makes the particle follow a circular path. The component of velocity which is along the magnetic field does not cause any force on the particle, hence the particle continues to move in a straight line path due to this component so, the resultant path will be helical.

The path is as shown.

Question 19.
Two parallel coaxial circular costs of equal radius ‘R’ and an equal number of turns ‘N’, carry equal currents ‘I’ in the same direction and are separated by a distance ‘2R’. Find the magnitude and direction of the net magnetic field produced at the mid-point of the line joining their centers.
Answer:
The magnetic field at a distance R from a circular coil is given by the expression

Both are directed in the same direction, therefore the resultant magnetic field at the center is
B = \(\frac{\mu_{0} N l R^{2}}{\left(2 R^{2}\right)^{1 / 2}}\)

Question 20.
(a) State Biot-Savart’s law. Using this law, derive the expression for the magnetic field due to a current-carrying circular loop of radius ‘R’, at a point which is at a distance ‘x’ from Its center along the axis of the loop.
Answer:
It states that for a small current element the small magnetic field is given by the expression Biot-Savart’s law for a small current element \(\overrightarrow{d B}=\frac{\mu_{0}}{4 \pi} \frac{I \overline{(d L} \times \hat{r})}{r^{2}}\).

(b) Two small identical circular loops, marked (1) and (2), carrying equal currents, are placed with the geometrical axes perpendicular to each other as shown in the figure. Find the magnitude and direction of the net magnetic field produced at point O.

Answer:
The magnetic field at O due to the circular loop 1 is B₁ = \(\frac{\mu_{0} l R^{2}}{2\left(x^{2}+R^{2}\right)^{1 / 2}}\) directed towards left.

The magnetic field at O due to the circular loop is B₁ = \(\frac{\mu_{0} l R^{2}}{2\left(x^{2}+R^{2}\right)^{1 / 2}}\) directed upwards.

The net magnetic field is therefore
B = \(\sqrt{B_{1}^{2}+B_{2}^{2}}=\sqrt{2} B_{1}=\frac{\mu_{0} l R^{2}}{\sqrt{2}\left(x^{2}+R^{2}\right)^{3 / 2}}\)

The direction of the net magnetic field is 45° with the axis of the loop as shown in the figure below.

Question 21.
Derive an expression for the magnetic field along the axis of an air-cored solenoid, using Ampere’s circuital law. Sketch the magnetic field lines for a finite solenoid. Explain why the field at the exterior mid-point Is weak while at the interior it is uniform and strong.
Answer:
A solenoid is a coil of wire with a Length, which is Large as compared with its diameter.

Consider an ideaL soLenoid carrying current l and having n turns per unit length.
(a) Consider a rectangular path abed of length L as shown in the figure below. Let us apply Ampere’s circuital law to this rectangular path so that we have



But by Ampere’s circuital law we have
\(\oint \vec{B} \cdot \vec{d} L\) = μ₀Nl = μ₀(nL)l …(2)

From equations (1) and (2) we have
BL = μ₀nLl
or
B = μ₀ n l

But n = N/L, therefore we have
B = μ₀ \(\frac{N l}{L}\)

This gives the value of a magnetic field inside a solenoid.

(b) The sketch Is as shown below.

The magnetic field gets added inside the solenoid whereas it is not added outside the solenoid.

Question 22.
(a)Using Ampere’s circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis.
Answer:
Consider a rectangular path abed of length L as shown in the figure below. Let us apply Ampere’s circuital law to this rectangular path so that we have



But by Ampere’s circuital law we have
\(\oint \vec{B} \cdot \vec{d} L\) = μ₀Nl = μ₀(nL)l …(2)

From equations (1) and (2) we have
BL = μ₀nLl
or
B = μ₀ n l

But n = N/L, therefore we have
B = μ₀ \(\frac{N l}{L}\)

This gives the value of a magnetic field inside a solenoid.

(b) In what respect is a toroid different from a solenoid? Draw and compare the pattern of the magnetic field lines in the two cases.
Answer:
(b) In a toroid magnetic field is uniform whereas in a solenoid it is different at the two ends and the center.
The magnetic field lines around the two are as shown.


(c) How is the magnetic field inside a given solenoid made strong? (CBSE Al 2011)
Answer:
(i) By inserting a ferromagnetic substance inside the solenoid.
(ii) By increasing the amount of current through the solenoid.

Question 23.
(a) Explain, giving reasons, the basic difference in converting a galvanometer into
(i) a voltmeter
Answer:
A voltmeter is always connected in parallel with the section of the circuit whose potential difference has to be measured. Further, it must draw a small current, otherwise, the voltage measurement will disturb the original setup by an amount that is very large. Thus a large resistance is connected to the galvanometer in series so as to minimize this effect.

(ii) an ammeter
Answer:
An ammeter measures current and is to be connected in series in a circuit. A galvanometer has a large resistance, therefore a shunt is connected to it in order to decrease its resistance such that the current in the circuit is not altered.

(b) Two long straight parallel conductors carrying steady currents l₁ and l₂ are separated by a distanced’. Explain briefly, with the help of a suitable diagram, how the magnetic field due to one conductor acts on the other. Hence deduce the expression for the force acting between the two conductors. Mention the nature of this force. (CBSE At 2012)
Answer:

Consider two long, straight parallel wires separated by a distance ‘a’ and carrying currents l₁ and l₂ in the same direction as shown. We can easily determine the force on one wire due to the magnetic field set-up by the other wire. Wire 2, which carries a current l₂, sets up a magnetic field B₂ at the position of wire 1.

The direction of B₂ is perpendicular to the wire, as shown in figure. Now the magnetic force on length L of wire 1 is
\(\vec{F}_{1}=l_{1}\left(\vec{L} \times \vec{B}_{2}\right)\)

Since L is perpendicular to B₂, the magnitude of F, is given by F₁ = l₁ L B₂ …(1)
But the field due to wire 2 is given by the relation
B₂ = \(\frac{\mu_{0} l_{2}}{2 \pi a}\) ….(2)

Therefore, from equations 1 and 2 we have
F₁= l₁ L B₂ = l₁ L\(\left(\frac{\mu_{0} l_{2}}{2 \pi a}\right)=\frac{L \mu_{0} l_{1} l_{2}}{2 \pi a}\)

We can rewrite it in terms of force per unit length as
\(\frac{F_{1}}{L}=\frac{\mu_{0} l_{1} l_{2}}{2 \pi a}\) ….(3)

The direction of F₁ is downward, towards wire 2, since \(\vec{L} \times \vec{B}\) is downwards. If one considers the field set up at wire 2 due to wire 1, the force F₂ is found to be equal and opposite to F₁ which is in accordance with Newton’s third law of motion. When the currents are in opposite directions, the forces are reversed and the wires repel each other. Hence we find that the force per unit length of wire between two parallel current-carrying wires is given by
F = \(\frac{\mu_{0} l_{1} l_{2}}{2 \pi a}\)

Therefore, two conductors carrying current in the same direction attract each other whereas two conductors carrying current in the opposite directions repel each other.

The above expression can be used to define ampere, the SI unit of current.
Let l₁ = l₂ = 1 ampere, a = 1 m then F = 2 × 10^-7 Nm^-1

Thus we have
One ampere is that much current which when flowing through each of the two infinitely long straight conductors held 1 meter apart in space, produce a force of F = 2 × 10^-7 N per meter of their length.

Question 24.
(a) State the underlying principle of a moving coil galvanometer.
Answer:
A current-carrying coil placed in a magnetic field experiences a torque.

(b) Give two reasons to explain why a galvanometer cannot as such be used to measure the value of the current in a given circuit.
Answer:

1. A galvanometer is a very sensitive device; it gives a full-scale deflection for a current of the order of a few pA.
2. The resistance of the galvanometer is not very small, hence it will change the value of current in the circuit branch when connected in series in that branch.

(c) Define the terms:
(i) voltage sensitivity and
(ii) current sensitivity of a galvanometer. (CBSE Delhi 2019)
Answer:

• Voltage sensitivity: Voltage sensitivity is defined as the deflection per unit potential difference applied.
• Current sensitivity: It is defined as the deflection per unit current.

Question 25.
(a) State Biot-Savart law. Deduce the expression for the magnetic field due to a circular current carrying loop at a point lying on its axis.
Answer:
It states that for a small current element the small magnetic field is given by the expression Biot-Savart’s law for a small current element \(\overrightarrow{d B}=\frac{\mu_{0}}{4 \pi} \frac{I \overline{(d L} \times \hat{r})}{r^{2}}\).

(b) Two long parallel wires carry currents l₁ and l₁ flowing in the same direction. When a third current-carrying wire is placed parallel and coplanar in between the two, find the condition when the third wire would experience no force due to these two wires. (CBSE AI 2012C)
Answer:
Two current-carrying wires carrying current in the same direction attract and those carrying current in the opposite direction repel. The current in the third wire at the center should be opposite to the current in the two wires.

The conditions should be

• The center wire should carry current in the opposite direction to the two wires and
• The center wire should be closer to the wire carrying lesser current.

Question 26.
(a) Derive the expression for the torque on a rectangular current-carrying loop suspended in a uniform magnetic field.
Answer:
The figure below shows a rectangular loop of wire with length ‘a’ and breadth ‘b’. A line perpendicular to the plane of the loop (i.e. a normal to the plane) makes an angle Φ with the direction of the magnetic field B, and the loop carries a current / as shown. Let the forces acting on the various sides of the loop be \(\vec{F}_{1}\), \(\vec{F}_{2}\), \(\vec{F}_{3}\) and \(\vec{F}_{4}\) as shown.

It follows from the expression for the force experienced by a conductor in a magnetic field that force on arm AB is
\(\vec{F}_{1}=I(\overrightarrow{\mathrm{AB}} \times \vec{B})\) …(1)
(in the plane of the paper and is directed upwards as shown).

The force on arm CD is given by \(\vec{F}_{2}=l(\overrightarrow{\mathrm{CD}} \times \vec{B})\) …(2)
(in the plane of the paper and is directed downwards as shown.)

Since these two forces are equal and opposite and have the same line of action therefore they cancel out each other’s effect and their resultant effect on the coil is zero.

Now the force on arm BC is
\(\vec{F}_{3}=l(\overrightarrow{\mathrm{BC}} \times \vec{B})\) …(3)
(acts perpendicular to the plane of the paper and is directed outwards as shown) Finally the force on arm DA is
\(\vec{F}_{4}=I(\overrightarrow{\mathrm{DA}} \times \vec{B})\) …(4)
(acts perpendicular to the plane of the paper and is directed inwards as shown in figure).

Both forces F₃ and F4 make an angle of 90° with the direction of the magnetic field. Therefore, in magnitude, these forces are given by
F₃ = F₄ = l a B sin90° = l a B …(5)

The lines of action of both these forces are perpendicular to the plane of the paper. The two forces F₃ and F₄ lie along different lines and each give rise to a torque about the X-axis. The two torques produce a resultant torque in the + X direction. The arm of the couple (perpendicular distance between the lines of action of the two forces) from the figure below is given by
The arm of couple = b Sin Φ …(6)

Therefore, by the definition of torque we have
Torque = either force × arm of couple Using equations (5) and (6) we have
Torque = l B a × b Sin Φ
But a b = A, area of the coil, therefore τ = l B A Sin Φ

(b) A proton and a deuteron having equal momenta enter in a region of the uniform magnetic field at a right angle to the direction of the field. Depict their trajectories in the field. (CBSE Delhi 2013)
Answer:
The radius of the curved path which will be followed by the two particles is given by the expression r = \(\frac{m v}{B q}\)

As m_d > m_p it will follow the path of a bigger radius.

Question 27.
State Biot-Savart law, expressing it in the vector form. Use it to obtain the expression for the magnetic field at an axial point, distance ‘d’ from the center of a circular coil of radius ‘a’ carrying current ‘l’. Also, find the ratio of the magnitudes of the magnetic field of this coil at the center and at an axial point for which d = a \(\sqrt{3}\) (CBSE Delhi 2013C)
Answer:
(a) It states that the magnetic field due to a current element dl at a distance r from it is given by the expression

dB = \(\frac{\mu_{0}}{4 \pi} \frac{l d l \sin \theta}{r^{2}}\). In vector form it is written as
\(\overrightarrow{d B}=k_{m} \frac{l mid \overrightarrow{d L} \times \hat{r}}{r^{2}}\)

Consider a circular loop of wire of radius R located in the YZ plane and carrying a steady current / as shown in the figure below. Let us calculate the magnetic field at an axial point P a distance x from the center of the loop. From the figure it is clear that any element dL is perpendicular to r, furthermore, all the elements around the loop are at the same distance r from P, where r² = X² + R². Hence by Biot-Savart’s law, the magnetic field at point P due to the current element dL is given by


The direction of the magnetic field dB due to the eLement dL is perpendicular to the plane formed by r and dL as shown in the figure above. The vector dB can be resolved into components dB aLong the X-axis and dB which is perpendicular to the X-axis. When the components perpendicular to the X-axis are assumed over the whole loop, the result is zero. That is, by symmetry any element on one side of the loop will set up a perpendicular component that cancels the component setup by an element diametrically opposite it. Therefore, it is obvious that the resultant magnetic field at P will be along the X-axis. This result can be obtained by integrating the components dB_x = dB cos θ.
Therefore, we have

where the integral is to be taken over the entire loop since θ,x and R are constants for all elements of the loop and since
cos θ = \(\frac{R}{\sqrt{x^{2}+R^{2}}}\)

B = \(\frac{\mu_{0} l R}{4 \pi\left(x^{2}+R^{2}\right)^{3 / 2}} \oint d L=\frac{\mu_{0} l R^{2}}{2\left(x^{2}+R^{2}\right)^{1 / 2}}\)

To find the magnetic field at the centre of the loop setting x = O in equation the above equation we have B = \(\frac{\mu_{0} l}{2 R}\)

Now magnetic field at the centre of a circular coil is
\(B_{C}=\frac{\mu_{0} l}{2 a}\) …(1)

Also magnetic field on the axial line when
d = a\(\sqrt{3}\) is

From (1) and (2) we have
\(\frac{B_{C}}{B_{\text {axial }}}=\frac{\mu_{0} l}{2 a} \times \frac{16 a}{\mu_{0} l}\) = 8

Question 28.
A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2.0 cm as shown in figure (a) Consider the magnetic field B at the center of the arc
(a) What is the magnetic field due to the straight segments?
(b) In what way the contribution to B from the semicircle differs from that of a circular loop and in what way does it resemble?
(c) Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in figure (b)? (NCERT)

Answer:
(a) dl and r for each element of the straight segments are parallel. Therefore \(d \vec{L} \times \hat{r}\) =0. Straight segments do not contribute to the magnetic field at the center of the semicircular arc.

(b) For all segments of the semicircular arc \(d \vec{L} \times \hat{r}\) are all parallel to, each other (into the plane of the paper). All such contributions add up in magnitude. Hence the direction of B for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop. Thus
B = \(\frac{1}{2} \frac{\mu_{0} l}{2 r}=\frac{4 \pi \times 10^{-7} \times 12}{4 \times 2 \times 10^{-2}}\) = 9 × 10^-4 T normal to the plane of the paper going into it.

(c) Same magnitude of B but opposite in direction to that in (b)

Question 29.
A long straight wire carrying a current of 25 A rests on a table as shown in the figure. Another wire PQ of length 1 m, mass 2.5 g carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise? (NCERT Exemplar)

Answer:
Given l₁ = l₁ = 25 A, L = 1 m, m = 2.5 g = 2.5 × 10^-3 kg
The repulsive force between PQ and the wire on the table will balance the weight of wire PQ. Let this happen when PQ is at a height h above the wire, then
F = mg

Question 30.
Two straight parallel conductors carry steady currents l₁ and l₂ separated by a distance d. if currents are flowing in the same direction, show how the magnetic field set-up in one produces an attractive force on the other. Obtain the expression for force. Hence define one ampere. (CBSE Delhi 2016)
Answer:
Consider two long, straight parallel wires separated by a distance ‘a’ and carrying currents l₁ and l₂ in the same direction as shown. We can easily determine the force on one wire due to the magnetic field set-up by the other wire. Wire 2, which carries a current l₂, sets up a magnetic field B₂ at the position of wire 1.

The direction of B₂ is perpendicular to the wire, as shown in figure. Now the magnetic force on length L of wire 1 is
\(\vec{F}_{1}=l_{1}\left(\vec{L} \times \vec{B}_{2}\right)\)

Since L is perpendicular to B₂, the magnitude of F, is given by F₁ = l₁ L B₂ …(1)
But the field due to wire 2 is given by the relation
B₂ = \(\frac{\mu_{0} l_{2}}{2 \pi a}\) ….(2)

Therefore, from equations 1 and 2 we have
F₁= l₁ L B₂ = l₁ L\(\left(\frac{\mu_{0} l_{2}}{2 \pi a}\right)=\frac{L \mu_{0} l_{1} l_{2}}{2 \pi a}\)

We can rewrite it in terms of force per unit length as
\(\frac{F_{1}}{L}=\frac{\mu_{0} l_{1} l_{2}}{2 \pi a}\) ….(3)

The direction of F₁ is downward, towards wire 2, since \(\vec{L} \times \vec{B}\) is downwards. If one considers the field set up at wire 2 due to wire 1, the force F₂ is found to be equal and opposite to F₁ which is in accordance with Newton’s third law of motion. When the currents are in opposite directions, the forces are reversed and the wires repel each other. Hence we find that the force per unit length of wire between two parallel current-carrying wires is given by
F = \(\frac{\mu_{0} l_{1} l_{2}}{2 \pi a}\)

Therefore, two conductors carrying current in the same direction attract each other whereas two conductors carrying current in the opposite directions repel each other.

The above expression can be used to define ampere, the SI unit of current.
Let l₁ = l₂ = 1 ampere, a = 1 m then F = 2 × 10^-7 Nm^-1

Thus we have
One ampere is that much current which when flowing through each of the two infinitely long straight conductors held 1 meter apart in space, produce a force of F = 2 × 10^-7 N per meter of their length.

Question 31.
(a) In a moving coil galvanometer, why is the magnetic field required to be radial?
Answer:
The radial field is always normal to the surface of the coil in all positions of the coil, i.e. θ = 0. The radial field ensures a linear relation between 0 and l in a moving coil galvanometer.

(b) A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A.
Calculate (i) the magnetic field at the centre of the coil, and
(ii) its magnetic moment. (CBSE2019C)
Answer:
Here N = 100, r = 10 cm = 0.1 m,
l = 3.2 A, B = ?, p = ?
(i) At the centre of the coil

(ii)Magnetic moment
p = NIA
= 100 × 3.2 × π × (0.1 )² ≃ 10 Am2

Question 32.
State Ampere’s circuital law. Use this law to find a magnetic field due to a straight infinite current-carrying wire. How are the magnetic field lines different from the electrostatic field lines? (CBSE Al 2016)
Answer:
Statement: “The line integral of B around any closed path equals μ₀ l, where l is the total steady current passing through any surface bounded by the closed path.”

Consider a long circular wire of radius ‘a’ carrying a steady current (dc) that is uniformly distributed along the cross-section of the wire as shown in the figure. Let us calculate the magnetic field in the regions r ≥ a and r < a. In region 1 let us choose a circular path of radius r centered at the wire. From symmetry, we find that B is perpendicular to dL at every point on the circular path.

Since total current linked with the circular path is l₀, therefore by Ampere’s law we have

for r ≥ a
Magnetic field lines form closed loops while electrostatics field lines do not.

Question 33.
(a) State Biot-Savart law in vector form.
Answer:
It states that for a small current element the small magnetic field is given by the expression Biot-Savart’s law for a small current element \(\overrightarrow{d B}=\frac{\mu_{0}}{4 \pi} \frac{I \overline{(d L} \times \hat{r})}{r^{2}}\).

(b) Deduce the expression for the magnetic field at a point on the axis of a current-carrying circular loop of radius ‘R’, distant V from the center, hence write the magnetic field at the center of a loop. (CBSE AI 2015, Delhi 2018C)
Answer:
Consider a circular loop of wire of radius R located in the YZ plane and carrying a steady current / as shown in the figure below. Let us calculate the magnetic field at an axial point P a distance x from the center of the loop. From the figure it is clear that any element dL is perpendicular to r, furthermore, all the elements around the loop are at the same distance r from P, where r² = X² + R². Hence by Biot-Savart’s law, the magnetic field at point P due to the current element dL is given by


The direction of the magnetic field dB due to the element dL is perpendicular to the plane formed by r and dL as shown in the figure above. The vector dB can be resolved into components dB along the X-axis and dB which is perpendicular to the X-axis. When the components perpendicular to the X-axis are assumed over the whole loop, the result is zero. That is, by symmetry any element on one side of the loop will set up a perpendicular component that cancels the component setup by an element diametrically opposite it. Therefore, it is obvious that the resultant magnetic field at P will be along the X-axis. This result can be obtained by integrating the components dB_x = dB cos θ.
Therefore, we have

where the integral is to be taken over the entire loop since θ,x and R are constants for all elements of the loop and since
cos θ = \(\frac{R}{\sqrt{x^{2}+R^{2}}}\)

B = \(\frac{\mu_{0} l R}{4 \pi\left(x^{2}+R^{2}\right)^{3 / 2}} \oint d L=\frac{\mu_{0} l R^{2}}{2\left(x^{2}+R^{2}\right)^{1 / 2}}\)

To find the magnetic field at the centre of the loop setting x = O in equation the above equation we have B = \(\frac{\mu_{0} l}{2 R}\)

Question 34.
Derive an expression for the maximum force experienced by a straight conductor of length 1, carrying current I and kept in a uniform magnetic field, B.
Answer:
Consider a straight segment of a conducting wire; with length L and cross-section at area A, the current is from bottom to top as shown In the figure below. The wire is in a uniform magnetic field B perpendicular to the plane of the diagram and directed into the plane. Let us assume that the moving charges are positive.

The drift velocity is upward, perpendicular to B. The average force experienced by each charge is
\(\vec{f}=q\left(\vec{v}_{d} \times \vec{B}\right)\) …..(1)

Directed to the left as shown in the figure

Since \(\vec{V}_{d}\) and \(\vec{B}\) are perpendicular, the magnitude of the force is given by
f = q v_d B …(2)

Let n be the number density of charges, i.e. number of charges per unit volume. A segment of the conductor with length L has volume V = A L and contains a number of charges N given by
N = n A L …(3)

Now the total force F on all the charges moving in this segment is
F = N f = (n A L) q v_d B = (n q v_d A) (L B)

But n q v_d A = l, therefore the above equation becomes
F = B l L …(4)

Question 35.
Derive an expression for the torque on a rectangular coil of area A, carrying a current l and placed in a magnetic field B, the angle between the direction of 8 and the vector perpendicular to the plane of the coil is θ. (CBSE Delhi 2019)
Answer:
The figure below shows a rectangular loop of wire with length ‘a’ and breadth ‘b’. A line perpendicular to the plane of the loop (i.e. a normal to the plane) makes an angle Φ with the direction of the magnetic field B, and the loop carries a current / as shown. Let the forces acting on the various sides of the loop be \(\vec{F}_{1}\), \(\vec{F}_{2}\), \(\vec{F}_{3}\) and \(\vec{F}_{4}\) as shown.

It follows from the expression for the force experienced by a conductor in a magnetic field that force on arm AB is
\(\vec{F}_{1}=I(\overrightarrow{\mathrm{AB}} \times \vec{B})\) …(1)
(in the plane of the paper and is directed upwards as shown).

The force on arm CD is given by \(\vec{F}_{2}=l(\overrightarrow{\mathrm{CD}} \times \vec{B})\) …(2)
(in the plane of the paper and is directed downwards as shown.)

Since these two forces are equal and opposite and have the same line of action therefore they cancel out each other’s effect and their resultant effect on the coil is zero.

Now the force on arm BC is
\(\vec{F}_{3}=l(\overrightarrow{\mathrm{BC}} \times \vec{B})\) …(3)
(acts perpendicular to the plane of the paper and is directed outwards as shown) Finally the force on arm DA is
\(\vec{F}_{4}=I(\overrightarrow{\mathrm{DA}} \times \vec{B})\) …(4)
(acts perpendicular to the plane of the paper and is directed inwards as shown in figure).

Both forces F₃ and F₄ make an angle of 90° with the direction of the magnetic field. Therefore, in magnitude, these forces are given by
F₃ = F₄ = l a B sin90° = l a B …(5)

The lines of action of both these forces are perpendicular to the plane of the paper. The two forces F₃ and F₄ lie along different lines and each give rise to a torque about the X-axis. The two torques produce a resultant torque in the + X direction. The arm of the couple (perpendicular distance between the lines of action of the two forces) from the figure below is given by
The arm of couple = b Sin Φ …(6)

Therefore, by the definition of torque we have
Torque = either force × arm of couple Using equations (5) and (6) we have
Torque = l B a × b Sin Φ
But a b = A, area of the coil, therefore τ = l B A Sin Φ

Question 36.
Draw a schematic sketch of a cyclotron. State its working principle and write its two uses. (CBSE 2019C)
Answer:
Cyclotron

Working Principle: It is based on the principle that a positively charged particle can acquire very large energy with the small alternating potential difference if the particle is made to cross again and again the electric field produced by alternating potential difference applied and a strong perpendicular magnetic field is applied.

Uses:
(a) It is used to accelerate positively charged particles to very high energies.
(b) Cyclotrons are a source of high-energy beams for nuclear physics experiments.

Question 37.
(a) Draw a schematic sketch of a moving coil galvanometer and describe briefly its working.
Answer:
Principle: It is based on the principle that the positive ions can be accelerated to high energies with a comparatively smaller alternating potential difference by making them cross the electric field again and again, by making use of a strong magnetic field.

Construction: It consists of two D-shaped, hollow metal dees D₁ and D₂. The dees are placed with a small gap in between them. The dees are connected to a source of high frequency alternating potential difference. The dees are evacuated in order to minimize energy losses resulting from collisions between the ions and air molecules. The whole apparatus is placed between the poles of an electromagnet as shown in the figure below. The magnetic field is perpendicular to the plane of the dees.

Working: In a cyclotron, particles of mass m and charge q move inside an evacuated chamber in a uniform magnetic field B that is perpendicular to the plane of their paths. The alternating potential difference applied between the hollow electrodes D₁ and D₂ create an electric field in the gap between them which changes precisely twice in each revolution so that the particles get a push each time they cross the gap.

The pushes increase their speed and kinetic energy, boosting them into paths of larger radius. The electric field increases the speed and the magnetic field makes the particles move in circular paths. Due to the presence of the perpendicular magnetic field the particle moves in a circle of radius r given by
r = \(\frac{m v}{q B}\) …(1)

The path of the particles in the dees is a semicircle and the time the particle spends in each dee is semi-circular
τ = \(\frac{\text { length of the semi-circular path }}{\text { velocity }}=\frac{\pi r}{v}=\frac{\pi m}{B q}\) by using equation (1)

The above time is independent of the radius of the path and the velocity of the charged particle. Hence the charged particle spends the same time in each dee and encounters a dee of opposite polarity whenever it comes into the gap between the dees and thus is continuously accelerated.

(b) “Increasing the current sensitivity of a galvanometer does not necessarily increase the voltage sensitivity.” Justify’ this statement. (CBSE Delhi 2014C)
Answer:
Voltage and current sensitivities are related as VS = \(\frac{CS}{R}\) .

An increase in current sensitivity may lead to an increase in the resistance of the coil. Thus the factor CS/R may not be affected.

Question 38.
Explain with the help of a labeled diagram the underlying principle, construction, and working of a moving coil galvanometer.
Answer:
It is an instrument used to detect weak currents in a circuit.

Principle: It is based on the principle that, whenever a loop carrying current is placed in a magnetic field, it experiences a torque, which tends to rotate it.

Construction: It consists of a rectangular or circular coil made by winding a fine insulated copper wire on an aluminum frame. A thin phosphor bronze strip from a torsion head, which is connected to a terminal screw, suspends this coil. The lower end of the coil is connected to a fine spring which is connected to another terminal screw. The coil hangs in space between the pole pieces of a powerful horseshoe magnet NS as shown in the figure below. The pole pieces are made concave cylindrical.

This provides a radial magnetic field. Since the field is radial, therefore the plane of the coil remains parallel to the magnetic field in all the orientations of the coil. In between the pole pieces, within the coil, lies a soft iron cylindrical piece called ‘core’. The core does not touch the coil anywhere. The whole arrangement is enclosed in a non-magnetic box to protect it from air currents. Three leveling screws are provided at the base.

Question 39.
Explain with the help of a labeled diagram the underlying principle, construction, and working of a cyclotron. (CBSE Delhi 2019)
Answer:
Principle: It is based on the principle that the positive ions can be accelerated to high energies with a comparatively smaller alternating potential difference by making them cross the electric field again and again, by making use of a strong magnetic field.

Construction: It consists of two D-shaped, hollow metal dees D₁ and D₂. The dees are placed with a small gap in between them. The dees are connected to a source of high frequency alternating potential difference. The dees are evacuated in order to minimize energy losses resulting from collisions between the ions and air molecules. The whole apparatus is placed between the poles of an electromagnet as shown in the figure below. The magnetic field is perpendicular to the plane of the dees.

Working: In a cyclotron, particles of mass m and charge q move inside an evacuated chamber in a uniform magnetic field B that is perpendicular to the plane of their paths. The alternating potential difference applied between the hollow electrodes D₁ and D₂ create an electric field in the gap between them which changes precisely twice in each revolution so that the particles get a push each time they cross the gap. The pushes increase their speed and kinetic energy, boosting them into paths of larger radius. The electric field increases the speed and the magnetic field makes the particles move in circular paths. Due to the presence of the perpendicular magnetic field the particle moves in a circle of radius r given by
r = \(\frac{m v}{q B}\) …(1)

The path of the particles in the dees is a semicircle and the time the particle spends in each dee is semi-circular
τ = \(\frac{\text { length of the semi-circular path }}{\text { velocity }}=\frac{\pi r}{v}=\frac{\pi m}{B q}\) by using equation (1)

The above time is independent of the radius of the path and the velocity of the charged particle. Hence the charged particle spends the same time in each dee and encounters a dee of opposite polarity whenever it comes into the gap between the dees and thus is continuously accelerated.

Question 40.
The figure below shows a long straight wire of circular cross-section (radius a) carrying steady current l. The current is uniformly distributed across this cross-section. Calculate the magnetic field in the region r < a and r > a. Draw a graph showing the variation of a magnetic field for the above two cases.

Answer:
Consider a long circular wire of radius ‘a’ carrying a steady current (dc) that is uniformly distributed along the cross-section of the wire as shown in the figure. Let us calculate the magnetic field in the regions r ≥ a and r < a. In region 1 let us choose a circular path of radius r centered at the wire. From symmetry, we find that B is perpendicular to dL at every point on the circular path. Since total current linked with the circular path 1 is l0, therefore by Ampere’s law we have

for r ≥ a

Now consider the interior of the wire, i.e. region 2 where r < a. In this, the current l enclosed by the path is less than l₀. Since the current is assumed to be uniform over the area of the wire,
Therefore

Now applying Ampere’s circuital rule to region 2 we have

The magnetic field versus r for this system is as shown in the figure below.

Question 41.
Draw the magnetic field lines due to a circular loop of area \(\vec{A}\) carrying current I. Show that it acts as a bar magnet of magnetic moment \(\vec{m}=I \vec{A}\). (CBSE Al 2015)
Answer:
The magnetic field lines are as shown

Magnetic field due to circular loop on its axis at far off points
B = \(\frac{\mu_{0}}{4 \pi} \frac{2 I A}{X^{3}}\)

Magnetic field due to a bar magnet at its axial point is
B = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{x^{3}}\)

Comparing the above two we have m = IA

Question 42.
(a) State Ampere’s circuital law. Use this law to obtain the expression for the magnetic field inside an air-cored toroid of average radius ‘r, having ‘n’ turns per unit length and carrying a steady current.
(b) An observer to the left of a solenoid of N turns each of cross-section area ‘A’ observes that a steady current I in it flows in the clockwise direction. Depict the magnetic field lines due to the solenoid specifying its polarity and show that it acts as a bar magnet of magnetic moment m = NIA. (CBSE Delhi 2015)

Answer:
(a) “The line integral of 8 around any closed path equal μ₀l, where l is the total steady current passing through any surface bounded by the closed path.” Consider a toroidal solenoid. Let N be the number of turns and l be the current passed through it. For a solenoid whose coils are closely spaced, the field inside the solenoid is tangent to the dotted circular path as shown in the figure and is the same at all points lying on the dotted line.

