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Properties Of Bulk Matter

Expansion Of Solid And Liquids Short Question And Answers

Short Answer Questions on Thermal Expansion

Question 1. The liquid with a co-efficient of volume expansion (γ) is filled in a container of a material having the coefficient of linear expansion a. If the liquid overflows on heating, then

1. γ = 3α
2. γ > 3α
3. γ <3α
4. γ > 3α³

Answer:

Given

The liquid with a co-efficient of volume expansion (γ) is filled in a container of a material having the coefficient of linear expansion a.

Apparent coefficient of volume expansion, γ’= γ-3α

Now, γ’ is positive when the liquid overflows on heating.

So the condition is γ – 3α > 0 or, ϒ > 3α

The option 2 is correct.

Question 2. When a copper sphere is heated, which physical quantity of the sphere will show maximum percentage change?
Answer:

When a copper sphere is heated, its volume will show maximum change in percentage.

Question 3. A metal rod is fixed rigidly at two ends so as to prevent its thermal expansion. If L, a and Y respectively denote the length of the rod, coefficient of linear thermal expansion and Young’s modulus of its material, then for an increase in temperature of the rod by ΔT, the longitudinal stress developed in the rod is

1. Inversely proportional to α
2. Inversely proportional to Y
3. Directly proportional to ΔT/Y
4. Independent of L

Answer:

Given

A metal rod is fixed rigidly at two ends so as to prevent its thermal expansion. If L, a and Y respectively denote the length of the rod, coefficient of linear thermal expansion and Young’s modulus of its material, then for an increase in temperature of the rod by ΔT,

Longitudinal thermal stress = YαΔT.

The option 4 is correct.

Key Concepts in Solid and Liquid Expansion

Question 4. A solid rectangular sheet has two different coefficients of linear expansion α₁ and α₂ along its length and breadth respectively. The coefficient of surface expansion is (for α₁ t<<1 1, α₂ t<<l)

1. \(\frac{\alpha_1+\alpha_2}{2}\)
2. \(2\left(\alpha_1+\alpha_2\right)\)
3. \(\frac{4 \alpha_1 \alpha_2}{\alpha_1+\alpha_2}\)
4. \(\alpha_1+\alpha_2\)

Answer:

Given

A solid rectangular sheet has two different coefficients of linear expansion α₁ and α₂ along its length and breadth respectively.

Let, initial length and breadth of rectangular sheet are a₁ and b₁ respectively.

If the final length and breadth are a₂ and b₂ respectively in increase in temperature t° C, then

⇒ \(a_2=a_1\left(1+a_1 t\right) \text { and } b_2=b_1\left(1+\alpha_2 t\right)\)

∴ \(a_2 b_2=a_1 b_1\left(1+\alpha_1 t\right)\left(1+\alpha_2 t\right)\)

or, \(a_1 b_1(1+\beta t)=a_1 b_1\left\{1+\left(\alpha_1+\alpha_2\right) t+\alpha_1 \alpha_2 t^2\right\}\)

or, \(1+\beta t=1+\left(\alpha_1+\alpha_2\right) t\)

[neglecting the term \(\alpha_1 \alpha_2 t^2\)]

∴ \(\beta=\alpha_1+\alpha_2\)

The option 4 is correct

Real-Life Applications of Thermal Expansion

Question 5. The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100 °C is (For steel, Young’s modulus is 2 x 10¹¹ N • m^-2 and coefficient of linear expansion is 1.1 x 10^-5 K^-1)

1. 2.2x 10⁸ Pa
2. 2.2x 10⁹ Pa
3. 2.2 x 10⁷ Pa
4. 2.2 x 10⁶ Pa

Answer: As length is constant, strain =ΔL/L = αΔQ

Now, pressure = stress = Yx strain

= 2x 10¹¹ x 1.1 x 10^-5 x 100

= 2.2 x 10⁸ Pa

The option 1 is correct.

Question 6. A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the coefficient of linear expansion (α) of the metal of the pendulum shaft are respectively

1. 25°C and 1.85 x 10^-5 l °C
2. 60°C and 1.85 x 10^-4 l °C
3. 30°C and 1.85 x 10^-3 l°C
4. 55°C and 1.85 x 10^-2 l °C

Answer:

Given

A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time,

Increase or decrease in time for change in temperature ΔT,

⇒ \(\Delta t =\frac{1}{2} \alpha \Delta T t\)

∴ 12 = \(\frac{1}{2} \alpha\left(40-T_0\right) \times 1\)

T₀ = temperature at which clock gives accurate breading, t = 1 day]

and 4 = \(\frac{1}{2} \alpha\left(T_0-20\right) \times 1\)….(2)

(1)+(2) we get,

3 = \(\frac{40-T_0}{T_0-20} \quad \text { or, } 3 T_0-60=40-T_0\)

or, \(T_0=\frac{100}{4}=25^{\circ} \mathrm{C}\)

Putting the value of T₀ in equation (1) we get,

⇒ \(15 \alpha=\frac{24}{24 \times 60 \times 60}\)

or, \(\alpha=\frac{1}{15 \times 60 \times 60}=1.85 \times 10^{-5} /{ }^{\circ} \mathrm{C}\)

The option 1 is correct.

Coefficient of Expansion Short Answer Questions

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Question 7. An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by:

1. \(3 P K \alpha\)
2. \(\frac{P}{3 \alpha K}\)
3. \(\frac{P}{\alpha K}\)
4. \(\frac{3 a}{P K}\)

Answer:

Given

An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating

Bulk modulus \(K=\frac{P}{\left(\frac{\Delta V}{V}\right)}\)

∴ \(\frac{\Delta V}{V}=\frac{P}{K}[\Delta V= change in volume ]\)

If increase in temperature Δt brings the cube to its original size, then

∴ \(\Delta V=V \cdot \gamma \Delta t \quad \text { or, } \Delta t=\frac{\Delta V}{V} \cdot \frac{1}{\gamma}=\frac{\Delta V}{V} \cdot \frac{1}{3 \alpha}=\frac{P}{3 K \alpha}\)

The option 2 is correct.

Question 8. Coefficient of linear expansion of brass and steel rods are α₁ and α₂. Lengths of brass and steel rods are l₁ and l₂ respectively. If (l₂ – l₁) is maintained same at all temperatures, which one of the following relations holds good?

1. \(\alpha_1 l_2^2=\alpha_2 l_1^2\)
2. \(\alpha_1^2 l_2=\alpha_2^2 l_1\)
3. \(\alpha_1 l_1=\alpha_2 l_2\)
4. \(\alpha_1 l_2=\alpha_2 l_1\)

Answer:

Given

Coefficient of linear expansion of brass and steel rods are α₁ and α₂. Lengths of brass and steel rods are l₁ and l₂ respectively. If (l₂ – l₁) is maintained same at all temperatures,

Let, lengths of brass rod at temperatures t₁ and t₂ are l₂ and l’₂ respectively.

∴ \(l_1^{\prime}=l_1\left\{1+\alpha_1\left(t_2-t_1\right)\right\}\)

Again, let lengths of steel rod at temperatures t₁ and t₂ are l₂ and l’₂ respectively.

∴ \(l_2^{\prime}=l_2\left\{1+\alpha_2\left(t_2-t_1\right)\right\}\)

According to the question, \(l_2-l_1=l_2^{\prime}-l_1^{\prime}\)

or, \(l_2-l_1=l_2\left\{1+\alpha_2\left(t_2-t_1\right)\right\}-l_1\left\{1+\alpha_1\left(t_2-t_1\right)\right\}\)

or, \(l_2-l_1=l_2+l_2 a_2\left(t_2-t_1\right)-l_1-l_1 a_1\left(t_2-t_1\right)\)

or, \(l_2-l_1=\left(l_2-l_1\right)+\left(l_2-t_1\right)\left(l_2 a_2-l_1 a_1\right)\)

or, \(\left(t_2-t_1\right)\left(l_2 a_2-l_1 a_1\right)=0\)

as \(\left(t_2-t_1\right) \neq 0\)

∴ \(\left(l_2 \alpha_2-l_1 \alpha_1\right)=0 \quad or, l_1 \alpha_1=l_2 a_2\)

The option 3 is correct.

Thermal Expansion Problems with Solutions

Question 9. Show that the coefficient of superficial expansion of a rectangular sheet of the solid is twice its coefficient of linear expansion.
Answer:

Let a₁,b₁ be the length and breadth, respectively, of the rectangular sheet at the initial temperature t₁.

Then, S₁ = a₁b₁ = initial area

Now, the temperature is raised to t₂, where t₂  – t₁ =t

The new values of the length and breadth are, respectively, a₂ and b₂; area S₂ = a₂b₂.

If α be the coefficient of linear expansion, then

⇒ \(a_2=a_1(1+\alpha t) \text { and } b_2=b_1(1+\alpha t)\)

∴ \(S_2=a_2 b_2=a_1 b_1(1+\alpha t)^2=S_1(1+2 \alpha t)\)

(neglecting α²t² as α<<1)

On the other hand, if β be the coefficient of superficial expansion, then

S₂ = S₁(1 +βt)

On comparison, we have β = 2α.

WBCHSE Class 12 Physics MCQs

Electric Potential Multiple Choice Question And Answers

Question 1. A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge

1. Remains constant because the electric field is uniform
2. Increases because the charge moves along the electric field
3. Decreases because the charge moves along the electric field
4. Decreases because the charge moves opposite to the electric field

Answer: 3. Decreases because the charge moves along the electric field

The positive charge will move along the electric field, i.e., it will move from higher to lower potential. Therefore, the electric potential energy will decrease.

