Two charges, each equal to , are kept at and on the . A particle of mass m and charge is placed at the origin. If charge is given a small displacement along the y-axis, the net force acting on the particle is proportional to
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2
3
4
Official Solution
Correct Option: (1)
02
PYQ 2014
medium
physicsID: jee-main
A spherically symmetric charge distribution is characterised by a charge density having the following variation : for for Where is the distance from the centre of the charge distribution and is a constant. The electric field at an internal point is
1
2
3
4
Official Solution
Correct Option: (2)
03
PYQ 2015
medium
physicsID: jee-main
A long cylindrical shell carries positive surface charge in the upper half and negative surface charge in the lower half. The electric field lines around the cylinder will look like figure given in:
(figures are schematic and not drawn to scale)
1
2
3
4
Official Solution
Correct Option: (4)
Tangent to the electrical field lines will give us the direction at a given point.
04
PYQ 2015
hard
physicsID: jee-main
A thin disc of radius has a concentric hole of radius ' ' in it (see figure). It carries uniform surface charge ' ' on it. If the electric field on its axis at height ' ' from its centre is given as ' ' then value of ' ' is :
1
2
3
4
Official Solution
Correct Option: (3)
05
PYQ 2016
medium
physicsID: jee-main
The region between two concentric spheres of radii and , respectively (see figure), has volume charge density , where is a constant and is the distance from the centre. At the centre of the spheres is a point charge . The value of such that the electric field in the region between the spheres will be constant, is :
1
2
3
4
Official Solution
Correct Option: (1)
06
PYQ 2017
medium
physicsID: jee-main
An electric dipole has a fixed dipole moment , which makes angles with respect to x-axis. When subjected to an electric field , it experiences a
torque . When subjected to another electric field it experiences a torque . The angle is
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2
3
4
Official Solution
Correct Option: (3)
......(i)
....(ii) From (i) and (ii)
07
PYQ 2018
medium
physicsID: jee-main
A body of mass and charge is connected to a spring of spring constant . It is oscillating along x-direction about its equilibrium position, taken to be at , with an amplitude . An electric field is applied along the x-direction. Which of the following statements is correct ?
1
The new equilibrium position is at a distance from
2
The total energy of the system is
3
The total energy of the system is
4
The new equilibrium position is at a distance from
Official Solution
Correct Option: (2)
Total energy of the system = kinetic energy + potential energy Given that amplitude of oscillation is . Therefore, energy at extreme point is . When electric field is applied, new mean position is
Thus, new total energy after electric field is applied is
Total energy
08
PYQ 2019
easy
physicsID: jee-main
Let a total charge be distributed in a sphere of radius , with the charge density given by , where r is the distance from the centre. Two charges and , of each, are placed on diametrically opposite points, at equal distance, a, from the centre. If and do not experience any force, then :
1
2
3
4
Official Solution
Correct Option: (3)
09
PYQ 2019
hard
physicsID: jee-main
Charge is distributed within a sphere of radius with a volume charge density where and are constants. If is the total charge of this charge distribution, the radius is :
1
2
3
4
Official Solution
Correct Option: (4)
The correct answer is (D) :
10
PYQ 2019
medium
physicsID: jee-main
An electric dipole is formed by two equal and opposite charges with separation . The charges have same mass . It is kept in a uniform electric field . If it is slightly rotated from its equilibrium orientation, then its angular frequency is :
1
2
3
4
Official Solution
Correct Option: (3)
moment of inertia (I) Now by
for small
Angular frequency
11
PYQ 2019
medium
physicsID: jee-main
For a uniformly charged ring of radius , the electric field on its axis has the largest magnitude at a distance from its centre. Then value of is :
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2
3
4
Official Solution
Correct Option: (3)
Electric field on axis of ring
for maximum electric field
12
PYQ 2020
medium
physicsID: jee-main
A particle of charge and mass is subjected to an electric field in the -direction, where a and are constants. Initially the particle was at rest at . Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :
1
2
3
4
Official Solution
Correct Option: (4)
For Hence
13
PYQ 2020
medium
physicsID: jee-main
Consider a sphere of radius which carries a uniform charge density . If a sphere of radius Ls carved out of it, as shown, the ratio of magnitude of electric field and respectively, at points A and B due to the remaining portion is :
1
2
3
4
Official Solution
Correct Option: (4)
Fill the empty space with and charge density.
14
PYQ 2020
medium
physicsID: jee-main
In finding the electric field using Gauss law the formula is applicable. In the formula is permittivity of free space, is the area of Gaussian surface and is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation ?
1
For any choice of Gaussian surface.
2
Only when the Gaussian surface is an equipotential surface.
3
Only when the Gaussian surface is an equipotential surface and is constant on the surface.
4
Only when constant on the surface.
Official Solution
Correct Option: (3)
should be constant on the surface and the surface should be equipotential.
15
PYQ 2022
medium
physicsID: jee-main
In the figure, a very large plane sheet of positive charge is shown. and are two points at distance and from the charge distribution. If is the surface charge density, then the magnitude of electric fields and and and respectively are
1
2
3
4
Official Solution
Correct Option: (3)
For an infinite charged plane
for each value of distance l
Thus,
So, the correct option is (C):
16
PYQ 2023
medium
physicsID: jee-main
A 10 μC charge is divided into two equal parts and kept at 1 cm distance. Find repulsion between charges?