Therefore we have
\(\oint \vec{B} \cdot d \vec{L}\) = B ∮dL = B (2 π r) …(1)

By Ampere’s circuital law we have ∮\(\overrightarrow{\mathbf{B}} \cdot d \vec{L}\) = μ₀Nl …(2)

From equations 1 and 2 we have
B (2 π r) = μ₀NI
or
B = \(\frac{\mu_{0} N I}{2 \pi r}\) but
\(\frac{N}{2 \pi r}\) = n

i. e. number of turns per unit length Therefore we have B = μ₀ n I

This gives the field inside a toroidal solenoid.

(b) The magnetic field lines and the polarity of the solenoid is as shown.

Magnetic field due to the coil on its axis at far off points
B = \(\frac{\mu_{0}}{4 \pi} \frac{2 N I A}{x^{3}}\)

Magnetic field due to a bar magnet at its axial point is
B = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{x^{3}}\)

Comparing the above two we have m = NIA

Question 43.
(a) Use Biot-Savart law to derive the expression for the magnetic field due to a circular coil of radius R having N turns at a point on the axis at a distance ‘x’ from its center.
Draw the magnetic field lines due to this coil.
(b) A current l enters a uniform loop of radius R at point M and flows out at point N as shown in the figure. Obtain the net magnetic field at the center of the loop. (CBSE Delhi 2015C)

Answer:
(a) According to Biot-Savart law:
The magnitude of magnetic field d \(\vec{B}\) , is due to a current element d \(\vec{l}\), is proportional to current l and element length, dl.

inversely proportional to the square of the distance r.

Its direction is perpendicular to the plane containing d \(\vec{l}\) and \(\vec{r}\) .

In vector notation,
\(\overrightarrow{d B}=\frac{\mu_{0}}{4 \pi}, \frac{d \vec{l} \times \vec{r}}{r^{3}}\)

Using Biot-Savart law, obtain the expression for the magnetic field due to a circular coil of radius r, carrying a current I at a point on its axis distant x from the centre of the coil. (CBSE Delhi 2018C)
Answer:

we have \(\overrightarrow{d B}=\frac{\mu_{0}}{4 \pi} I \frac{|d \vec{i} \times \vec{r}|}{r^{3}}\)
r² = x² + R²

∴ dB = \(\frac{\mu_{0} I}{4 \pi} \frac{d l}{\left(X^{2}+R^{2}\right)^{3 / 2}}\)

The magnetic fieLd Lines are as shown.

(b) At point O, the net magnetic field is the sum of fields due to two current segments carrying currents /1 and l2. If the potential difference between points A and B be V, then
V = I₁ R₁ = I₂R₂
where R, is the resistance of segment subtending an angle 90° at O and R₂ is the resistance of segment subtending an angle (360° – 90°) at O.

Now R₁ = \(\frac{\pi / 2}{2 \pi} R=\frac{R}{4}\)

Now field B₁, at O due to smaller segment is

directed inwards.

Now field B₂ at 0 due to larger segment is

directed outwards

Hence net magnetic field in the center is
B₁ – B₂ = \(\frac{3 \mu_{0} I}{32 r}-\frac{3 \mu_{0} I}{32 r}\) = 0

(ii) The velocity of the particle inside a cyclotron is given by v = \(\frac{B q r}{m}\) , which again depends upon the q/m ratio. The q/m ratio of an a particle is less than that of a proton, therefore a proton will come out with higher velocity.

Question 44.
(a) Define the SI unit of current in terms of the force between two parallel current-carrying conductors.
(b) Two long straight parallel conductors carrying steady currents l_a and l_b along the same direction are separated by a distance d. How does one explain the force of attraction between them? If a third conductor carrying a current lc in the opposite direction is placed just in the middle of these conductors, find the resultant force acting on the third conductor. (CBSE Delhi 2018C)
Answer:
(a) The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross¬section, and placed one meter apart in vacuum, would produce on each of these conductors a force equal to 2 × 10^-7 newton per meter of length.
(b) The wire (ii) experiences a force due to the magnetic field caused by the current flowing in wire (i).

The magnetic field at any point on the wire (b) due to the current in the wire (a) is perpendicular to the plane of two wires and pointing inwards and hence force on it will be towards wire (a). Similarly, the force on the wire (a) will be towards wire (b). Hence two wires carrying currents in the same direction attract each other.

Force on wire (3) due to wire (1)
\(\frac{\mu_{0} l_{a} l_{c}}{2 \pi\left(\frac{d}{2}\right)}\) towards right

Force on wire 3 due to wire 2
\(\frac{\mu_{0} l_{b} l_{c}}{2 \pi\left(\frac{d}{2}\right)}\) towards Left

Net force on wire 3
\(\frac{\mu_{0} I_{c}}{\pi d}\left[I_{a}-I_{b}\right]\) towards right nd °

Also accept
\(\frac{\mu_{0} I_{c}}{\pi d}\left[I_{b}-I_{a}\right]\) towards left

Question 45.
(a) State Biot-Savart law and express it in the vector form.
Answer:
(a) According to Biot-Savart law:
The magnitude of magnetic field d \(\vec{B}\) , is due to a current element d \(\vec{l}\), is proportional to current l and element length, dl.

inversely proportional to the square of the distance r.

Its direction is perpendicular to the plane containing d \(\vec{l}\) and \(\vec{r}\) .

In vector notation,
\(\overrightarrow{d B}=\frac{\mu_{0}}{4 \pi}, \frac{d \vec{l} \times \vec{r}}{r^{3}}\)

(b) Using Biot-Savart law, obtain the expression for the magnetic field due to a circular coil of radius r, carrying a current I at a point on its axis distant x from the centre of the coil. (CBSE Delhi 2018C)
Answer:

we have \(\overrightarrow{d B}=\frac{\mu_{0}}{4 \pi} I \frac{|d \vec{i} \times \vec{r}|}{r^{3}}\)
r² = x² + R²

∴ dB = \(\frac{\mu_{0} I}{4 \pi} \frac{d l}{\left(X^{2}+R^{2}\right)^{3 / 2}}\)

Question 46.
Derive an expression for the velocity y0 of a positive ion passing undeflected through a region where crossed uniform electric field E and magnetic field B are simultaneously present. Draw and justify the trajectory of identical positive ions whose velocity has a magnitude less than \(\left|v_{c}\right|\).
OR
A particle of mass m and charge q is in motion at speed Y parallel to a long straight conductor carrying current I as shown below.

Find the magnitude and direction of the electric field required so that the particle goes undefeated. (CBSE Sample Paper 2018-19)
Answer:

If the ion passes undeflected, therefore the magnetic and electric forces acting on the ion must be equal and opposite.
Therefore
qE = Bqv_c
v_c = \(\frac{E}{B}\)

The trajectory would be as shown. Justification: For positive ions with speed v < v_c. Force due to electric field wilt remains the same as It does not depend upon Vc.

But force due to the magnetic field will become Less than the initial value. This unbalances the two, electric and magnetic, forces hence, the ion will experience a net electric force. This will accelerate the ion along the direction of the electric field. Since initiaL velocity is perpendicular to E, the trajectory would be parabolic.
OR
For the charged particle to move undeflected

Electric force = magnetic force
qE = Bq v
or
E=Bv

Now magnetic field at a distance r from the long straight conductor is
B = \(\frac{\mu_{0} I}{2 \pi r}\)

This magnetic force Will act towards the wire.

Hence electric field is

E = \(\frac{\mu_{0} l v}{2 \pi r}\)

This electric field should act away from the wire.

Numerical Problems:

Formulae for solving numerical problems

• Magnetic field due to a small current element dB = \(\frac{\mu_{0}}{4 \pi} \frac{l mid d L \sin \theta}{r^{2}}\)
• Magnetic field due to an infinitely long straight conductor B = \(\frac{\mu_{0}}{4 \pi} \frac{2l}{a}\)
• Magnetic field at the centre of a circular coil B = \(\frac{\mu_{0}}{4 \pi} \frac{2 \pi l}{r}\)
• Force on a charge moving in a magnetic field F = Bq v Sin θ
• The magnetic field inside a solenoid B = µ₀ n I
• Force on a current-carrying conductor placed in a magnetic field F = BIL Sin θ
• Force between two current carrying conductors \(\frac{F_{1}}{L}=\frac{\mu_{0} l_{1} l_{2}}{2 \pi a}\)
• Torque on a current loop τ = BI n A sin θ
• Current in a galvanometer l = \(\frac{C}{n B A}\) sin θ
• Shunt required S = \(\frac{l_{8} G}{l-I_{g}}\)
• Resistance required R = \(\frac{v}{l_{g}}\) – G
• Radius of a charged particle in a magnetic field r = \(\frac{m v}{B q}=\frac{\sqrt{2 m E}}{B q}\)

Question 1.
A proton and an alpha particle having the same kinetic energy are in turn allowed to pass through a uniform magnetic field perpendicular to their direction of motion. Compare the radii of the paths of the proton and the alpha particle.
Answer:
Given E_α = E_p, m_α = 4 mp, q_α = 2q_p, B is same for both. Now the radius of the path followed is given by the expression

Question 2.
A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal to the coil. Calculate the magnitude of the counter-torque that must be applied to prevent the coil from turning.
Answer:
Given n = 30, r = 8.0 cm = 8 × 10^-2 m,
l = 6.0 A, B = 1.0 T, θ = 60°, τ = ?
Using the formula for torque
τ = BlnA

we have τ = Blnπr² sin θ
τ = 1 × 6 × 30 × 3.14 × (8 × 10^-2)² × sin 60°
= 3.1 N m

Question 3.
How can a moving coil galvanometer be converted into an ammeter? To increase the current sensitivity of a moving coil galvanometer by 50%, its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does its voltage sensitivity change?
Answer:
Voltage sensitivity
= \(\frac{\alpha}{V}=\frac{\alpha}{I R}=\frac{\text { current sensitivity }}{R}\)

When the current sensitivity increases by 50 % resistance becomes 2R.

New voltage sensitivity
= \(\frac{1+0.5}{2}\) = 0.75

Hence the voltage sensitivity decreases and becomes 75% of its original value.

Question 4.
A long straight conductor PQ carrying a current of 75 A is fixed horizontally. Another long conductor XY is kept parallel to PQ at a distance of 5 mm, in air. Conductor XY is free to move and carries a current l. Calculate the magnitude and direction of current l for which the magnetic repulsion just balances the weight of conductor XY (Mass per unit length for conductor XY is 10^-2 kg m^-1.)

Answer:
Given l₁ = 75 A, r = 5 mm = 5 × 10^-3 m, l₂ = ? mass per unit length = 10^-2kg m^-1.

The force between the two wires should be repulsive and should balance the weight of the wire XY. Thus the current in wire XY will be opposite to that in wire PQ.

The force between the two current-carrying conductors is given by

Question 5.
A galvanometer has a resistance of 30 Ω. It gives full-scale deflection with a current of 2 mA. Calculate the value of the resistance needed to convert it into an ammeter of range 0-0.3 A.
Answer:
Given G = 30 Ω, lg = 2 mA= 2 × 10^-3 A, S = ?, I = 0.3 A
Using the formula S = \(\frac{I_{g} G}{I-I_{g}}\)

we have
S = \(\frac{I_{g} G}{I-I_{g}}\) = \(\frac{30 \times 2 \times 10^{-3}}{0.3-2 \times 10^{-3}}\) = 0.20 Ω

Question 6.
An infinitely long straight current-carrying wire produces a magnetic field 8, at a point distant ‘a’ from it. What must be the radius of a circular loop, so that, for the same current through it, the magnetic field at (i) its center equals B/2 and (ii) an axial point, distant equal to the radius of the loop, equals B?
Answer:
The magnetic field at a distance ‘a’ from an infinitely long straight conductor is
B = \(\frac{\mu_{0} I}{2 \pi a}\)

(i) Given B_c = B/2. Let r be the radius of the circular coil for which the magnetic field is B/2.

Magnetic field at the centre of a circular coil is \(B=\frac{\mu_{0} I}{2 \pi a}\) Therefore
\(\frac{\mu_{0} l}{2 r}=\frac{1}{2} \times \frac{\mu_{0} l}{2 \pi a}\) or r = 2 π a

(ii) Magnetic field at the axial line of a loop is
B = \(\frac{\mu_{0} l R^{2}}{2\left(x^{2}+R^{2}\right)^{1 / 2}}\)

Here x = R radius of the loop.
Therefore \(\frac{\mu_{0} I}{2 \pi a}=\frac{\mu_{0} I R^{2}}{2\left(R^{2}+R^{2}\right)^{3 / 2}}\)

solving for R we have R = \(\frac{\pi a}{\sqrt{8}}\)

Question 7.
Calculate the value of resistance needed to convert a galvanometer of resistance 120 ft, which gives a full-scale deflection for a current of 5 mA, into a voltmeter of 0 – 50 V range.
Answer:
Given G = 120 Ω, l_g = 5 × 10^-3A, V= 50 V, R =?

Using the relation R = \(\frac{v}{l_{g}}\) – G we have
R = \(\frac{50}{5 \times 10^{-3}}\) -120 = 9880 Ω

Question 8.
Two infinitely long straight wires A₁ and A₂ carrying currents l₁ and l₂ flowing in the same directions are kept distance apart. Where should a third straight wire A₃ carrying current 1.5 l be placed between A₁ and A₂ so that it experiences no net force due to A₁ and A₂? Does the net force act on A3 depend on the current flowing through it? (CBSE Delhi 2019)
Answer:
The diagram is as shown.


If no force is experienced by the conductor A₃, then

The net force on A₃ does not depend upon the current flowing through it.

Question 9.
An ammeter of resistance 0.80 Ω can measure current up to 1.0 A.
(a) What must be the value of shunt resistance to enable the ammeter to measure current up to 5.0 A?
Answer:
Given G = 0.80 Ω, /g = 1.0 A, l = 5.0 A, S = ?, R_A = ?
(a) Using the expression
S = \(\frac{I_{g} G}{I-I_{0}}=\frac{1 \times 0.8}{5-1}=\frac{0.8}{4}\) = 0.2 Ω

(b) What is the combined resistance of the ammeter and the shunt? (CBSE Delhi 2013)
Answer:
Now R_A = \(\frac{G S}{G+S}=\frac{0.8 \times 0.2}{0.8+0.2}\) = 0.016 Ω

Question 10.
A wire AB is carrying a steady current of 12 A and is lying on the table. Another wire CD carrying 5A is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. (Take the value of g = 10 m s^-2) (CBSE AI 2013)
Answer:
Given l₁ = 12 A, r = 1 mm = 1 × 10^-3 m, l₂ = 5 A, mass per unit length = ?

The force between the two wires should be repulsive and should balance the weight of the wire CD. Thus the current in wire CD will be opposite to that in wire AB.

The force between the two current¬carrying conductors is given by

Question 11.
A square loop of side 20 cm carrying a current of I A is kept near an infinitely long straight wire carrying a current of 2 A in the same plane as shown in the figure.

Calculate the magnitude and direction of the net force exerted on the loop due to the current carrying conductor. (CBSEAI 2015C)
Answer:
Here, PQ = 20 cm = 20 × 10^-2 m,
PS = 20 cm = 10 × 10^-2 m

Distance of PQ from AB,
r₁ = 10 cm = 10 × 10^-2 m

Distance of RS from AB,
r₂ = (10 + 20) = 30 cm = 30 × 10^-2 m

Current through long wire AB, l₁ = 2 A
Current through rectangular loop, l₂ = 1 A

Force on the arm PQ,

F₁ = \(\frac{\mu_{0} I_{1} I_{2}}{2 \pi a}\) × length PQ

F₂ = \(\frac{2 \times 10^{-7} \times 2 \times 1 \times 20 \times 10^{-2}}{10 \times 10^{-2}}\) = 8 × 10^-7 N
= 8 × 10^-7 N (towards AB)

Force on the arm RS,
F₁ = \(\frac{\mu_{0} I_{1} I_{2}}{2 \pi a}\) × length RS

F₂ = \(\frac{2 \times 10^{-7} \times 2 \times 1 \times 20 \times 10^{-2}}{30 \times 10^{-2}}\) =2.66 × 10^-7

= 2.66 × 10^-7 N (away from AB)

Effective force on the Loop,
F= F₁ – F₂
=8 × 10^-7 – 2.66 × 10^-7
= 5.34 × 10^-7 N (towards AB)

Question 12.
A square-shaped plane coil of area 100 cm2 of 200 turns carries a steady current of 5A. It is placed in a uniform magnetic field of 0.2 T acting perpendicular to the plane of the coil. Calculate the torque on the coil when its plane makes an angle. of 60° with the direction of the field. In which orientation will the coil be in stable equilibrium? (CBSE Al 2015C)
Answer:
Given A= 10-2 m², n = 200, l = 5 A, θ = 60°, B = 0.2 T, τ = ?
Using the expression τ = B I n A sin θ we have
τ = 0.2 × 5 × 200 × 10^-2 × sin 60° = 20 Nm
Stable equilibrium, when the magnetic field is in the direction of the coil.

Question 13.
A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B (figure). What is the magnitude of the magnetic field? (NCERT)

Answer:
For the wire to be suspended in mid-air, it must experience an upward force F of magnitude F = B I L to balance its weight W= mg
Therefore, BI L = mg
B = \(\frac{m g}{1 L}=\frac{0.2 \times 9.8}{2 \times 1.5}\) = 0.65 T

Question 14.
A cyclotron’s oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating proto is? If the radius of its dees is 60 cm what is the kinetic energy (in m MeV) of the proton beam produced by the accelerator (e =1.60 × 10^-19 C, mp = 1.67 × 10^-27 kg, 1 MeV = 1.6 × 10^-13 J) (NCERT)
Answer:
The oscillator frequency should be the same as the proton’s cyclotron frequency, i.e. 10 MHz = 107 Hz.
Therefore
B = \(\frac{2 \pi m v}{q}\)

= \(\frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 10^{7}}{1.6 \times 10^{-19}}\) = 0.66 T

Final velocity of protons is
V = r × 2πv = 0.6 × 6.3 × 10⁷ = 3.78 × 107 m s^-1.

E = 1/2mv² = 1.67 × 10^-27 × 14.3 × 10¹⁴ / (2 × 1.6 × 10^-13) = 7 MeV

Question 15.
A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. Does it carry a current of 5 A. What is the magnitude of the magnetic field inside the solenoid? (NCERT)
Answer:
Given n = 500/0.5 = 1000 turns per unit length. l = 5 A
The length L = 0.5 m and radius r = 0.01 m. Thus, L/a = 50,
Hence we have
B = μ_onl = 4π x 10^-7 × 1000 × 5 = 6.28 × 10^-3 T

Question 16.
A circular coil of wire consisting of 100 turns, each of a radius 8.0 cm, carries a current of 0.40 A. What is the magnitude of the magnetic field B at the center of the coil? (NCERT)
Answer:
Given n = 100, r = 8.0 cm = 8.0 × 10^-2 m,
l = 0.40 A,B = ?

Using the expression B = \(\frac{\mu_{0} n l}{2 r}\)
we have
B = \(\frac{\mu_{0} n l}{2 r}=\frac{4 \pi \times 10^{-7} \times 100 \times 0.40}{2 \times 8.0 \times 10^{-2}}\)

Question 17.
A horizontal overhead power line carries a current of 90 Ain the east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line? (NCERT)
Answer:
Given l = 90 A, r = 1.5 m, B = ?
Using the expression B = \(\frac{\mu_{0} l}{2 \pi r}\) we have

B = \(\frac{\mu_{0} l}{2 \pi r}=\frac{4 \pi \times 10^{-7} \times 90}{2 \times \pi \times 1.5}\) = 1.2 × 10^-5 T
The magnetic field will be towards the south.

Question 18.
A square coil of side 10 cm consists of 20 turns and carries a current of 12 A The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil? (NCERT)
Answer:
Given L= 10 cm=0.1 m, A = (0.1)² = 0.01 m², n = 20 , l = 12 A, θ = 30°, B = 0.80T, τ = ?

Using the expression τ = B l n A sin θ we have
τ = 0.80 × 12 × 20 × 0.01 × sin 30° = 0.96 Nm

Question 19.
A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(i) What magnetic field should be set up normally to the conductor in order that the tension in the wires is zero?
(ii) What will be the total tension in the wires if the direction of the current is reversed keeping the magnetic field the same as before? (Ignore the mass of the wires.) g = 9.8 ms^-2 (NCERT)
Answer:
Given L = 0.45 m, m = 60 g, I = 5.0 A, B = ?
(i) The tension in the wires will be zero if the weight of the rod is balanced by the force on it due to the magnetic field.
Therefore we have
B I L = mg
B = \(\frac{m g}{I L}=\frac{0.06 \times 9.8}{5.0 \times 0.45}\) = 0.26 T

Thus a horizontal magnetic field of magnitude 0.26 T normal to the conductor should be applied in such a direction that Fleming’s left-hand rule gives a magnetic force in the upward direction.

(ii) The tension will become twice the weight of the wire i.e.,
T = B / L + mg = mg + mg = 2 mg
Or
T= 2 × 0.06 × 9.8 = 1.176 N

Question 20.
A galvanometer coil has a resistance of 12 ohms and the meter shows full-scale deflection for a current of 3 mA. How will you convert the meter into a voltmeter of range 0 to 18 V? (NCERT)(CBSE 2019C)
V= 18 V, R = ?
We will connect a resistance R = \(\left(\frac{V}{l_{g}}-G\right)\) in series with the galvanometer.

Therefore R = \(\frac{18}{3 \times 10^{-3}}\) – 12 = 5988 Ω

Question 21.
A galvanometer coil has a resistance of 15 ohms and the meter shows full-scale deflection for a current of 4 mA. How will you convert the meter into an ammeter of range 0 to 6A? (NCERT)
Answer:
Given G = 15 Ω, l_g = 4 mA = 4 × 10^-3 A, l = 6 A, S = ?

We will connect a resistance S = \(\frac{l_{s} G}{l-l_{s}}\) in parallel with the galvanometer.
Therefore S = \(\frac{4 \times 10^{-3} \times 15}{6-\left(4 \times 10^{-3}\right)}\) = 0.01 Ω
Or
S = 10 mΩ

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 3 Current Electricity. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 3 Important Extra Questions Current Electricity

Current Electricity Important Extra Questions Very Short Answer Type

Question 1.
A wire or resistivity ρ is stretched to double its length. What will be its new resistivity?
Answer:
The resistivity remains the same as it does not depend upon the length of the wire.

Question 2.
What is the effect of temperature on the relaxation time of electrons in a metal?
Answer:
The relaxation time of electrons decreases with the rise in temperature of the metal.

Question 3.
Which physical quantity does the voltage versus current graph for a metallic conductor depict? Give its SI unit.
Answer:
It represents resistance. It is measured in ohm.

Question 4.
Define drift velocity of electrons.
Answer:
The mean velocity acquired by electrons in a conductor when an external electric field is applied to it.

Question 5.
A resistance R is connected across a cell of emf ε and internal resistance r. A potentiometer now measures the potential difference between the terminals, of the cell as V., Write the expression for ‘r’ in terms of ε, V and R. (CBSE Delhi 2011)
Answer:
The required relation is r = \(\left(\frac{\varepsilon}{V}-1\right)\)R

Question 6.
How is the drift velocity in a conductor affected by the rise in temperature? (CBSE Delhi 2019)
Answer:
It decreases.

Question 7.
Two students A and B were asked to pick a resistor of 15 kΩ from a collection of carbon resistors. A picked a resistor with bands of colours brown, green, orange, while B chose a resistor with bands of black, green, red. Who picked the correct resistor? (CBSE AI 2013C)
Answer:
A

Question 8.
Define the term ‘Mobility’ of charge carriers in a conductor. Write its S.l. unit. (CBSE Delhi 2014, AI 2015)
Answer:
Mobility of charge carriers in a conductor is defined as the magnitude of their drift velocity per unit applied electric field. Its SI unit is m² V^-1 s^-1.

Question 9.
How does the mobility of electrons in a conductor change, if the potential difference applied across the conductor is doubled, keeping the length and temperature of the conductor constant? (CBSE Delhi 2019)
Answer:
No change.

Question 10.
Graph showing the variation of current versus voltage for a material GaAs is shown in the figure. Identify the region of
(i) negative resistance (ii) where Ohm’s

Answer:
(i) DE (ii) AB

Question 11.
Define mobility of a charge carrier. What is its relation with relaxation time? (CBSEAI 2016)
Answer:
It is defined as the drift velocity per unit electric field. The relation is μ = \(\frac{-e \tau}{m}\).

Question 12.
Nichrome and copper wires of the same length and same radius are connected in series. Current l is passed through them. Which wire gets heated up more? Justify your answer. (CBSEAI 2017)
Answer:
Nichrome, as it has more resistivity.

Question 13.
When a potential difference is applied across the ends of a conductor, how is the drift velocity of the electrons related to the relaxation time? (CBSE Delhi 2019)
Answer:
Drift velocity is directly proportional to the relaxation time.
V_d = \(\frac{e E}{m} \tau\)

Question 14.
Two bulbs are marked 60 W, 220 V and 100 W, 220 V. These are connected in parallel to 220 V mains. Which one out of the two will glow brighter?
Answer:
In parallel combination, the bulb having, more power glows more. Therefore the bulb marked 100 W, 220 V glows brighter.

Question 15.
A heater joined in parallel with a 60 W bulb is connected to the mains. If the 60 W bulb is replaced by a 100 W bulb, will the rate of heat produced by the heater be more, less or remain the same?
Answer:
The rate of heat produced in the heater will be the same as the two are connected in parallel.

Question 16.
Two conductors, one having resistance R and another 2R, are connected in turn across a dc source. If the rate of heat produced in the two conductors is Q₁ and Q₂ respectively, what is the value of Q₁/ Q₂?
Answer:
We know that Q = \(\frac{V^{2}}{R}\)t, therefore
\(\frac{Q_{1}}{Q_{2}}=\frac{R_{2}}{R_{1}}=\frac{2 R}{R}\) = 2

Question 17.
When electrons drift in metal from lower to higher potential, does it mean that all the free electrons of the metal are moving in the same direction? (CBSE Delhi 2012)
Answer:
No.

Question 18.
How does one explain the increase in resistivity of metal with an increase in temperature? (CBSE AI 2014C)
Answer:
With the increase in temperature, the average relaxation time decreases; this causes an increase in resistivity.

Question 19.
A carbon resistor is marked in red, yellow and orange bands. What is the approximate resistance of the resistor?
Answer:
24 × 10³  ohm ± 20%

Question 20.
If potential difference V applied across a conductor is increased to 2 V, how will the drift velocity of the electrons change?
Answer:
The drift velocity is given by the expression V_d = \(\frac{e V}{m L}\)τ Therefore if the potential is doubled, drift velocity is also doubled.

Question 21.
What is the largest voltage you can safely put across a resistor marked 98 ohm – 0.5 W?
Answer:
Using the expression
V = \(\sqrt{P R}=\sqrt{0.5 \times 98}\) = 7 V

Question 22.
Two wires A and B are of the same metal, have the same area of cross-section and have their lengths in the ratio 2:1. What will be the ratio of currents flowing through them respectively, when the same potential difference is applied across the length of each of them?
Answer:
\(\frac{I_{\mathrm{A}}}{I_{\mathrm{B}}}=\frac{R_{\mathrm{B}}}{R_{\mathrm{A}}}=\frac{l_{\mathrm{A}}}{l_{\mathrm{B}}}=\frac{2}{1}\)

Question 23.
How does the heat produce in a resistor depend on its resistance when
(i) a constant current is passed through it
Answer:
For l = constant, heat produced H ∝ R

(ii) a constant potential difference is applied across its ends?
Answer:
For constant potential difference V, the heat produced, H ∝ \(\frac{1}{R}\)

Question 24.
Two wires of equal length, one of copper and the other of manganin have the same resistance. Which wire is thicker? (CBSE AI 2012)
Answer:
Manganin.

Question 25.
Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through this resistance? (CBSE AI 2013)
Answer:
The current is l = E/R

Question 26.
Define temperature coefficient of resistivity. (CBSE Sample Paper 2018-2019)
Answer:
It is defined as the fractional change in resistivity per unit change in temperature.

Question 27.
A cell of emf E and internal resistance r is connected across an external resistance R. Plot a graph showing the variation of P.D. across R, versus R. (NCERT Exemplar)
Answer:
The graph is as shown.

Current Electricity Important Extra Questions Short Answer Type

Question 1.
Why is a potentiometer preferred over a voltmeter for determining the emf of a cell? (CBSE Delhi 2016C)
Answer:
The potentiometer is based on the null method or it does not draw any (net) current from the cell and measures emf However, the voltmeter draws some current from the cell when connected across it, hence measures terminal voltage.

Question 2.
The figure shows the V – l graph for a parallel and series combination of two resistors A and B. Which line represents the parallel combination?

Answer:
For the same potential, the current is less in series combination than parallel combination. Therefore from the graph, it is apparent that the same potential current is less in A. Therefore B represents the parallel combination.
As, R = \(\frac{V}{l}\)

The slope of B > Slope of A

Question 3.
V – l graph for a given metallic wire at two temperatures is shown. Which of these is at a higher temperature?

Answer:
At higher temperature resistance of a metallic wire is more or its conductance is low. Hence, graph (2) is at a higher temperature, i.e. T₂ > T₁.

Question 4.
In an experiment on a metre bridge, if the balancing length AC is ‘x’, what would be its value, when the radius of the metre bridge wire AB is doubled? Justify your answer. (CBSE AI 2011C)

Answer:
In a metre bridge, at the balance point we have
\(\frac{R_{1}}{R_{2}}=\frac{x}{100-x}\)

As R1 and R2 remain the same, x will also remain the same. It does not depend upon the diameter of the wire.

Question 5.
The emf of a cell is always greater than its terminal voltage. Why? Give reason. (CBSE Delhi 2013)
Answer:
When current passes through a cell, there is a drop in potential across it due to its internal resistance. This is called the lost volt. Thus terminal voltage is less than the emf of the cell.

Question 6.
Draw a graph showing the variation of resistivity with temperature for nichrome. Which property of nichrome is used to make standard resistance coils? (CBSEAI 2013C)
Answer:
The graph is as shown.

The property has a low-temperature coefficient of resistance.

Question 7.
Define the term ‘mobility’ for a charge carrier and state its SI unit.
Name the mobile charge carriers in
(i) an electrolyte,
(ii) a semiconductor and
(iii) an ionised gas. (CBSE Al 2015)
Answer:
Mobility is defined as the ratio of the drift velocity of the charge to the applied electric field.
(i) Anions and cations.
(ii) Electrons and holes
(iii) Free electrons.

Question 8.
Define the term current density of a metallic conductor. Deduce the relation connecting current density (J) and the conductivity (a) of the conductor, when an electric field E is applied to it. (CBSE AI 2015)
Answer:
Current density is defined as the current flowing per unit area of the conductor.

Mathematically current density is given by the expression J = \(\frac{l}{A}\)

But l = V/R and R = \(\frac{ρl}{A}\) = \(\frac{L}{σA}\). Substituting in the above relation, we have

J = \(\frac{l}{A}\) = \(\frac{V}{AL}\) × σA= \(\frac{V}{L}\) × σ = Eσ

Question 9.
(a) Define the terms ‘drift velocity’ and ‘relaxation time’ giving their physical significance.
Answer:
Drift Velocity: It is the average velocity of the free electrons with which they get drifted towards the positive terminal under the influence of the external field.

Significance: The net current flowing through any cross-section is controlled by drift velocity and there is no transport of charges in a direction perpendicular to the applied field. Relaxation Time (T): The average time between successive collisions of electrons or ions in a conductor is called the relaxation time.

Significance: It determines the drift velocity acquired by the electrons under the given applied electric force and also determines the electrical conductivity of a conductor at different temperatures.