Question 2. The electrostatic potential on the surface of a charged conducting sphere is 100 V. Two statements are made in this regard

S₁: At any point inside the sphere, electric intensity is zero.

S₂: At any point inside the sphere, the electrostatic potential is 100 V.

Which of the following is a correct statement?

1. S₁ is true but S₂ is false
2. Both Sj and S₂ are false
3. S₁ is true, S₂ is also true and S₁ is the cause of S₂
4. S₁ is true, and S₂ is also true but the statements are independent

Read and Learn More Class 12 Physics Multiple Choice Questions

Answer: 3. S₁ is true, S₂ is also true and S₁ is the cause of S₂

The field E inside a conductor is zero.

⇒ \(E=-\frac{d V}{d r}\)

or, -Edr = dV = 0

∴ V is constant

Question 3. Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately

1. Sphere
2. Planes
3. Paraboloidsÿ
4. Ellipsoids

Answer: 1. Sphere

Question 4. Equipotential surfaces

1. Are closer in regions of large electric fields compared to regions of lower electric fields
2. Will be more crowded near the sharp edges of a conductor
3. Will be more crowded near regions of large charge densities
4. Will always be equally spaced

Answer:

1. Are closer in regions of large electric fields compared to regions of lower electric fields

2. Will be more crowded near sharp edges of a conductor

3. Will be more crowded near regions of large charge densities

Question 5. The work done to move a charge along an equipotential from A to B

1. Cannot be defined as \(-\int_A^B \vec{E} \cdot d \vec{l}\)
2. Must be defined as \(-\int_A^B \vec{E} \cdot d \vec{l}\)
3. Is zero
4. Can have a non-zero value

Answer:

2. Must be defined as \(-\int_A^B \vec{E} \cdot d \vec{l}\)

3. Is zero

⇒ \(W=-\int_A^B \vec{E} \cdot d \vec{l}=-\int_A^B E d l \cos \theta\)

For the equipotential surface, θ = 90°.

So, W = 0

Electric Potential MCQs for WBCHSE Class 12

Question 6. In a region of constant potential

1. The electric field is uniform
2. The electric field is zero
3. There can be no charge inside the region
4. The electric field shall necessarily change if a charge is placed outside the region

Answer:

2. The electric field field is zero

3. There can be no charge inside the region

⇒ \(E=-\frac{d V}{d r}\)

If V is constant, then E = 0

Again, \(\oint_S \vec{E} \cdot d \vec{S}=\frac{q}{\epsilon_0}\)

Since E= 0.

So, q = 0

Question 7. Which of the following quantities do not depend on the choice of zero potential or zero potential energy?

1. Potential at a point
2. The potential difference between the two points
3. The potential energy of a two-charge system
4. Change in potential energy of a two-charge system

Answer:

2. Potential difference between two points

4. Change in potential energy of a two-charge system

WBCHSE class 12 physics MCQs

Question 8. p is a point on an equipotential surface S. The field at P is E

1. E is perpendicular to S in all cases
2. E is perpendicular to S only if S is a plane surface
3. E cannot have a component along a tangent to S
4. E may have a non-zero component along a tangent to S if S is a curved surface

Answer:

1. E is perpendicular to S in all cases

3. E cannot have a component along a tangent to S

Question 9. When a proton is accelerated from rest through a potential difference of 1000 V, its kinetic energy becomes

1. 1.6 x 10^-16 J
2. 1.6 x 10^-13 eV
3. 1000J
4. 1000 eV

Answer:

1. 1.6 x 10^-16 J

4. 1000 eV

Question 10. In a uniform electric field, equipotential surfaces must

1. Be Plane Surfaces
2. Be Normal To The Direction Of The Field
3. Be Spaced Such That The Surfaces Having Equal Differences In Potential Are Separated By Equal Distances
4. Have Decreasing Potentials In The Direction Of The Field

Answer:

1. Be Plane Surfaces

2. Be Normal To The Direction Of The Field

3. Be Spaced Such That The Surfaces Having Equal Differences In Potential Are Separated By Equal Distances

4. Have Decreasing Potentials In The Direction Of The Field

Multiple Choice Questions on Electric Potential

Question 11. An ellipsoidal cavity is cut within a perfect conductor. A positive charge Q is placed at the centre of the cavity. If points A and B are on the cavity surface then which among the following choices are correct?

1. Electric field near A = Electric field near B
2. Potential at A = potential at B
3. The total electric flux through the entire surface of the cavity \(\frac{Q}{\epsilon_0}\)
4. Charge density at A = charge density at B

Answer:

2. Potential at A = potential at B

3. Total electric flux through the entire surface of the cavity \(\frac{Q}{\epsilon_0}\)

Question 12. A conductor A is given a charge of around +Q and then placed inside a deep metal can B, without touching it.

1. The potential of A does not change when it is dipped inside B.
2. If B is earthed, +Q amount of charge flows from it to the earth
3. If B is earthed, the potential of A is reduced
4. Either 2 or 3 is, true or both are true only if the outer surface of B is connected to the media. i surface

Answer:

1. The potential of A does not change when it is dipped inside B.

2. If B is earthed, +Q amount of charge flows from it to the earth

3. If B is earthed, the potential of A is reduced

Electric potential class 12 MCQs

Question 13. For a spherical symmetrical charge distribution, variation of electric potential with distance from the centre Is given in the diagram. Given that \(V = \frac{1}{4 \pi \epsilon_0} \frac{q}{R_0} ; \text { for } r ≤ R_0 \text { and } V=\frac{1}{4 \pi \epsilon_0} \frac{q}{r}\) r ≥ R₀. Choose the correct options.

1. The total charge within 2R₀ is q
2. Total electrostatic energy for r≤R₀ is zero
3. At r = R₀ electric field is discontinuous
4. There will be no charge anywhere except at r = R₀

Answer:

1. Total charge within 2R₀ is q

2. Total electrostatic energy for r≤R₀ is zero

3. At r = R₀ electric field is discontinuous

4. There will be no charge anywhere except at r = R₀

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Electric Potential Practice MCQs for Students

Question 14. An elliptical cavity is carved within a perfect conductor. A positive charge q is placed at the centre of the cavity. Points A and B are on the cavity surface. Then,

1. The electric field near A in the cavity = electric field near B In the cavity
2. Charge density at A = charge density at B
3. Potential at A = potential at B
4. The total electric flux through the surface of the cavity is \(\frac{q}{\epsilon_0}\)

Answer:

3. Potential at A = potential at B

4. Total electric flux through the surface of the cavity is \(\frac{q}{\epsilon_0}\)

Question 15. Four charges, all of the same magnitude, are placed at the four corners of a square. At the centre of the square, the potential is V and the field is E. Which of the following is possible?

1. V = 0, E = 0
2. V= 0, E ≠ O
3. V ≠ 0, F = 0
4. V ≠ 0, E ≠ 0

Answer:

1. V = 0, E = 0

2. V= 0, E ≠ O

3. V ≠ 0, F = 0

4. V ≠ 0, E ≠ 0

Electric potential class 12 MCQs

Question 16. Which of the following statement (s) is/are correct?

1. If the electric Hold duo lo a point charge varies aa r-2.5 Instead of r-2, then Gatins’ law will still ho valid
2. Gatina’s law can ho linked to calculating holding around an electric dipole
3. If the electric field at some point between two points charges In zero, then the algo of two charged la the name
4. The work done by the external force In moving a unit positive charge from point A at potential V_A to point B at potential V_B is (V_B– V_A)

Answer:

3. If the electric field at some point between two points charges In zero, then the algo of two charged la the name

4. The work done by the external force In moving a unit positive charge from point A at potential VA to point B at potential V_B is (V_B – V_A)

Question 17. Six-point charges are kept at the vertices of a regular hexagon of side L, and center O. Given that \(K=\frac{1}{4 \pi \epsilon_0} \frac{q}{L^2}\). Which of the following statements is correct?

1. The electric field at O is 6K along OD
2. Potential at O Is zero
3. Potential at all points on the line PR Is the same
4. Potential at all points on the Lino ST Is the same

Answer:

1. Electric field at O is 6K along OD

2. Potential at O Is zero

3. Potential at all points on the line PR Is the same

Question 18. Some of the field lines correspond to an electric field. If E = electric field and V = potential, then

1. E_A > E_B
2. B_A < E_B
3. V_A > V_B
4. V_A < V_B

Answer:

1. E_A > E_B

4. V_A < V_B

Question 19. The number of statvolt corresponding to 1 volt is

1. \(\frac{1}{100}\)
2. 10⁹
3. \(\frac{1}{300}\)
4. 300

Answer: 3. \(\frac{1}{300}\)

Question 20. 1V m^-1 =?

1. 1N.C^-1
2. 3 x 10¹⁰ N.C^-1
3. 10⁷N.C^-1
4. 10¹⁰ N.C^-1

Answer: 1. 1N.C^-1

Question 21. When a body is connected to the earth, electrons flow from the earth to the body. The body is

1. Negatively charged
2. Insulator
3. Uncharged
4. Positively charged

Answer: 4. Positively charged

Electric potential class 12 MCQs

Question 22. In a uniform electric field of intensity E, a small charge q is carried once along the elliptical path ABCD. The amount of work in the paths AB, BC, CD and DA are respectively W₁, W₂, W₃, W₄, then

1. W₁ = W₂ = W₃ = W₄ ≠ 0
2. W₁ = W₂ = W₃ = W₂ = 0
3. W₁ + W₂ + W₃ + 4 = 0
4. W₁ = W₂+ W₃-W₄ = 0

Answer: 3. W₁ + W₂ + W₃ + W₄ = 0

Electric Potential Revision MCQs for Class 12

Question 23. Four point changes each + q is placed on the circumference of a circle of diameter 2d in such a way that they form a square. The potential at the centre of the circle (in CGS) is

1. 0
2. \(\frac{4q}{d}\)
3. \(\frac{4d}{q}\)
4. \(\frac{q}{4d}\)

Answer: 2. \(\frac{4q}{d}\)

Question 24. Work done in taking a point charge from P to A is W_A, from P to B is W_B and from P to C is W_C. Which of the following options is correct?

1. W_A < W_B < W_C
2. W_A>W_B>W_C
3. W_A = W_B = W_C
4. None Of These

Answer: 3. W_A = W_B = W_C

Question 25. The radius of a soap bubble whose potential is 16 V is doubled. The new potential of the bubble is

1. 2 V
2. 4 V
3. 8 V
4. 16 V

Answer: 3. 8 V

Question 26. Four electric charges +q, +q,- q, and- q are placed at the corners of a square of side 2L. The electric potential at point A midway between the two charges + q and +q is

1. \(\frac{1}{4 \pi \epsilon_0} \frac{2 q}{L}(1+\sqrt{5})\)
2. \(\frac{1}{4 \pi \epsilon_0} \frac{2 q}{L}\left(1+\frac{1}{\sqrt{5}}\right)\)
3. \(\frac{1}{4 \pi \epsilon_0} \frac{2 q}{L}\left(1-\frac{1}{\sqrt{5}}\right)\)
4. Zero

Answer: 3. \(\frac{1}{4 \pi \epsilon_0} \frac{2 q}{L}\left(1-\frac{1}{\sqrt{5}}\right)\)

Question 27. A charge Q is placed at the center of the circle ABCDE. Work done to bring another charge q from A to B, C, D, and E is

1. Minimum along the path AB
2. Maximum along the path AD
3. Zero along each of the paths AB, AC, AD, AE
4. Positive but equal along all the paths

Answer: 3. Zero along each of the paths AB, AC, AD, AE

Electric potential class 12 MCQs

Question 28. Work done to carry a charge q once along the circular path of radius r with charge q’ at its centre, will be

1. 0
2. \(\frac{q q^{\prime}}{4 \pi \epsilon_0}\left(\frac{1}{\pi r}\right)\)
3. \(\frac{q q^{\prime}}{4 \pi \epsilon_0}\left(\frac{1}{2 \pi r}\right)\)
4. \(\frac{q q^{\prime}}{4 \pi \epsilon_0 r}\)

Answer: 1. \(\frac{q q^{\prime}}{4 \pi \epsilon_0}\left(\frac{1}{\pi r}\right)\)

Question 29. There is a uniform electric field of intensity E. How many labelled points do have the same potential as the fully shaded point? ‘

1. 2
2. 3
3. 8
4. 11

Answer: 2. 3

Question 30. If a charge is displaced against the Coulomb force in an electric field

1. Work is flooded by the electrical force
2. Work is done by an electric agency
3. The energy of the electric, field decreases
4. The energy of the system decreases

Answer: 2. Work is done by an electric agency

Question 31. Three point charges each equal to q are placed on the vertices of an equilateral triangle of side 1. The potential energy of the system is

1. \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{q^2}{l}\)
2. \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{2 q^2}{l}\)
3. \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{3 q^2}{l}\)
4. \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{4 q^2}{l}\)

Answer: 3. \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{3 q^2}{l}\)

Question 32. There is a charge of 10μC at the centre of a circle of radius 10 m. The work done in moving a unit positive charge once around the circle is

1. 0
2. 100 J
3. 10 J
4. 150 J

Answer: 1. 0

Class 12 physics electric potential questions 

Question 33. An electron of mass m and charge e is accelerated from rest through a potential difference V in a vacuum. Its final speed will be

1. \(\sqrt{\frac{2 e V}{m}}\)
2. \(\sqrt{\frac{2 V}{m}}\)
3. \(\frac{e V}{m}\)
4. \(\frac{e V}{2m}\)

Answer: 1. \(\sqrt{\frac{2 e V}{m}}\)

Question 34. The radius of a charged hollow metal sphere is 5 cm and its surface potential is 10 V. Potential at its centre is

1. 0
2. 10V
3. Equal to the potential at k distance 5 cm away from its surface
4. Equal to the potential at a distance 25 cm away from its surface

Answer: 2. 10V

Question 35. The variation of electric potential with distance from a fixed point. What is the value of electric field intensity at x = 2 m?

1. 0
2. 3 V.m^-1
3. 6 V.m^-1
4. 2 V.m^-1

Answer: 2. 6 V.m^-1

Question 36. If the equation of electric field is \(\vec{E}=(y \hat{i}+x \hat{j})\), then the equation of electric potential is represented by

1. V = -(x + y) + constant
2. V = constant
3. V = -(x² + y²) + constant
4. V = – xy + constant

Answer: 4. V = – xy + constant

Question 37. A charge of 3C experiences a force of 3000 N in a uniform electric field. The potential difference between two points situated 1 cm apart along the electric lines of force will be

1. 300 V
2. 100 V
3. 30 V
4. 10 V

Answer: 4. 10V

Class 12 physics electric potential questions 

Question 38. The electric potential V at any point (x, y, z) all in metres in space is given by V = 4 x 2V. The electric field at the point (1, 0, 2) in V.m^-1 is

1. 8 along the negative X-axis
2. 8 along the positive X-axis
3. 16 along the negative X-axis
4. 16 along the positive X-axis

Answer: 1. 8 along the negative X-axis

Question 39. Two points A and B are 2 cm apart and a uniform electric field 1? acts along the straight line AB directed from A to B with E = 200 N.C^-1. A particle of charge +10^-6C is taken from A to B along AB.

1. The force on the charge is

1. 2 x 10^-4 N
2. 3N
3. 2N
4. 2x 10^-2 N

Answer: 1. 2 x 10^-4 N

2. The potential difference between A and B is

1. 4 x 10^-6 V
2. 1 V
3. 4V
4. 2 V

Answer: 3. 4V

3. The work done on the charge by \(\vec{E}\) is

1. 4J
2. 1 x 10^-6 J
3. 2 x 10^-6 J
4. 4 x 10^-6 J

Answer: 4. 4 x 10^-6 J

Question 40. Two point charges q₁ = 10 x 10^-6C and q₂ = -2 X 10^-8C are separated by a distance of 6 cm in air.

1. The distance from q1 of the point of zero electric potential is

1. 5 cm
2. 2.5 cm
3. 8 cm
4. 10 cm

Answer: 1. 5 cm

2. The electrostatic potential energy of the system is

1. 15 X 10^-3J
2. 24 x 10^-4 J
3. 18 X 10^-3 J
4. -18 X 10^-4 J

Answer: 4. -18 X 10^-4 J

WBCHSE Physics Electric Potential Question Bank

Question 41. A positively charged oil drop is in equilibrium in the electric field existing in the space between two horizontal plates separated by a distance of 1 cm. The charge of the oil drop is 3.2 x 10^-19C and its mass is 10^-17 g.

1. The potential difference between the plates is

1. 1020 V
2. 1531 V
3. 3062 V
4. 2454 V

Answer: 3. 3062 V

2. The instantaneous acceleration of the oil drop, when the polarity of the two plates is reversed, is

1. 24.54 m.s^-2
2. 19.60 m s^-2
3. 29.70 m s^-2
4. 34.52 m.s^-2

Answer: 2. 19.60 m.s^-2

Question 42. Each of the two concentric spheres of radii 5 cm and 10 cm are given a charge of 10μC.

1. The electric potential at a point situated at a distance of 2.5 cm from the centre is

1. 27 x 10⁵ V
2. 0V
3. 16.5 x 10⁵ V
4. 24.2 X 10⁴ V

Answer: 1. 27 x 10⁵ V

2. The electric potential at a point situated at a distance of 8 cm from the centre is

1. 10.52 X 10⁴V
2. 20.25 x 10⁵ V
3. 32.24 X 10⁵ V
4. 42.28 X 10⁵V

Answer: 2. 20.25 x 10⁵ V

3. The electric potential at a point situated at a distance of 20 cm from the centre is

1. 4 x 10⁵ V
2. 32 x 10⁴ V
3. 9 X 10⁵ V
4. 16 X 10⁵ V

Answer: 3. 9 X 10⁵ V

Question 43. The plot of the x component of the electric field as a function of x in a certain region. The y and z components of the electric field are zero in this region. The electric potential at the origin is 10 V

1. Electric potential at x = 2 m is

1. 10V
2. 20V
3. 30V
4. 40V

Answer: 3. 30V

2. The greatest positive value of electric potential for any point in the region 0 ≤ x ≤ 6 m on the x-axis is

1. 10V
2. 20V
3. 30V
4. 40V

Answer: 4. 40V

3. The value of x for which potential is zero is

1. 2m
2. 3m
3. 4m
4. 5.5 m

Answer: 4. 5.5 m

Question 44. 64 tiny drops of water having the same radius and same charge are combined to form one large drop. The ratio of the potential of the large drop to the small drop is

1. 4:1
2. 1:4
3. 16:1
4. 1:16

Answer: 3. 16:1

Potential of one tiny drop, \(V_1=\frac{1}{4 \pi \epsilon_0} \frac{q}{r}\)

Now, a charge of the large water drop = 64q

The ratio of the volume of the tiny water drop to the volume of the large water drop = 1: 64

Then, ratio of their radii = \(1: \sqrt[3]{64}\)