(b) A conductor of length L is connected across a dc source of emf E. If the conductor is replaced by another of the same material and area of cross-section but of length 5L, by what factor will the drift velocity change? (CBSE2019C)
Answer:
In the first case:
V_d= \(\frac{e V}{m L}\)τ

In the second case:

Thus we find that the drift velocity becomes \(\frac{1}{5}\) of its original value.

Question 10.
A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable resistor. ‘R’. Plot a graph showing the variation of terminal voltage ‘ V’ of the cell versus the current ‘l’. Using the plot, show how the emf of the cell and its internal resistance can be determined. (CBSE AI 2014)
Answer:
The graph is as shown.

For l = 0,V= E
The internal resistance can be found by using the expression
V = E – l_r

For V = 0
E = l_r
r = E/l

Question 11.
Two electric bulbs P and Q have their resistances in the ratio of 1: 2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs. (CBSE AI, Delhi 2018)
Answer:
Given \(\frac{R_{\mathrm{p}}}{R_{\mathrm{Q}}}\) = \(\frac{1}{2}\)

Power dissipated \(\frac{P_{\mathrm{p}}}{P_{\mathrm{Q}}}=\frac{I^{2} R_{\mathrm{p}}}{I^{2} R_{\mathrm{Q}}}=\frac{1}{2}\)

Question 12.
A 10 V cell of negligible Internal resistance Is connected In parallel across a battery of emf 200 V and Internal resIstance 38 š as shown in the figure. Find the value of current in the circuit. (CBSE AI, Delhi 2018)

Answer:
Given ε₁ = 10 V, r₁ = 0, ε₂ = 200 V, r₂ = 38 Ω
The two cells send current in the opposite direction; therefore, net emf of the combination
E = 200 – 10 = 190V

Hence current in the circuit
l = \(\frac{E}{R}=\frac{190}{38}\) = 5 A

Question 13.
In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell In an open circuit is 350 cm. When the resistance of 9 Ω Is used In the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell. (CBSE AI, Delhi 2018)
Answer:
Given L₁ = 350 cm, L₂ – 300 cm, R = 9 Ω, r = ? Using the expression
r = \(\left(\frac{L_{1}-L_{2}}{L_{2}}\right) R=\left(\frac{350-300}{300}\right)\) × 9 = 1.5 Ω

Question 14.
TWo bulbs are rated (P₁ V) and (P₂, V). If they are connected (i) in series and (ii) in parallel across a supply V, find the power dissipated in the two combinations in terms of P₁ and P₂. (CBSE Delhi 2019)
Answer:
Resistances of the two bulbs R₁ = \(\frac{V^{2}}{P_{1}}\) and R₂ = \(\frac{V^{2}}{P_{2}}\)

(i) Net resistance in series
R_s = R₁ + R₂ = \(\frac{V^{2}}{P_{1}}+\frac{V^{2}}{P_{2}}=V^{2}\left(\frac{P_{1}+P_{2}}{P_{1} P_{2}}\right)\)

(ii) Net resistance in parallel

Question 15.
What is the advantage of using thick metallic strips to join wires in a potentiometer? (NCERT Exemplar)
Answer:
The metal strips have low resistance and need not be counted in the potentiometer Length l of the null point. One measures only their lengths along with the straight segments (of lengths 1 metre each). This is easily done with the help of centimetre rulings or metre ruler and leads to accurate measurements.

Question 16.
AB is a potentiometer wire (figure). If the value of R is increased, in which direction will the balance point J shift? (NCERT Exemplar)

Answer:
If R is increased, the current through the wire will decrease. This will decrease the potential gradient across the potentiometer wire (k = V/L = l_pR_p/L). Hence more length of the wire will be required to balance the same cell. Therefore J will shift towards B.

Question 17.
While doing an experiment with a potentiometer (figure) it was found that the deflection is one-sided and
(i) the deflection decreased while moving from one end A of the wire to the end B;
(ii) the deflection increased while the jockey was moved towards the end B.
(iii) Which terminal + or -ve of the cell E1 is connected at X in case (i) and how is E1 related to E?
(iv) Which terminal of the cell E1 is connected at X in case (ii)? (NCERT Exemplar)

Answer:
(i) Positive terminal of E₁ is connected at Xand E₁ > E.
(ii) Negative terminal of E₁ is connected at X.

Question 18.
The circuit in the figure shows two cells connected in opposition to each other. Cell E₁ is of emf 6 V and internal resistance 2 Ω; the cell E₂ is of emf 4 V and internal resistance 8 Ω. Find the potential difference between the points A and B. (NCERT Exemplar)

Answer:
The cells are connected in the opposite direction, therefore net emf in the circuit is
E = E₁ – E₂ = 6 – 4 = 2 V

Hence current in the circuit is
l = \(\frac{E}{R+r}\) = \(\frac{2}{10}\) = 0.2 A

P.D. across E₁ = 6 – 0.2 × 2 = 5.6 V
P.D. across E₂ = VAB = 4 + 0.2 × 8 = 5.6 V
Point B is at a higher potential than A

Current Electricity Important Extra Questions Long Answer Type

Question 1.
Explain the term ‘drift velocity’ of electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of ‘drift velocity’. (CBSE AI 2013, 2013C)
Answer:
Drift velocity (V_d) is defined as the average velocity with which the free electrons get drifted inside a conductor under the effect of the electric field, opposite to the direction of the field.

Let n be the electrons per unit volume in the conductor. Here n is called the number density of electrons. Assume that all electrons move with the same drift velocity V_d. In a time interval dt, each electron moves a distance vdt. Now the volume of the cylinder covered by the electrons in time dt is
V = A v_ddt …(1)

and the number of electrons in this volume is
N = nV = nA v_ddt …(2)

If e is the charge on the electron, then charge flowing through the conductor in small time dt is
dQ = e(nA v_ddt) …(3)

Hence the current through the conductor is
l = \(\frac{dQ}{dt}\) = nAev_d

Question 2.
Draw a plot showing the variation of resistivity of an (i) conductor and (ii) semiconductor, with the increase in temperature.
How does one explain this behaviour in terms of the number density of charge carriers and the relaxation time? (CBSE Delhi 2014C)
Answer:
The plots are as shown.


With a rise in temperature the average relaxation time for a conductor decreases and resistivity increases, while for a semiconductor the number density of charge carriers increases, hence the resistivity decreases.

Question 3.
State the two Kirchhoff’s rules used in electric networks. How are these rules justified? (CBSE Delhi 2015)
Answer:
The two rules are

Kirchhoff’s junction rule: The algebraic sum of currents meeting at a junction is zero.
Kirchhoff’s loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero.

The two rules are justified as they are based on the law of conservation of charge and law of conservation of energy respectively.

Question 4.
(a) Deduce the relation between current l flowing through a conductor and drift velocity \(\vec{v}_{\mathrm{d}}\) of the electrons.
(b) Figure shows a plot of current ‘l’ flowing through the cross-section of a wire versus the time ‘t’. Use the plot to find the charge flowing in 10 s through the wire. (CBSE AI 2015C, 2016C)

Answer:
Suppose there are n electrons per unit volume, where n is called the number density of electrons. Assume that all electrons move with the same drift velocity V_d. In a time interval dt, each electron moves a distance v_ddt. Now the volume of the cylinder covered by the electrons in time dt is
V = A v_ddt ….(1)

and the number of electrons in this volume is
N = nV=nA v_ddt …(2)

If e is the charge on the electron, then charge flowing through the conductor in small time dt is
dQ= e(nA v_ddt) …(3)

Hence the current through the conductor is
l = \(\frac{dQ}{dt}\) = vdenA

The area under the l – t curve gives the value of the charge
Q = 1/2 × 5 × 5 + (10 – 5) × 5 = 12.5 + 25
= 37.5 C

Question 5.
Answer the following:
(a) Why are the connections between the resistors in a metre bridge made of thick copper strips?
Answer:
The metal strips have low resistance;
hence their resistance does not come into play while calculating the unknown resistance.

(b) Why is it generally preferred to obtain the balance point in the middle of the metre bridge wire?
Answer:
It is done to reduce the index correction.

(c) Which material is used for the metre bridge wire and why? (CBSE Al 2014)
Answer:
Alloys are used due to their high resistivity.

Question 6.
(a) State the underlying principle of a potentiometer. Why is it necessary to (i) use a long wire, (ii) have a uniform area of cross-section of the wire and (iii) use a driving cell whose emf is taken to be greater than the EMFs of the primary cells?
Answer:
Principle: It is based on the principle that if a wire of uniform area of cross-section carries a constant current, the potential drop across any portion of the wire is directly proportional to the length of that portion of the wire.

The diagram is as shown.
First, the key K is inserted. This brings the cell of emf E₁ in the circuit. The jockey is moved on the wire to obtain a balance point, i.e. a point on the wire where the galvanometer gives zero deflection. Let the balancing length be L₁. Therefore by the potentiometer principle, we have
E₁ ∝ L₁ …(1)

Now, the key K₂ is inserted. This brings the cell of emf E₂ into the circuit. The jockey is again moved on the wire to obtain the balance point. Let the balancing length be L₂. Then by potentiometer principle, we have
E₂ ∝ L₂ …(2)

Dividing equations (1) by (2) we have
\(\frac{E_{1}}{E_{2}}=\frac{L_{1}}{L_{2}}\) …. (3)

Knowing the values of L₁ and L₂, the EMFs can be compared.

1. The least count of the potentiometer is given by k = V/l, where l is the length of the potentiometer wire. The larger the value of l, the smaller is the least count. Hence a long wire is used so that there is more accuracy.
2. If the area of cross-section is not uniform, the value of the potential ‘ gradient will not be the same along
   the length of the wire.
3. The balance point will be obtained on the wire.

(b) In a potentiometer experiment, if the area of the cross-section of the wire increases uniformly from one end to the other, draw a graph showing how potential gradient would vary as the length of the wire increases from one end. (CBSE AI 2014C)
Answer:
k = \(\frac{V}{l}\) = \(\frac{lR}{l}\)

⇒ k ∝ \(\frac{1}{l}\)

Question 7.
Plot a graph showing the variation of current density (j) versus the electric field (E) for two conductors of different materials. What information from this plot regarding the properties of the conducting material, can be obtained which can be used to select suitable materials for use in making (i) standard resistance and (ii) connecting wires in electric circuits?
Electron drift speed is estimated to be of the order of mm s^-1. Yet large current of the order of few amperes can be set up in the wire. Explain briefly. (CBSEAI 2015)
Answer:
We know that J = σE
Therefore the graph between J and E will be a straight line passing through the origin.

This is as shown.

The slope of the graph = conductivity (σ)

Hence materiaL with Less slope (smaller conductivity) is used for making standard resistors and materiaL with greater slope (higher conductivity) for making connecting wires.

A large current can be set up in the wire because the electron number density is enormous, 10²⁹ m^-3.

Question 8.
The reading of the (ideal) ammeter, In the circuit shown here, equals:

(i) l when key K₁ is closed but key K₂ is open.
Answer:
(i) Current l when K₂ is open
l = \(\frac{ε}{R+X}\)

(ii) l/2 when both keys K₁ and K₂ are closed.
Find the expression for the resistance of X In terms of the resistances of R and S. (CBSE Delhi 2016C)
Answer:
Let current be l when both K₁ and K₂ are closed

Current through X

Question 9.
Define the current sensitivity of a galvanometer. Write Its S.I. unit. The figure shows two circuits each having a galvanometer and a battery of 3 V. When the galvanometers in each arrangement do not show any deflection, obtain the ratio R₁/R₂. (CBSE AI 2013)

Answer:
The current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer per unit current passing through It. It is measured in div per ampere. In both cases the Wheatstone bridge is in the balanced state. Therefore we have
\(\frac{4}{R_{1}}=\frac{6}{9}\) Or R₁ = 6 ohm

\(\frac{6}{12}=\frac{R_{2}}{8}\) Or R2 = 4 ohm

Therefore \(\frac{R_{1}}{R_{2}}=\frac{6}{4}=\frac{3}{2}\)

Question 10.
The following circuit shows the use of a potentiometer to measure the internal resistance of a cell, (i) When the key K is open, how does the balance point change, if the current from the driver cell decreases? (ii) When the key K is closed, how does the balance point change if R is increased, keeping the current from the driver cell constant?

Answer:
(i) When the key K is open, a decrease in the current from the driver cell decreases the potential drop across the potentiometer wire. Therefore to balance the same emf again, more length of the wire will be required. Thus the balance point will shift towards point B.
(ii) When key K is closed and R is increased, it increases the terminal potential difference of the cell. Thus to balance the new terminal potential difference, more length of the wire will be required. Thus the balance point will shift towards point B.

Question 11.
(a) State, with the help of a circuit diagram, the working principle of a metre bridge. Obtain the expression used for determining the unknown resistance.
Answer:
(a) The experiment is based on the Wheatstone bridge principle.

The connections are made as shown in the figure. A resistance R is introduced from the resistance box and the key K is closed. The jockey is moved on the wire to the point where there is no deflection in the galvanometer. In such case points, B and D are at the same potential. Point B is called the “null” point.

Let in this position AB = L cm and BC = (100 – L) cm. Therefore resistance of AB, i.e.
P ∝ L
and resistance of BC, i.e. Q ∝ (100 – L)

hence
\(\frac{P}{Q}=\frac{L}{100-L}\) …..(1)

In the balanced state by the Wheatstone bridge Principle we have
\(\frac{P}{Q}=\frac{R}{X}\) … (2)

Substituting equation (1) in equation (2) we have
\(\frac{R}{x}=\frac{L}{100-L}\) ….(3)

rewriting equation (3) we have
X = \(\left(\frac{100-L}{L}\right)\)R

(b) What happens if the galvanometer and cell are interchanged at the balance point of the bridge?
Answer:
There is no change in the position of the balance point if the galvanometer and the cell are interchanged.

(c) Why is it considered important to obtain the balance point near the mid-point of the wire? (CBSE Delhi 2011C)
Answer:
It is important to get the balance point near the mid-point of the wire because then the resistances in the four arms of the bridge are of the same order. The sensitivity of the bridge is maximum and the resistance is determined most accurately.

Question 12.
(a) Which material is used for potentiometer wire and why?
Answer:
The potentiometer wire is usually of constantan or manganin.
The material of the wire should have

• high specific resistance and
• low-temperature coefficient of resistance.

(b) How can the sensitivity of a potentiometer be increased? (CBSE Delhi 2011C)
Answer:
For greater sensitivity of the measurement, the fall of potential per cm should be less. The smaller the value of K, the greater will be the balancing length and the greater will be the accuracy of the measurement. For this, a wire of longer length should be taken or the current in the wire should be less.

Question 13.
Write two possible causes for one-sided deflection in a potentiometer experiment. (CBSE Delhi 2013)
Answer:

1. The Emf of the auxiliary battery may be less than the emf of the test cell.
2. The positive terminal of the test cell and the auxiliary cell may not be connected to the same point.

Question 14.
(a) Derive a relation between the internal resistance, emf and terminal potential difference of a cell from which current l is drawn. Draw V vs l graph for a cell and explain its significance.
Answer:
Consider the circuit shown.

By Kirchhoff’s rules we have
E – lR – rl = 0
E – V – lr = 0
E = V + lr

The V-l graph is as shown.

Significance of Graph: To find emf and internal resistance of the cell.

(b) A voltmeter of resistance 998 0 is connected across a cell of emf 2 V and internal resistance 2Q. Find the potential difference across the voltmeter and also across the terminals of the cell. Estimate the percentage error in the reading of the voltmeter.
Answer:
The diagram is as shown.

V= E – lr
998 × l = 2 – 2l
1000 l = 2
l = 0.002 A

Therefore,
V= 0.002 × 998= 1.996 V

Percentage error
\(\frac{2-1.996}{2}\) × 100 = \(\frac{0.004}{2}\) = 0.2%

Question 15.
Two cells of the same emf E but internal resistance r₁ and r₂ are connected in series to an external resistor R (figure). What should be the value of R so that the potential difference across the terminals of the first cell becomes zero? (NCERT Exemplar)

Answer:
Since the cells are connected in series,
total emf is
E_T = E + E = 2E

Now current in the circuit Is
l = \(\frac{2 E}{R+r_{1}+r_{2}}\)

For terminal potential difference across the first cell to be zero we have

Question 16.
Two cells of EMFs ε₁ and ε₂ and Internal resistances r1 and r2 respectively are connected in parallel. Obtain expressions for the equivalent
(i) resistance and
(ii) emf of the combination (CBSE Delhi 2018C, 2019C)
Answer:

Question 17.
Using Kirchhoff’s rules, calculate the potential difference between B and D In the circuit diagram as shown in the figure. (CBSE Delhi 2018C)

Answer:
Using Kirchhoff’s voLtage ruLe, we have:
For Loop DABD
l₁ × 1 + (1) + (-2) + 2l₁ + 2(l₁ + l₂) = 0
Or 5l₁ + 2l₂ = 1 …. (i)

For Loop DCBD
l₂ × 3 + (3) + (-1) + l₂ + 2(l₁ + l₂) = 0
Or 2l₁ + 6 l₂ = -2 …. (ii)

Solving (i) and (ii), we get
l₁ = \(\frac{5}{13}\)A

l₂ = \(\frac{-6}{13}\)A

∴ Current through DB = l₁ + l₂ = \(\frac{-1}{13}\)A
∴ P.D. between B and D = 0.154 V

Question 18.
Define the term resistivity of a conductor. Give its SI unit. Show that the resistance R of a conductor Is given by \(\frac{m L}{n e^{2} A \tau}\)where the symbols have their usual meaning.
Answer:
The resistivity of a conductor is the resistance of the conductor of unit length and unit cross-sectional area.

Let V_d be the drift velocity of the electrons, then its relation with the electric field is
V_d = \(-\frac{e E}{m}\)τ …..(1)

Let V be the potential difference applied across the two ends of a conductor of length L and area of cross-section A, then
E = \(\frac{-V}{L}\) …..(2)

Substituting equation (1) in (2) we have
V_d = \(\frac{\mathrm{eV}}{\mathrm{mL}}\)τ ….(3)

Also l = neAV_d ….(4)
Substituting equation (3) in equation (4),
we have

l = neA\(\left[\frac{e V}{m L} \tau\right]=\left[\frac{n e^{2} A \tau}{m L}\right]\)V ….(5)

Comparing with Ohm’s “Law”, i.e
l = \(\frac{V}{R}\) …. (6)

we have
R = \(\left[\frac{m L}{n e^{2} A \tau}\right]\) ….(7)

Question 19.
The given figure shows a network of resistances R₁, R₂, R₃ and R₄.

Using Kirchhoff’s laws, establish the balance condition for the network. (CBSEAI 2015)
Answer:
Consider the diagram as shown below.

Now applying Kirchhoff’s Loop rule to the closed Loop ABDA, we have
– l₁R₁ – l_gG + (l – l₁) R₃ = 0 ….(1)

Again applying Kirchhoff’s Loop rule to closed-loop BCDB, we have
– (l₁ – l_g)R₂ + (l – l₁ + l_g)R₄ + l_gG = 0 …(2)

The values of R₁, R₂, R₃ and R₄ are so adjusted that the galvanometer gives zero deflection. This means that both B and D will be at the same potential and hence no current will flow through the galvanometer,

i.e. l_g = 0. In this situation, the Wheatstone bridge is said to be balanced. Putting l_g = 0 in equations (1) and (2) we have
– l₁R₁ + (l – l₁) R₃ = 0
and
-l₁R₂ + (l – l₁)R₄ = 0 …(4)

Rewriting the above two equations we have
– l₁R₁ = (l – l₁) R₃ ….(5)
and
– l₁R₂ + (l – l₁)R₄ ….(6)

Dividing the above equations we have
\(\frac{R_{1}}{R_{2}}=\frac{R_{3}}{R_{4}}\) … (7)

The above expression gives the condition for the balance of a Wheatstone bridge.

Question 20.
Draw a circuit diagram of a potentiometer. State its working principle. Derive the necessary formula to describe how it is used to compare the EMFs of the two cells. (CBSE AI 2015C)
Answer:
PRINCIPLE: It is based on the principle that if a wire of uniform area of cross-section carries a constant current, the potential drop across any portion of the wire is directly proportional to the length of that portion of the wire.

The diagram is as shown.
First, the key K is inserted. This brings the cell of emf E₁ into the circuit. The jockey is moved on the wire to obtain a balance point, i.e. a point on the wire where the galvanometer gives zero deflection. Let the balancing length be L₁. Therefore by the potentiometer principle, we have
E₁ ∝ L₁ …(1)

Now, the key K₂ is inserted. This brings the cell of emf E₂ into the circuit. The jockey is again moved on the wire to obtain the balance point. Let the balancing length be L₂. Then by potentiometer principle, we have
E₂ ∝ L₂ …(2)

Dividing equations (1) by (2) we have
\(\frac{E_{1}}{E_{2}}=\frac{L_{1}}{L_{2}}\) …. (3)

Knowing the values of L₁ and L₂, the EMFs can be compared.

Question 21.
(i) Define the term ‘conductivity’ of a metallic wire. Write its SI unit.
Answer:
The property of a material that allows a flow of electrons between two points of the material when a potential difference is applied between these two points is called the conductivity of a wire. SI unit of conductivity: siemens per metre.

(ii) Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E. (CBSE At 2018, Delhi 2018)
Answer:
Drift velocity of electrons in a conductor is given as
V_d = eEτ/m …(1)

Small distance covered in time
Δt = V_dΔt

Amount of charge passing through the area A in time Δt, q = lΔt
Δt = neAV_dΔt
or l = neAV_d …(2)
where
n → Number of free electrons per unit volume or number density

Now from equations (1) and (2), we get
l = ne₂AτE/m …(3)

Since resistivity of a conductor is given as
ρ = m/ne₂τ

Now, we know that the conductivity of a conductor is mathematically defined as the reciprocal of resistivity of the conductor. Thus,
σ = \(\frac{1}{ρ}\) …(4)

where σ = conductivity of the conductor. Thus, from equations (3) and (4), we get
σ = ne₂τ/m ….(5)

Now, from equations (3) and (5), we have
lA = σE …(6)
and current density is given as
J = lA
Thus, J = σE

Question 22.
(a) You are required to select a carbon resistor of resistance of 56 kΩ ± 10% from a shopkeeper. What would be the sequence of colour bands required to code the desired resistor?
Answer:
Number 5 corresponds to green, No. 6 corresponds to blue, 10³ corresponds to orange and 10% corresponds to silver.
∴ The sequence of Colours is
Green, blue, orange, silver

(b) Write two characteristic properties of the material of a metre bridge wire.
Answer:
Metre bridge wire must have

• High resistivity
• Low-temperature coefficient of resistivity

(c) What precautions do you take to minimise the error in finding the unknown resistance of the given wire? (CBSE2019C)
Answer:
To minimise the error in determining the resistance of a wire, the

• The wire should be of uniform thickness
• The balance point should be near the midpoint of the wire

Question 23.
The diagram below shows a potentiometer set-up. On touching the jockey near the end X of the potentiometer wire, the galvanometer pointer deflects to left. On touching the jockey near to end Y of the potentiometer, the galvanometer pointer again deflects to left but now by a larger amount. Identify the fault in the circuit and explain, using appropriate equations or otherwise, how it leads to such a one¬sided deflection. (CBSE Sample Paper 2018-19)

Or
The following circuit was set up in a metre bridge experiment to determine the value X of unknown resistance.

(i) Write the formula to be used for finding X from the observations. 39.5 cm from end A, when the resistor Y is 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or metre bridge made of thick copper strips?
(ii) If the resistance R is increased, what will happen to balance length?
Answer:
The positive of E₁ is not connected to terminal X.
In Loop XGNX,
E₁ – V_G + E = 0
V_G = E₁ + E_XN
V_G = E₁ + kl

So, V_G (or deflection) Will be maximum when l is maximum, i.e. when Jockey Is touched near the end Y. Also, VG (or deflection) Will, be minimum when l is minimum, i.e. when a jockey is touched near end X.
Or
(i) X = (100 – l) R/l
(ii) The balancing Length wilt increase.

Question 24.
(i) In a metre bridge as shown, the balance point is found to be at 39.5 cm from the end A, when the resistor Y is 12.5. Determine the resistance of X. Why are the connections between resistors In a Wheatstone or metre bridge made of thick copper strips?
(ii) Determine the balance point of the bridge above if X and Y are interchanged.
(iii) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current? (NCERT)

Answer:
(i) Given L₁ = 39.5 cm and L₂ =100 – 39.5 = 60.5 cm, Y=12.5Q, X=?
Using the Wheatstone bridge principle we have
\(\frac{39.5}{60.5}=\frac{X}{12.5}\) or X = 8.2 Ω

The connections between resistors in a Wheatstone or metre bridge are made of thick copper strips to minimise the resistance of the connection which are not accounted for in the bridge formula.

(ii) If X and Y are interchanged, then
\(\frac{L}{(100-L)}=\frac{12.5}{8.2}\); solving for L we have
L = 60.4 cm from point A.

(iii) When the galvanometer and cell are interchanged, the condition for the balance of the bridge remains satisfied. Hence galvanometer will not show any current.

Question 25.
The figure below shows a potentiometer with a cell of 2.0 V and internal resistance of 0. 40 Ω maintaining a potential drop across the resistor wire AB. A standard cell that maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf e and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
1. What is the value of e?
2. What purpose does the high resistance of 600 kΩ have?
3. Is the balance point affected by this high resistance?
4. Is the balance point affected by internal resistance?
5. Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
6. Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

Answer:
1. Given ε₁ = 1.02 V, L₁ = 67.3 cm, L₂ = 82.3 cm, ε₂ = ?
By potentiometer principle we have

2. This resistance reduces the current through the galvanometer when the movable contact (a jockey) is far away from the balance point. This, in turn, protects the galvanometer from the damage likely to be caused by the high current.
3. There is no effect of this resistance on the balance point.
4. There is no effect.
5. In case, the driving emf is smaller than the emf to be measured, the balance point cannot be found on the potentiometer wire.
6. In such a case, the balance point will be very close to the end giving a very large error in the measurement of e. If resistance is placed in series with the wire AB such that the potential drop across AB is only slightly larger than the emf to be measured, the balance point will be found on a larger length. This will make the error very small.

Numerical Problems:

Formulae for solving numerical problems.

• The mobility of electrons is given by μ = \(\frac{V_{d}}{E}=\frac{e \tau}{m}\)
• By Ohm’s law, l = \(\frac{V}{R}\) and J = σE
• The current density is J = \(\frac{l}{A}\) = nevd
• Theresistanceofawireis R = ρ\(\frac{L}{A}\) = \(\left[\frac{m L}{n e^{2} A \tau}\right]\)
• The resistivity of a wire is ρ = \(\frac{m}{n e^{2} \tau}\)
• The resistance of a wire at t°C is R = R₀ (1 + αt)
• In parallel combination \(\frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\)
• In series combination R_s = R₁ + R₂ + R₃ + ……..
• Ratio of resistances in series and parallel \(\frac{R_{\mathrm{s}}}{R_{\mathrm{p}}}\) = n²
• The Wheatstone bridge principle is \(\frac{P}{Q}=\frac{R}{S}\)
• V = E – lr, where V is the terminal potential difference, E is emf and r is the internal resistance of the cell.
• The internal resistance of a cell is r = \(\frac{(E-V) R}{V}=\frac{\left(L_{1}-L_{2}\right) R}{L_{2}}\)

Question 1.
A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus (i) R and (ii) the current l.
It is found that when R = 4 Ω the current is 1 A and when R is increased to 9 Ω, the current reduces to 0.5 A. Find the values of the emf E and internal resistance r. (CBSE Delhi 2015)
Answer:
The plots are as shown

Here l₁ = 1.0 A, R₁ =4 ohm, l₂ = 0.5 A, R₂ =9 ohm
Using the equation l = \(\frac{E}{(R+r)}\) Or E = l(R + r)

we have
1.0 × (4 + r) = 0.5 × (9 + r)

Solving the above equation for r we have r = 1 ohm
Also E = 0.5 (9 + 1) = 5 V

Question 2.
A wire of resistance R, length l and area of cross-section A is cut into two parts, having their lengths in the ratio 1:2. The shorter wire is now stretched till its length becomes equal to that of the longer wire. If they are now connected in parallel, find the net resistance of the combination. (CBSE Al 2019)
Answer:
Since the wires are cut in the ratio of 1:2 therefore,
Resistance of the shorter wire R₁ = \(\frac{R}{3}\) and

Resistance of the longer wire R₂ = \(\frac{2R}{3}\)

Since the shorter wire is stretched to make it equal to the longer wire therefore, it is stretched by n = 2 times its length. Hence New resistance of the shorter wire

Question 3.
In the figure, a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of EMFs ε₁ and ε₂ connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) ε₁/ε₂ and (ii) position of null point for the cell ε₁. How is the sensitivity of a potentiometer increased? (CBSE Delhi 2012)

Answer:
From the diagram we have
\(\frac{\varepsilon_{1}-\varepsilon_{2}}{\varepsilon_{1}+\varepsilon_{2}}=\frac{120}{300}=\frac{2}{5}\)
Or
5ε₁ – 5ε₂ = 2ε₁ + 2ε₂

Solving we have
\(\frac{\varepsilon_{1}}{\varepsilon_{2}}=\frac{7}{3}\) ….(1)

Also let L be the balancing length for cell of emf ε1, then
\(\frac{\varepsilon_{1}}{\varepsilon_{1}+\varepsilon_{2}}=\frac{L}{300}\)

Using equation (1) we have
\(\frac{\varepsilon_{1}}{\varepsilon_{1}+\frac{3}{7} \varepsilon_{1}}=\frac{L}{300}\)

Solving for L we have L = 210 cm
The sensitivity of a potentiometer can be increased by increasing the length of the potentiometer wire.

Question 4.
The network PQRS, shown in the circuit diagram, has batteries of 4 V and 5 V and negligible internal resistance. A milli- ammeter of 20 Ω resistance is connected between P and R. Calculate the reading in the milliammeter. (CBSE AI 2012C)

Answer:
Using Kirchhoff’s junction rule to distribute current we have

Consider the loop SRPS, by Kirchhoff’s loop rule we have
200 l₂ + 20 (l₁ + l₂ ) – 5 = 0 …(1)
Or
220 l₂ + 20 l₁ = 5 …(2)

Consider the loop PRQP, by Kirchhoff’s loop rule we have
– 60 l₁ + 4 – 20 (l₁ + l₂) = 0 …(3)
80 l₁ + 20 l₂ =4 …(4)

Multiplying equation (2) by (4) we have
880 l₂ + 80 l₁ = 20 …(5)

Subtracting equation (4) from equation (5)
we have
860 l₂ = 16 or l₂ = 4/215 A

Substituting in equation (4) we have
l₁ = \(\frac{39}{860}\)A

Therefore reading of the milliammeter is
l₁+ l₂ = \(\frac{4}{215}+\frac{39}{860}\) = 0.063 A = 63 mA

Question 5.
A set of ‘n’ identical resistors, each of resistance ‘R’ when connected in series have an effective resistance ‘X’. When they are connected in parallel, their effective resistance becomes ‘Y’. Find out the product of X and Y. (CBSEAI2019) Answer:
In series
R_s = R₁ + R₂ + R₃ + ……
R_s = X = R + R + R + …. upto n
X=nR

In Parallel

Question 6.
In the following circuit, a metre bridge is shown in its balanced state. The metre bridge wire has a resistance of 1 ohm per centimetre. Calculate the value of the unknown resistance X and the current drawn from the battery of negligible internal resistance.

Answer:
Using the Wheatstone bridge principle we have
\(\frac{40}{60}=\frac{X}{3}\)
or X = 2 Ω

Now total resistance of the combination is
R = \(\frac{5 \times 100}{5+100}=\frac{500}{105}\) = 4.76 Ω

Current drawn is
l = V/R = 6/4.76 = 1.26 A

Question 7.
Calculate the electrical conductivity of the material of a conductor of length 3 m, area of cross-section 0.02 mm² having a resistance of 2 ohms.
Answer:
Given L = 3 m,
A = 0.02 mm2 = 0.02 × 10-6 m2
R = 2 ohm.