= 1: 4

∴ The potential of the large water drop

⇒ \(V_2=\frac{1}{4 \pi \epsilon_0} \frac{64 q}{4 r}=16 V_1\)

i.e., \(\frac{V_2}{V_1}=\frac{16}{1}\)

= 16: 1

The option 3 is correct

WBCHSE physics electric potential MCQs 

Question 45. An infinite sheet carrying a uniform surface charge density σ lies on the xy-plane. The work done to carry a charge q from point \(\vec{A}=a(\hat{i}+2 \hat{j}+3 \hat{k})\) to the point \(\vec{B}=a(\hat{i}-2 \hat{j}+6 \hat{k})\)(where a is a constant with the dimension of length and ∈₀ is the permittivity of free space) is

1. \(\frac{3 \sigma a q}{2 \epsilon_0}\)
2. \(\frac{2 \sigma a q}{\epsilon_0}\)
3. \(\frac{5 \sigma a q}{2 \epsilon_0}\)
4. \(\frac{3 \sigma a q}{\epsilon_0}\)

Answer: 1. \(\frac{3 \sigma a q}{2 \epsilon_0}\)

Distance between the points, \(\vec{r}=\vec{B}-\vec{A}=a(-4 \hat{j}+3 \hat{k})\)

Field intensity at a point due to an infinite sheet carrying a uniform surface charge density cr lies on the xy-plane,

⇒ \(\vec{E}=\frac{\sigma}{2 \epsilon_0} \hat{k}\)

∴ Electric force acting on the charge q, \(q, \vec{F}=q \vec{E}=\frac{q \sigma}{2 \epsilon_0} \hat{k}\)

∴ Work done, \(W=\vec{F} \cdot \vec{r}=\frac{q \sigma}{2 \epsilon_0} \hat{k} \cdot a(-4 \hat{j}+3 \hat{k})\)

= \(\frac{3 q \sigma a}{2 \epsilon_0}\)

The option 1 is correct

Question 46. The angle between an equipotential surface and electric lines of force is

1. 0°
2. 90°
3. 180°
4. 270°

Answer: 2. 90°

The option 2 is correct.

Question 47. A point charge -q is carried from point A to point B on the axis of a charged ring of radius r carrying a charge +q. If point A is at a distance|r from the centre of the ring and point B is|r from the centre but on the opposite side, what is the network that needs to be done for this?

1. \(-\frac{7}{5} \frac{q^2}{4 \pi \epsilon_0 r}\)
2. \(-\frac{1}{5} \frac{q^2}{4 \pi \epsilon_0 r}\)
3. \(\frac{7}{5} \frac{q^2}{4 \pi \epsilon_0 r}\)
4. \(\frac{1}{5} \frac{q^2}{4 \pi \epsilon_0 r}\)

Answer: 2. \(-\frac{1}{5} \frac{q^2}{4 \pi \epsilon_0 r}\)

Net work done (W) = electric potential energy at point B – electric potential energy at point A

or, \(W=\frac{1}{4 \pi \epsilon_0} \cdot \frac{(+q)(-q)}{\sqrt{r^2+\left(\frac{3}{4} r\right)^2}}-\frac{1}{4 \pi \epsilon_0} \cdot \frac{(+q)(-q)}{\sqrt{r^2+\left(\frac{4}{3} r\right)^2}}\)

⇒ \(-\frac{1}{5} \cdot \frac{q^2}{4 \pi \epsilon_0 r}\)

The option 2 is correct.

Question 48. Two positive charges Q and 4Q are placed at points A and B respectively, where B is at a distance of d units to the right of A. The total electric potential due to these charges is minimum at P on the line through A and B. What is (are) the distance(s) of P from A?

1. \(\frac{d}{3}\) units to the right of A
2. \(\frac{d}{3}\) units to the left of A
3. \(\frac{d}{5}\) units to the right of A
4. d units to the left of A

Answer: 1. \(\frac{d}{3}\) units to the right of A

Let a unit positive charge be placed at point P between points A and B and at a distance x from A

Total potential at point P due to charges Q and 4Q,

⇒ \(V=\frac{Q}{x}+\frac{4 Q}{(d-x)}\)

For the potential to be minimum at point P

⇒ \(\frac{d V}{d x}=0 \quad\)

or, \(\frac{d}{d x}\left[\frac{Q}{x}+\frac{4 Q}{d-x}\right]=0 \quad\)

or, \(x=\frac{d}{3}\)

∴ The electric potential is minimum at a distance \(\frac{d}{3}\) from point A on its right.

The option 1 is correct.

Question 49. Assume that an electric field \(\vec{E}=30 x^2 \hat{i}\) exists in space. Then the potential difference V_A– V_O, where V_O is the potential at the origin and VA the potential at x – 2 m is

1. 80 J
2. 120 J
3. -120 J
4. -80 J

Answer: 4. -80 J

We know, \(\vec{E}=-\frac{d V}{d x} \hat{i} \quad\)

or, \(\int_{V_O}^{V_A} d V=-\int_0^2 30 x^2 d x\)

∴ \(V_A-V_O=-30\left[\frac{x^3}{3}\right]_0^2\)

= -80 J

The option 4 is correct.

Electric Potential Assessment Questions for Class 12

Question 50. A uniformly charged solid sphere of radius R has potential V_O (measured with respect to ∞ ) on its surface. For this sphere the equipotential surfaces with potentials \(\frac{3 V_0}{2}\), \(\frac{5 V_0}{4}, \frac{3 V_0}{4} \text { and } \frac{V_0}{4}\) have radius R1, R2, R3 and R4 respectively. Then

1. R₁= 0 and R₂ > (R₄– R₃)
2. R₁ ≠ 0 and (R₂-R1)>(R₄-R₃)
3. R₁ = 0 and R₂ < (R₄– R₃)
4. 2R < R₄

Answer: 3. R₁ = 0 and R₂ < (R₄– R₃)

⇒ \(V_0=\frac{1}{4 \pi \epsilon_0} \frac{q}{R}\) [q = charge of tire sphere]

Now, \(\frac{3 V_0}{4}, \frac{V_0}{4}<V_0\)

So, these equipotential surfaces will be situated outside the sphere

⇒ \(R_3, R_4>R\)

In that case,

⇒ \(\frac{3 V_0}{4}=\frac{1}{4 \pi \epsilon_0} \frac{q}{R_3} \quad\)

or, \(\frac{3}{4} \frac{1}{4 \pi \epsilon_0} \frac{q}{R}=\frac{1}{4 \pi \epsilon_0} \frac{q}{R_3}\)

i,e., \(R_3=\frac{4}{3} R\)

Againg \(\frac{V_0}{4}=\frac{1}{4} \frac{1}{4 \pi \epsilon_0} \frac{q}{R}\)

= \(\frac{1}{4 \pi \epsilon_0} \frac{q}{R_4} \quad\)

i.e., \(R_4=4 R\)

∴ \(R_4-R_3=\frac{8}{3} R\)

On the other hand, \(\frac{3 V_0}{2}, \frac{5 V_0}{4}>V_0\); these equipotential surfaces are situated inside the sphere R1, R2<R. Potential at any point inside the sphere at a distance r(r<R) from the center of the sphere,

⇒ \(V=\frac{1}{4 \pi \epsilon_0} \frac{q}{2 R^3}\left(3 R^2-r^2\right)\)

= \(\frac{V_0}{2 R^2}\left(3 R^2-r^2\right)\)

∴ In the first case,

⇒ \(\frac{3}{2} V_0=\frac{V_0}{2 R^2}\left(3 R^2-R_1^2\right) \quad \text { or, } R_1=0\)

In the second case,

⇒ \(\frac{5}{4} V_0=\frac{V_0}{2 R^2}\left(3 R^2-R_2^2\right) \quad \text { or, } R_2=\frac{1}{\sqrt{2}} R\)

The option 3 is correct.

Question 51. Three concentric metal shells A, B and C of respective radii a, b and c (a<b<c) have surface charge densities +σ, -σ and +σ respectively. The potential of shell B is

1. \(\frac{\sigma}{\epsilon_0}\left[\frac{b^2-c^2}{b}+a\right]\)
2. \(\frac{\sigma}{\epsilon_0}\left[\frac{b^2-c^2}{c}+a\right]\)
3. \(\frac{\sigma}{\epsilon_0}\left[\frac{a^2-b^2}{a}+c\right]\)
4. \(\frac{\sigma}{\epsilon_0}\left[\frac{a^2-b^2}{b}+c\right]\)

Answer: 4. \(\frac{\sigma}{\epsilon_0}\left[\frac{a^2-b^2}{b}+c\right]\)

The potential of shell B,

⇒ \(V_B=\frac{1}{4 \pi \epsilon_0}\left[\frac{(+\sigma) 4 \pi a^2}{b}+\frac{(-\sigma) 4 \pi b^2}{b}+\frac{(+\sigma) 4 \pi c^2}{c}\right]\)

⇒ \(\frac{\sigma}{\epsilon_0}\left[\frac{a^2-b^2}{b}+c\right]\)

The option 4 is correct.

Question 52. A conducting sphere of radius R is given a charge Q . The electric potential and the electric field at the centre of the sphere respectively are

1. \(\text { zero and } \frac{Q}{4 \pi \epsilon_0 R^2}\)
2. \(\frac{Q}{4 \pi \epsilon_0 R} \text { and zero }\)
3. \(\frac{Q}{4 \pi \epsilon_0 R} \text { and } \frac{Q}{4 \pi \epsilon_0 R^2}\)
4. Both are Zreo

Answer: 2. \(\frac{Q}{4 \pi \epsilon_0 R} \text { and zero }\)

WBCHSE physics electric potential MCQs 

Question 53. In a region, the potential is represented by V(x, y, z) = 6x-8xy-8y+6yz where V is in volts and x, y, z are in metres. The electric force experienced by a charge of 2 coulomb situated at point (1, 1, 1) is

1. 6√5N
2. 30 N
3. 24 N
4. 4√35N

Answer: 4. 4√35N

We know,

⇒ \(\vec{E}=E_x \hat{i}+E_y \hat{j}+E_z \hat{k}\)

= \(-\frac{d V}{d x} \hat{i}-\frac{d V}{d y} \hat{j}-\frac{d V}{d z} \hat{k}\)

Now, V(x, y,z) = 6x – 8xy – 8y + 6yz

∴ \(\frac{d V}{d x}=6-8 y ; \frac{d V}{d y}=-8 x-8+6 z ; \frac{d V}{d y}=6 y\)

∴ \(\vec{E}=-(6-8 y) \hat{i}-(-8 x-8+6 z) \hat{j}-6 y \hat{k}\)

∴ Electric field at point (1, 1, 1),

⇒ \(\vec{E}=-(6-8 \times 1) \hat{i}-(-8 \times 1-8+6 \times 1) \hat{j}-(6 \times 1) \hat{k}\)

⇒ \(2 \hat{i}+10 \hat{j}-6 \hat{k}\)

⇒ \(E=\sqrt{2^2+10^2+(-6)^2}=\sqrt{140}=2 \sqrt{35}\)

∴ Force acting on charge 2 C,

F = qE

= 2 x 2√35

= 4√35 N

The option 4 is correct

Question 54. A molecule of a substance has permanent dipole moment p. A mole of this substance is polarised by applying a strong electrostatic field E. The direction of the field is suddenly changed by an angle of 60°. If (V is Avogadro’s number the amount of work done by the field is

1. 2NpE
2. \(\frac{1}{2}\)
3. NpE
4. \(\frac{3}{2}\)

Answer: 2. \(\frac{1}{2}\)

The potential energy of electric dipole (\(\vec{p}\)) in an electric field \((\vec{E})=-\vec{p} \cdot \vec{E}=-p E \cos \theta\)

So, the work done by the field on N molecules

= change in potential energy

= -NpE( cos60°- cosθ°) = \(\frac{1}{2}\) NpE

The option 2 is correct.

Unit 1 Electrostatics Chapter 3 Electric Potential Long Question and Answers

Question 1. If the intensity at a point in an electric field is zero, will the electric potential be also zero at that point? If the electric potential be zero at a point, will the Intensity be also zero at that point?
Answer:

If the intensity of the electric field is E and the potential is V, then the relation between them is, E = – \(\frac{dV}{dx}\)

So, if E = 0 at any point, we have \(\frac{dV}{dx}\)= 0 or, V = constant.

Thus, the potential has a constant value, not necessarily zero, around that point. Now, for example, we consider an electric dipole (q,-q) placed at AB along the x-axis.

The variation of the potential V with the dis- Answer: We know that charges of a conductor reside on its outer stance x is shown in the figure. At the surface.

Placing the smaller conductor inside the bigger one and mid-point O, clearly, V = 0; but the connecting them with a wire, we can prepare a single conductor.

The slope of the V-x curve is not zero, i.e., \(\frac{dV}{dx}\) not = 0. As E = – \(\frac{dV}{dx}\), we Thus charge from the smaller conductor will then flow to the can conclude that E not = 0 at O.

Therefore, if the potential is zero at a point, but it is not a constant in the immediate neighbourhood of that point, the electric field intensity is non-zero there.

Read And Learn More WBCHSE Class 12 Physics Long Question And Answers

In particular, the potential due to a dipole at any point on the perpendicular bisector is zero, but the field at that point is certainly non-zero being directed opposite to the direction of the dipole moment.

Question 2. Two hollow conductors are charged positively. The potential of the smaller conductor is 50 V and that of the larger conductor is 100 V. How are these two conductors to be placed so that when connected by a wire, charges will flow from the smaller conductor to the larger one?
Answer:

Two hollow conductors are charged positively. The potential of the smaller conductor is 50 V and that of the larger conductor is 100 V.

We know that the charges of a conductor reside on its outer surface. Placing the smaller conductor inside the bigger one and connecting them with a wire, we can prepare a single conductor. Thus charges from the smaller conductor will then flow to the outer surface of the larger one.

Question 3. Explain the variations of potential and field Intensity with distance due to a hollow charged spherical conductor (radius a and charge q ), both inside and outside the sphere, with a graph.
Answer:

Electric potential: Potential at all points inside a hollow charged spherical conductor and on its surface are equal. Here this potential is, V = ⇒ \(\frac{q}{a}\). So, from the centre to the surface of the tire conductor, the potential is constant and it has been shown by the line AB, parallel to the x-axis.

If the distance of an external point from the centre of the conductor be r, then electric potential at that point is given by, So as r increases, V decreases. This decrease has been shown by the curved line BC. The inverse variation is illustrated by the moderate slope of the curve.

Electric intensity: As the potential at all points inside a hollow charged spherical conductor is constant, intensity at those points, i.e., inside the hollow spherical conductor is zero.

As the radius of the conductor is an intensity at a point on its surface, \(E=\frac{q}{a^2}\). If r is the distance of an external point from the centre of the sphere, the intensity at that point will be \(E=\frac{q}{a^2}\)

So, intensity on the surface of the conductor is maximum and it decreases as we move away from the conductor. This variation has been shown by the curved line AB.

The inverse square variation has been shown by the steeper slope, as compared to the V-r curve.

Electric Potential Theory Long Questions WBCHSE

Question 4. The potential difference between two conductors is very large. What will happen under the following three conditions?

1. The conductors are connected by a metallic wire,
2. Both positive and negative ions are present in the air medium in between the conductors,
3. The conductors are placed in a vacuum.

Answer:

1. Charge will continue to flow till the potentials of the two conductors become equal.

2. Positive ions will move to the conductor having lower potential and negative ions will move to the conductor having higher potential.

3. Charges on the conductors will not move and the potential difference between them will be maintained.

Question 5. Give an Example of a line In the electric field of an electric dipole along which a positive charge may be moved without any work being done.
Answer:

There are numerous equipotential surfaces in the electric field of a dipole. Any movement of a charge, on any of these surfaces, would involve no work. For Example, the plane perpendicular to the dipole and passing through its mid-point is an equipotential, actually zero-potential surface. So if a positive charge is moved along any line on this plane, no work is to be done.

Electric Potential Long Answer Questions WBCHSE

Question 6. Can two different equipotential surfaces intersect each other?
Answer:

If two different equipotential surfaces intersected each other, the potential would have two different values at the point of intersection. Moreover, there would be two normals on the two surfaces at the point of intersection. That would mean two elec-/ trie fields in two different directions at the same point. These situations are absurd. So two different equipotential surfaces cannot intersect each other.

So, the intensity on the surface of the conductor is maximum and it decreases as we move away from the conductor. This variation has been shown by the curved line AB. The inverse square variation has been shown by the steeper slope, as compared to the V-r curve.

Question 7. Two pairs of planes A, B and A’, B’ are kept in two uniform electric fields. The potential dx difference between the planes In the two cases are equal. Which pair of planes Is situated in a stronger electric field?

Answer:

Since E = – \(\frac{dV}{dx}\); therefore, dx = – \(\frac{dV}{E}\)

So it is seen that dx will be smaller if E is stronger.

Therefore, planes A and B are in a stronger electric field.

Question 8. A positively charged and a negatively charged body are connected to the earth separately. What will be their potential before and after the connection?
Answer:

A positively charged and a negatively charged body are connected to the earth separately.

Before connection with the earth, the positively charged body will have positive potential and the negatively charged body will have negative potential. But after connection with the earth both the bodies will have zero potential.

Question 9. When the body Is connected with the earth, electrons are found to move from the earth to the body. What Is Your Idea about the nature of the charge of the body?
Answer:

When the body Is connected with the earth, electrons are found to move from the earth to the body.

Since electrons are moving from the earth to the body, Its potential Is positive. As the body has positive potential, it Is positively charged.

Question 10. In a vacuum, equal charges of the same nature are placed at the vertices of an equilateral triangle. What will be the electric field Intensity and potential at the centroid of the triangle
Answer:

In a vacuum, equal charges of the same nature are placed at the vertices of an equilateral triangle.

Let three charges, each equal to +q, be placed at the vertices of the equilateral triangle. The length of the side of the triangle is r.

We know that the centroid of the triangle G divides the median of the n triangle In (the ratio 2:1).

Height of the equilateral triangle

⇒ \(\frac{\sqrt{3}}{2} r\)

So, the potential at G Is,

⇒ \(V=\frac{q}{4 \pi \epsilon_0\left(\frac{2}{3} \cdot \frac{\sqrt{3} r}{2}\right)} \times 3\)

= \(\frac{3 q}{4 \pi \epsilon_0 \cdot \frac{r}{\sqrt{3}}}\)

= \(\frac{1}{4 \pi \epsilon_0} \frac{3 \sqrt{3} q}{r}\)

If E be the electric field intensity at G due to each charge +q (HI and if they are resolved into two perpendicular components, we see that the two components, ‘Ecos30° are equal and opposite balance each other.

Again 2sin30° \(\left(=\frac{2 E}{2}=E\right)\) and F being equal and opposite, balance each other. So the electric field Intensity at the centroid of the angle is zero.

In-depth Questions on Electric Potential for Class 12

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 11. Draw three equipotential surfaces corresponding to a field that uniformly Increases In magnitude but remains constant along the z-direction. How are these surfaces different from those of u constant electric field along the z-direction?
Answer:

The separation between the equipotential surfaces will gradually decrease along the direction of the Increment of the electric field (in this case z-axis).