Using the equation
R = \(\frac{ρL}{A}\)
Or
ρ = \(\frac{RA}{L}\)

σ = \(\frac{L}{AR}\) = \(\frac{3}{0.02 \times 10^{-6} \times 2}\) = 7.5 × 107 Sm-1

Question 8.
A potential difference of 2 volts is applied between points A and B has shown in the network drawn in the figure. Calculate (i) equivalent resistance of the network across the points A and B and (ii) the magnitudes of currents in the arms AFCEB and AFDEB.

Answer:
The circuit can be redrawn as shown below.

As seen the circuit is a balanced Wheatstone bridge; therefore the resistance in the arm CD is superfluous.
(i) Resistance of arm FCE = 2 + 2 = 4 Ω
Resistance of arm FDE = 2 + 2 = 4 Ω
Hence net resistance of the circuit between A and B is
R = \(\frac{4 \times 4}{4+4}=\frac{16}{8}\) = 2 Ω

(ii) current in the arm AFCEB
l = V/R = 2/4 = 0.5 A

Current in the arm AFDEB
l = V/R = 2/4 = 0.5 A

Question 9.
A cell of emf E and internal resistance ‘r’ gives a current of 0.8 A with an external resistor of 24 ohms and a current of 0.5 A with an external resistor of 40 ohms.
Calculate
(i) emf E and
(ii) internal resistance ‘r’ of the cell.
Answer:
Given l₁ = 0.8 A, R₁ = 24 ohm l₂ = 0.5 A, R₂ = 40 ohm
Using the equation
E = l(R + r) we have
0.8 × (24 + r) = 0.5 × (40 + r)

Solving for r we have r = 2.67 ohm
Also E = 0.5( 40 + 2.67) = 21.3 V

Question 10.
In the circuit diagram of the metre bridge given below, the balance point is found to be at 40 cm from A. The resistance of X is unknown and Y is 10 ohms.
(i) Calculate the value of X;
(ii) if the positions of X and Y are interchanged in the bridge, find the position of the new balance point from A; and
(iii) if the galvanometer and the cell are interchanged at the balance point, would the galvanometer show any current.

Answer:
(i) Using the Wheatstone bridge principle we have
\(\frac{40}{60}=\frac{X}{10}\)
Or
X = 6.67 Ω

(ii) If X and Y are interchanged then
\(\frac{L}{(100-L)}=\frac{10}{6.67}\) solving for L we have
L = 59.9cm

(iii) The ga(vanometer will not show any current.

Question 11.
The length of a potentiometer wire is 600 cm and it carries a current of 40 mA. For a cell of emf 2.0 V and internal resistance 10 ohms, the null point is found to be at 500 cm. If a voltmeter is connected across the cell, the balancing length is decreased by 10 cm. Find (i) the resistance of the whole wire, (ii) reading of the voltmeter and (iii) resistance of the voltmeter.
Answer:
Here total length of potentiometer wire L = 600 cm, current flowing
l = 40 mA = 0.04 A,
Emf of cell E = 2 V,

Internal resistance of cell r = 10 Ω, balancing length for cell l = 500 cm
(i) Since ε = kl
k = \(\frac{\varepsilon}{l}=\frac{2}{500}\) , V = kl

∴ V = \(\frac{2}{500}\) × 600 = 2.4 V

Therefore resistance of the potentiometer wire =\(\frac{V}{l}=\frac{2.4}{0.04}\) = 60 Ω.

(ii) With voltmeter connected across the cell, balancing Length
L’ = l – 10 = 500 – 10 = 490cm

Therefore, potential difference, i.e. reading of voltmeter,
= \(\frac{L^{\prime}}{L} \times V=\frac{490}{600}\) × 2.4 = 1.96 V

(iii) Let resistance of the voltmeter be R,
R = \(\frac{V}{l}=\frac{1.96}{0.04}\) = 49 Ω
R = 49 ohm

Question 12.
Find the value of the unknown resistance X in the following circuit, if no current flows through the section AO. Also, calculate the current drawn by the circuit from the battery of emf 6 V and negligible internal resistance.

Answer:
If no current flows through the section AO, the given circuit is an example of a balanced Wheatstone bridge. The circuit can be redrawn as

Then \(\frac{2}{4}=\frac{3}{X}\)
Or
X = \(\frac{3 \times 4}{2}\) = 6 Ω

Now 2 Ω and 4 Ω are in series and 3 Ω and 6 Ω are also in series; so the circuit becomes

Hence total resistance of the circuit
R = 2.4 + \(\left(\frac{9 \times 6}{15}\right)\) = 6 Ω

Hence current drawn from the battery
l = V/R = 6/6 = 1 A

Question 13.
In the given circuit, calculate the value of current in a 4.5-ohm resistor and indicate its direction. Also, calculate the potential difference across each cell.

Answer:
The effective resistance of 3 ohms and 6-ohm resistors in parallel is
R = \(\frac{3 \times 6}{3+6}\) =2Q

Hence the resistance of the entire circuit
R = 2 + 4.5 + 0.5 + 1 = 8 ohm

Effective emf of the circuit E = 8 – 4 = 4 V.

Therefore current through the circuit and hence through the 4.5-ohm resistor is
l = E/R = 4/8 = 0.5 A.

The direction of the current is from cell E₂ towards cell E₁.

The potential difference across 8 V battery
V = E – lr = 8 – 0.5 × 1 = 7.5 V

The potential difference across 4 V battery
V = E + lr = 4 + o.5 × 0.5 = 4.25 V

Question 14.
Potentiometer wire PQ of 1-metre length is connected to a standard cell E₁. Another cell, E₂ of emf 1.02 V, is connected as shown In the circuit diagram with a resistance ‘r’ and a switch, S. with the switch open, the null position is obtained at a distance of 51 cm from P.
Calculate (i) the potential gradient of the potentiometer wire and
(ii) the emf of the cell E₁.
(iii) When switch S is closed, will the null point move towards P or towards Q? Give a reason for your answer.

Answer:
Here E₂ = 1.02V and with switch open L = 51 cm

(i) Potential gradient k = E/L
= 1.02 /51 = 0.02 V cm^-1

(ii) As total length of potentiometer wire L = 1 m = 100 cm
Therefore E₁ = kL = 0.02 × 100 = 2 V

(iii) When switch S is closed, the null point will remain unaffected because cell E₂ is even now in an open circuit and no current is being drawn from it.

Question 15.
Three identical resistors when connected in series to a dc source dissipate a power of X watt. If these resistors are connected in parallel to the same dc source, what will be the power dissipation in this case?
Answer:
Given P_s = X, R_s = 3R, R_p = R/3

Power in series Ps = \(\frac{V^{2}}{R_{s}}\) or X = \(\frac{V^{2}}{3 R}\) or 3X = \(\frac{V^{2}}{R}\)

When resistors are connected In parallel, we have
P_p = \(\frac{V^{2}}{R_{p}}=\frac{V^{2}}{R / 3}=\frac{3 V^{2}}{R}\) = 3 × 3X = 9X

Question 16.
A heater coil is rated 100W, 200V. It is cut into two Identical parts. Both parts are connected In parallel, to the same source of 200 V. Calculate the energy liberated per second in the new combination.
Answer:
The resistance of the coil
R = \(\frac{V^{2}}{P}=\frac{(200)^{2}}{100}\) = 400 Ω

When the coil is cut into two equal halves, the resistance of each half becomes 200 ohms.

When these two parts are connected in parallel, the resultant resistance is
R_p = \(\frac{200 \times 200}{200+200}\) = 100 Ω

Now energy Liberated
H = \(\frac{V^{2}}{R_{\mathrm{p}}}=\frac{(200)^{2}}{100}\) = 400 Ω

Question 17.
AB is a 1 m long uniform wire of 10 Ω resistance. The other data are as shown in the circuit diagram given in the figure below. Calculate (i) potential gradient along with AB and (ii) length AO of the wire, when the galvanometer shows no deflection.

Answer:
Emf of battery E₁ = 2 V.
Resistance of potentiometer R = 10 Ω,
Resistance joined in series R₁ = 15 Ω and
Length of potentiometer wire L = 1 m = 100 cm
(i) Potential gradient
k = \(\frac{E_{1} R}{\left(R+R_{1}\right) L}=\frac{2 \times 10}{(10+15) \times 100}\) = 0.008 Vcm^-1

(ii) Current through 0.3 Ω resistance due
to cell E₁, l = \(\frac{1.5}{1.2+0.3}\) = 1 A

Potential difference across 0.3 Ω resistance = l × 0.3 = l × 0.3 = 0.3 V

∴ Length AO = \(\frac{\text { Potential difference }}{\text { Potential gradient }}\) = 37.5 cm

Question 18.
In a potentiometer, a standard cell of emf 5V and of negligible resistance maintains a steady current through the potentiometer wire of length 5 m. Two primary cells of EMFs E₁ and E₂ are joined in series with (a) the same polarity and (b) opposite polarity. The combination Is connected through a galvanometer and a jockey to the potentiometer. The balancing lengths In the two cases are found to be 350 cm and 50 cm, respectively.
(i) Draw the necessary circuit diagram.
(ii) Find the value of the EMFs of the two cells.
Answer:
(i) The circuit diagram is as shown below.

(ii) With cells E₁ and E₂ joined in series with the same polarity, the resultant emf
E = E₁ + E₂ and

Balancing Length L = 350 cm.
With celLs joined in series with opposite polarity, the resuLtant emf
E’ = E₁ – E₂ and

Balancing Length L = 50 cm.

Moreover, the potentiaL gradient

k = \(\frac{V}{L}=\frac{5 \mathrm{~V}}{5 \mathrm{~m}}=\frac{5 \mathrm{~V}}{500 \mathrm{~cm}}\) = 0.01 Vcm^-1

Therefore
E₁ + E₂ = kL = 0.01 × 350 = 3.5 V and
E₁ – E₂ = kL = 0.01 × 50 = 0.5 V

Solving for the two EMFs we get E₁ = 2V and E₂ = 1.5 V

Question 19.
Using Kirchhoff’s rules, calculate the current through the 40 Q and 200 resistors in the following circuit:

Or
What is the end error in a metre bridge? How is it overcome? The resistances in the two arms of the metre bridge are R = 5 Ω and S respectively. When the resistance S is shunted with equal resistance, the new balance length found to be 1.5 L₁, where L₁ is the initial balancing length. Calculate the value of S. (CBSE Delhi 2019)

Answer:
The distribution of current is as shown

In loop ABCDA
+ 80 – 20 l₂ + 40 l₁ = 0
4 = l₂ – 2 l₁

In loop DCFED
– 40 l₁ – 10(l₁ + l₂) + 40 = 0
-50 l₁ – 10 l₂ + 40 = 0
5 l₁ + l₂ = 4

Solving these two equations
l₁ = 0 A and
l₂ = 4 A
Or
The end error, in a meter bridge, is the error arising due to
(i) Ends of the wire not coinciding with the 0 cm/100 cm mark on the meter scale.
(ii) Presence of contact resistance at the joints of the meter bridge wire with the metallic strips.

It can be reduced/overcome by finding balance length with two interchanged positions of R and S and taking the average value of ‘S’ from these two readings.

For a metre bridge, we have
\(\frac{R}{S}=\frac{L}{100-L}\)

For the two given conditions, we have

Putting this value in equation (1) we have S = 10 Ω.

Question 20.
A 10 m long wire of uniform cross-section and 20-ohm resistance is used in a potentiometer. The wire is connected in series with a battery of 5 V along with an external resistance of 480 ohms. If an unknown emf E is balanced at 6.0 m length of the wire, calculate:
(i) the potential gradient of the potentiometer wire
(ii) the value of unknown emf E.
Answer:
Total resistance of the circuit
R = 20 + 480 = 500 ohm

Therefore current through the potentiometer
l = \(\frac{V}{R}=\frac{5}{500}\) = 0.01 A

Now potential drop across the potentiometer wire of 20 ohm = 20 × 0.01 = 0.2 V

(i) Potential gradient
\(\frac{V}{L}=\frac{0.2}{10}\) = 0.02 Vm^-1

(ii) Unknown emf = balancing length × potential gradient = 6 × 0.02 = 0.12 V

Question 21.
In a potentiometer arrangement, a cell of emf 1.20 volt gives a balance point at 30 cm length of the wire. This cell is now replaced by another cell of unknown emf. If the ratio of the EMFs of the two cells is 1.5, calculate the difference in the balancing length of the potentiometer wire in the two cases.
Answer:
Given E₁ = 1.20 V, L₁ = 30 cm, E₂ = ?,
E₁/E₂ = 1.5, L₁ – L₂ = ?

Now \(\frac{E_{1}}{E_{2}}=\frac{L_{1}}{L_{2}}\) or 1.5 = \(\frac{30}{L_{2}}\) or L₂ = 20 cm

Therefore L₁ – L₂ = 30 – 20 = 10 cm

Question 22.
A uniform wire of resistance 12 Ω is cut into three pieces so that the ratio of the resistances R₁: R₂: R₃ = 1: 2: 3 and the three pieces are connected to form a triangle across which a cell of emf 8 V and internal resistance 1 Ω is connected as shown. Calculate the current through each part of the circuit. (CBSE A! 2013C)

Answer:
Value of the three resistances are
R₁ = \(\frac{1}{6}\) × 12 = 2 Ω,
R₂ = \(\frac{2}{6}\) × 12 = 4 Ω
and R₃ = \(\frac{3}{6}\) × 12 = 6 Ω

Now R1 and R2 are connected in series, therefore we have
R = R₁ + R₂ = 2 + 4 = 6 Ω

Now R and R3 are connected in parallel, therefore net resistance
R_p = \(\frac{R R_{3}}{R+R_{3}}=\frac{6 \times 6}{6+6}\) = 3 Ω

Now Rp and the internal resistance of the cell are connected in series, therefore net resistance of the circuit is
R_N = R_p + 1 = 3 + 1 = 4 Ω

Hence current in the circuit is
l = V/R_N = 8/4 = 2 A

This is equally divided amongst R₁, R₂ and R₃. Therefore current through all the three resistors is 1 A.

Question 23.
A battery of emf E and internal resistance r when connected across an external resistance of 12 ohms produces a current of 0.5 A. When connected across a resistance of 25 ohms it produces a current of 0.25 A. Determine the (i) emf and (ii) internal resistance of the cell. (CBSEAI2013C)
Answer:
Here l₁ = 0.5 A, R₁ =12 ohm, l₂ = 0.25 A, R2₂ = 25 ohm

Using the equation l = \(\frac{E}{(R+r)}\) or E = l(R + r)

we have
0. 5 × (12 + 1) = 0.25 × (25 + r)

Solving the above equation for r we have
r = 1 ohm

Also E = 0.5 (12 + 1) = 6.5 V

Question 24.
The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5 Ω and at steam, the point is 5.23 Ω. When the thermometer is inserted into a hot bath, the resistance of the platinum wire is 5.795 Ω. Calculate the temperature of the bath. (NCERT)
Answer:
R_o = 5 Ω, R₁₀₀ = 5.23 Ω, R_t = 5.795 Ω

Question 25.
A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1 Ω. Determine the equivalent resistance of the network and the current along each edge of the cube. (NCERT)

Answer:
The distribution of current is as shown.

Next take a closed-loop, say, ABCC’EA and apply Kirchhoff’s second rule:
lR – (1/2)lR – lR + E = 0
where R is the resistance of each edge and E is the emf of the battery.
Thus E = 5lR/2

Now if R_eq is the equivalent resistance, then
R = E/31 = 5lR/6l = 5/6 R

Now current in the circuit is 3l × 5/6 R = 10 or l = 4 A.

Hence the other currents can also be found.

Question 26.
Determine the current in each branch of the network shown. (NCERT)

Answer:
Each branch of the network is assigned an unknown current to be determined by the application of Kirchhoff’s rules. To reduce the number of unknowns at the outset, the first rule of Kirchhoff is used at every junction to assign the unknown current in each branch. We then have three unknowns l₁, l₂ and l₃ which can be found by applying the second rule of Kirchhoff to three different closed loops.

Kirchhoff’s second rule for the closed loop ADCA gives,
10 – 4(l₁ – l₂) + 2(l₂ + l3 – l₁) – l₁ = 0 …(1)
that is, 7 l₁ – 6l₂ – 2l₃ = 10

For the closed loop ABCA, we get
10 – 4l₂ – 2 (l₂ + l₃) – l₁ = 0 …(2)

That is, l₁ + 6l₂ + 2l₃ = 10

For the closed loop BCDEB, we get
5 – 2(l₂ + l₃) – 2(l₂ + l₃– l₁) = 0 ….(3)

That is 2 l₁ – 4l₂ – 4l₃ = – 5

Equations (1), (2) and (3) are three simultaneous equations in three unknowns. These can be solved by the usual method to give
l₁ = 2.5 A, l₂ = 5/8 A, l₃ = 15/8 A

The currents in the various branches of the network are
AB = 5/8 A, CA = 5/2 A, DEB = 15/8 A, AD = 15/8 A, CD = 0 A, BC = 5/2 A

Question 27.
In a metre bridge, the null point is found at a distance of 33.7 cm from A. If now resistance of 12 Ω is connected in parallel with 5, the null point occurs at 51.9 cm. Determine the values of R and S. (NCERT)
Answer:
From the first balance point we get
\(\frac{R}{S}=\frac{33.7}{66.3}\)

After S is connected in parallel with a resistance of 12 Ω, the resistance across the gap changes from S to S_eq where and hence the new balance condition now gives
\(\frac{R}{S_{e q}}=\frac{51.9}{48.1}=\frac{R(S+12)}{12 S}\)

Substituting for R/S we have
\(\frac{51.9}{48.1}=\frac{(5+12)}{12} \times \frac{33.7}{66.3}\)
which gives S = 13.5 Ω and R = 6.86 Ω

Question 28.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell? (NCERT)
Answer:
Given ε₁ = 1.25 V, L₁ = 35.0 cm and L₂ = 63.0 cm, ε₂ = ?
ε₂ = \(\frac{L_{2} \times \varepsilon_{1}}{L_{1}}=\frac{1.25 \times 63.0}{35.0}\) = 2.25 V

Question 29.
The figure below shows a potentiometer circuit for the comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε? (NCERT)

Answer:
Given R= 10.0 Ω, L₁ = 58.3 cm, L₂ = 68.5 cm, X = 1

Let k be the potential gradient along the wire AB, then
\(\frac{X}{R}=\frac{k L_{2}}{k L_{1}}\)
or
X = \(\frac{R \times L_{2}}{L_{1}}=\frac{10 \times 68.5}{58.3}\)
or
X= 11.75 Ω

If we fail to find a balance point with the given cell of emf e, it means the potential drop across R or X is greater than the potential drop across the potentiometer wire AB. Therefore to obtain the balance point on the wire, reduce the current in the circuit by putting a resistor in series with the circuit.

Question 30.
The figure below shows a 2.0 V potentiometer used for the determination of the internal resistance of a 1.5 V cell. The balance point of the cell in the open circuit is 76.3 cm. When a resistor of 9.5 Q is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell. (NCERT)

Answer:
Given 4 = 76.3 cm, ε = 1.5 V, R = 9.5 Ω, L₂ = 64.8 cm
Internal resistance of a cell is given by the expression
r = \(\frac{\left(L_{1}-L_{2}\right) R}{L_{2}}\)
or
r = \(\frac{(76.3-64.8) \times 9.5}{64.8}\)
= 1.7 Ω

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 1 Electric Charges and Fields. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 1 Important Extra Questions Electric Charges and Fields

Electric Charges and Fields Important Extra Questions Very Short Answer Type

Question 1.
What is the value of the angle between the vectors \(\vec{p}\) and \(\vec{E}\) for which the potential energy of an electric dipole of dipole moment \(\vec{p}\), kept in an external electric field \(\vec{E}\), has maximum value.
Answer:
P.E. = –pEcos θ
P.E. is maximum when cos θ = – 1, i.e.
θ = 180°

Question 2.
Define electric field intensity at a point.
Answer:
Electric field intensity at a point is defined as the force experienced by a unit test charge placed at that point. Mathematically
we have

Question 3.
Two equal point charges separated by 1 m distance experience force of 8 N. What will be the force experienced by them, if they are held in water, at the same distance? (Given: K_water = 80) (CBSE Al 2011C)
Answer:
The force in water is given by
F_w = \(\frac{F_{\text {air }}}{K}=\frac{8}{80}\) = 0.1 N

Question 4.
A charge ‘q’ is placed at the center of a cube of side l. What is the electric flux passing through each face of the cube? (CBSE AI 2012) (CBSE Sample Paper 2019)
Answer:
Φ = q/6ε₀

Question 5.
Why do the electric field lines not form closed loops? (CBSE Al 2012C)
Answer:
It is due to the conservative nature of the electric field.

Question 6.
Two equal balls having equal positive charge ‘q’ coulomb are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two? (CBSE AI 2014)
Answer:
It decreases because force ∝= \(\frac{1}{k}\) and k > 1.

Question 7.
What is the electric flux through a cube of side l cm which encloses an electric dipole? (CBSE Delhi 2015)
Answer:
Zero

Question 8.
Why are electric field lines perpendicular at a point on an equipotential surface of a conductor? (CBSE Al 2015C)
Answer:
So that no net force acts on the charge at the equipotential surface and it remains stationary.

Question 9.
What is the amount of work done in moving a point charge Q. around a circular arc of radius ‘r’ at the centre of which another point charge ‘q’ is located? (CBSE Al 2016)
Answer:
Zero.

Question 10.
How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased? (CBSE Delhi 2016)
Answer:
No change, as flux does not depend upon the size of the Gaussian surface.

Question 11.
Draw the pattern of electric field lines, when a point charge –Q is kept near an uncharged conducting plate. (CBSE Delhi 2019)
Answer:
The pattern is as shown.

Question 12.
Draw a plot showing the variation of the electric field with distance from the center of a solid conducting sphere of radius R, having a charge + Q on its surface. (CBSE Delhi 2017C)
Answer:
The plot is as shown.

Question 13.
Draw a graph to show the variation of E with perpendicular distance r from the line of charge (CBSE Delhi 2018)
A
Answer:
E = \(\frac{\lambda}{2 \pi \varepsilon_{0} r}\)
E ∝ \(\frac{1}{r}\)

The graph between E and r is as shown.

Question 14.
Define the term ‘electric flux’. Write its S.I. unit. (CBSE Delhi 2018)
Answer:
Electric flux: It is the measure of the number of electric field lines crossing a given area normally.

Mathematically the electric flux passing through an area \(\vec{dS}\) is given by
dΦ = \(\vec{E}\) . \(\vec{dS}\)

SI unit of electric flux is Nm²C^-1 or Vm.

Question 15.
Why can the interior of a conductor have no excess charge in the static situation? (CBSE Ai 2019)
Answer:
Since the electric field inside the conductor is zero, electric flux through the closed surface is also zero. Hence by Gauss’s law, there is no net charge enclosed by the closed surface.

Question 16.
Two field lines never cross each other. Why?
Answer:
It is because at the point of intersection two perpendiculars can be drawn. Thus there will be two directions of the electric field at that point which is not possible.

Question 17.
In an electric field, an electron is kept freely. If the electron is replaced by a proton, what will be the force experienced by the proton?
Answer:
The magnitude of force will be the same but the direction will be reversed.

Question 18.
Consider the situation shown in the figure given below. What are the signs of q₁ and q₂?
Answer:

q₁ is negative and q₂ is positive.

Question 19.
In the figure given below, at which point electric field is maximum?
Answer:

The electric field is maximum at point C.

Question 20.
An uncharged insulated conductor A is brought near a charged insulated conductor B. What happens to the charge and potential of B?
Answer:
On bringing uncharged conductor A near a charged conductor B, charges are induced on A as shown in the figure below. As a result, the potential of conductor B is slightly lowered but the charge on it remains unchanged.

Question 21.
In a medium the force of attraction between two point electric charges, distance ‘d’ apart, is F. What distance apart should these be kept in the same medium so that the force between them becomes 3F?
Answer:
Let the new distance be ‘d’, since F ∝\(\frac{1}{r^{2}}\) ,
therefore \(\frac{F}{3 F}=\frac{r^{2}}{d^{2}}=\frac{1}{3} \Rightarrow r=\frac{d}{\sqrt{3}}\)

Question 22.
Find the value of an electric field that would completely balance the weight of an electron.
Answer:
mg = eE ⇒ E = \(\frac{m g}{e}\)
= \(\frac{9.1 \times 10^{-31} \times 9.8}{1.6 \times 10^{-19}}=5.57 \times 10^{-11} \mathrm{Vm}^{-1}\)

Question 23.
Two charges, one +5 µC, and the other -5 µC are placed 1 mm apart. Calculate the electric dipole moment of the system.
Answer:
p = q × 2a = 5 × 10 ^-6 × 10^-3 = 5 × 10^-9 Cm

Question 24.
Two-point charges of+3 µC each are 100 cm apart. At what point on the line joining the charges will the electric intensity be zero?
Answer:
At the mid-point of the line joining the two point charges.

Question 25.
Four charges of came magnitude and same sign are placed at the corners of a square, of each side 0.1 m. What is electric field intensity at the center of the square?
Answer:
Zero.

Question 26.
Why should the electrostatic field be zero ‘ inside a conductor? (CBSE Delhi 2012)
Answer:
Because it does not contain any charge.

Question 27.
A metallic spherical shell has an inner radius R₁ and outer radius R₂. A charge Q is placed at the center of the shell. What will be the surface charge density on the (i) inner surface, and (ii) the outer surface of the shell? (CBSE Delhi 2018)
Answer:
On inner surface

On the outer surface,

Question 28.
An arbitrary surface encloses a dipole. What is the electric flux through this surface? (NCERT Exemplar)
Answer:
Zero.

Electric Charges and Fields Important Extra Questions  Short Answer Type

Question 1.
(a) Electric field inside a conductor is zero. Explain.
(b) The electric field due to a point charge at any point near it is given as
E  =

what is the physical significance of this limit?
Answer:
(a) By Gauss theorem \(\phi \vec{E} \cdot \overrightarrow{d S}=\frac{q}{\varepsilon_{0}}\). Since there is no charge inside a conductor therefore in accordance with the above equation the electric field inside the conductor is zero.
(b) It indicates that the test charge should be infinitesimally small so that it may not disturb the electric field of the source charge.

Question 2.
Define the electric line of force and give its two important properties.
Answer:
It is a line straight or curved, a tangent to which at any point gives the direction of the electric field at that point.
(a) No two field lines can cross, because at the point of intersection two tangents can be drawn giving two directions of the electric field which is not possible.
(b) The field lines are always perpendicular to the surface of a charged conductor.

Question 3.
Draw electric field lines due to (i) two similar charges, (ii) two opposite charges, separated by a small distance.
Answer:
(a) The diagram is as shown.

(b) The diagram is as shown.

Question 4.
An electric dipole is free to move in a uniform electric field. Explain what is the force and torque acting on it when it is placed
(i) parallel to the field
Answer:
When an electric dipole is placed parallel to a uniform electric field, net force, as well as net torque acting on the dipole, is zero and, thus, the dipole remains in equilibrium.

(ii) perpendicular to the field
Answer:
When the dipole is placed perpendicular to the field, two forces acting on the dipole form a couple, and hence a torque acts on it which aligns its dipole along the direction of the electric field.

Question 5.
A small metal sphere carrying charge +Q. is located at the center of a spherical cavity in a large uncharged metallic spherical shell. Write the charges on the inner and outer surfaces of the shell. Write the expression for the electric field at the point P₁(CBSE Delhi 2014C)
Answer:

Charge on inner surface – Q.
Charge on outer surface + Q,
Electric field at point P = E = k\(\frac{Q}{r_{1}^{2}}\)

Question 6.
Two-point charges q and –2q are kept ‘d’ distance apart. Find the location of the point relative to charge ‘q’ at which potential due to this system of charges is zero. (CBSE Al 2014C)
Answer:
Let the potential be zero at point P at a distance x from charge q as shown

Now potential at point P is
V = \(\frac{k q}{x}+\frac{k(-2 q)}{d+x}\) = 0

Solving for x we have
x = d

Question 7.
Two small identical electrical dipoles AB and CD, each of dipole moment ‘p’ are kept at an angle of 120° as shown in the figure. What is the resultant dipole moment of this combination? If this system is subjected to the electric field (\(\vec{E}\)) directed along +X direction, what will be the magnitude and direction of the torque acting on this? (CBSE Delhi 2011)

Answer:
The resultant dipole moment of the combi-nation is
P_R = \(\sqrt{p^{2}+p^{2}+2 p^{2} \cos 120^{\circ}}\) = p

since cos 120° = -1/2
This will make an angle of 30° with the X-axis, therefore torque acting on it is
τ=PE sin 30° = pE/2 (Along Z-direction)

Question 8.
A metallic spherical shell has an inner radius R₁ and outer radius R₂. A charge Q is placed at the center of the spherical cavity. What will be surface charge density on (i) the inner surface, and (ii) the outer surface? (NCERT Exemplar)
Answer:
The induction of charges is as shown.

Therefore surface charge density on the inner and the outer shell is on the outer surface is

Question 9.
If the total charge enclosed by a surface is zero, does it imply that the electric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that the net charge inside is zero. (NCERT Exemplar)
Answer:
No, the field may be normal to the surface. However, the converse is true i.e. when the electric field everywhere on the surface be zero then the net charge inside it must be zero.

Electric Charges and Fields Important Extra Questions Long Answer Type

Question 1.
(a) State Gauss theorem in electrostatics. Using it, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
Answer:
It states, “The net electric flux through any Gaussian surface is equal to \(\frac{1}{\varepsilon_{0}}\) times the net electric charge enclosed by the surface.

Mathematically, Φ =\( \phi \vec{E} \cdot d \vec{A}=\frac{q_{i n}}{\varepsilon_{0}}\)

Consider an infinite plane sheet of charge. Let a be the uniform surface charge density, i.e. the charge per unit surface area. From symmetry, we find that the electric field must be perpendicular to the plane of the sheet and that the direction of E on one side of the plane must be opposite to its direction on the other side as shown in the figure below. In such a case let us choose a Gaussian surface in the form of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A.

The charged sheet passes through the middle of the cylinder’s length so that the cylinder’s ends are equidistant from the sheet. The electric field has a normal component at each end of the cylinder and no normal component along the curved surface of the cylinder. As a result, the electric flux is linked with only the ends and not the curved surface.

Therefore by the definition of eLectric flux, the flux Linked with the Gaussian surface is given by
Φ = \(\oint_{A} \vec{E} \cdot \vec{\Delta} A\)
Φ = E_A + E_A = 2E_A … (1)

But by Gauss’s Law
Φ = \(\frac{q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}\) [∵ q = σA] … (2)

From equations (1) and (2), we have
2E_A = \(\frac{\sigma A}{\varepsilon_{0}}\) … (3)
E = \(\frac{\sigma}{2 \varepsilon_{0}}\) …. (4)

This gives the electric field due to an infinite plane sheet of charge which is independent of the distance from the sheet.

(b) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged? (C8SE Delhi 2012)
Answer:
(a) directed outwards
(b) directed inwards.

Question 2.
Use Gauss’s law to derive the expression for the electric field (\(\vec{E}\) ) due to a straight uniformly charged infinite line of charge λ Cm^-1. (CBSE Delhi 2018)
Answer:

Consider an infinitely Long, thin wire charged positively and having uniform Linear charge density λ. The symmetry of the charge distribution shows that must be perpendicular to the tine charge and directed outwards. As a result of this symmetry, we consider a Gaussian surface in the form of a cylinder with arbitrary radius r and arbitrary Length L. with its ends perpendicular to the wire as shown in the figure. Applying Gauss’s theorem to curved surface ΔA₁ and circular surface ΔA₂.

Φ EΔA1 cos 0°+ EΔA2 cos 90° = \(\frac{q}{\varepsilon_{0}}\) = \(\frac{\lambda l}{\varepsilon_{0}}\) [∵ λ = \(\frac{q}{e}\)]
Or
E . 2πrl = \(\frac{\lambda l}{\varepsilon_{0}}\) ⇒ E = \(\frac{1}{2 \pi \varepsilon_{0}} \frac{\lambda}{r}\)

This is the expression for the electric field due to an infinitely long thin wire.
The graph is as shown.