In the case of a constant uniform electric field, the equipotential surfaces are equally spaced.

Question 12. A unit positive charge is moving along PQR in an electric field of intensity E. What is the potential difference between the points P and R?

Answer:

A unit positive charge is moving along PQR in an electric field of intensity E.

The work done to move a test charge from one point to another is Independent of the path followed.

⇒ \(V_{P R}=W_{P R}=-\int \vec{E} \cdot d \vec{l}=-\int_P^R E d l \cos 180^{\circ}\)

or, V_VR = Er

i.e., the potential difference is r times of the electric field intensity.

Question 13. The variation of electric potential V for two charges Q₁ and Q₂ (where r Is the distance from a point charge).

1. What are the natures of the charges Q₁ and Q₂?
2. Which of the two charges Is greater In magnitude? Justify

Answer:

1. Electric potential due to a point charge Q is given by

⇒ \(V=\frac{1}{4 \pi \epsilon_0} Q\)

V is positive if Q is positive and V is negative if Q is negative.

Therefore, Q₁ is a positive charge and Q₂ is a negative charge.

2. Since V₂ > V₁, so charge Q₂ Is greater In magnitude as compared to charge Q₁

Long Form Questions about Electric Potential WBCHSE

Question 14. Two electric charges q and -2q ore placed 6 m apart on a horizontal plane. Find the locus of any point on this plane where the potential has a value zero.
Answer:

Two electric charges q and -2q ore placed 6 m apart on a horizontal plane.

The potential at P due to the charges q and -2q,

⇒ \(V=\frac{1}{4 \pi \epsilon_0}\left(\frac{q}{r_1}-\frac{2 q}{r_2}\right)\)

But V = 0 (given)

∴ \(\frac{1}{r_1}=\frac{2}{r_2} \quad\)

or, \(r_2=2 r_1 \quad\)

or, \(r_2^2=4 r_1^2\)

or, y² + (6-x)² = 4(x²+y²)

or, y² + 36 + x²- I2x = 4x² + 4y²

or, 3x² + 12x + 3y² = 36

or, x² + 4x+ y² = 12

or, (x + 2)² + y² = 42

or, [x-(-2)]² + (y-0)² = 42

Therefore, the locus of point P is a circle whose centre is at (-2, 0) and whose radius is 4 m.

Question 15. A charge Q Is uniformly distributed over a long rod YA AB of length L as shown In the figure. Determine the electric potential at the point O lying at a distance L from the end A.

Answer:

A charge Q Is uniformly distributed over a long rod YA AB of length L as shown In the figure.

Potential at O,

⇒ \(V=\frac{1}{4 \pi \epsilon_0} \int_0^L \frac{d q}{L+x}\)

⇒ \(\left.\frac{1}{4 \pi \epsilon_0} \int_0^L \frac{\lambda d x}{L+x} \text { [linear charge density, } \lambda=\frac{d q}{d x}=\frac{Q}{L}\right]\)

⇒ \(\frac{\lambda}{4 \pi \epsilon_0} \int_0^L \frac{d x}{L+x}\)

= \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{L} \ln \frac{L+L}{L}\)

= \(\frac{Q \ln 2}{4 \pi \epsilon_0 L}\)

Electric Potential Concepts and Long Answers WBCHSE

Question 16. A conducting of radius a, thickness l(<<a) has a potential V. Now the bubble transforms into a droplet. Find the potential on the surface of the droplet. Answer t Let q be the charge on the bubble. The potential on the surface of the bubble, V =
Answer:

A conducting of radius a, thickness l(<<a) has a potential V. Now the bubble transforms into a droplet.

Let q be the charge on the bubble

The potential on the surface of the bubble, \(V=\frac{1}{4 \pi c_0} \frac{q}{a}\)

∴ \(q=\frac{V a}{k}\left[\text { where } k=\frac{1}{4 \pi \epsilon_0}\right]\)

Let R be the radius of the droplet.

By comparing their volumes, we get

⇒ \(\left(4 \pi a^2\right) t=\frac{4}{3} \pi R^3\)

∴ \(R=\left(3 a^2 t\right)^{1 / 3}\)

Now potential on the surface of the droplet becomes

⇒ \(V_1=\frac{k q}{R}\)

or, \(V_1=k \frac{\frac{V a}{k}}{\left(3 a^2 t\right)^{1 / 3}}\) [from equation (1)]

⇒ \(V_1=V\left(\frac{a}{3 t}\right)^{1 / 3}\)

Question 17. Some equipotential surfaces. What can you say about the magnitude and the direction of the electric field A 30 cm Intensity?

Answer:

The electric field is always perpendicular to the equipotential surface. Also, potential decreases along the direction of electric field intensity.

The electric field makes an angle of 120° with the X-axis.

The magnitude of the electric field along X -axis,

⇒ \(E \cos 120^{\circ}=-\frac{(20-10)}{(20-10) \times 10^{-2}}\)

or, \(-E \cdot \frac{1}{2}=-\frac{10}{0.10}\)

E = 200 V/m

The direction of the electric field is radially outward.

Here, rV = constant = 6 V.m hypotenuse,

Thus potential at any distance r from the centre,

V(r) = \(\frac{6}{r}\)

Hence,\(E=-\frac{d V}{d r}=\frac{6}{r^2} \mathrm{~V} / \mathrm{m}\)

WBCHSE Electric Potential Explanations and Solutions

Question 18. The radii of two concentric metal spheres are a and b (b>a). The outer sphere Is charged with a charge q. If the Inner sphere is connected to the earth, what will be its charge?
Answer:

The radii of two concentric metal spheres are a and b (b>a). The outer sphere Is charged with a charge q.

Let the charge of the inner sphere be Q.

Given that the charge of the outer sphere is q.

Radii of the inner and outer spheres are a and b respectively.

As the inner sphere is earthed, the potential is zero.

Thus tire net potential is

⇒ \(\frac{1}{4 \pi \epsilon_0} \frac{Q}{a}+\frac{1}{4 \pi \epsilon_0} \frac{q^{\circ}}{b}=0\) [∵ \(\frac{1}{4 \pi \epsilon_0}\) is constant, so it is not equal to zero.]

∴ \(\frac{Q}{a}=-\frac{q}{b}\)

or, \(Q=-q \frac{a}{b}\)

Therefore charge of the inner sphere is \(-q \frac{a}{b}\).

Question 19. +Q, +q, +q charge θ arc placed on the vertices of an Isosceles right-angled triangle. If a the electric potential energy of the system of charges Is zero, what will +q be the value of Q?

Answer:

Electric potential energy due to Q.and q through the hypotenuse,

⇒ \(U_1=\frac{1}{4 \pi \epsilon_0}\left[\frac{Q \cdot q}{\sqrt{2 a}}\right]\)

Similarly, potential energy due to Q and q through the perpendicular

⇒ \(U_2=\frac{1}{4 \pi \epsilon_0}\left[\frac{Q \cdot q}{a}\right]\)

and potential energy due to q and q through the base,

⇒ \(U_3=\frac{1}{4 \pi \epsilon_0}\left[\frac{q \cdot q}{a}\right]\)

Thus the net potential energy,

⇒ \(U=U_1+U_2+U_3\)

= \(\frac{1}{4 \pi \epsilon_0}\left[\frac{Q q}{\sqrt{2} a}+\frac{Q q}{a}+\frac{q q}{a}\right]=0\)

or, \(\frac{Q q}{\sqrt{2} a}[1+\sqrt{2}]=-\frac{q q}{a}\)

or, \(Q=-\frac{\sqrt{2} q}{1+\sqrt{2}}\)

∴ \(Q=-\frac{2 q}{\sqrt{2}+2}\)

Long Answer Format for Electric Potential Topics

Question 20. Discuss the variation of electric potential due to positive and negative point charges with distance from a charge.
Answer:

Electric potential at a distance r from a charge +q ,

⇒ \(V=\frac{+q}{4 \pi \epsilon_0 r}\)

Electric potential at a distance r from a charge -q, \(V=\frac{-q}{4 \pi \epsilon_0 r}\)

The variation of V with r. From the nature of the curves it can be said that the electric potential gradually decreases as the distance from charge +q increases and it gradually increases as the distance from charge -q increases.

Unit 1 Electrostatics Chapter 3 Electric Potential Short Answer Questions

Question 1. A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of -2 x 10^-9 C from a point. P(0, 0, 3 cm) to a point Q(0, 4 cm,0), via a point R (0, 6 cm, 9 cm).

Answer:

A charge of 8 mC is located at the origin.

Work done depends only on the linear distance between the initial and final position of the particle.

∴ Work done

⇒ \(=\frac{q_1 q_2}{4 \pi \epsilon_0}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\)

⇒ \(9 \times 10^9 \times\left(8 \times 10^{-3}\right) \times\left(-2 \times 10^{-9}\right) \times\left(\frac{1}{0.04}-\frac{1}{0.03}\right)\)

= 1.2 J

Read And Learn More WBCHSE Class 12 Physics Short Question And Answers

Question 2. In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 A.

1. Estimate the potential energy of the system in eV, taking the zero of the potential energy at an infinite separation of the electron from the proton.
2. What is the minimum work to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
3. What are the answers to (a) and (b) above if the zero of the potential energy is taken at 1.06 A separation?

Answer:

In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 A.

1. Potential energy

⇒ \(E_p=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r}\)

⇒ \(\frac{\left(9 \times 10^9\right) \times\left(-1.6 \times 10^{-19}\right) \times\left(1.6 \times 10^{-19}\right)}{0.53 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}\)

= -27.17 eV

2. Kinetic energy of the electron,

⇒ \(E_K=\frac{1}{2} E_P=13.585 \mathrm{eV}\)[Ek is always positive]

∴ The total energy of the electron,

E = Ep + EK = (-27.17 + 13.585)eV

=-13.585 eV

Minimum work done to free an electron from the atom

= [0-(-13.585)]eV

= 13.585 eV

Potential energy at a separation of 1.06 A

⇒\(E_P=\frac{\left(9 \times 10^9\right) \times\left(-1.6 \times 10^{-19}\right)\left(1.6 \times 10^{-19}\right)}{1.06 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}\)

= -13.585 eV

and kinetic energy, \(E_K=\frac{E_P}{2}\)

= 6.792 eV

∴ The total energy of the electron

= [6.792 + (-13.585)]eV

=-6.792 eV

∴ Minimum work done to free the electron from the atom

= [0-(-6.792)]eV

= 6.792 eV

Conceptual Questions on Voltage and Work Done

Question 3. Two charges -q and +q are located at points (0, 0, -a) and (0, 0, a) respectively.

1. What is the electrostatic potential at the points (0, 0, z) and (x, y, 0)?
2. Obtain the dependence of potential on the distance f of a point from the origin when r >> a.
3. How much work is done in moving a small test charge from the point (5, 0, 0) to (-7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

Answer:

Two charges -q and +q are located at points (0, 0, -a) and (0, 0, a) respectively.

1. The point (0, 0, z) lies on the axis of the dipole.

∴ Potential on a point on the axis of the dipole,

⇒ \(V= \pm \frac{1}{4 \pi \epsilon_0} \cdot \frac{p}{x^2-a^2}\)

where p = dipole moment = q.2a

and x²-a² = z²-a²

∴ \(V= \pm \frac{1}{4 \pi \epsilon_0} \cdot \frac{2 q a}{z^2-a^2}\)

The point (x,y,0) is on the line perpendicular to m, x the axis of the dipole, the potential at that point is zero.

2. Let P be the point at a distance r from the dipole. OP makes an angle θ with the axis of the dipole. The length of the dipole = 2a.

r₂ = BP == CP = r – a cosθ

r₁ = AP == DP = r + a cosθ

Potential at P,

⇒ \(V=V_1+V_2=\frac{q}{4 \pi \epsilon_0}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\)

⇒ \(\frac{q}{4 \pi \epsilon_0}\left(\frac{1}{r-a \cos \theta}-\frac{1}{r+a \cos \theta}\right)\)

∴ \(V=\frac{q}{4 \pi \epsilon_0} \frac{2 a \cos \theta}{r^2-a^2 \cos ^2 \theta}\)

⇒ \(\frac{2 a q \cos \theta}{4 \pi \epsilon_0}\left\{\frac{1}{r^2}\left(1-\frac{a^2}{r^2} \cos ^2 \theta\right)^{-1}\right\}\)

⇒ \(\approx \frac{p \cos \theta}{4 \pi \epsilon_0 r^2}\) ∵\( \frac{a^2}{r^2} \ll 1\)

∴ \(V \propto \frac{1}{r^2}\)

3. The work done is zero because the potential along the x-axis is zero.

The work done is always zero between these two points because work done does not depend on the path taken, it only depends on the position of the endpoints.

WBBSE Class 12 Electric Potential Short Q&A

Question 4. A charge array is known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for \(\frac{r}{a}\) >> 1, and contrast your results with that due to an electric dipole and an electric monopole (i,e., single charge).

Answer:

A charge array is known as an electric quadrupole.

Potential at P,

⇒ \(V_1=\frac{1}{4 \pi \epsilon_0}\left[\frac{q}{r-a}-\frac{2 q}{r}+\frac{q}{r+a}\right]\)

⇒ \(\frac{q}{4 \pi \epsilon_0} \cdot \frac{2 a^2}{r\left(r^2-a^2\right)} \approx \frac{q}{4 \pi \epsilon_0} \cdot \frac{2 a^2}{r^3}\)

∴ \(V_1 \propto \frac{1}{r^3}\)

Potential due to dipole,

⇒ \(V_2=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{r^2} \quad\)

∴ \(V_2 \propto \frac{1}{r^2}\)

Potential due to monopole,

⇒ \(V_3=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{r} \quad\)

∴ \(V_3 \propto \frac{1}{r}\)

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Question 5. If Coulomb’s law involved \(\frac{1}{r^3}\) dependence (instead of \(\frac{1}{r^2}\) ), would Gauss’ law be still true?
Answer:

Gauss’ law and Coulomb’s law depend on each other. So if Coulomb’s law changes its form, Gauss’ law would also change its present form.

Question 6. Describe schematically the equipotential surfaces corresponding to

1. A constant electric field in the z-direction,
2. A field that uniformly increases in magnitude but remains in a constant (say, z ) direction,
3. A single positive charge at the origin and
4. A uniform grid consisting of long equally spaced parallel charged wires in a plane.

Answer:

1.  A plane parallel to xy-plane.

2. A plane parallel to the xy-plane. These planes will come closer as the field intensity increases.

3. Concentric spheres with the charge at the centre.

4. Nearer the grid the equipotential surfaces will change their shape periodically slowly becoming planer and parallel to the grid at far-off distances from the grid.

Question 7. The top of the atmosphere is at about, 440 kV with respect to the surface of the earth, corresponding, to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 V.m^-1. Why then we do not get an electric shock as we step out of our house into the open? (Assume the house to be a still cage so there is no field inside.)
Answer:

The top of the atmosphere is at about, 440 kV with respect to the surface of the earth, corresponding, to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 V.m^-1.

Our body and the earth’s surface become equipotential i.e., there is no potential difference between the earth and our body. Hence no current flows through our body and therefore we do not experience a shock

Question 8. Write down the name of the physical quantity whose unit is joule/coulomb.
Answer: Electric potential

Question 9. Determine how much work is to be done to move a 10 C positive charge 1 m along the y-axis in a uniform electric field ⇒ \(\vec{E}=5(\hat{i}+\hat{j}) \mathrm{V} \cdot \mathrm{m}^{-1}\)
Answer:

Force acting on a charge q placed in a uniform electric field \(\vec{E}\) is given by,

⇒ \(\vec{F}=q \vec{E}=10 \times 5(\hat{i}+\hat{j}) \mathrm{N}\)

Again, displacement along y-axis, \(\vec{s}=1 \hat{j} \mathrm{~m}\)

∴ \(\vec{F} \cdot \vec{s}=50(\hat{i}+\hat{j}) \cdot \hat{j}\)

Short Answer Questions on Electrostatic Potential

Question 10. The electric field strength at a point in an electric field is zero. Is the electric potential also zero at that point? Answer with reason.
Answer:

The electric field strength at a point in an electric field is zero.

If the intensity of the electric field be E and the potential be V, thenrthear&atitinjbetween them is, E = \(\frac{dV}{dx}\)

So, if E = 0 at any point, we have \(\frac{dV}{dx}\) = 0

or, V = constant

Thus, the potential has a constant value, not necessarily zero, around that point.

Question 11. The electric potential at a point (x, y, z) is given by V = -x²y- xz³ + 4 . Find the intensity of electric field \(\vec{E}\) at that point
Answer:

The electric potential at a point (x, y, z) is given by V = -x²y- xz³ + 4 .

⇒ \(E_x=-\frac{d V}{d x}=2 x y+z^3, E_y=-\frac{d V}{d y}=x^2\)

⇒ \(E_z=-\frac{d V}{d z}=3 x z^2\)

∴ \(\vec{E}=E_x \hat{i}+E_y \hat{j}+E_z \hat{k}\)

= \(\left(2 x y+z^3\right) \hat{i}+x^2 \hat{j}+3 x z^2 k\)

Question 12. A test charge q is moved without acceleration along the path from A to B and from D to C in electric field E.

1. Calculate the potential difference between A and C

2. At which point (of the two) is the electric potential more and why?

Answer:

1. \(V_A-V_C=-\int_2^6 \vec{E} \cdot d \vec{x}=-\int_2^6 E d x \cos 0^{\circ}=-\int_2^6 E d x\)

⇒ \(-E \int_2^6 d x=-4 E\)

∴ \(V_C-V_A=4 E\)

2. In the direction of the electric field potential decreases. So, point C is a a higher potential.

Question 13. Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero
Answer:

Some equipotential surfaces of a dipole

All the points on the equatorial plane of a dipole, being equidistant from +q and -q, have zero potential.

Question 14. Obtain an expression for the work done to dissociate the system of three charges placed at the vertices of an equilateral triangle of side a as shown below.
Answer:

Work done to keep the system bound Is

⇒ \(W=\frac{1}{4 \pi \epsilon_0}\left[\frac{q_1 q_2}{a}+\frac{q_2 q_3}{a}+\frac{q_1 q_3}{a}\right]\)

⇒ \(\frac{1}{4 \pi \epsilon_0}\left[\frac{q(-4 q)}{a}+\frac{(-4 q) 2 q}{a}+\frac{q(2 q)}{a}\right]\)

⇒ \(\frac{1}{4 \pi \epsilon_0}\left[\frac{-4 q^2}{a}+\frac{8 q^2}{a}+\frac{2 q^2}{a}\right]\)

⇒ \(\frac{q^2}{4 \pi \epsilon_0 a}(-4-8+2)=\frac{-10 \cdot q^2}{4 \pi \epsilon_0 a}\)

Therefore, the work done to dissociate the system is,

⇒ \(W_d=-W=\frac{10 q^2}{4 \pi \epsilon_0 a}\)

Common Short Questions on Potential Energy

Question 15. 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.
Answer:

An infinitely large thin plane sheet has a uniform surface charge density +σ.