Question 3.
Obtain the expression for the potential energy of an electric dipole placed with its axis at an angle (θ) to an external electric field (E). What is the minimum value of the potential energy? (CBSE 2019C)
Answer:
The torque x acting on an electric dipole of dipole moment p placed in a uniform electric field E is given by:
τ = pEsinθ …(i)
where θ is the angle made by the dipole with the electric field E. The torque tends to align the dipole along the direction of the field. If the dipole is rotated through a small angle dθ against the torque, work has to be done, which is stored in the form of the potential energy of the dipole.

Work done in rotating the dipole through the angle dθ against the torque τ is given by dw = τ dθ = pE sin θ dθ

If the dipole is rotated from θ₁ to θ₂, then
Total work is done,

W = – pE(cos θ₂ – cos θ₁)

This work done is stored as potential energy U.

U = – pE(cos θ₂ – cos θ₁)
If the dipole is rotated from θ₁ = \(\frac{\pi}{2}\) to θ₂ =θ, then

U = – pE(cos θ – cos \(\frac{\pi}{2}\) )
= – pE(cos θ – 0)

U = – pE cos θ
minimum value of potential energy
U = – pE When θ = 0°

Question 4.
Why does the electric field inside a dielectric decrease when it is placed in an external electric field?
Answer:
When a dielectric is placed in an electric field (E₀), it gets polarised, i.e. within the dielectric, an electric field (E) is induced in a direction opposite to that of the external field. Therefore, the net field within the dielectric decreases to \(\vec{E}_{0}\) – \(\vec{E}\)

Question 5.
Two-point charges +q and -2q are placed at the vertices ‘B’ and ‘C’ of an equilateral triangle ABC of side ‘a’ as given in the figure. Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric field at vertex A due to these two charges. (CBSE Al 2014 C)

Answer:
The electric fields due to the two charges placed at B and C are inclined at an angle θ = 120° as shown
Now in magnitude, we have

E_B = k\(\frac{q}{a^{2}}\) and
E_C = k\(\frac{2q}{a^{2}}\) = 2 E_B

Hence E = \(\sqrt{E_{B}^{2}+E_{c}^{2}+2 E_{B} E_{c} \cos \theta}\)
Or E = \(\sqrt{E_{B}^{2}+\left(2 E_{B}\right)^{2}+2 E_{B}\left(2 E_{B}\right) \cos 120^{\circ}}\)

On Solving we have
E = \(\sqrt{3} E_{B}\) = \(\sqrt{3} \frac{k q}{a^{2}}\)

Direction

Therefore B = 90°, the resultant is inclined at an angle of 90° with EB.

Question 6.
Four-point charges Q, q, Q, and q are placed at the corners of a square of side ‘a’ as shown in the figure.
Find the
(a) the resultant electric force on a charge Q and
(b) the potential energy of this system. (CBSEAI, Delhi 2018)
Answer:
(a) Let us find the force on charge Q at point C.

Force due to charge Q placed at point A is
F_A = k\(\frac{Q^{2}}{(a \sqrt{2})^{2}}\) =k\(\frac{Q^{2}}{2 a^{2}}\) along AC

Force due to the charge q placed at D
F_D = k\(\frac{q Q}{a^{2}}\) alongDA

Force due to the charge q placed at B
F_B = k \(\frac{q Q}{a^{2}}\) along BC

The resuLtant of F_D and F_B is
F_BD = K\(\frac{q Q \sqrt{2}}{a^{2}}\) along AC

∴ net force of charge Q placed at point C is

(b) PotentiaL energy of the system

Question 7.
A charge +Q is uniformly distributed within a sphere of radius R. Find the electric field, distant r from the center of the sphere where: (1) r < R and (2) r > R. (CBSEAI 2016C)
Answer:
For a solid sphere p = \( \frac{q}{\frac{4}{3} \pi R^{3}}\) = \(\frac{q}{\text { volume }}\)

Case 1. 0 < r < R The point Lies within the sphere.

Using Gauss’s theorem
Let Q’ be the charge encLosed by Gaussian’s surface of radius r < R.

E(4πr²)=\(\frac{Q^{\prime}}{\varepsilon_{0}}=\frac{Q^{\prime}}{4 \pi \varepsilon_{0} r^{2}}\)

From (i) and (ii)
E = \(\frac{Q r^{3}}{4 \pi \varepsilon_{0} r^{2} R^{3}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q r}{R^{3}}\)

Case 2. For r > R
\(\oint \vec{E} \cdot \overrightarrow{d s}\) = \(\frac{q_{\text {enclosed }}}{\varepsilon_{0}}\)

E(4πr²) = \(\frac{Q}{\varepsilon_{0}}\)
E = \(\frac{Q}{4 \pi \varepsilon_{0} r^{2}}\)

Question 8.
(a) An electric dipole is kept first to the left and then to the right of a negatively charged infinite plane sheet having a uniform surface charge density. The arrows p₁ and p₂ show the directions of its electric dipole moment in the two cases. Identify for each case, whether the dipole is in stable or unstable equilibrium. Justify each answer.
Answer:
p₁: stable equilibrium, p₂: unstable equilibrium. The electric field, on either side, is directed towards the negatively charged sheet and its magnitude is independent of the distance of the field point from the sheet. For position P₁ dipole moment and electric field are parallel. For position p₂, they are antiparallel.

(b) Next, the dipole is kept in a similar way (as shown), near an infinitely long straight wire having uniform negative linear charge density. Will the dipole be in equilibrium at these two positions? Justify your answer. (CBSE Sample Paper 2018-2019)

Answer:
The dipole will not be in equilibrium in any of the two positions.
The electric field due to an infinite straight charged wire is non-uniform (E ∝ 1/r).
Hence there will be a net non-zero force on the dipole in each case.

Question 9.
Two large parallel plane sheets have uniform charge densities +σ and -σ. Determine the electric field (i) between the sheets, and (ii) outside the sheets. (CBSE Delhi 2018)
Answer:
Let us consider two parallel planes charged conductors A and B carrying +ve and -ve charge density σ (charge per unit area). According to Gauss’ theorem, the electric intensity at P due to the charge on sheet A is

EA = \(\frac{\sigma}{2 \varepsilon_{0}}\) (From A and B)

The electric field at P due to charge on sheet B is
E= \(\frac{\sigma}{2 \varepsilon_{0}}\) (From A to B)

The electric field at P is
E = EA + EB
= \(\frac{\sigma}{\varepsilon_{0}}\)

Question 10.
Define electric flux and write its SI unit. The electric field components in the figure shown are : E_x = αx, E_y = 0, E_z = 0 where α = 100N/cm. Calculate the charge within the cube, assuming a = 0.1 m. (CBSE Sample Paper 2019)
Or
An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 10⁴ N/C (figure a)

(a) Calculate the time it takes to fall through this distance starting from rest.

(b) If the direction of the field is reversed (figure b) keeping its magnitude unchanged, calculate the time taken by a proton to fall through this distance starting from rest. (CBSE Delhi 2018C)
Answer:
Electric Flux is the dot product of the electric field and area vector.
Φ = \(\oint \vec{E} \cdot \overrightarrow{d s}\)

SI Unit: Nm²/C or Vm

For a given case
Φ = Φ₁ – Φ₂ = [Ex(atx= 2a) – Ex (atx = a)]a²
= [α(2a)-α(a)]a² = αa³
= 10⁴ × (0.1 )³ = 10 Nm²/C
But
Φ = \(\frac{q}{\varepsilon_{0}}\)

∴ q = ε₀Φ = 8.854 × 10^-12 × 10 C = 8.54 pC
Or
We have
F = qE

Acceleration, a = \(\frac{q E}{m}\)

Also
s = \(\frac { 1 }{ 2 }\) at² [u = 0]
∴ t = \(\sqrt{\frac{2 s}{a}}\)

(i) For the electron
a = \(\frac { eE}{ m }\)

∴ t = \(\sqrt{\frac{2 s m}{e E}}\)

∴ t = \(\sqrt{\frac{3 \times 10^{-2} \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19} \times 2.0 \times 10^{4}}}\) = 2.92 ns

(ii) for proton
t = \(\sqrt{\frac{2 \times 1.5 \times 10^{-2} \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 2 \times 10^{4}}}\)
= -.125 μs

Question 11.
What will be the total flux through the faces of the cube (figure) with the side of length ‘a’ if a charge q is placed at
(a) A: a corner of the cube.
(b) B: mid-point of an edge of the cube.
(c) C: center of the face of the cube.
(d) D: mid-point of B and C. {NCERJ Exemplar)

Answer:
(a) The charge wilt is shared by eight cubes if it has to be enclosed. Therefore the flux through the cube will be one-eighth of the total flux. Φ = q/8ε₀
(b) The charge will be shared by four cubes if it has to be enclosed. Therefore the flux through the cube will be one-fourth of the total flux. Φ = q/ 4ε₀
(c) The charge will be shared by two cubes if it has to be enclosed. Therefore the flux through the cube will be one-half of the total flux. Φ = q/ 2ε₀
(d) The charge will be shared by two cubes if it has to be enclosed. Therefore the flux through the cube will be one-half of the total flux. Φ = q/ 2ε₀

Question 12.
Two charges q and -3q are placed fixed on the x-axis separated by distance ‘d’. Where should a third charge 2q be placed such that it will not experience any force? (NCERT Exemplar)
Answer:
The situation is as shown.

Let the charge 2q be placed at a distance ‘x’ from charge q. For the charge 2q to experience zero force we have
\(\frac{2 q^{2}}{4 \pi \varepsilon_{0} x^{2}}=\frac{6 q^{2}}{4 \pi \varepsilon_{0}(d+x)^{2}}\)
(d + x)² = 3x²

Solving for x we have
x = \(\frac{d}{2} \pm \frac{\sqrt{3} d}{2}\)
(-ve sign would be between q and -3q and hence is unacceptable.)

Therefore, we have
x = \(\frac{d}{2}+\frac{\sqrt{3} d}{2}=\frac{d}{2}(1+\sqrt{3})\) to the left of q.

Question 13.
(a) State Gauss’s law. Use it to deduce the expression for the electric field due to a uniformly charged thin spherical shell at points (i) inside and (ii) outside the shell. Plot a graph showing the variation of the electric field as a function of r > R and r < R. (r being the distance from the center of the shell) (CBSE Delhi 2011, Al 2013)
Answer:
Gauss’s law states that the net outward flux through any closed surface is equal to 1 /ε₀ times the charge enclosed by the closed surface.

Consider a thin spherical shell of radius R and center at O. Let σ be the uniform surface charge density (charge per unit surface area) and q be the total charge on it. The charge distribution is spherically symmetric. Three cases arise

Case 1: at a point outside the spherical shell
In order to find the electric field at a point P outside the shell let us consider a Gaussian surface in the form of a sphere of radius r (r >>R).

By symmetry, we find that the electric field acts radially outwards and has a normal component at alt points on the Gaussian sphere. Therefore by definition of electric flux we have

Φ = E × A, where A is the surface area of the Gaussian sphere therefore
Φ = E × 4πr² …(1)

But by Gauss’s law
Φ = \(\frac{Q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}=\frac{\sigma \times 4 \pi R^{2}}{\varepsilon_{0}}\) … (2)

from equations (1) and (2) it follows that
E × 4πr² = \(\frac{Q}{\varepsilon_{0}}\) Or E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r^{2}}\)

E × 4πr² = \(\frac{\sigma \times 4 \pi R^{2}}{\varepsilon_{0}}\)

Ē = \(\frac{\sigma R^{2}}{\varepsilon_{0} r^{2}}\) … (3)

It follows that the electric field due to a spherical shell outside it is same as that due to a point charge. Therefore for points Lying outside the spherical shell the shell behaves as if the entire charge is concentrated at the centre.

Case 2: at a point inside the spherical shell
In this case, the Gaussian surface Lies inside the shell. Since no charge is enclosed In this surface therefore we have

E × 4πr²=q₀/ε₀ [∵ q=0]
Therefore E = 0

(b) Two identical metallic spheres A and B having charges +4Q. and -10 Q are kept a certain distance apart. A third identical uncharged sphere C is first placed in contact with sphere A and then with sphere B. Spheres A and B are then brought in contact and then separated. Find the charges on the spheres A and B. (CBSEAI 2011C)
Answer:
The initial charge on the sphere A = + 4 Q.
The initial charge on the sphere B = -10 Q.

Since all the three spheres are identical, i.e. they have the same capacity, therefore when uncharged sphere C is placed in contact with A, the total charge is equally shared between them.

Charge on C after contact with A
= \(\frac{0+4 Q}{2}\) = 2Q

Charge on A after contact with C is 2Q.
When sphere C carrying a charge 2Q is placed in contact with B, again charges are equally shared between C and B equally.

Charge on C after it is in contact with B
= \(\frac{2 Q-10 Q}{2}\) = -4Q

Now when sphere A with a charge of 2Q. is placed in contact with B, with charge -4Q.

Charge are again shared
∴ charge on A or B = \(\frac{2 Q-4 Q}{2}\) = -Q.

Question 14.
(a) Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.
Answer:
It is defined as the product of the magnitude of either of the two charges and the distance between them.
For derivation see sol. 9(a) of LA-II.

(b) Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero. (CBSE Al 2013)
Answer:
The diagram is as shown

The zero potential points lie on the equatorial line.

Question 15.
(a) Using Gauss’s law obtain expressions for the electric field (i) inside, and (ii) outside a positively charged spherical shell.
(c) A Square plane sheet of side 10 cm is inclined at an angle of 30° with the direction of a uniform electric field of 200 NC“1. Calculate the electric flux passing through the sheet. (CBSE 2019C)
Answer:
(a) Spherical Shell
Consider a spherical shell of radius R. Let q be a charge on the shell. Let us find the electric field at a point P at a distance r from the center 0 of the spherical shell.

Case (i): When point P lies inside the spherical shell
From the point, P draws a Gaussian surface which will be a sphere of radius r.

From the Gauss’s Theorem,
\(\oint_{s} \vec{E} \cdot \vec{d} S=\frac{0}{\varepsilon_{0}}\) [∵ No charge exists inside the spherical shell]

Or
E = 0
i. e. electric field inside the charged spherical shell is zero.

Case (ii): When point P is lying outside the shell (i.e. r > R)
From point P, draw a Gaussian surface which will be a spherical shell of radius r. Let dS be a small area element on the Gaussian surface P.


Or

i. e. the electric field outside the spherical shell behaves as if the whole charge is concentrated at the center of the spherical shell.

(b) Show graphically variation of the electric field as a function of the distance r from the center of the sphere.
Answer:
Variation of electric field E with distance
The given figure shows the variation of an electric field with distance from the center of the charged spherical shell.

(c) Here E = 200 N/C, S = 0.1 × 0.1 = 0.01 m²
And θ = 90° – 30° = 60°

The electric flux linked with the square sheet
Φ = E S cos 60°
= 200 × 0.01 × \(\frac { 1 }{ 2 }\) = 1.0 Nm² C^-1

Question 16.
A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at a large distance from the ring, it behaves like a point charge. (CBSE Delhi 2016)
Answer:
Consider a uniformly charged ring of radius ‘a’. Let the total charge on the ring be Q, Let us find the electric field on the axis of the ring at point P distance x from the center of the ring. Consider a segment of charge dQ as shown in the figure.

The magnitude of etectñc field at P due to the segment is
dE=k\(\frac{d Q}{r^{2}}\) …(1)

This field can be resolved into its components: x component dE_x = dE cos α an along the axis of the ring and y component dE perpendicular to the axis. Since these perpendicular components, due to alt the charge segments, are equal and opposite, therefore they cancel out each other. From the diagram we have r = \(\sqrt{x^{2}+a^{2}}\) and cos α = x/r, therefore we have
…(2)

In this case, all the segments of the ring give the same contribution to the field at P since they are all equidistant from this point. Thus we can easily sum over all segments to get the total electric field at point P

If the point of observation is far away, i.e. x >> a, then E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{x^{2}}\). This is the same as that for a point charge. Thus at far-off axial points, a charged ring behaves as if a point charge is situated at the center of the ring.

Question 17.
Two thin concentric and coplanar spherical shells, of radii ‘a’ and ‘b’ (b > a), carry charges, q, and Q respectively. Find the magnitude of the electric field, at a point distant x, from their common center for
(i) O < x < a
(ii) a ≤ x < b
(iii) b ≤ x < ∞ (CBSE Delhi 2016C)
Answer:
The diagram is as shown.

(i) For 0 < x < a
Point Lies inside both the spherical shells.

∴ charge enclosed = 0
Hence, E(x) = 0

(ii) For a < x < b
Point is outside the spherical shell of radius ‘a’ but inside the spherical Shell of radius ‘b’.
Therefore

(iii) For b ≤ x< ∞
The point is outside of both the spherical shells. The total effect we charge at the center equals (Q + q).

Question 18.
(a) Define electric flux. Is It a scalar or a vector quantity?

A point charge q is at a distance of d/2 directly above the center of a square of side d, as shown In the figure. Us. Gauss’s law to obtain the expression for the electric flux through the square.
Answer:
The electric flux through a given surface is defined as the dot product of the electric field and area vector over that surface. It is a scalar quantity.

Constructing a cube of side ‘d’ so that charge ‘q’ gets placed within this cube (Gaussian surface)
According to Gauss’s law the electric flux
Φ = \(\frac{q}{\varepsilon_{0}}\)

This is the total flux through all the six faces of the cube.

Hence electric flux through the square (one face of the cube)
Φ = \(\frac{1}{6} \frac{q}{\varepsilon_{0}}\)

(b) If the point charge is now moved to a distance ‘d’ from the center of the square and the side of the square is doubled, explain how the electric flux will be affected. (CBSE Al, Delhi 2018)
Answer:
If the charge is moved to a distance ‘d’ and the side of the square is doubled the cube will be constructed to have a side 2d but the total charge enclosed in it will remain the same. Hence the total flux through the cube and therefore the flux through the square will remain the same as before.

Question 19.
(a) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
Answer:
The required graph is

(b) Find the work done in bringing a charge q from perpendicular distance r₁ to r₂ (r₂ > r₁). (CBSEAI, Delhi 2018)
Answer:
Work done in moving the charge “q”. through a smaLL displacement ‘dr’

dW = \(\vec{F}\) . \(\vec{dr}\) =q\(\vec{E}\) .\(\vec{dr}\)
dW = qE dr cos 0 = qEdr
dW= q x \(\frac{\lambda}{2 \pi \varepsilon_{0} r}\) dr

Hence work done in moving the charge from r₁ to r₂ (r₂ > r₁)

Question 20.
Derive an expression for the torque acting on an electric dipole of dipole moment \(\vec{p}\) placed in a uniform electric field \(\vec{E}\). Write the direction along which the torque acts.
OR
Derive an expression for the electric field at a point on the axis of an electric dipole of dipole moment \(\vec{p}\). Also, write its expression when the distance r >> a the length ‘a’ of the dipole. (CBSEAI 2019)
Answer:
Consider an electric dipole consisting of charges -q and +q and dipole length d is placed in a uniform electric field E as shown in the figure.

Let the dipole moment makes an angle θ with the direction of the electric field. The two charges experience force qE each.

These forces are equal, parallel, and opposite. Therefore, the net force acting on the dipole is
F_n = qE – qE = 0

Thus the net force acting on the dipole is zero.

These two forces constitute a couple. This applies a torque on the dipole given by
τ = either force × arm of the couple
τ = qE × d sin θ, where d sin θ is the arm of the couple.
τ = q d Esin θ, where p = qd, dipole moment
τ = pE sin θ.

Consider an electric dipole consisting of -q and +q charges separated by a distance 2a as shown in the figure. Let P be the point of observation on the axial Line where the electric field has to be found. Let it be at a distance r from the center O of the dipole. Let us suppose that the dipole is placed in a vacuum.

Let E and E_B be the electric fields at point P due to the charges at A and B respectively. Therefore,

The two fields at P are in opposite directions. Thus the resultant electric field at P is given by
E = \(\sqrt{E_{A}^{2}+E_{B}^{2}-2 E_{A} E_{B} \cos \theta}=\sqrt{\left(E_{B}-E_{A}\right)^{2}}\) = E_B – E_A

since θ = 180⁰ Therefore, the resultant electric field is

Solving we have

where p = q 2a.
If the dipole is short then r >> a, therefore, ‘a’ is neglected as compared to r, hence

This field is in the direction of the dipole moment.

Question 21.
(a) Derive the expression for the electric field at a point on the equatorial line of an electric dipole. (CBSE 2019C)
Answer:
Consider an electric dipole consisting of charges -q and +q separated by a distance 2a as shown in the figure. Let the point of observation P lie on the. right bisector of the dipole AB at a distance r from its midpoint 0. Let E_A and E_B be the electric field intensities at point P due to charges at A and B, respectively.

The two electric fields have magnitudes

in the direction of AP

in the direction of PB.

The two fields are equal in magnitude but have different directions. Resolving the two fields E_A and E_B into their rectangular components, i.e. perpendicular to and parallel to AB. The components perpendicular to AB, i.e E_A sinG and E_B sinG being equal and opposite cancel out each other while the components parallel to AB, i.e. E_A cos θ and E_B cos θ being in the same direction add up as shown in the figure. Hence the resultant electric field at point P is given by

E = E_Acosθ + E_Bcosθ

∵ cos θ = \(\frac{a}{\left(r^{2}+a^{2}\right)^{1 / 2}}\) and q 2a = p

For a short dipole r²,>>a² therefore
E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\)

(b) Discuss the orientation of the dipole in (a) stable, (b) unstable equilibrium in a uniform electric field. (CBSE Delhi 2017)
Answer:
For stable equilibrium \(\vec{P}\) is along \(\vec{E}\)
θ = 0°, τ = PE sin 0°, Torque alligns the dipole in the direction of field

For unstable equilibrium \(\vec{P}\) is antiparallel to \(\vec{E}\)
∵ θ = 180°, τ = PE sin 180° = 0, Torque alligns the dipole in a direction opposite to \(\vec{E}\).

Question 22.
(a) An electric dipole of dipole moment \(\vec{p}\) consists of point charges +q and -q separated by a distance 2a apart. Deduce the expression for the electric field due to the dipole at a distance x from the center of the dipole on its axial line in terms of the dipole moment. Hence show that in

the limit r >> a, \(\vec{E}\) → \(\frac{2 \vec{p}}{4 \pi \varepsilon_{0} r^{3}}\)
Answer:

Consider an electric dipole consisting of charges -q and +q and dipole length d is placed in a uniform electric field E as shown in the figure.

Let the dipole moment makes an angle θ with the direction of the electric field. The two charges experience force qE each.

These forces are equal, parallel, and opposite. Therefore, the net force acting on the dipole is
F_n = qE – qE = 0

Thus the net force acting on the dipole is zero.

These two forces constitute a couple. This applies a torque on the dipole given by
τ = either force × arm of the couple
τ = qE × d sin θ, where d sin θ is the arm of the couple.
τ = q d Esin θ, where p = qd, dipole moment
τ = pE sin θ.

Consider an electric dipole consisting of -q and +q charges separated by a distance 2a as shown in the figure. Let P be the point of observation on the axial Line where the electric field has to be found. Let it be at a distance r from the center O of the dipole. Let us suppose that the dipole is placed in a vacuum.

Let E and EB be the electric fields at point P due to the charges at A and B respectively. Therefore,

The two fields at P are in opposite directions. Thus the resultant electric field at P is given by
E = \(\sqrt{E_{A}^{2}+E_{B}^{2}-2 E_{A} E_{B} \cos \theta}=\sqrt{\left(E_{B}-E_{A}\right)^{2}}\) = E_B – E_A

since θ = 1800 Therefore, the resultant electric field is

Solving we have

where p = q 2a.
If the dipole is short then r >> a, therefore, ‘a’ is neglected as compared to r, hence

This field is in the direction of the dipole moment.

(b) Given the electric field in the region E = 2xî, find the net electric flux through the cube and the charge enclosed by it. (CBSE Delhi 2015)

Answer:
Since the electric field has only an x component, for faces perpendicular to the x-direction, the angle between E and ΔS is ± π/2.

Therefore, the flux Φ = E.ΔS is separately zero for each face of the cube except the two faces along the X-axis.

Now the magnitude of the electric field at the left face is
E_L= 0 (x = 0 at the left face)

The magnitude of the electric field at the right face is
E_R = 2x = 2a (x = a at the right face)

The corresponding fluxes are
Φ_L = E_L.ΔS = ΔS (\(\vec{E}_{L} \cdot \hat{n}_{\mathrm{L}}\)) = E_LΔS COS θ
= – E_L ΔS = 0, since θ =180°

Φ_R = E_R.ΔS = E_RΔS cos θ = E ΔS = (2a)a², since θ = 0°

Net flux through the cube
Φ = -Φ_R + Φ_L = 2a³ -0 = 2a³

We can use Gauss’s law to find the total charge q inside the cube.
We have Φ = q/ε₀ or q = Φε₀. Therefore, q = 2a³ × 8.854 × 10^-12C

Question 23.
(a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.
Answer:
Consider an electric dipole consisting of charges -q and +q separated by a distance 2a as shown in the figure. Let the point of observation P lie on the. right bisector of the dipole AB at a distance r from its midpoint 0. Let EA and EB be the electric field intensities at point P due to charges at A and B, respectively.

The two electric fields have magnitudes

in the direction of AP

in the direction of PB.

The two fields are equal in magnitude but have different directions. Resolving the two fields E_A and E_B into their rectangular components, i.e. perpendicular to and parallel to AB. The components perpendicular to AB, i.e E_A sinG and E_B sinG being equal and opposite cancel out each other while the components parallel to AB, i.e. E_A COS θ and E_B cos θ being in the same direction add up as shown in the figure. Hence the resultant electric field at point P is given by

E = E_A cos θ + E_B cos θ

∵ cos θ = \(\frac{a}{\left(r^{2}+a^{2}\right)^{1 / 2}}\) and q 2a = p

For a short dipole r²,>>a² therefore
E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\)

(b) Two identical point charges, q each, are kept 2 m apart in the air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q. (CBSE Delhi 2019)
Answer:
The third charge Q. wilt is in equilibrium if it experiences zero net force. Let it be placed at a distance x meter from the charge q.

Solving for x, we have x= 1 m

For the equilibrium of charges “q”, the nature of charge Q must be opposite to the nature of charge q and should be placed at the center of two charges.

Question 24.
(a) Define electric flux. Write its S.I. unit.
Answer:
It is defined as the total number of electric field lines crossing a given area. The electric flux can be found by multiplying the component of the electric field in the direction of the area vector (or perpendicular to the area) with the area of the closed surface. It is measured in Nm²C-^-1.

(b) A small metal sphere carrying charge +Q is located at the center of a spherical cavity inside a large uncharged metallic spherical shell as shown in the figure. Use Gauss’s law to find the expressions for the electric field at points P₁ and P₂.

Answer:
For point P, using Gauss law we have

Since E and dA are in the same direction therefore we have E = \(\frac{Q}{\varepsilon_{0} A}\)

Point P₂ lies inside the metal, therefore the Gaussian surface drawn at P₂ does not include a charge, hence the electric field at P₂ is zero.

(c) Draw the pattern of electric field lines in this arrangement. (CBSEAI 2012C)
Answer:
The electric field lines are as shown.

Question 25.
(a) Deduce the expression for the torque acting on a dipole of dipole moment \(\vec{P}\) in the presence of a uniform electric field \(\vec{E}\)
Answer:

Consider an electric dipole consisting of charges -q and +q and dipole length d is placed in a uniform electric field E as shown in the figure.

Let the dipole moment makes an angle θ with the direction of the electric field. The two charges experience force qE each.

These forces are equal, parallel, and opposite. Therefore, the net force acting on the dipole is
F_n = qE – qE = 0

Thus the net force acting on the dipole is zero.

These two forces constitute a couple. This applies a torque on the dipole given by
τ = either force × arm of the couple
τ = qE × d sin θ, where d sin θ is the arm of the couple.
τ = q d Esin θ, where p = qd, dipole moment
τ = pE sin θ.

Consider an electric dipole consisting of -q and +q charges separated by a distance 2a as shown in the figure. Let P be the point of observation on the axial Line where the electric field has to be found. Let it be at a distance r from the center O of the dipole. Let us suppose that the dipole is placed in a vacuum.

Let E and EB be the electric fields at point P due to the charges at A and B respectively. Therefore,

The two fields at P are in opposite directions. Thus the resultant electric field at P is given by
E = \(\sqrt{E_{A}^{2}+E_{B}^{2}-2 E_{A} E_{B} \cos \theta}=\sqrt{\left(E_{B}-E_{A}\right)^{2}}\) = E_B – E_A

since θ = 180° Therefore, the resultant electric field is

Solving we have

where p = q 2a.
If the dipole is short then r >> a, therefore, ‘a’ is neglected as compared to r, hence

This field is in the direction of the dipole moment.

(b) Consider two hollow concentric spheres S₁ and S₂ and enclosing charges 2Q and 4Q. respectively as shown in the figure, (i) Find out the ratio of the electric flux through them, (ii) How will the electric flux through the sphere S₁ change if a medium of dielectric constant ‘er’ is introduced in the space inside S1 in place of air? Deduce the necessary expression. (CBSE Al 2014)

Answer:
Φ₁ = \(\frac{2 Q}{\varepsilon_{0}}\) and Φ2 = \(\frac{2 Q+4 Q}{\varepsilon_{0}}=\frac{6 Q}{\varepsilon_{0}}\)

Hence ratio

When a dielectric of dielectric constant εr is introduced then we have
Φ₁ = \(\frac{2 Q}{\varepsilon}=\frac{2 Q}{\varepsilon_{r} \varepsilon_{0}}\)

Question 26.
(a) Use Gauss’s theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density σ.
Answer:
It states, “The net electric flux through any Gaussian surface is equal to \(\frac{1}{\varepsilon_{0}}\) times the net electric charge enclosed by the surface.

Mathematically, Φ =\( \phi \vec{E} \cdot d \vec{A}=\frac{q_{i n}}{\varepsilon_{0}}\)

Consider an infinite plane sheet of charge. Let a be the uniform surface charge density, i.e. the charge per unit surface area. From symmetry, we find that the electric field must be perpendicular to the plane of the sheet and that the direction of E on one side of the plane must be opposite to its direction on the other side as shown in the figure below. In such a case let us choose a Gaussian surface in the form of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A.

The charged sheet passes through the middle of the cylinder’s length so that the cylinder’s ends are equidistant from the sheet. The electric field has a normal component at each end of the cylinder and no normal component along the curved surface of the cylinder. As a result, the electric flux is linked with only the ends and not the curved surface.

Therefore by the definition of eLectric flux, the flux Linked with the Gaussian surface is given by
Φ = \(\oint_{A} \vec{E} \cdot \vec{\Delta} A\)

Φ = E_A + E_A = 2E_A … (1)

But by Gauss’s Law

Φ = \(\frac{q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}\) [∵ q = σA] … (2)

From equations (1) and (2), we have

2E_A = \(\frac{\sigma A}{\varepsilon_{0}}\) … (3)

E = \(\frac{\sigma}{2 \varepsilon_{0}}\) …. (4)

This gives the electric field due to an infinite plane sheet of charge which is independent of the distance from the sheet.

(b) An infinitely large thin plane sheet has a uniform surface charge density +σ. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point, distant r, in front of the charged plane sheet. (CBSE Al 2017)
Answer:
Work done

Numerical Problems:

Formulae for solving numerical problems

• q = ±ne
• F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\)
• Fmed = \(\frac{1}{4 \pi \varepsilon_{0} K} \frac{q_{1} q_{2}}{r^{2}}\) where K is dielectric constant.
• 
• E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 p}{r^{3}}\) for an electric dipole on its axial line.
• E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\) for an electric dipole on its equitorial line.
• Torque on an electric dipole in a uniform electric field, τ = PE sin θ.
• U = -pE cos θ, the potential energy of an electric dipole.
• Φ = \(\oint \vec{E} \cdot d \vec{A}=\frac{q_{\text {in }}}{\varepsilon_{0}}\)
• E = \(\frac{\sigma}{2 \varepsilon_{0}}\) , electric field due to an infinite plane sheet of charge.