Electric field due to the infinitely large plane sheet = \(\frac{\sigma}{2 \epsilon_0}\)

Work done to bring a point charge q from infinity to a point at a distance r =q(Vr-Vco)

⇒ \(-q \int_{\infty}^r E d r=-q \int_{\infty}^r \frac{\sigma}{2 \epsilon_0} d r=\infty\)

Question Four point charges Q, q, Q and q are placed at the comers of a square of side.

Find the

1. Resultant electric force on a charge Q, and

2. Potential energy of this system

Answer:

The electric force on charge Q due to charge q

⇒ \(F_q=\frac{1}{4 \pi \epsilon_0} \times \frac{q Q}{a^2}\)

The electric force on charge Q due to the other charge Q

⇒ \(F_Q=\frac{1}{4 \pi \epsilon_0} \times \frac{Q^2}{(a \sqrt{2})^2}=\frac{1}{4 \pi \epsilon_0} \frac{Q^2}{2 a^2}\)

The net force on charge Q

⇒ \(F_{\text {net }}=F_Q+\sqrt{F_q^2+F_q^2}=F_Q+F_q \sqrt{2}\)

⇒ \(\frac{1}{4 \pi \epsilon_0} \times \frac{Q^2}{2 a^2}+\frac{1}{4 \pi \epsilon_0} \times \frac{q Q}{a^2} \sqrt{2}\)

⇒ \(\frac{Q^2}{4 \pi \epsilon_0 a^2}\left[\frac{Q}{2}+\sqrt{2} q\right]\) along diagonal

2. Potential energy of the system,

⇒ \(\dot{U}=U_{q Q}+U_{Q q}+U_{q Q}+U_{Q q}+U_{q q}+U_{Q Q}\)

⇒ \(4 U_{q Q}+U_{q q}+U_{Q Q}\)

⇒ \(\frac{4 q Q}{4 \pi \epsilon_0 a}+\frac{q^2}{4 \pi \epsilon_0(\sqrt{2} a)}+\frac{Q^2}{4 \pi \epsilon_0(\sqrt{2} a)}\)

⇒ \(\frac{1}{4 \pi \epsilon_0 a}\left[4 q Q+\frac{q^2}{\sqrt{2}}+\frac{Q^2}{\sqrt{2}}\right]\)

Practice Short Questions on Capacitors and Potential

Question 16.

1. Three point charges q, -4q, and 2q are placed at the vertices of an equilateral triangle ABC of side. Obtain the expression for the magnitude of the resultant electric force acting on the charge q.

2. Hind out this amount of work done to separate the charges at Infinite distance.

Answer:

1.

⇒ \(F_{A B}=\frac{1}{4 \pi \epsilon_0} \frac{q(4 q)}{l^2}=\frac{1}{4 \pi \epsilon_0} \frac{4 q^2}{l^2}\)

⇒ \(F_{A C}=\frac{1}{4 \pi \epsilon_0} \frac{q(2 q)}{l^2}=\frac{1}{4 \pi \epsilon_0} \frac{2 q^2}{l^2}\)

The angle between forces \(\vec{F}_{A B} \text { and } \vec{F}_{A C}\) is 120°.

The magnitude of the resultant force,

⇒ \(F=\sqrt{F_{A B}^2+F_{A C}^2+2 F_{A B} F_{A C} \cos 120^{\circ}}\)

⇒ \(\frac{1}{4 \pi \epsilon_0}\left(\frac{q^2}{l^2}\right) \sqrt{(4)^2+(2)^2+2 \times 4 \times 2 \times\left(\frac{-1}{2}\right)}\)

⇒ \(\frac{1}{4 \pi \epsilon_0} \frac{q^2}{l^2}(2 \sqrt{3})\)

2. Amount of work done = change in potential energy of the system

= U_f – U_i

= 0 – (U_AB + U_BC + U_CA)

⇒ \(\frac{-1}{4 \pi \epsilon_0 l}[q(-4 q)+(-4 q)(2 q)+(q)(2 q)]\)

⇒ \(\frac{10 q^2}{4 \pi \epsilon_0 l}\)

Unit 6 Optics Chapter 4 Dispersion Of Light Very Short Question And Answers

Question 1. Does dispersion of light take place in a vacuum?
Answer: No

Question 2. If white light is allowed to pass through a prism filled with water, will a spectrum be produced?
Answer:  No

Question 3. If rays of light pass through a glass slab which colour of light will have greater velocity—red or blue?
Answer: Red

Question 4. For which colours of light, the refractive index ofthe mate¬
rial of the prism is maximum and minimum?
Answer:

Maximum for violet light and minimum for Red light

Question 5. What colour of light will bend maximum and what colour will bend minimum while passing through a prism?
Answer:

Violet will bend maximum and red will bend minimum

Very Short Questions on Dispersion of Light WBCHSE

Question 6. Arrange in descending order μ_b ,μ_v ,μ_r, Here μ_b, μ_r and μ_v are the refractive indices of a medium for blue, red and violet colours respectively.
Answer: μ_b >μ_v>μ_r

Question 7. Which one is fundamental—frequency or wavelength, in the analysis of the spectrum?
Answer: Frequency

Question 8. Does a prism produce colour?
Answer: No

Question 9. Will a spectrum be produced if the light of a particular colour is allowed to pass through a prism?
Answer: No

Question 10. Which range of the visible spectrum is most sensitive for our eyes
Answer: Yellow

Question 11. What colour of light Is used while driving a car through fog?
Answer: Yellow

Question 12. Out of red and blue lights for which colour Is the refractive Index of glass greater?
Answer: [μ_b>μ_r]

Question 13.

1. Which has a larger wavelength, blue or red colour
Answer: Red colour has a larger wavelength as compared to blue

2. Between red and violet light whose refractivity Is higher?
Answer: Violet

3. n case of dispersion of light, the angle of deviation for red colour is maximum and it is minimum for violet colour. Is the statement true ot false?
Answer: False

Question 14. What is the range of wavelength of visible light? Can the dispersive power of any medium be negative?
Answer: No

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Question 15. On what factors does the dispersion power of the medium depend?
Answer: Nature of the medium and colour of light

Question 16. For which two colours of light is angular dispersion the
Answer:  Violet and Red

Question 17. What is the unit of angular dispersion?
Answer: Degree or radio

WBCHSE Physics Quick Q&A on Dispersion

Question 18. Does the dispersive power of a prism depend on the angle of the prism?
Answer: No

Question 19. Which one has the greater dispersive power: crown glass or flint glass?
Answer: Flint glass

Question 20.

1. What are the factors on which angular dispersion of a not? prism depends
Answer: Angle of prism and nature of the material of the prism

2. Is it true that for a thin prism the angular dispersion of of scattered light? violet and red light rays is equal to the product of the dispersive power of the prism and the deviation of the yellow light ray?
Answer: Yes

Question 21. What are the reasons the formation of a rainbow?
Answer: Dispersion, refraction and reflection of light

Question 22. What is the most essential condition to observe a rainbow?
Answer:

The sun must be situated behind the observer after rainfall or, the observer must stand with his back towards the sun

Question 23.  Under what condition should an object be able to produce a band spectrum?
Answer: Radioactive condition

Question 24. Under what condition should an object be able to produce a band spectrum?
Answer: Molecular state

Question 25. What will be the nature of the spectrum produced by a light bulb?
Answer: Line absorption spectrum

Question 26. What kind or type of spectrum solar spectrum is?
Answer: Line absorption spectrum

Question 27. What is the name of the black lines present in the solar spectrum? Answer: Fraunhofer lines

Question 28. What Is the ratio of velocity of ultraviolet rays and infrared rays in a vacuum?
Answer: 1:1

Brief Q&A on Dispersion of Light Concepts

Question 29.

1. How will a blue object look in sodium vapour flame? What kind of spectrum is solar spectrum?
Answer:  Black, line absorption spectrum

2. Write down the nature of spectra produced from an atom, molecule and an incandescent body respectively.
Answer:  Line spectrum, band spectrum, continuous spectrum

3. What is the nature of the spectrum that a red hot heater produces?
Answer: Continuous

4. For the identification of a gas which spectrum is used?
Answer: Line spectrum

5. What characteristics of a body is indicated by the band spectrum?
Answer:  Atomic characteristics

6. A bunch of red roses kept in a room illuminated with yellow light looks black.—Is the statement correct or not
Answer: Correct

Question 30. What is the relation between the intensity and wavelength of scattered light
Answer: \(\left[I \propto \frac{1}{\lambda^4}\right]\)

Question 31. ‘In addition to the liquid, solid and gaseous media also exhibit Raman effect’—true or false.
Answer:  True

Question 32. In the Raman spectrum, what are the names of the spectral lines of longer and shorter wavelengths, apart from the low light ray?
Answer:  Stokes line and anti-stokes line

Question 33. Why is red light used as a danger signal?
Answer:

Because its wavelength is maximum and scattering is minimum]

Simple Q&A Format on Dispersion of Light

Question 34. The intensity of Stokes’ lines is more than that of Anti-stokes’ line Is the statement true or false?
Answer: True

Question 35. Why can’t we see clearly through the fog? Name the phenomenon responsible for it.
Answer:

1. During fog, the light coming from an object is partly
   deflected by the particles of the fog and does not reach the
   eye of the observer.
2. For that reason, we cannot see the objects clearly through fog. This responsible phenomenon is called the scattering of light.