Question 1.
An electric dipole of length 2 cm is placed with its axis making an angle of 60° to a uniform electric field of 105 NC^-1. If it experiences a torque of 8\(\sqrt{3}\) Nm, calculate the (i) magnitude of the charge on the dipole and (ii) potential energy of the dipole.
Answer:
(i) Using
q = τ / L E sin Φ

we have
q = 8N\(\sqrt{3}\) × 2 / 2 × 10^-2 × 10⁵ × \(\sqrt{3}\) = 8 × 10^-3 C

(ii) Using U = -pE cos θ, we have
U = – 8 × 10^-3 × 0.02 × 10⁵ × 0.5 = – 8 J

Question 2.
A charge of 17.7 × 10^-4 C is distributed uniformly over a large sheet of area 200 m². Calculate the electric field intensity at a distance of 20 cm from it in the air.
Answer:
Given q = 17.7 × 10^-4C, A = 200 m², r = 20cm = 0.2 m

Using the relation

Question 3.
Two fixed point charges +4e and +e units are separated by a distance ‘a’. Where should the third point charge be placed for it to be in equilibrium?
Answer:
The third charge will be in equilibrium if it experiences zero net force. Let it be placed at a distance x from the charge 4e.

4(a – x)² = x² solving for x we have
2 (a – x) = x or x = a/3

Question 4.
An electric field along the x-axis is given by \(\vec{E}\) = 100 îN/C for x > 0 and \(\vec{E}\) = -100 îN/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm lies parallel to the x-axis with its centre at the origin and one face at x = +10 cm, the other face at x = -10 cm. Calculate the net outward flux through the cylinder. (CBSE 2019C)

r = 5 cm = 0.05 m
Φ = ?

Net outward flux through the cylinder
Φ = Φ₁ + Φ₂ + Φ₃
= \(\vec{E} \cdot \overrightarrow{d s}_{1}+\vec{E} \cdot \overrightarrow{d s}_{2}+\vec{E} \cdot \overrightarrow{d s}_{3}\)

= Eds₁ cos 180° + Eds₂ cos 90° + Eds3 cos 0°
= – Eds₁ + Eds₂ (0) + Eds cos 0°
= – (-100) ds + 100 ds
= (100 +100) ds
= 200 × πr² = 200 × 3.14 × (0.05)²
= 1.57 Nm² C^-1

Question 5.
A hollow cylindrical box of length 1m and area of cross-section 25 cm² is placed in a three-dimensional coordinate system as shown in the figure. The electric field
in the region is given by \(\vec{E}\) = 50x î, where E is in NC^-1 and x is in meters. Find

(i) Net flux through the cylinder.
Answer:
We can see from the figure that on the left face E and dS are antiparallel. Therefore, the flux is
ΦL = \(\vec{E}\).\(\vec{dS}\) = E dS cos 180°
= -50 × 1 × 25 × 10^-4 = -0.125 N m² C^-1

On the right face, E and dS are parallel and therefore
ΦR = \(\vec{E}\).\(\vec{ds}\) = EdS cos 0°
= 50 × 2 × 25 × 10^-4 = 0.250 N m² C^-1

Therefore net flux is – 0.125 + 0.250 = + 0.125 N m² C^-1

(ii) Charge enclosed by the cylinder. (CBSE Delhi 2013)

Answer:
q = Φε₀ = 0.125 × 8.854 × 10^-12 = 1.1 × 10^-12C

Question 6.
(i) Define electric flux. Write its SI units,
Answer:
(i) It is defined as the number of electric field lines crossing a unit area perpendicular to the given area. It is measured in N m² C^-1

(ii) Consider a uniform electric field \(\vec{E}\) = 5 × 10³î NC^-1. Calculate the flux of this field through a square surface of area 12 cm² when
(a) its plane is parallel to the Y – Z plane, and
(b) the normal to its plane makes a 60° angle with the x-axis. (CBSE Delhi 2013C)
Answer:
Given \(\vec{E}\) = 5 × 10³î N C^-1, A = 12 cm² = 12 × 10^-4 m²

(a) Here \(\vec{E}\) = 5 × 103î N C^-1
Area of square = 12 × 10^-4 m²

The plane of surface area being parallel to YZ plane, hence
A = 12 × 10^-4 î m²

Electric flux of the field
Φ = \(\vec{E}\) \(\vec{A}\) = (5 × 10³ î). (12 × 10^-4î) = 6 N C^-1 m²
(b) When normal to the plane of surface area makes an angle of 60° with the X-axis, the flux is given by
Φ = EA cos 0 = 5 × 10³ × 12 × 10^-4 × 0.5
= 3 NC^-1 m²

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 2 Electrostatic Potential and Capacitance. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 2 Important Extra Questions Electrostatic Potential and Capacitance

Electrostatic Potential and Capacitance Important Extra Questions Very Short Answer Type

Question 1.
Express dielectric constant in terms of the capacitance of a capacitor.
Answer:
It is given by the expression K = \(\frac{c}{c_{0}}\) where C is the capacitance of the capacitor with dielectric and C₀ is the capacitance without the dielectric.

Question 2.
On what factors does the capacitance of a parallel plate capacitor depend?
Answer:

1. Area of plates,
2. The separation between the plates and
3. Nature of dielectric medium between the plates.

Question 3.
What is the ratio of electric field intensities at any two points between the plates of a capacitor?
Answer:
The ratio is one, as the electric field is the same at all points between the plates of a capacitor.

Question 4.
Write a relation between electric displacement vector D and electric field E.
Answer:
\(\vec{D}\) = ε₀ \(\vec{E}\) + \(\vec{P}\)

Question 5.
Write the relation between dielectric constant (K) and electric susceptibility χ_e
Answer:
K = 1 + χ_e

Question 6.
A hollow metal sphere c radius 5 cm is charged such that the potential on its surface is 10 V. What is the potential at the center of the sphere? (CDSE AI 2011)
Answer:
10 V

Question 7.
What is the geometrical shape of equipotential surfaces due to a single isolated charge? (CBSE Delhi 2013)
Answer:
Concentric circles.

Question 8.
Draw the equipotential surfaces due to an isolated point charge. (CBSE Delhi 2019)
Answer:
These areas are shown.

Question 9.
‘For any charge configuration, equipotential surface through a point is normal to the electric field’. Justify. (CBSE Delhi 2014)
Answer:
This is because work done in moving a charge on an equipotential surface is zero. This is possible only if the equipotential surface is perpendicular to the electric field.

Question 10.
The given graph shows the variation of charge ‘q’ versus potential difference ‘V for two capacitors C₁ and C₂. Both the capacitors have the same plate separation but the plate area of C₂ is greater than that of C_y Which line (A or B) corresponds to C₁ and why? (CBSEAI 2014C)

Answer:
Since C = ε₀ A/d, since the area for C₂ is more, therefore capacitance of C₂ is more. From the graph greater the slope greater is than the capacitance, therefore, graph A belongs to capacitor C₂. While graph B belongs to capacitance C_v

Question 11.
Write a relation for polarisation P of dielectric material in the presence of an external electric field E . (CBSE AI 2015)
Answer:
P = χ_e ε₀ \(\vec{E}\).

Question 12.
Why are electric field lines perpendicular at a point on an equipotential surface of a conductor? (CBSE AI 2015C)
Answer:
So that no net force acts on the charge on the equipotential surface and it remains stationary.

Question 13.
Why does a given capacitor store more charge at a given potential difference when a dielectric is filled in between the plates?
Answer:
Because the capacitance of the capacitor increases on filling the dielectric medium in between the plates.

Question 14.
Why is the electrostatic potential inside a charged conducting shell constant throughout the volume of the conductor? (CBSE AI 2019)
Answer:
We know that E = – \(\frac{d V}{d r}\)
Inside a conductor, the electric field is zero and no work is done in moving a charge inside a conductor. Therefore, V is constant.

Question 15.
Does the charge given to a metallic sphere depend on whether it is hollow or solid? (CBSE Delhi 2017)
Answer:
If the capacitance of the two spheres, solid and hollow, is the same, then they will hold the same quantity of charge.

Question 16.
Is the electric potential necessarily zero at a place where the electric field is zero?
Answer:
No, it is not necessary. The electric field inside a hollow metallic conductor is zero but the electric potential is not zero.

Question 17.
In a parallel plate capacitor, the potential difference of 10² V is maintained between the plates. What will be the electric field at points A and B as shown in the figure below?
Answer:
The electric field between the plates of a capacitor is uniform; therefore the electric field at points A and B will be the same.

Question 18.
If the plates of a charged capacitor are suddenly connected to each other by a wire, what will happen?
Answer:
The capacitor is discharged immediately.

Question 19.
Can you place a parallel plate capacitor of one farad capacity in your house?
Answer:
Ordinarily, it is not possible because the surface area of such a capacitor will be extra-large.

Question 20.
Is there any conductor which can be given almost unlimited charge?
Answer:
Yes, the earth.

Question 21.
What would be the work done if a point charge + q is taken from a point A to a point B on the circumference of a circle drawn with another point charge + q at the center?

Answer:
Zero.

Question 22.
Sketch a graph to show how a charge Q, acquired by a capacitor of capacitance, C, varies with the increase in the potential difference between the plates.
Answer:

Question 23.
In a parallel plate capacitor, the capacitance increases from 4 μF to 80 μF, introducing a dielectric medium between the plates. What is the dielectric constant of the medium?
Answer:
The dielectric constant is given by
K = \(\frac{80}{4}\) = 20

Electrostatic Potential and Capacitance Important Extra Questions Short Answer Type

Question 1.
Draw a plot showing the variation of (i) electric field (E) and (ii) electric potential (V) with distance r due to a point charge Q. (CBSE Delhi 2012)
Answer:
The plot is as shown.

Question 2.
Two identical capacitors of 10 pF each are connected in turn (i) in series and (ii) in parallel across a 20 V battery. Calculate the potential difference across each capacitor in the first case and the charge acquired by each capacitor in the second case. (CBSE AI 2019)
Answer:
(i) Since the two capacitors have the same capacitance, therefore, the potential will be divided amongst them. Hence V = 10 V each
(ii) Since the capacitors are connected in parallel, therefore, potential difference = 20 V
Hence charge Q = CV = 10 × 20 = 200 pC

Question 3.
A point charge ‘q’ is placed at O as shown in the figure. Is V_A – V_B positive, negative, or zero, if ‘q’ is an (i) positive, (ii) negative charge? (CBSE Delhi 2011, 2016)

Answer:
If V_A – V_B = \(\frac{q}{4 \pi \varepsilon_{0}}\left(\frac{1}{\mathrm{OA}}-\frac{1}{\mathrm{OB}}\right)\)

As OA < OB
∴ If q is positive then V_A– V_B is positive and
if q is negative V_A – V_B is also negative.

Question 4.
The graph shows the variation of voltage V across the plates of two capacitors A and B versus charge Q stored on them. Which of the two capacitors has higher capacitance? Give a reason for your answer.

Answer:
Capacitor A has higher capacitance. We know that capacitance C = Q/V.

For capacitor A
\(c_{A}=\frac{Q}{V_{A}}\)

For capacitor B
\(c_{B}=\frac{Q}{V_{B}}\)

As V_B > V_A
∴ C_B < C_A
Thus capacitance of A is higher.

Question 5.
A test charge ‘q’ is moved without acceleration from A to C along the path from A to B and then from B to C in electric field E as shown in the figure,

(i) Calculate the potential difference between A and C
Answer:
(i) dV = – E dr = – E (6 – 2) = – 4E

(ii) At which point (of the two) is the electric potential more and why? (CBSE AI 2012)
Answer:
Electric potential is more at point C as dV = – Edr, i.e. the electric potential decreases in the direction of the electric field.

Question 6.
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor. (CBSE AI 2013)
Answer:
Given t = d/2, C = ?
We know that when a dielectric of thickness ‘t’ is inserted between the plates of a capacitor, its capacitance is given by
C = \(\frac{\varepsilon_{0} A}{d-t+\frac{t}{K}}\)

Hence we have
C = \(\frac{\varepsilon_{0} A}{d-\frac{d}{2}+\frac{d}{2 K}}=\frac{2 K \varepsilon_{0} A}{d(1+K)}\)

Question 7.
Two-point charges q and -2q are kept ‘d’ distance apart. Find the location of the point relative to charge ‘q’ at which potential due to this system of charges is zero. (CBSE Al 2014C)
Answer:
Let the potential be zero at point P at a distance x from charge q as shown

Now potential at point P is
V = \(\frac{k q}{x}+\frac{k(-2 q)}{d+x}\) = 0

Solving for x we have
x = d

Question 8.
Four-point charges Q, q, Q., and q are placed at the corners of a square of side ‘a’ as shown in the figure.

Find the potential energy of this system. (CBSEAI, Delhi 2018)
Answer:
The potential energy of the system
U = \(\frac{1}{4 \pi \varepsilon_{0}}\left(4 \frac{q Q}{a}+\frac{q^{2}}{a \sqrt{2}}+\frac{Q^{2}}{a \sqrt{2}}\right)\)

U = \(\frac{1}{4 \pi \varepsilon_{0} a}\left(4 q Q+\frac{q^{2}}{\sqrt{2}}+\frac{Q^{2}}{\sqrt{2}}\right)\)

Question 9.
Three-point charges q, -4q, and 2q are placed at the vertices of an equilateral triangle ABC of side T as shown in the figure. (CBSE Delhi 2018)

Find out the amount of the work done to separate the charges at infinite distance. (CBSE AI, Delhi 2018)
Answer:
Net potential energy of the system
= \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{l}[-4+2-8]=\frac{5 q^{2}}{2 \pi \varepsilon_{0} l}\)

Electrostatic Potential and Capacitance Important Extra Questions Long Answer Type

Question 1.
Two-point charges 2 μC and —2 μC are placed at points A and B 6 cm apart.
(a) Draw the equipotential surfaces of the system.
Answer:
The diagram is as shown.

(b) Why do the equipotential surfaces get closer to each other near the point charges? (CBSEAI2O11C)
Answer:
We know that E = – dV/dr
Therefore, dr =- dV/E
Since near the charge, electric field E is large, dr will be less.

Question 2.
(a) Obtain the expressions for the resultant capacitance when the three capacitors C₁, C_2, and C₃ are connected (i) in parallel and then (ii) in series.
Answer:
(i) Parallel combination of three capacitors.
Let three capacitors of capacitances C₁, C₂, and C₃ be connected in parallel, and potential difference V be applied across A and B. If q be total charge flowing in the circuit and q₁ q₂ and q₃ be charged flowing across C₁, C_2, and C₃ respectively, then
q = q₁+ q₂ + q₃
or q = C₁v + C₂V + C₃V …(i)

If C_P is the capacitance of the arrangement in parallel, then
q = C_PV

So equation (i) becomes
C_PV = C₁V + C₂V + C₃V
Or
C_P = C₁ + C₂ + C₃

(ii) Series combination of three capacitors Let three capacitors C₁, C₂, and C₃ be connected in series. Let q charge be flowing through the circuit.
If V₁, V₂, and V₃ be potential differences across the plates of the capacitor and V be the potential difference across the series combination, then

V = V₁ + V₂ + V₃
Or
V = \(\frac{q}{C_{1}}+\frac{q}{C_{2}}+\frac{q}{C_{3}}\) … (i)

If Cs is the capacitance of series combination, then V = \(\frac{q}{\mathrm{C}_{\mathrm{s}}}\).

So the equation (i) becomes
\(\frac{q}{\mathrm{C}_{\mathrm{s}}}=\frac{q}{\mathrm{C}_{1}}+\frac{q}{\mathrm{C}_{2}}+\frac{q}{\mathrm{C}_{3}}\)
Or
\(\frac{1}{\mathrm{C}_{\mathrm{s}}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}\)

(b) In the circuit shown in the figure, the charge on the capacitor of 4 μF is 16 μC. Calculate the energy stored in the capacitor of 12 μF capacitance. (CBSE 2019C)

Answer:
Charge q across 4 μF Capacitor is 10 μc Potential difference across the capacitor of capacitance 4 μF will be
V= \(\frac{q}{C}=\frac{16 \mu C}{4 \mu F}=\frac{16 \times 10^{-6} \mathrm{C}}{4 \times 10^{-6} \mathrm{~F}}\)=4V

∴ Potential across 12 μF Capacitors
= 12V – 4V = 8V

Energy stored in the capacitors of capacitance C = 12 μF

U = \(\frac{1}{2}\) CV² = \(\frac{1}{2}\) × 12 × 10^-6 × 8² joule
= 384 × 10^-6 J = 384 μJ

Question 3.
Two identical plane metallic surfaces A and B are kept parallel to each other in the air, separated by a distance of 1 cm as shown in the figure.

A is given a positive potential of 10 V and the outer surface of B is earthed.
(i) What is the magnitude and direction of the uniform electric field between Y and Z?
Answer:
The electric field between the plates is
E = \(\frac{V}{d}\) = 10³ V m^-1
directed from plate A at the higher potential to plate B at a lower potential, i.e. from Y to Z

(ii) What is the work done in moving a charge of 20 µC from X to Y?
Answer:
Since X and Y are on the same plate A, which is an equipotential surface, work done in moving a charge of 20 µC from X to Y on the equipotential surface is zero.

Question 4.
(i) Draw the equipotential surfaces corresponding to a uniform electric field in the z-direction.
Answer:
The equipotential surfaces are as shown.

Equipotential surfaces

(ii) Derive an expression for the electric potential at any point along the axial line of an electric dipole. (CBSE Delhi 2019)
Answer:
Consider an electric dipole of length 2a and having charges +q and -q. Let us find the potential on the axial line at point P at a distance OP = x from the center of the dipole.

Now potential at point P is

Question 5.
A network of four capacitors, each of capacitance 15 µF, is connected across a battery of 100 V, as shown in the figure. Find the net capacitance and the charge on the capacitor C₄. (CBSE Al 2012C)

Answer:
Capacitors C1 C2 and C3 are in series, therefore their equivalent capacitance is
\(\frac{1}{C_{\mathrm{s}}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}=\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\frac{3}{15}\)

Hence C_S = 5 µF
Now C_S and C₄ are in parallel, hence
C_P = C_S + C₄ = 5 + 15 = 20 µF

Now C4 is connected to 100 V, therefore charge on it is
Q = CV=15 × 10^-6 × 100=15 × 10^-4 C

Question 6.
A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor. (CBSE AI 2014)
Answer:
U_i = \(\frac{1}{2}\) CV²

When the capacitors are connected then the energy stored is
U_F = \(\frac{1}{2}\left(C_{1}+C_{2}\right)\left(\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}\right)^{2}\)

Since C₁ = C₂ = C, and V₂ = 0, we have
Uf = \(\frac{1}{2}(C+C)\left(\frac{C V}{C+C}\right)^{2}=\frac{1}{2}(2 C) \times \frac{V^{2}}{4}=\frac{1}{4} C V^{2}\)

Hence we have
\(\frac{U_{f}}{U_{i}}=\frac{1}{2}\)

Question 7.
An isolated air capacitor of capacitance C₀ is charged to a potential V₀. Now if a dielectric slab of dielectric constant K is inserted between its plates, completely filling the space between the plates, then how to do the following change, when the battery remains connected
(i) capacitance,
Answer:
When the battery remains connected, the potential on the capacitor does not change.
The capacitance of the capacitor becomes K times the original value, i.e. C = K C₀.

(ii) charge
Answer:
Now new charge is Q = CV = K C₀ V = K Q₀.

(iii) the field between the plates
Answer:
The field between the plates becomes
E = \(\frac{V}{d}=\frac{V_{0}}{d}\) = E₀, i.e. no change.

(iv) energy stored by the capacitor?
Answer:
The energy stored becomes
U = \(\frac{1}{2} C V^{2}=\frac{1}{2} K C_{0} V^{2}\) = KU₀.

Question 8.
An isolated air capacitor of capacitance C0 is charged to a potential V0. Now if a dielectric slab of dielectric constant K is inserted between its plates, completely filling the space between the plates, then how to do the following change, when the battery is disconnected
(i) charge
Answer:
If battery is disconnected then charge remains same, Q = Q₀

(ii) electric field between the plates,
Answer:
V = \(\frac{V_{0}}{K}\) ,
∴E = \(\frac{E_{0}}{K}\) ,

(iii) capacitance
C = KCO [∵ C = \(\frac{Q_{0}}{V}=\frac{Q_{0}}{V_{0} / K}=\frac{K Q_{0}}{V_{0}}\)]

(iv) energy stored by the capacitor
U = \(\frac{U_{0}}{K}\left[U=\frac{1}{2} C V^{2}=\frac{1}{2}\left(K C_{0}\right)\left(\frac{V_{0}}{K}\right)^{2}\right]\)

Question 9.
The figure shows two identical capacitors, C₁ and C₂, each of 1 µF capacitance connected to a battery of 6 V. Initially switch ‘S’ is closed. After some time ‘S’ is left open and dielectric slabs of dielectric constant K = 3 are inserted to fill completely the space between the plates of the two capacitors. How will (i) the charge and (II) potential difference between the plates of the capacitors be affected after the slabs are inserted? (CBSE Delhi 2011)

Answer:
When switch S is opened then capacitor C₁ remains connected to the battery white capacitor C₂ is disconnected. Thus for the two capacitors, we have

Physical quantity	Capacitor C1	Capacitor C2
Charge	Becomes K times, i.e. Q = CV = K C0V = 3 × 1 × 10-6 × 6 = 1.8 × 10-5 C	Remains same Q = Q0  = CV = 1 × 10-6 × 6 = 6 × 10-6 C
Potential difference	Remains same V = V0 = 6 V	Becomes 1/K times V =V0/K = 6/3 = 2V

Question 10.
A particle, having a charge +5 µC, is initially at rest at the point x = 30 cm on the x axis. The particle begins to move due to the presence of a charge Q that is kept fixed at the origin. Find the kinetic energy of the particle at the instant it has moved 15 cm from its initial position if (i) Q = +15 µC and (ii) Q = -15 µC (CBSE Sample Paper 2018-19)
Answer:
From energy conservation, U_i + K_i = U_f + K_f
kQq/r_i + 0 = kQq/r_f + K_f
K_f = kQq (1/r_i — 1/r_f)

When Q is +15 µC, q will move 15 cm away from it. Hence r_f = 45 cm
K_f = 9 × 10⁹ × 15 × 10^-6 × 5 × 10^-6 [1/(30 × 10^-6) – 1/(45 × 10^-2)] = 0.75 J

When Q is -15 µC, q will move 15 cm towards it. Hence r_f = 15 cm K_f = 9 × 10⁹ × (-15 × 10^-6) × 5 × 10⁶ [1/(30 × 10^-2) – 1/(15 × 10^-2)] = 2.25 J

Question 11.
Two metal spheres, one of radius R and the other of radius 2R, both have same surface charge density σ. They are brought in contact and separated. What will be new surface charge densities on them? (NCERT Exemplar)
Answer:
Let the two spheres have charges Q₁ and Q₂ respectively. Since σ₁ = σ₂, before contact, we have
\(\frac{Q_{1}}{4 \pi R^{2}}=\frac{Q_{2}}{4 \pi(2 R)^{2}}\)
Or
Q₂ = 4 Q₁

After contact
Let q1 and q2 be the charges on them, then
q₁ + q₂ = Q₁ + Q₂= Q₁ + 4Q₁ = 5Q₁ = 5(σ × 4πr²)

The two will exchange charge till their potentials are equal, therefore we have
\(\frac{q_{1}}{R}=\frac{q_{2}}{2 R}\)
Or
q₂ = 2 q₁

Therefore 3q₁ = 5(σ × 4πr²)
Or
q₁ = \(\frac{5}{3}\)(σ x 4πr²) and q₂ = 2q₁ = \(\frac{10}{3}\)5(σ x 4πr²)

Therefore
σ₁ = \(\frac{q_{1}}{4 \pi R^{2}}=\frac{5}{3}\left(\frac{\sigma \times 4 \pi R^{2}}{4 \pi R^{2}}\right)=\frac{5 \sigma}{3}\)
And
σ₂ = \(\frac{q_{2}}{4 \pi(2 R)^{2}}=\frac{10}{3}\left(\frac{\sigma \times 4 \pi R^{2}}{4 \pi(2 R)^{2}}\right)=\frac{5 \sigma}{6}\)

Question 12.
(a) Derive an expression for the capacitance of a parallel plate capacitor when the space between the plates is partially filled with a dielectric medium of dielectric constant ‘K’.
Answer:
Consider a parallel plate capacitor having each plate of area A and separated by a distance d. When there is a vacuum between the two plates, the capacitance of the parallel plate capacitor is given by
C₀ = ε₀ A/d.

Suppose that when the capacitor is connected to a battery, the electric field of strength E₀ is produced between the two plates of the capacitor. Further, suppose that when a dielectric slab of thickness t (t < d) is introduced between the two plates of the capacitor as shown in the figure, the electric field reduces to E due to the polarisation of the dielectric. Therefore between the two plates of the capacitor, over a distance of t, the strength of the electric field is E, and over the remaining distance (d – f)

the strength is E₀. If V is the potential between the plates of the capacitor, then

V = E_t + E₀(d – t)
Since E = E₀/K where K is the dielectric constant, the above equation becomes

V = \(\frac{E_{0}}{K} t+E_{0}(d-t)=E_{0}\left(d-t+\frac{t}{K}\right)\)

The electric field between the plates of the capacitor is given by
E₀ = σ / ε₀ = Q/ A ε₀

Hence the potential between the two plates becomes.
V = \(E_{0}\left(d-t+\frac{t}{K}\right)=\frac{Q}{\varepsilon_{0} A}\left(d-t+\frac{t}{K}\right)\)

Hence the capacitance of the parallel plate capacitor is given

∴ clearly, C > C₀. Therefore, capacitance increases in the presence of a dielectric medium.

(b) Explain why the capacitance decreases when the dielectric medium is removed from between the plates.
Answer:
On removing the dielectric, the capacitance will decrease.

Question 13.
(a) Obtain the expression for the energy stored per unit volume in a charged parallel plate capacitor.
(b) The electric field inside a parallel plate capacitor is E. Find the amount of work done in moving a charge q over a closed rectangular loop a b c d a. (CBSE Delhi 2014)

Answer:
(a) Suppose the capacitor is charged fully, its final charge is Q. and a final potential difference is V. These are related as Q = CV

Let q and V be the charge and potential difference, respectively.

At an intermediate stage during charging process q = CV. At this stage the small work done dW to transfer an additional charge dq is

dW = Vdq = \(\frac{q d q}{C}\)

The total work W needed to increase the capacitor’s charge q from zero to its final value Q is given by
W = \(\int_{0}^{w} d W=\int_{0}^{Q} \frac{q d q}{C}=\frac{1}{C} \int_{0}^{Q} q d q\)
Or
W = \(\frac{1}{C}\left|\frac{q^{2}}{2}\right|_{0}^{Q}=\frac{Q^{2}}{2 C}\)

This work is stored in the capacitor in the form of its electric potential energy. Hence,
U = \(\frac{Q^{2}}{2 C}\) … (1)

substituting Q = CV in equation (1) we have
U = \(\frac{1}{2}\) CV²

Energy density = Energy stored per unit volume

(b) Since the surface is an equipotential surface, work done is zero.

Question 14.
(a) Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.
Answer:
Suppose Q. is the charge on the capacitor, and c is the uniform surface charge density on each plate as shown in the figure. Therefore by Gauss’s theorem, the electric field between the plates of the capacitor (neglecting fringing of electric field at the edges) is given by

E = \(\frac{\sigma}{\varepsilon_{0}}=\frac{Q}{\varepsilon_{0} A}\) …. (1)

The field is uniform, and the distance between the plates is d, so the potential difference between the two plates is
V = Ed = \(\frac{1}{\varepsilon_{0}} \frac{Q d}{A}\) ….. (2)

Therefore by the definition of capacitance we have
C = \(\frac{Q}{V}=\frac{\varepsilon_{0} A}{d}\) …. (3)

This gives the capacitance of a parallel plate capacitor with a vacuum between plates.

(b) Two charged spherical conductors of radii R₁ and R₂ when connected by a conducting wire acquire charges q₁ and q₂ respectively. Find the ratio of their surface charge densities in terms of their radii. (CBSE Delhi 2014)
Answer:

The ratio of the surface charge densities is given by
\(\frac{\sigma_{1}}{\sigma_{2}}=\frac{q_{1}}{4 \pi R_{1}^{2}} \times \frac{4 \pi R_{2}^{2}}{q_{2}}\) ….(4)

Now when they are connected with a wire, their potentiaLs wilt be same: therefore, from the expression.
V = \(\frac{Q}{c}=\frac{Q}{R}\)

∵ C = 4πε₀R for a spherical body
∴\(\frac{q_{1}}{q_{2}}=\frac{R_{1}}{R_{2}}\)

Substituting in equation (4) we have
\(\frac{\sigma_{1}}{\sigma_{2}}=\frac{R_{1}}{R_{1}^{2}} \times \frac{R_{2}^{2}}{R_{2}}=\frac{R_{2}}{R_{1}}\)

Question 15.
(a) Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.
Answer:
The electrons are transferred to the positive terminal of the battery from the metallic plate connected to the positive terminal, leaving behind a positive charge on it. Similarly, the electrons move on to the second plate from the negative terminal, hence it gets negatively charged. This process continues till the potential difference between the two plates becomes equal to the potential of the battery.

Suppose the capacitor is charged fully; its final charge is Q and the final potential difference is V.
These are related as Q = CV

Let q and V be the charge and potential difference respectively, after some time during the charging of the capacitor, then q = CV. At this time, the small work done dW required to transfer an additional charge dq is given by

dW = Vdq = \(\frac{q d q}{C}\)

The total work W needed to increase the capacitor’s charge q from zero to its final value Q is given by

This work is stored in the capacitor in the form of its electric potential energy. Hence,
U = \(\frac{Q^{2}}{2 C}\) …(1)

Substituting Q= CVin equation (1) we have
U = \(\frac{1}{2}\) CV²

(b) A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor. (CBSE Delhi 2019)
Answer:
The charge stored on the capacitor q = CV, when it is connected to the uncharged capacitor of same capacitance, sharing of charge, takes place between the two capacitors till the potential of both the capacitors becomes V/2.

Energy stored on the combination (U₂)
= \(\frac{1}{2} C\left(\frac{V}{2}\right)^{2}+\frac{1}{2} C\left(\frac{V}{2}\right)^{2}=C\left(\frac{V}{2}\right)^{2}=\frac{C V^{2}}{4}\)

Energy stored on single capacitor before connecting
U₁ = –\(\frac{1}{2}\) CV²

Ratio of energy stored in the combination to that in the single capacitor.
\(\frac{U_{2}}{U_{1}}=\frac{C V^{2} / 4}{C V^{2} / 2}=\frac{1}{2}\)

Question 16.
(a) Obtain the expression for the potential due to an electric dipole of dipole moment p at a point ‘x’ on the axial line.
(b) Two identical capacitors of plate dimensions l × b and plate separation d have dielectric slabs filled In between the space of the plates as shown In the figures.

Obtain the relation between the dielectric constants K, K_1, and K₂. (CBSE Al 2013C)
Answer:
(a) Consider an electric dipole of length 2a and having charges + q and — q. Let us find the potential on the axial Une at point P at a distance OP = x from the center of the dipole.

Now potential at point P is

(b) When there is no dieLectnc then
C = \(\frac{\varepsilon_{0} l b}{d}\)

For the first capacitor
C’ = \(\frac{K \varepsilon_{0} l b}{d}\) = KC

The second case is a case of two capacitors connected in paralleL, therefore
C₁ = \(\frac{K_{1} \varepsilon_{0} l b}{2 d}\) and
C₂ = \(\frac{K_{2} \varepsilon_{0} l b}{2 d}\)

These two are connected in parallel, therefore we have
C” = C₁ + C₂ = \(\frac{K_{1} \varepsilon_{0} l b}{2 d}\) + \(\frac{K_{2} \varepsilon_{0} l b}{2 d}\)
= C\(\left(\frac{K_{1}+K_{2}}{2}\right)\)

If the capacitance in each case be same, then C’ = C”

Hence K= \(\left(\frac{K_{1}+K_{2}}{2}\right)\)

Question 17.
(a) Explain using suitable diagrams the difference in the behavior of a
(i) conductor:
Answer:
The diagram is as shown.

When a conductor is placed in an external electric field, the free charge carriers move and charge distribution in the conductor adjusts itself in such a way that the electric field due to induced charges opposes the external field within the conductor. This happens until in the static situation, the two fields cancel each other and the net electrostatic field in the conductor is zero.

(ii) dielectric In the presence of the external electric field. Define the term polarisation of a dielectric and write its relation with susceptibility.
Answer:

In a dielectric, this free movement of charges is not possible. It turns out that the external field induces dipole moment by stretching or re-orienting molecules of the dielectric. The collective effect of all the molecular dipole moments is that the net charges on the surface of the dielectric produce a field that opposes the external field. However, the opposing field so induced does not exactly cancel the external field. It only reduces it. The extent of the effect depends on the nature of the dielectric.

Dielectric polarization is defined as the dipole moment per unit volume of a dielectric. It is related to susceptibility as P = χ_eε₀\(\vec{E}\)

(b) A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge is placed at its center C and another charge +2Q. outside the shell, at a distance r from the center as shown in the figure. Find (i) the force on the charge at the center of the shell and at the point A and

Answer:
(i) F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q^{2}}{2 R^{2}}\) and
F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(Q+Q / 2) \times 2 Q}{r^{2}}\)

(ii) the electric flux through the shell. (CBSE Delhi 2015)
Answer:
Flux = \(\frac{Q}{2 \varepsilon_{0}}\)

Question 18.
(a) Define the SI unit of capacitance.
Answer:
The capacity of a capacitor is said to be one farad when a charge of 1 coulomb is required to raise the potential difference by 1 volt.

(b) Obtain the expression for the capacitance of a parallel plate capacitor.
Answer:
Let Q. be the charge on the capacitor, and o be the uniform surface charge density on each plate as shown in the figure.

d = distance between plates of the capacitor.

Therefore by Gauss’ theorem, the electric field between the plates of the capacitor is given by
E = \(\frac{\sigma}{\varepsilon_{0}}=\frac{Q}{\varepsilon_{0} A}\)

The field is uniform, so the potential difference between the two plates is
V = Ed = \(\frac{1}{\varepsilon_{0}} \frac{Q d}{A}\)

Therefore, by the definition of capacitance we have
C = \(\frac{Q}{V}=\frac{\varepsilon_{0} A}{d}\)

This gives the capacitance of a parallel plate capacitor.

(c) Derive the expression of the effective capacitance of a series combination of n capacitance (CBSE Delhi 2016C)
Answer:
Capacitors in series. Capacitors are said to be connected in series if the second plate of one capacitor is connected to the first plate of the next and so on as shown in the figure.

Resultant capacitance will be
\(\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\ldots \ldots \ldots \ldots=+\frac{1}{C_{n}}\)

Question 19.
A charge Q is distributed over the surfaces of two concentric hollow spheres of radii r and R (R >> r), such that their surface charge densities are equal. Derive the expression for the potential at the common center.
Or
Three concentric metallic shells A, B, and C of radii a, b, and c (a <b < C) have surface charge densities +a, -a, and + o respectively as shown. Obtain the expressions for the potential of three shells A, B, and C. If shells A and C are at the same potential, obtain the relation between a, b and c. (CBSE Al 2019)

Answer:
Let the charges on the spheres be q, and q2 such that
Q=q₁ + q₂
= 4πσ(r² + R²)
Or
σ = \(\frac{Q}{4 \pi\left(r^{2}+R^{2}\right)}\)

Now potential at the common centre V=V₁ + V₂
V = \(\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1}}{r}+\frac{q_{2}}{R}\right)\)
= \(\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{4 \pi r^{2} \sigma}{r}+\frac{4 \pi R^{2} \sigma}{R}\right)\)

V = \(\frac{(r+R) \sigma}{\varepsilon_{0}}\)

Substituting for o, we have

Numerical Problems :

Formulae for solving numerical problems

• V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\) electric potential at a point.
• Q = CV for a capacitor.
• C = 4πε₀R for a spherical conductor.
• C =\(\frac{Q}{V}=\frac{\varepsilon_{0} A}{d}\) capacitance of a parallel plate capacitor with air as dielectric.
• C = \(\frac{K \varepsilon_{0} A}{d}\) capacitance of a parallel plate capacitor with a dielectric.
• For capacitors in senes \(\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}\) and for capacitors in parallel C_p = C₁ + C₂
  
• U = \(\frac{1}{2} \frac{Q^{2}}{c}\) = \(\frac{1}{2}\)CV² = \(\frac{1}{2}\)QV Energy stored in a capacitor.
• Energy density u = \(\frac{1}{2}\) ε₀E₂
  
• Capacitance of a parallel plate capacitor with a conducting slab of thickness t between plates is C = \(\frac{\varepsilon_{0} A}{d-t}\)
• Capacitance of a capacitor with dielectric slab of thickness t << d , C = \(\frac{\varepsilon_{0} A}{d+t\left(\frac{1}{K}-1\right)}\)
• Common potential V = \(\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}\)
• Loss of energy when two conductors are combined, U₁ – U₂ = \(\frac{C_{1} C_{2}}{2\left(C_{1}+C_{2}\right)}\left(V_{1}-V_{2}\right)^{2}\)
• If n small drops each having a charge Q, capacitance C, and potential V coalesce to form a big drop, then

1. The charge on the big drop = nQ
2. The capacitance of the big drop = n^1/3 C
3. Potential of the big drop = n^2/3 V
4. The potential energy of the big drop = n^5/3 U

Question 1.
Electric field intensity at point B due to a point charge Q kept at point A is 24 N C^-1 and the electric potential at point B due – to the same charge is 12 J C^-1. Calculate the distance AB and also the magnitude of the charge Q.
Answer:
Given E = 24 N C^-1 , V = 12 J C^-1 , r = ? and Q = ?

Using the relation
E = \(\frac { V }{ r }\)
Or
\(\frac { V }{ E }\) = \(\frac { 12 }{ 24 }\) = 0.5 m

Also V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}\)

Therefore 12 = 9 × 10⁹ × \(\frac { Q }{ 0.5 }\) , solving for Q

we have Q = 6.67 × 10^-10C

Question 2.
A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 360 μC. When the potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 μC. Calculate:
(i) The potential V and the unknown capacitance C.
(ii) What will be the charge stored in the capacitor, if the voltage applied had increased by 120 V? (CBSE Delhi 2013)
Answer:
Given Q₁ = 360 μC, Q₂ = 120 μC,
V₂ = (V- 120) volt, V₁ = V

(i) We know that C = \(\frac { Q }{ V }\)

Since capacitance is same, we have
\(\frac{Q_{1}}{V_{1}}=\frac{Q_{2}}{V_{2}}\)
Or
\(\frac{360 \times 10^{-6}}{V}=\frac{120 \times 10^{-6}}{V-120}\)

Solving for V we have V= 180 volt

Also C = Q₁/ V₁ = 360 × 10^-6 / 180 = 2 × 10^-6 C = 2 pF

(ii) V= 180 + 120 = 300 V
Therefore Q = CV = 2 x 10^-5 × 300 = 600 x 10^-6 = 600 pC

Question 3.
Two identical capacitors of 12 pF each are connected in series across a 50 V battery. Calculate the electrostatic energy stored in the combination. If these were connected in parallel across the same battery, find out the value of the energy stored in this combination. (CBSE Al 2019)
Answer:
Given C = 12 pF = 12 × 10^-12 F, V= 50 V,
In series
C_s = \(\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{12 \times 12}{12+12}\) = 6 pF

Hence energy stored
U_s = \(\frac{1}{2}\) C_sV₂ = \(\frac{1}{2}\) × 6 × 10^-12× (50)²
U_s = 7.5 × 10^-9 J

In parallel
C_P = C₁ + C₂ = 12 + 12 = 24 pF
U_P = \(\frac{1}{2}\) C_PV² = \(\frac{1}{2}\) × 24 × 10^-12 × (50)²
U_P = 3 × 10^-8 J

Question 4.
The figure shows a network of three capacitors C₁ = 2 μF; C₂ = 6 μF and C₃ = 3 μF connected across a battery of 10 V. If a charge of 6 μC is acquired by the capacitor C₃, calculate the charge acquired by C₁ (CBSE Al 2019)

Answer:
Capacitors C₂ and C₃ are connected in parallel, therefore, the net capacitance of the combination.
C₂₃ = (6 + 3) = 9 μF

Let V1, be the potential across C1 and V2 be the potential across C₂₃
Now C₁ = Q/V, or V₁ = Q/C₁ = Q/2

Also V₂ = Q/9
But V = V₁ + V₂

Solving for Q we get
Q= 16.4 μC

Question 5.
(a) Find equivalent capacitance between A and B in the combination given below. Each capacitor is of 2 μF capacitance.

Answer:
This can be redrawn as

Like C₂, C₃ and C₄ are in parallel,
C₂₃₄ = C₂ + C₃ + C₄ = 6 μF

Further, C₁, C₂₃₄ and C₅ are in series
\(\frac{1}{C_{\text {net }}}=\frac{1}{2}+\frac{1}{6}+\frac{1}{2}=\frac{3+1+3}{6}=\frac{7}{6}\)
\(C_{\text {net }}=\frac{6}{7}\)μF

(b) If a DC source of 7 V is connected across AB, how much charge is drawn from the source and what is the energy stored in the network? (CBSE Delhi 2017)
Answer:
∴ C_net = \(\frac { 6 }{ 7 }\) μF ,

q = C_net V = \(\frac { 6 }{ 7 }\) × 10^-6 × 7 = 10^-6C

∴ Energy stored = \(\frac{1}{2}\)qV

= \(\frac{1}{2}\) × 6 × 10^-6 × 7 = 21 × 10^-6J

Question 6.
Two identical capacitors of 12 pF each are connected in series across a battery of 50 V. How much electrostatic energy is stored in the combination? If these were connected in parallel across the same battery, how much energy will be stored in the combination now? Also, find the charge drawn from the battery in each case. (CBSE Delhi 2017)
Answer:
(i) Net capacitance C_net = \(\frac{C_{1} C_{2}}{C_{1}+C_{2}}\)
= \(\frac{12 \times 12}{12+12}\)pF = 6 pF

∴ Energy stored = \(\frac{1}{2}\)C_net V²
= \(\frac{1}{2}\) × 6 × 10^-12 × (50)² = 75 × 10^-10J
q = Chargedrawn = C_net V=6 × 10^-12 × 50 = 3 × 10^-10 C

(ii) C_net=12 + 12 = 24pF
∴ Energy stored = \(\frac{1}{2}\) C_net V²
= \(\frac{1}{2}\) × 24 × 10^-12 × (50)²
= 3 × 10^-8J

Charge drawn, q = C_netV
= 24 × 10^-12 × 50
= 1200 × 10^-12
= 12 × 10^-10 C

Question 7.
In the following arrangement of capacitors, the energy stored in the 6 μF capacitor is E. Find the value of the following. (Foreign 2016)
(a) Energy stored in 12 pF capacitor.
(b) Energy stored in 3 pF capacitor.
(c) Total energy drawn from the battery.

(a) Energy stored in 6 μ capacitor is E. Capacitors 6 μF and 12 μF are connected in parallel.
So, the voltage across 6 μF capacitor
= voltage across 12 μF capacitor
= voltage across 12 μF capacitor
= V (say)

As E = \(\frac{1}{2}\) × 6 × V²
∴ V = \(\sqrt{\frac{E}{3}}\)

Similarly energy U’ stored in 12 pF capacitor
= \(\frac{1}{2}\) × 12 × V²
= \(\frac{1}{2}\) ×12 × \(\frac{E}{3}\) = 2E

U’ = 2 E

(b) Equivalent capacitance of 6 µF and 12 µF is 6 + 12 = 18 µF

Charge on 18 µF and 3 µF is same as they are in series as they are in series
∴ Q = CV= 18 × V

∴ Energy in 3 µF capacitor
U” = \(\frac{1}{2} \frac{Q^{2}}{C}=\frac{1}{2} \frac{(18 V)^{2}}{3}\)

U” = \(\frac{1}{2} \times \frac{18 \times 18}{3} \frac{E}{3}\)
U” = 18E

(c) Total energy drawn from battery U = E + 2E + 18E = 21E

Question 8.
Two parallel plate capacitors X and Y have the same area of plates and the same separation between them, X has air between the plates, while Y contains a dielectric medium of ε_r = 4.
(a) Calculate the capacitance of each capacitor if the equivalent capacitance of the combination is 4 μF.

C_X = \(\frac{\varepsilon_{0} A}{d}\) = C(say)
C_Y= 4\(\frac{\varepsilon_{0} A}{d}\) = 4C

EquivaLent capacitance = 4 μF
C_eq = \(\frac{C_{X} C_{Y}}{C_{X}+C_{Y}}\) = 4
= \(\frac{C \times 4 C}{C+4 C}\)
Or
= \(\frac{4C}{5}\) = 4

C = 5 μF
∴ C_X = C = 5 μF and
C_Y = 4C = 20 μF

(b) Total charge, q = C_eq V = 4 × 12 = 48μC
C_X and C_Y are in series. Hence, charge on both is 48 μC each.

∴ The potential difference across C_X,
V_X = \(\frac{q}{C_{X}}=\frac{48}{5}\) = 9.6 V0lt

The potential difference across C_Y,
V_Y = \(\frac{q}{C_{Y}}=\frac{48}{20}\) = 2.4 Volt

(c) U_X=\(\frac{1}{2}\)CXVX² ; UY = \(\frac{1}{2}\)CYVy²

∴ \(\frac{U_{x}}{U_{Y}}=\frac{5 \times(9.6)^{2}}{20 \times(2.4)^{2}}\)
= \(\frac{1}{4} \times \frac{9.6 \times 9.6}{2.4 \times 2.4}\) = 4

\(\frac{U_{x}}{U_{r}}\) = 4

Question 9.
Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C₁ and C₂ with their capacitances in the ratio 1:2 so that the energy stored in these two cases becomes the same. (CBSE Al 2016)
Answer:
(i) Let C₁ = C, ∴ C₂ = 2C
For series combination equivalent capacitance is
\(C_{s}=\frac{C_{1} \times C_{2}}{C_{1}+C_{2}}=\frac{C \times 2 C}{3 C}=\frac{2}{3} C\) …. (1)
and energy stored,
\(U_{\mathrm{s}}=\frac{1}{2} C_{s} V_{\mathrm{s}}^{2}=\frac{1}{2} \times \frac{2}{3} C V_{\mathrm{s}}^{2}\) …. (2)

For parallel combination equivalent capacitance
C_p = C + 2C = 3C ….(3)
and energy stored
∴ \(U_{D}=\frac{1}{2} 3 C \times V_{p}^{2}\) …(4)

But U_s = U_p [Given]

Question 10.
Calculate the potential difference and the energy stored in the capacitor C₂ in the circuit shown in the figure. Given potential at A is 90 V, C₁ = 20 μF, C₂ = 30 μF, and C₃= 15 μF. (CBSEAI 2015)

Answer:

C = \(\frac{60}{9}\) μF = \(\frac{20}{3}\) μF

The potential at A = 90V

∴ Charge on each capacitor
Q = C x V= \(\frac{20}{3}\) x 90 =600 μC.

∴ Charge on C₂ = 600 μC

∴ V₂ = \(\frac{Q}{C_{2}}=\frac{600}{30}\) ; V₂ = 20 V

Energy stored in C₂ = \(\frac{1}{2}\)C₂V₂²

U₂ = \(\frac{1}{2}\) × 3o × 10^-6 × 20 × 20
U₂ = 6000 × 10^-6J = 6 × 10^-3J

Important Questions for Class 12 Physics Chapter Wise with Solution Pdf Download 2020-21: Here we are providing CBSE Important Extra Questions for Class 12 Physics Chapter Wise Pdf download in Hindi and English Medium. Students can get Class 12 Physics NCERT Solutions, Physics Class 12 Important Extra Questions and Answers designed by subject expert teachers.

CBSE Class 12th Physics Important Extra Questions and Answers Chapter Wise Pdf

1. Class 12 Physics Chapter 1 Important Questions Electric Charges and Fields
2. Class 12 Physics Chapter 2 Important Questions Electrostatic Potential and Capacitance
3. Current Electricity Class 12 Important Questions
4. Moving Charges and Magnetism Class 12 Physics Important Questions
5. Magnetism and Matter Physics Class 12 Important Questions
6. Electromagnetic Induction Imp Question of Physics Class 12
7. Alternating Current Important Questions of Physics Class 12
8. Electromagnetic Waves Physics Important Questions Class 12
9. Ray Optics and Optical Instruments Class 12 Important Questions
10. Wave Optics Class 12 Important Questions
11. Dual Nature of Radiation and Matter Class 12 Important Questions
12. Atoms 12th Physics Important Questions
13. Nuclei CBSE Class 12 Physics Important Questions
14. Semiconductor Electronics: Materials, Devices and Simple Circuits Important Questions
15. Communication Systems Class 12 Important Questions

We hope the given CBSE Important Questions of Physics Class 12 Chapter Wise Pdf download in Hindi and English Medium will help you. If you have any queries regarding NCERT Class 12 Physics Extra Important Questions and Answers, drop a comment below and we will get back to you at the earliest.

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 13 Organisms and Populations. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 13 Important Extra Questions Organisms and Populations

Organisms and Populations Important Extra Questions Very Short Answer Type

Question 1.
Define microclimate.
Answer:
Microclimate represents the climatic

Question 2.
Define habitat and niche.
Answer:
Habitat is the native environment of conditions that prevail at a local scale or in animal or place. area of limited size. Niche is the position or function of an organism in a community of plants and animals.

Question 3.
How is thermoregulation achieved in polar bears?
Answer:
Thermoregulation in the polar bear. By developing blubber in a subcutaneous zone prevents loss of body heat.

Question 4.
What are osmoconformers? Give one example.
Answer:
Osmoconformers. Animals that can change the osmolarity of their body fluids according to that of the surrounding medium are termed osmoconformers, e.g. Myxine.

Question 5.
What factors cause annual variations in the intensity and duration of temperature?
Answer:

1. Rotation of the earth around the Sun.
2. The tilt of the earth on its axis.

Question 6.
Species that can tolerate a narrow range of temperature are called ………………..
Answer:
Stenothermic.

Question 7.
What are eurythermic species?
Answer:
Species that can tolerate a wide range of temperature variations are called eurythermic.

Question 8.
Species that can tolerate a wide range of salinity are called ……………………
Answer:
Euryhaline.

Question 9.
Define stenohaline species.
Answer:
It is a species that lives within a narrow range of salinity.

Question 10.
Name two factors that cause the formation of major biomes.
Answer:

1. Variation in the intensity and duration of temperature.
2. Variation in precipitation.

Question 11.
What is the main cause of salinity?
Answer:
Salinity is due to the accumulation of soluble minerals on the surface or beneath the surface of the earth.

Question 12.
What does the stratification of community depict?
Answer:
Stratification of a community depicts vertical layering of vegetation.

Question 13.
From where are individual organisms derived?
Answer:
Individual organisms are always derived from pre-existing organisms through the mechanism of reproduction-may be vegetative, asexual or sexual.

Question 14.
List two negative interactions between two species.
Answer:
Parasitism, Predation.

Question 15.
What is the other term for facultative mutualism?
Answer:
Protocooperation.

Question 16.
Name the association in which one species produces a poisonous substance or a change in environmental conditions that is harmful to another species. (CBSE Delhi 2019)
Answer:
Amensalism.

Question 17.
What is mycorrhiza?
Answer:
It is a symbiotic or mutually beneficial association between a fungus and roots of higher plants.

Question 18.
Emergent land plants that can tolerate the salinity of the sea are called ………………..
Answer:
Mangrove plants.

Question 19.
Give an example of a commensal relationship.
Answer:

1. Epiphytes and tree,
2. Remora fish and shark.

Question 20.
Define community periodicity.
Answer:
The recurrence of daily or seasonal or lunar changes in a community is called community periodicity.

Question 21.
Mention any two significant roles predation plays in nature. (CBSE 2008)
Answer:
Role of predators:

1. Transferring energy to a higher trophic level.
2. They keep the prey population under control.

Question 22.
Why is the polar region not a suitable habitat for a tiny hummingbird? (CBSE 2008)
Answer:
Tiny animals like hummingbird have more surface area compared to their volume. So heat loss may occur which will not be good for the tiny bird in polar regions.

In polar regions, only animals with a thick layer of fat below the skin can survive.

Question 23.
Between birds and amphibians which will be able to cope with global warming? Give reason. (CBSE2008)
Answer:
Birds will be able to cope with global warming because they are homeotherms as body temperature remains constant irrespective to change in surroundings.

Question 24.
Which one of the two stenothermal or eurythermal shows a wide range of distribution on the earth? (CBSE 2008)
Answer:
Eurythermal shows a wide range of distribution on the earth because they can tolerate a wide range of temperature.

Question 25.
In a pond, there were 30 Hydrilla plants. Through reproduction 10 new Hydrilla plants were added in a year. Calculate the birth rate of the population. (CBSE 2010)
Answer:
A number of individuals added = 10 per 20 Hydrilla plants.

\frac{10}{30}= 0.33 offspring per hydrilla plant per year.

Question 26.
How is ‘stratification’ represented in a forest ecosystem? (CBSE Outside Delhi 2014)
Answer:
Stratification is a grouping of plants in a forest into two or more well-defined layers depending upon height like tall trees, medium-sized trees, small trees, bushes and herbs. In the rainforest, it is multi-storeyed.

Question 27.
Give an example of an organism that enters ‘diapause’ and why? (CBSE Delhi 2014)
Answer:
Under unfavourable conditions, many zooplanktonic species in lakes and ponds enter a stage of suspended development called diapause.

Organisms and Populations Important Extra Questions Short Answer Type

Question 1.
Certain species of wasps are seen to
frequently visit flowering fig trees. What type of interaction is seen between them and why? (CBSE 2008, 2014)
Answer:

1. Mutualism.
2. Flowers of a fig tree can only be pollinated by wasp species. In return, female wasp visits the fruit of fig as not only an egg-laying site but also uses the developing seed in fruit for nourishing its larvae.

Question 2.
Write a short note on microclimate.
Answer:
Microclimate: It represents the climatic conditions that prevail at a local scale, or in areas of limited size such as the immediate surroundings of plants and animals. Microclimate generally differs from the prevailing regional climatic conditions. For example, in a forest, dense foliage reduces the amount of light reaching the ground. This also results in a changed air temperature profile. The day-time air temperature inside the forests is lower than the outside temperature. Also, the interior of a forest may be more humid than a nearby non-forested area.

Question 3.
What are the effects of organisms on habitat?
Answer:
Effects of organisms on habitat:

1. Excessive growth may lead to the death of the habitat as in the case of water hyacinth- an increase in population had led to the death of Hussain Sagar lake of Hyderabad.
2. The predominance of a predator will reduce the population of prey.
3. Man manipulates the habitat and has affected the forest causing serious problems due to deforestation. More so man has created pollution of air, water and land.

Question 4.
How do you differentiate habitat from the environment?
Answer:
Differences between habitat and environment:

Habitat	Environment
(i) It is the living place of organisms.	(i) It Is a specific region surrounding the population of organisms.
(ii) It is a part of the total environment.	(ii) It is larger than a habitat.
(iii) It must offer food, shelter and climate condition suitable for the organism to live and flourish.	(iii) It has a special condition suitable for specific organisms. Example. Desert rats of Rajasthan are not found in plains of U.P and M.P

Question 5.
Differentiate microhabitat. between habitat and
Answer:
Differences between habitat and microhabitat:

Habitat	Microhabitat
(i) It is a living place of an organism.	(i) It is a more localised part of the habitat.
(ii) It is a part of the total environment of the region.	(ii) It is a part of the habitat.
(iii) It has a common climate for all organisms. Example. Sundarban forests are the habitat of Bengal tigers.	(iii) It is mostly suited for specific animals, e.g. sediments of pond or togs.

Question 6.
List the means by which organisms survive at freezing temperatures.
Answer:

1. The animals are usually white or light coloured. The light colouration of animals helps in camouflage with snow and thermal regulation.
2. Animals are thickly coated with fur.
3. Presence of a thick layer of fat below the skin.

Question 7.
Differentiate eurythermal, stenothermal and euryhaline animals.
Answer:
Differences between eurythermal, stenothermal and euryhaline animals:

Eurythermal Animals	Stenothermal Animals	Euryhaline Animals
These are the animals/ organisms which can tolerate a wide range of temperatures.	These are the animals/ organisms which can tolerate only: a narrow range of temperatures.	These are the animals that can tolerate a wide range of salinity of the medium.

Question 8.
Explain how is an orchid plant adapted to changes in temperature and humidity.
Answer:
An orchid plant, e.g.Venda, an epiphyte, is a native of tropical forests of India and South Asia.

Its adaptations are:

• Stem stores water and leaves are adapted to keep water in and dryness out.
• Roots of the plant secure their attachment to the bark of the tree and help in the absorption of moisture from humid air or rain.

Question 9.
Why do submerged plants receive weaker illuminations than exposed floating plants in a lake?
Answer:
Because light intensities of stronger illuminations are absorbed by the exposed floating plants of the lake/water-bodies, so only the weaker illuminations reach the submerged, Some light is reflected at the water surface and a part is absorbed by upper layers of water.

Question 10.
Categorise the following plants into hydrophytes, halophytes, mesophytes and xerophytes. Give reasons for your answers.
(i) Salvinia
Answer:
Hydrophyte. Because it is not able to tolerate deficiency of water. It is partially or completely submerged.

(ii) Opuntia
Answer:
Xerophyte. Because it prefers a dry and hot climate with low rainfall, thus has succulent leaves.

(iii) Rhizophora
Answer:
Halophyte. Because it grows in a saline habitat.

(iv) Marigifera
Answer:
Mesophyte. Because it prefers areas with high moisture content.

Question 11.
If a freshwater fish is placed in an aquarium containing seawater, will the fish be able to survive? Explain giving reasons. (CBSE Delhi 2015)
Answer:
No, a freshwater fish placed in the aquarium containing sea-water will not be able to survive. Because its body system is adapted to function normally in a narrow range of salinity and it cannot survive in the high salinity of sea-water.

Question 12.
Distinguish between population and community.
Answer:
Differences between population and community:

Population	Community
(i) Group of individuals of same species inhabiting the same area.	(i) Group of species lying in the same area.
(ii) Organism in a population undergoes the same life cycle.	(ii) Different species have a different life cycle.

Question 13.
How does population size increase or decrease?
Answer:

• Population (size/density) increases by birth, hatching, germination and immigration that add individuals.
• Population (size/density) decreases by death and emigration. The population is regulated by food, space, disease, natural calamities and environmental factors.

Question 14.
Differentiate between natality rate and death/mortality rate.
Answer:
Differences between natality rate and mortality rate:

Character	Natality rate	Morality rate
(i) Definition	A number of births per 1,000 individuals of a population per year.	A number of death per 1,000 individuals of a population per year.
(ii) Population size and population density.	Increases	Decreases

Question 15.
Discuss the role of predators in an ecosystem.
Answer:
Predators are of great importance as they play the following important roles in an ecosystem:

1. They act as ‘conduits’ for energy transfer to higher trophic levels.
2. They keep the prey population under control, which otherwise can reach very high population density and cause an imbalance in the ecosystem,
3. They help in maintaining species diversity in a community by reducing the intensity of competition among the competing prey species.

Question 16.
What is brood parasitism? Give an example.
Answer:
Brood parasitism. It refers to the phenomenon in which one (parasitic) bird species lays its eggs in the nest of another bird species.

Evolution has occurred in such a way that the eggs of the parasitic birds resemble those of the host bird in size, colour, etc. to avoid the host bird detecting the foreign eggs and ejecting them from the nest. Example. Cuckoo birds lay eggs in the nests of crows.

Question 17.
List a few adaptations that parasites have developed. (CBSE Delhi 2019)
Answer:
Parasites have evolved one or more of the following adaptations:

1. Loss of unnecessary sense organs.
2. Presence of hooks/adhesive organs and suckers.
3. Loss of digestive system.
4. High reproductive capacity.
5. Produces antitoxins to counter toxins to the host.

Question 18.
How do parasites harm the host?
Answer:
The parasites harm the host in the following ways:

1. Reduces the survival of the host.
2. Growth and reproductive rates of hosts are reduced.
3. Render the most vulnerable to its predators by making them physically weak.

Question 19.
Justify the statement “Predators and scavengers are markedly different.”
Answer:
Predators feed on another organism, i.e. prey, whereas scavenger feeds on dead animals or an animal killed by another animal. A predator can be a prey also, e.g. a frog eats insects and the frog may be eaten by a snake. But a scavenger such as jackals, hyenas and vultures cannot kill zebra or giraffe or deer but a lion kills them and leaves a part of it to be eaten by such scavengers.

Question 20.
Mention any two significant roles predators play in nature. (CBSE 2008, 2016)
Answer:
Role of predators:

1. Transferring energy to a higher trophic level.
2. Keeping the prey population under control.

Question 21.
How do organisms manage stressful conditions existing in their habitat for a short duration? Explain with the help of one example each. (CBSE (Delhi) 2008)
Answer:
Physiological adaptations allow some organisms to respond quickly to stressful conditions. At high altitude, mountain (altitude) sickness is experienced resulting in nausea, fatigue and heart palpitations. It is due to low atmospheric pressure at high altitude and the body does not receive the proper amount of oxygen. It is accommodated by increasing RBC production, decreasing the binding capacity of haemoglobin and increasing breathing rate.

Question 22.
Explain with the help of an example each of any three ways the ecologists use to measure the population density of different organisms rather than by calculating their absolute number. (CBSE Delhi 2019 C)
Answer:
Method to measure population density: Population density is the number of individuals of a species per unit area or volume.

P_D= N/S

Where N = The no. of the individual in the region
S = No. of the unit area in a region

1. Number of animals per square kilometre
2. Number of trees per hectare
3. A number of phytoplanktons per cubic litre of water. Sometimes the relative densities also serve the purpose, that is

• The number of fish caught per trap is good enough to measure the population of fish.
• Tiger census is based on pug marks and faecal pellets.

Question 23.
Name the interaction in each of the following:
(i) The cuckoo lays her eggs in the crow’s nest.
Answer:
Brood parasitism.

(ii) Orchid grows on a mango tree.
Answer:
Commensalism.

(ii) Ticks live on the skin of dogs.
Answer:
Parasitism.

(iv) Sea anemone is often found on the shell of a hermit crab. (CBSE Delhi 2008)
Answer:
Mutualism.

Question 24.
Name the interaction in each of the following:
(i) Cuscuta growing on a shoe flower plant.
Answer:
Parasitism.

(ii) Mycorrhizae living on the roots of higher plants.
Answer:
Mutualism.

(iii) Clownfish living among the tentacles of a sea anemone.
Answer:
CommensaLism.

(iv) Koel laying its eggs in the crow’s nest. (CBSE Delhi 2011)
Answer:
Brood parasitism.

Question 25.
Name the interaction In each of the following:
(i) Ascaris worms living in the intestine of human.
Answer:
ParasItism

(ii) Suckerfish attached to the shark.
Answer:
Commensalism

(iii) Smaller barnacles disappeared when Balanus dominated the coast of Scotland.
Answer:
Competition

(iv) Wasp pollinating fig inflorescence. (CBSE 2008)
Answer:
Mutualism.

Question 26.
Why do clownfish and sea anemone pair up? What is this relationship called? (CBSE 2012)
Answer:

1. Commensalism,
2. Interaction between sea-anemone that has stinging tentacles and the clownfish that lives among them in which fish gets protection from predators. The anemone does not appear to derive any benefit from the clownfish.

Question 27.
Some organisms suspend their metabolic activities to survive in unfavourable conditions. Explain with the help of any four examples. (CBSE 2012)
Answer:

1. Thick-walled spores of bacteria and fungi help them to overcome unfavourable conditions.
2. Frogs and lizards undergo hibernation during the winter season.
3. Snails and fish undergo aestivation during summer to avoid summer-related problems of heat and desiccation.
4. Species of zooplankton are known to enter a suspended state of development termed-diapause.

Question 28.
A moss plant is unable to complete its life cycle in a dry environment. State reason. (CBSE Delhi 2015)
Answer:

1. Due to the absence of vascular tissue, water and minerals cannot be transported to various parts.
2. Transfer of flagellated antherozooids to oospore of archegonium depends upon water.

Question 29.
Plants that inhabit a rain-forest are not found in wetlands. Explain. (CBSE Delhi 2016)
Answer:
A habitat is a specific place or area or locality having a combination of factors, physical features and barrier where a community resides. In rainforest plants adapt themselves to the specific conditions of their habitat and the characteristics adaptations related to a particular habitat are not found in those plants living in wetland and vice versa. Thus they can not inhabit a new habitat.

Question 30.
Answer:
In certain seasons we sweat profusely while in some other season we shiver. Explain. (CBSE Delhi 2016)
Answer:
Humans are a warm-blooded or homeothermic or endothermal organism. They maintain a fixed body temperature despite changes in the surrounding. The optimum temperature for maximum efficiency of enzymes is 37°C. Thus the human body maintains a body temperature near it.

Homoeostasis is the phenomenon of maintaining a constant internal environment. Thus during heavy exercise or summer, sweating occurs. It helps in cooling the body. During winter, shivering helps in retaining heat within the body.

Question 31.
How did David Tillman show that the “Stability of a community depends on its species richness”? Explain. (Outside Delhi 2019)
Answer:
The stability of a community depends on its species richness. David Tilman’s long-term ecosystem experiments using outdoor plots provided the answer for species richness. Tilman found that plots with more species showed less year-to-year variations in total biomass. He also showed that increased diversity contributed to higher productivity.

Thus he established that the stability of a community depends on its species richness.

Organisms and Populations Important Extra Questions Long Answer Type

Question 1.
(i) “Organisms may be conformers or regulators.” Explain this statement and give one example of each.
(ii) Why are there more conformers than regulators in the animal world? (CBSE 2017)
Or
Human is categorised as ‘regulators’. Explain how they maintain a constant body temperature. (CBSE Outside Delhi 2019)
Answer:
(i) Regulators are organisms that are able to maintain homoeostasis by physiological and behavioural means. All birds and mammals and few lower vertebrate and invertebrate species maintain homoeostasis by thermoregulation and osmoregulation, e.g. birds and mammals.

However, a majority (99%) of animals and nearly all plants cannot maintain a constant internal environment, i. e. their body temperature is not constant. They are conformers, e.g. fishes, frogs, etc.

(ii) There are more conformers than regulators in the animal world because conformers lack the capability to maintain a constant internal environment or homoeostasis. It is limited to birds, mammals and few lower vertebrate and invertebrate species only because it is a very energy-expensive process.

Question 2.
Explain the ecological hierarchy.
Answer:
Ecological hierarchy: It is a series of graded ecological categories.
Characteristics of ecological hierarchy:

1. A biological unit at each level has a specific structure and function.
2. In this hierarchy, smaller biological units coordinate to form the next higher level of organisation.
3. Only the organisms show free existence.
4. Organisms cannot live in isolation.
5. At each level, different units show interdependence.
6. At each level, the unit shows interaction with the physical environment (energy and matter).
7. The biological units are successfully adapted to their environment.

Question 3.
In a seashore, the benthic animals live in sandy, muddy and rocky substrata and accordingly developed the following adaptations.
Find suitable substratum against each adaptation.
(i) Burrowing
Answer:
In a seashore, the benthic animals adapt to their mode of life according to the nature of the seafloor, e.g. Benthic animals become fossorial (burrowing) in the sandy substratum, e.g. tube worm.

(ii) Building cubes
Answer:
Benthic animals build cubes in the muddy substratum.

(iii) Holdfasts/peduncle
Answer:
Benthic animals develop holdfast/ peduncle if the substratum is rocky.

Question 4.
In a pond, we see plants that are free. floating, rooted-submerged, rooted-emergent, rooted with floating leaves. Write the type of plant against the following examples:
(i) Hydrilla
Answer:
Hydrilla is Rooted submerged.

(ii) Typha
Answer:
Typha is Rooted emergent.

(iii) Nymphaea
Answer:
Nymphaea is Rooted with floating leaves.

(iv) Lemna
Answer:
Lemna is Free-floating.

(v) Vallisneria.
Answer:
Vallisneria is Rooted submerged.

Question 5.
Why do all the freshwater organisms have contractile vacuoles whereas the majority of marine organisms lack them?
Answer:
Contractile vacuole helps in maintaining salt and water level called osmoregulation. Because of the cellular environment of a freshwater organism such as Amoeba, Paramecium etc. being hypertonic, the water diffuses inside the cell constantly and gets collected in the contractile vacuole, which squeezes the extra water out of the cell periodically.

While in the case of marine protozoans organisms, this does not occur due to high salt concentration. These organisms live in isotonic conditions in seawater. Thus there is no need for contractile vacuole.

Question 6.
Explain the following terms:
(i) Mimicry
Answer:
Mimicry: It is a phenomenon in which a living organism modifies its form, appearance, structure or behaviour and looks like another living organism or some inanimate (non-living) object so as to defend from its predators, or to increase the chances of capturing the prey.

The individual which shows mimicry is called a mimic, while the animate or inanimate object with which a mimic resembles is called a model. The concept of mimicry was first observed by an English naturalist, Henry Bates (1862 A.D.), so the phenomenon is also called Batesian mimicry.

Types: Mimicry is of three types:
(a) Protective mimicry,
(b) Aggressive mimicry,
(c) Feigning death or Conscious mimicry.

(ii) Acclimatisation
Answer:
Acclimatisation: The gradual physiological adjustment to slowly changing new environmental conditions is known as acclimatisation. If some factors shift beyond the tolerance range, the organism can come to a tolerance range or migrate to acclimatise.

Question 7.
DepIct the temperature-based thermal stratification in lakes.
Answer:
Thermal stratification in Lakes:

Thermal stratification occurs in lakes, the seasonal mixing patterns of a lake are determined by its temperature profile.

Question 8.
1. How does age distribution help in the study of the population?
2. How does an age pyramid, for the human population at a given point of time helps the policymakers in planning the future? (CBSE Delhi 2016)
Or
Draw a stable human age pyramid. (CBSE Outside Delhi 2019)
Answer:
Age distribution (Age composition). The relative abundance of the organisms of various age groups in the population is called the age distribution of the population. With regard to age distribution, there are three kinds of populations:

1. Rapidly growing population. It has a high birth rate and low death rate, so there is more number of young individuals in the population. According to a recent survey, more than 42% of the Indian population consists of children below the age of 14 years, so the Indian population is called the young population.

2. Stationary population: It has equal birth and death rates, so the population shows zero population growth.

3. Declining population. It has a higher death rate than the birth rate, so the population of young members is lower than that of old members.
(a) For the human population, the age pyramids generally show the age distribution of males and females in a combined diagram.
(b) The shape of the age pyramid reflects the growth status of the population. Thus age pyramid for the human population at a given time helps the policymakers in planning for the future.

Question 9.
Discuss life-history traits of an organism that have evolved in relation to the constraints imposed by biotic and abiotic factors in their habitat.
Answer:
According to ecologists, life-history traits of an organism have evolved in relation to the constraints imposed by the biotic and abiotic factors in their habitats.

1. It can be illustrated with vast variations and life history.
2. The evolution of populations aims at improving reproductive fitness or Darwinian fitness to the maximum in their habitats.
3. They evolve towards the most efficient reproductive strategy.
4. Organisms like Pacific Salmon fish and bamboo breed only once in their lifetime.
5. Most birds and mammals breed many times during their lifetime.
6. Oysters and pelagic fishes produce a large number of small-sized offspring.
7. Birds and mammals produce a small number of large-sized offspring.

Question 10.
What is the predator-prey relationship? Give example. (CBSE 2016)
Answer:
Predation. It is an interspecific interaction, where one animal (called predator) kills Organisms and Populations and consumes the other weaker animal (called prey). Herbivores are predators of plants.

Predator-prey relationship. It is a kind of direct food relationship between two species of animals in which larger species, called predator, attacks, kills and feeds on the smaller species, called prey. It was proved by G.F. Gause (1934).

Although predation and competition appear to be very harmful processes, they are essential to keep a check on the size of the population of other species.

Predation is nature’s way of transferring the energy fixed by plants, to higher trophic levels.
Examples:

1. A tiger killing and eating a deer.
2. A snake eating a frog.
3. A sparrow eating a fruit/seed.

Question 11.
Many prey organisms have developed different defence mechanism. Give a few examples.
Answer:
Prey species have developed various defence mechanisms to reduce the impact of predation; some of them are listed below:

1. Certain insect species and frogs have camouflage (cryptic colouration) to avoid detection by their predators.
2. Some animals (e.g. monarch butterfly) are highly distasteful to their predators. This butterfly species accumulates a chemical by feeding on a poisonous weed during its caterpillar stage.
3. Some prey species are poisonous and hence are avoided by predators, e.g. Dart frogs like Phyllobates bicolour and Dendrobats.

Question 12.
“Herbivores are the predators of plants”. Discuss a few defence mechanisms of plants against herbivory.
Or
Write what do phytophagous insects feed on? (CBSE2012)
Answer:
Herbivores are the predators of plants. The problem of predation is more severe for plants than animals as the plants cannot move away from the predators. About 25% of the known insects are phytophagous and feed on the sap and other parts of plants.

Plants have developed certain morphological and chemical defence mechanisms against herbivores; a few of them are listed below:

1. Morphological: Thorns (Bougainvillaea) and spines (Acacia, Cactus, etc.) are the most common morphological means of defence.
2. Chemical: Plants produce and store certain chemicals which function in one or more of the following ways:
   (a) They make the animal feel sick.
   (b) They inhibit them from feeding.
   (c) They interfere with digestion.
   (d) They even directly kill them, e.g. Calotropis produces a highly poisonous glycoside, that is a cardiac poison.
   (e) Nicotine, strychnine, opium, quinine, etc. are the chemicals produced by plants for their defence against herbivores.

Question 13.
What is parasitism? Define parasite and host. What are the kinds of parasite?
Answer:
Parasitism: This is a relationship between two organisms in which one obtains its nourishment from the other and harms it at the same time.

Parasite: The organism which obtains its food from the other without directly killing it is known as a parasite. Parasites are host specific and parasite and host tend to co-evolve.

Host: The second organism which provides food to the parasite is named the host. Endoparasite. The parasite that lives inside the body of the host is known as endoparasite.

Ectoparasite. The parasite lives on the outside of the body of the host, e.g. Leech, Louse, Bedbug.

Question 14.
Define commensalism. Give examples.
Answer:
Commensalism: It is defined as the interspecific interaction where one species is benefitted while the other species is neither benefitted nor harmed.
Examples:

1. Orchids grow as epiphytes on mango or other fruit trees. Orchids are benefitted by getting shelter, while the tree is neither benefitted nor harmed.
2. The clownfish living among sea anemones get protection from their predators, which stay away from the stinging tentacles of the sea anemone.
3. Barnacles growing on the whale are benefitted to move to where food is available.
4. The cattle egrets always forage near where the cattle are grazing. As the cattle animals stir up, the insects are flushed out from the vegetation. The egrets are benefitted from this as otherwise, it might be difficult for the birds to detect and catch the insects.

Question 15.
Differentiate between the following:
(i) Mutualism commensalism.
Answer:
Differences between mutualism and commensalism:

Mutualism	commensalism
(i) It is the inter-specific interaction, in which both the interacting species are benefitted.	(i) It is the interspecific interaction, in which one species is benefitted white the other is neither benefitted nor harmed.
(ii) It may or may not involve physical association.	(ii) The two individuals come in close physical contact.

(ii) Commensalism and amensalism.
Answer:
Differences between commensalism and amensalism:

Commensalism	Amensalism
It is the interspecific interaction in which one species is benefitted while the other one is neither harmed nor benefitted.	It is the interspecific interaction in which one species is harmed I affected, while the other is neither benefitted nor harmed.

(iii) Predators and parasites.
Answer:
Differences between predators and parasites:

Predators	Parasites
(1) Predators are Larger and stronger animals that kill and consume the prey.	(i) Parasites are small or microscopic organisms that depend on the host.
(ii) They do not take shelter on the prey	(ii) They take shelter on the host.
(iii) Their biotic potential is Low.	(iii) They have higher biotic potential.
(iv) They are mobile to capture the prey.	(iv) They have poor means of dispersal.
(v) They are not specific for the Prey	(v) They are host-specific.

Question 16.
Study the graph given below and answer the questions that follow:

(i) Write the status of food and space in the curves (a) and (b).
Answer:
Food and space are unlimited in the curve (a), while they are limited in the curve (b).

(ii) In the absence of predators, which one of the two curves would appropriately depict the prey population?
Answer:
In the absence of a predator, any species will grow exponentially and show curve (b).

(iii) Time has been shown on X-axis and there is a parallel dotted line above it. Give the significance of this dotted line. (CBSE 2014)
Answer:
The dotted line represents (k). It is carrying capacity.

Question 17.
Explain parasitism and co-evolution with the help of one example of each. (CBSE Outside Delhi 2016)
Answer:

1. Many parasites have evolved to be host-specific. They can live as parasite only in a single species of host.
2. These organisms parasitize in such a way that both the host and the parasite tend to co-evolve; that is if the host evolves a special mechanism for rejecting or resisting the parasite, in such conditions parasite has to evolve a mechanism to counteract and neutralise them, in order to be successful with the same host species.
3. In accordance with their mode of living, parasites evolve special adaptations such as loss of unnecessary sense organs, presence of adhesive organs or sucker so as to cling to the host. It also loses the digestive system. Parasites have a high reproductive capacity.
4. The human liver fluke depends on the intermediate host (a snail and a fish ) to complete the life cycle.
5. The malarial parasite needs a female anopheles mosquito as a vector to spread to other hosts.

Question 18.
(i) In a pond, there were 200 frogs. 40 more were born in the year. Calculate the birth rate of the population.
Answer:
Birth rate = No. of individuals born/ Total no. of individuals = 40/200 = 0.2 = 0.2 frog per year.

(ii) Population in terms of number is not always a necessary parameter to measure population density. Justify with two examples. (CBSE Outside Delhi 2019)
Answer:
To measure population density, the number is not always a necessary parameter.

For example:
(a) If there are 200 Parthenium plants but only a single huge banyan tree with a large canopy, the population density of banyan is low relative to that of Parthenium which amounts to underestimating the enormous role of the banyan in that community. In such cases, the per cent cover or biomass is a more meaningful measure of the population size.
(b) In a dense laboratory culture of a microbial population in a Petri dish, the total number of microbes is again not an easily adaptable measure because as the population is huge, counting is impossible and time¬consuming.

Question 19.
How do organisms which cannot migrate tend to overcome adverse environ¬mental conditions? Explain taking one example each from vertebrates and angiosperms respectively. (CBSE Delhi 2009)
Answer:
Organisms that cannot migrate tend to overcome adverse environmental conditions by developing several methods/ features. For example, some vertebrates escape the stress caused by unfavourable environmental conditions by escaping in time like bears go into hibernation during the winter months.

In angiosperms, seeds and some other vegetative reproductive structures serve as means to tide over periods of stress. They reduce their metabolic activity and go into an inactive, i.e. ‘dormant’, state. They germinate to form new plant when the favourable conditions return.

Question 20.
Explain how tolerance to environmental factors determines the distribution of species.
Answer:
Range of tolerance: Biological species can show a range of tolerance to environmental factors. These factors show variation in their effects and anyone who is present in the least amount may become limiting. The response of an organism to a range of gradient of a single environmental factor such as temperature, sunlight or nutrient concentration forms a bell-shaped curve as shown in the figure.

The response of an organism to a range or gradient of an environmental factor, (temperature, Light, nutrient)

Question 21.
Discuss in detail various adaptations found in plants and animals in snowy winter of polar regions.
Answer:
Adaptations in plant and animals of polar regions:

1. Organisms living in polar regions have to face the severe snowy winter and show entirely different types of adaptations. Animals in polar regions are generally white or tight coloured. This light colouration helps in camouflage and in thermal regulation.
2. During extreme winter they undergo hibernation.
3. Animals eat a lot of food during summer and autumn and store a lot of energy in the form of fat.
4. During hibernation, their metabolic activities are reduced considerably,
5. The plants growing in these regions remain dormant during extreme winter months.
6. They bear narrow leaves, spines, etc. which are shed easily and rapidly,
7. Plants remain dwarf. Trees are usually not found in these regions as they cannot withstand the low temperatures.

Question 22.
How do desert plants prevent loss of water?
Answer:
Xeric adaptations of desert plants:

1. Some xerophytes remain dormant either as seeds or as roots and during rains, they sprout up. It is termed ephemerals, e.g. Cassia, Argemone.
2. Some desert plants develop succulent organs (stems, leaves and roots), e.g. Asparagus, Begonia, Bryophyilum.
3. Presence of extensive root system.
4. Either leaf is absent or small¬sized leaves present to minimise the rate of transpiration.
5. The stomata get sunken to reduce the rate of transpiration.
6. Presence of thick cuticle on stem and leaf surfaces. This reduces the rate of transpiration.
7. Shedding of leaves in some desert plants. This helps in reducing the surface area and water loss.

Question 23.
(i) What is the competition?
Answer:
Competition is an interaction between organisms for life requirements (nutrition, shelter, sunlight, etc.). It is of two types: intraspecific and interspecific. The effort of a tiger and a leopard for prey is an interspecific competition. In general, it is believed that competition occurs among closely related species when they compete for the same resources that are limited.

(ii) Why is it not true always?
Answer:
But it is found to be not true always for the following reasons:
(a) Completely unrelated species can also compete for the same sources. For example in certain shallow lakes of South America, the visiting flamingoes and the native fishes compete for the same zooplankton as their food.

(b) Resources need not be limiting for the competition to occur, the feeding efficiency of one species might be reduced due to the interfering and inhibitory presence of the other species, even if resources are plenty.

For example, the Abingdon tortoise in Galapagos island became extinct within a decade after goats were introduced into the island: this was due to the fact that the goats had greater browsing efficiency than the tortoise.

(iii) Explain competitive release and Gause’s competitive exclusion principle.
Answer:
Another evidence for the competition is competitive release, a phenomenon in which a species whose distribution is restricted to a small geographical area due to the presence of a competitively superior species, expands its distributional range when the competing species is experimentally removed.

Gause’s competitive exclusion principle states that two closely related species competing for the same resources cannot exist together as the competitively inferior one will be eliminated, but this is true only when the resources are limiting and not otherwise.

They have also pointed out that species facing competition might evolve mechanisms that promote co¬existence, rather than exclusion.

(iv) Write contribution of Mac Arthur.
Answer:
Mac Arthur had shown that five closely related species of warblers living on the same tree were able to avoid competition and co-exist to behavioural differences in their foraging activities.

Question 24.
Illustrate symbiosis with any four examples.
Answer:
Symbiosis: It is the relationship between two individuals where both partners are benefited.

That following are examples of symbiosis:

1. The Rhizobium bacteria present in the root nodules of leguminous plant roots are an example of symbiosis. Bacteria live in the roots in the form of nodules and avail carbohydrate and other food substances. In exchange, bacteria fix the nitrogen present in the atmosphere and make it useful for plants.
2. Trichonympha, a protozoan parasite, lives inside the intestine of termites. The presence of this protozoan helps the termites to digest the cellulose food and the parasite gets food and shelter.
3. Lichen plants. Lichen is the result of a symbiotic relationship between algae and fungi. In this, algae depend on fungi for water, minerals, salts and safety, whereas fungi get food material prepared by algae.
4. The human intestine contains a large number of symbiotic bacteria that help in the synthesis of vitamin B-complex. Bacteria get necessary nutritional substances from the human intestine and a safe place for living.

Question 25.
Define phenotypic adaptation. Give one example.
Answer:
Phenotypic adaptation involves non-genetic changes in individuals such as physiological modifications, acclimatisation or behavioural changes. Some organisms possess adaptations that are phenotypic, which allow them to respond quickly to a stressful situation.

Example. If a person had ever been to any high altitude place (Rohtang pass near Manali or Mansrover in Tibet), he or she must have altitude sickness because the body does not get enough oxygen at such a height and protect itself from atmospheric hypoxia. But after some time, the body acclimatises itself to the situation by increasing the number of RBC and decreasing the binding affinity of Hb and oxygen.

Question 26.
Name important defence mechanisms in plants against herbivores.
Answer:
Defence mechanisms of plants against herbivorous animals:

1. Formation of thick cuticle on their leaf surface.
2. Formation of leaf spines, e.g. leaf spines in opuntia
3. Modification of leaves into thorns, e.g. Bougainvillea and Duranta
4. Development of spiny margins on leaves.
5. Development of sharp silicated edges in leaves.
6. Many plants produce and store toxic chemicals which cause discomforts to herbivores, e.g. cardiac glucoside by Calotropis, nicotine by tobacco.

Question 27.
Distinguish between the following:
(i) Hibernation and aestivation
Answer:
Differences between hibernation and aestivation:

Hibernation	Aestivation
(i) Spending the winter in an inactive? dormant state, Is called hibernation (winter sleep).	(i) Spending dr hot period (summer) in an inactive period is called aestivation (summer sleep).
(ii) Example. Northern ground squirrels.	(ii) Example. Ground squirrels in the South West deserts.

(ii) Ectotherms and endotherms.
Answer:
Differences between ectotherms and endotherms:

Ectotherms	Endotherms
Ectotherms are those animals whose body temperature changes to match with that of the surroundings, in which they are living. They cannot maintain their internal environment constant.	Endotherms are those animals whose body temperature is maintained relatively constant by physiological regulation.

Question 28.
Write short notes on:
(i) adaptations of desert plants and animals. (CBSE2011)
Answer:
Adaptations of desert plants and animals:
(a) Adaptations of desert animals.

• Animals faced with water scarcity as found in arid or desert areas, show two types of adaptations, reducing water loss and the ability to tolerate arid conditions. Kangaroo/Desert rat seldom drinks water. It has a thick coat to minimise evaporative desiccation. The animal seldom comes out of its comparatively humid and cool burrow during the daytime. 90% of its water requirement is met from metabolic water (water produced by respiratory breakdown) while 10% is obtained from food.
• Loss of water is minimised by producing nearly solid urine and faeces.
• Spiny skin and highly cornified in Phrynosoma (horned toad) and Moloch horridus.
• Camels have long legs to stay away from the hot desert surface.

(b) Adaptations of desert plants.

• Plants have thick cuticle, succulent organs where water and mucilage are stored.
• Stomata are sunken to prevent water loss.
• They have a well-developed branched root system.
• They possess a waxy coating on the surface.
• Crassulacean pathway of photosynthesis.

(ii) behavioural adaptations in animals. (CBSE 2011)
Answer:
Adaptations of plants to water scarcities. They are called xerophytes. The above-mentioned adaptation of plants are applicable (see part (a)).

(iii) Behavioural adaptations in animals,
(a) Hibernation
(b) Aestivation
(c) Periodic activity
(d) Camouflage
(e) Migration.

(iv) importance of light to plants.
Answer:
Importance of light to plants.
(a) Source of energy for photosynthesis
(b) Photoperiodism
(c) Pigmentation
(d) Daily rhythm
(e) Plant movements and (f) Growth.

(v) effect of temperature or water scarcity and the adaptation of animals.
Answer:
Effect of temperature or water scarcity and adaptation of animals. Optimum temperature is necessary for animals, survival as all the metabolic activities are driven by enzymes. And enzymes work actively only in a certain optimum range of temperature. Regulators can regulate their body temperature in case of temperature fluctuations in the external environment. Conformers also try to maintain temperature by certain methods but not internally.

Similarly, water is also necessary for the metabolic activities of animals. Various adaptations seen in animals to deal with temperature fluctuations are

• thick fur,
• sweating, short limbs and ears.

Various adaptation to counter the scarcity of water is the ability to use metabolic water, store water, reduced water loss, etc.

Question 29.
With the help of a suitable diagram describe the logistic population growth curve. (CBSE Delhi 2019 C)
Answer:
Logistic popuLation growth curve.

S-shaped Population growth curve (Verhulst-Pearl logistic growth curve)

Unlimited resources result in exponential growth. Many countries have realised this fact and introduced various restraints to limit human population growth. In nature, a given habitat has enough resources to support a maximum possible number, beyond which no further growth is possible. This limit is called carrying capacity (K) for that species in that habitat.

A population growing in a habitat with limited resources shows initially a lag phase, followed by phases of increase and decrease and finally the population density reaches the carrying capacity. A plot of N in relation to time (t) results in a sigmoid curve. This type of population growth is called Verhulst-Pearl Logistic Growth as explained by the following equation:

\(\frac{d \mathrm{~N}}{d t}=r \mathrm{~N}\left(\frac{\mathrm{K}-\mathrm{N}}{\mathrm{K}}\right)\)

where N = Population density at a time t; r = Intrinsic rate of natural increase and K = Carrying capacity,

\(\left[\frac{k-N}{k}\right]\) = Environmental resistance.

Since resources for growth for most animal populations become limiting sooner or later, the logistic growth model is considered a more realistic one.

Question 30.
List any three important characteristics of a population and explain. (CBSE Delhi 2019)
Answer:
Three important characteristics of a population are:

1. Population size and population density
2. Birth or natality rate
3. Death or mortality rate.

• Population size: It is the actual number of individuals in the population. The size of the population keeps changing with time depending on the factors like:
  (a) Food availability
  (b) Weather
  (c) Predation pressure and
  (d) Competition.

Population density is a measure of population size per unit area. The population density in a given habitat during a given period changes due to four basic processes, namely

1. Natality,
2. Mortality,
3. Immigration and
4. Emigration.

While natality and immigration contribute to an increase in the size and density of a population, mortality and emigration contribute to a decrease in them.

So the equation for population growth is: Nt + 1 = Nt + [(B + I) – (D + E)],
where Nt = population density at time t.
B = No of birth I = No of Immigration D = No of death E = No of emigration

• If B + I is more than D – E, the population density increases.
• If B + I is less than D – E, the population density decreases.

Birth or Natality rate. It is generally expressed as the number of births per 1,000 individuals of a population per year. It increases the population size (total number of individuals of a population) and population density. Natal or Birth Rate =

Death or Mortality rate. It is the opposite of the natality rate. It is commonly expressed as the number of deaths per 1,000 individuals of a population per year.

Question 31.
Define mutualism. Give examples.
Answer:
Mutualism: It is a symbiotic relationship between the members of two species in which the two partners are mutually benefited. There is a complete dependence of the partners on each other, and one cannot survive in the absence of the other.

Sometimes the term symbiosis is used as a synonym with mutualism.
Examples:
(a) Mutualism between animal and bacteria. Symbiotic bacteria like Ruminococcus are found in the rumen part of the compound stomach of cud-chewing mammals like cattle, sheep, goat, camel, etc. and secrete cellulase enzyme to digest the cellulose of plant food eaten by the ruminants which provide food and shelter to the bacteria.

(b) Mutualism between crab and sea anemone. In this case, a sea anemone gets attached to the back of the hermit crab.

(c) The Mediterranean orchid, Ophrys, employs sexual deceit to get its flowers pollinated. In this orchid, one petal of the flower resembles the female of a bee species in size, colour, markings, etc. The male bee perceives it as a female and pseudo copulate with it. During the process, the pollen grains get dusted on its body.

Question 32.
It is observed that plant-animal interactions often involve co-evolution. Explain with the help of a suitable example. (CBSE Delhi 2019 C)
Or
Mention four significant services that a healthy forest ecosystem provides. (CBSE Delhi 2019)
Answer:
Plants and animals interact for mutual ben¬efit. Plant-animal interactions often involve co-evolution of mutualism, that is, the evolution of flower and its pollinator species are tightly linked with one another. In many species of fig trees, there is a tight one to one relationship with the pollinator species of wasp. It means that a given fig species can be pollinated only by its partner wasp species and no other species.

The female wasp uses the fruit not only at the oviposition site but also uses the developing seeds within the fruit for nourishing its larvae. The wasp pol¬linates the fig inflorescence while searching for suitable egg-laying sites. In return for the favour of pollination, the fig offers the wasp some of its developing seeds as food for the developing wasp larvae.

Question 33.
(i) Compare, giving reasons, the J-shaped and S-shaped models of population growth of a species.
Answer:
There are two models of population growth-exponential growth and logistic growth.
(a) Exponential growth: This growth occurs where the resources, i.e. food and space, are unlimited. The equation can be represented as follows:
\(\frac{\mathrm{dN}}{\mathrm{dt}}\) = (b-d) x N

Let (b-d) = r
\(\frac{\mathrm{dN}}{\mathrm{dt}}\) = rN or N_t=N_oe^rt

N = population size
N_t = population density after time t.
N_o = population density at time zero
r = growth rate
e = base of natural log (2.71828)
b = birth rate
d = death rate
In this growth, when N in relation to time is plotted on a graph, the curve becomes J-shaped.

(b) Logistics growth model: This is a realistic approach as the resources become limited at a certain point in time.

(a) J-shaped curve exponential growth
(b) S-shaped curve logistics growth Every ecosystem has limited resources to support a particular maximum carrying capacity (K). When N is plotted in relation to time t, a sigmoid S-shaped curve is obtained. It is also called Verhulst-Pearl logistic growth.

The equation is:
\(\frac{\mathrm{dN}}{\mathrm{dt}}\)=rN\(\left(\frac{K-N}{K}\right)\)

N = population density at time t
r = growth rate
K = carrying capacity.

(ii) Explain “fitness of a species” as mentioned by Darwin. (CBSE 2017)
Answer:
According to Darwin “Fitness of a species” means reproductive fitness. All organisms after reaching reproductive age have a varying degree of reproductive potential. Some organisms produce more offspring, whereas some organisms produce few offspring only. This phenomenon is also called differential reproduction. The species which produces more offspring are selected by nature.

Question 34.
(i) The following are the responses of different animals to various abiotic factors. Describe each one with the help of an example.
(a) Regulate
Answer:
Regulate: Some organisms are able to maintain homoeostasis by regulating their body temperatures. The mechanisms used by most mammals to regulate their body temperature are similar to what we humans use.

For example, our body temperature remains constant at 37°C. In summer, when the outside temperature is more than our body temperature, we sweat profusely to cool down and when the temperature is much lower than 37° C, we shiver to generate heat. Thus body temperature remains constant.

(b) Conform
Answer:
Confirm: Most of the animals cannot maintain a constant internal environment. Their body temperature changes with the ambient temperature. These are called conformers. Heat loss or heat gain is a function of surface area. Since small animals have a larger surface area relative to their volume, they tend to lose body heat very fast when it is cold outside, for example, shrews and hummingbirds.

(c) Migrate
Answer:
Migrate: Organism can move away temporarily from the stressful habitat to a more hospitable area and then return when the stressful period is over.

For example, every winter, the famous Keoladeo National Park in Bharatpur hosts thousands of migratory birds coming from Siberia and other northern regions,

(d) Suspend
Answer:
Suspend: Bacteria, fungi and some lower plants survive unfavourable conditions by forming thick-walled spores. In higher plants, seeds and some other vegetative reproductive structures serve as means to tide over periods of stress. They do so by reducing their metabolic activity and going into a state of dormancy.

(ii) If 8 individuals in a population of 80 butterflies die in a week, calculate the death rate of the population of butterflies during that period. (CBSE 2018)
Answer:
Death Rate = \(\frac{8}{80}\) = 0.1 per week

Question 35.
(a) Identify the features of a stable biological community.
Answer:
(a) Features of the stable biological community.

• It is stable.
• A stable biological community is not replaced by any other community.
• The environment becomes moister and shadier.
• Communities should have greater biodiversity for greater stability.
• It should be able to prevent invasion by alien species.
• It should be able to restore itself in a short period of time.

(b) How did David Tilman’s findings link the stability of a biological community to its species richness? (CBSE Delhi 2019 C)
Answer:
David Tilman’s with his experiments established that the stability of a community depends on its species richness.