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2
3
4
Official Solution
Correct Option: (3)
Let the total charge be divided into two parts, and . The force between the two charges is given by Coulomb's law:
where is Coulomb's constant and is the distance between the charges. To maximize , differentiate with respect to and set the derivative equal to zero: Simplify: This gives: Solve for : The other part of the charge is: Thus, the charges of the two parts are and .
17
PYQ 2023
hard
physicsID: jee-main
A point source of emits light with efficiency At a distance of from the source, the intensity produced by the electric field component is:
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2
3
4
Official Solution
Correct Option: (1)
The correct option is(A):
18
PYQ 2023
easy
physicsID: jee-main
Given below are two statements : One is labelled as Assertion and the other is labelled as Reason R Assertion A : Two metallic spheres are charged to the same potential One of them is hollow and another is solid, and both have the same radii Solid sphere will have lower charge than the hollow one. Reason R : Capacitance of metallic spheres depend on the radii of spheres. In the light of the above statements, choose the correct answer from the options given below.
1
Both and are true and is the correct explanation of
2
Both and are true but is not the correct explanation of
3
A is true but is false
4
A is false but is true
Official Solution
Correct Option: (4)
Step 1: Analyze Assertion A
The potential of a conducting sphere (solid or hollow) is given by: where is the potential, is a constant, is the charge, and is the radius.
If two spheres have the same potential ( ) and the same radius ( ), then: Since , it follows that . Therefore, the assertion that the solid sphere will have a lower charge is false. The charges must be equal.
Step 2: Analyze Reason R
The capacitance of a spherical conductor is given by: where is the capacitance, is the permittivity of free space, and is the radius. This shows that the capacitance of a metallic sphere depends only on its radius. Therefore, the reason is true.
Conclusion: Assertion A is false, but Reason R is true. Therefore, the correct option is (4).
19
PYQ 2023
hard
physicsID: jee-main
A uniform electric field of is created between two parallel charged plates (as shown in figure) An electron enters the field symmetrically between the plates with a kinetic energy The length of each plate is The angle of deviation of the path of electron as it comes out of the field is ___ (in degree)
Official Solution
Correct Option: (1)
1. Kinetic energy of the electron: Convert to joules: Solve for : 2. Vertical displacement: Substituting: Thus, the angle of deviation is 45°. The deviation of the path in a uniform electric field depends on the balance between horizontal and vertical forces on the electron.
20
PYQ 2023
easy
physicsID: jee-main
Electric field in a certain region is given by The unit of and are :
1
2
3
4
Official Solution
Correct Option: (1)
Equations with Proper Units
Equation 1:
Units of A and B:
The unit of is derived as:
The unit of is derived as:
21
PYQ 2023
medium
physicsID: jee-main
64 identical balls made of conducting material each having a potential of 10 mV are joined to form a bigger ball. The potential of a bigger ball is?
Official Solution
Correct Option: (1)
Since the balls are made of conducting material, the potential will be distributed uniformly over the surface of each ball. When the balls are joined to form a bigger ball, they will form a parallel combination, and the total charge will remain the same. Let the potential of the bigger ball be V. Then, by the principle of conservation of charge, we have: total charge of the 64 balls = charge of the bigger ball q = CV = 64C(0.01V) where C is the capacitance of each ball and 0.01V is the potential of each ball. Also, capacitance of each ball, C = 4πε₀r where r is the radius of each ball and ε₀ is the permittivity of free space. Since all the balls are identical, r is the same for each ball. So, we have: q = 64(4πε₀r)(0.01V) And, the capacitance of the bigger ball, C' is given by: C' = 4πε₀R where R is the radius of the bigger ball. Since the balls are joined to form a bigger ball, the radius of the bigger ball, R, is equal to the radius of one ball multiplied by the cube root of 64. So, we have: R = r∛64 = 4r Therefore, the capacitance of the bigger ball is: C' = 4πε₀(4r) = 16πε₀r Now, equating the expressions for q and C', we get: 64C(0.01V) = 16πε₀rV or, V = Substituting the value of capacitance C, we get: V = Simplifying, we get: V = 0.16 V Therefore, the potential of the bigger ball is 0.16 V. Answer. 0.16V
22
PYQ 2023
medium
physicsID: jee-main
Electric field due to a dipole at a equatorial point depends upon r-n . Value of n is
Official Solution
Correct Option: (1)
The correct answer is : 3 At an equatorial point of a dipole, the electric field due to the dipole is given by:
where k is the Coulomb constant, p is the dipole moment, and r is the distance from the dipole. For an equatorial point, the distance r is equal to the distance between the two charges of the dipole, which is 2a, where a is the separation between the charges. Therefore, the electric field at the equatorial point can be written as:
where θ is the angle between the axis of the dipole and the line joining the dipole to the point where the electric field is to be calculated, and ε0 is the permittivity of free space. Since sinθ = 1 at the equatorial point, we can simplify the above equation as:
Therefore, the dependence of the electric field on the distance r at an equatorial point is , which is equivalent to n = 3. Hence, the value of n is 3.
23
PYQ 2023
medium
physicsID: jee-main
A condenser of capacitance is charged steadily from to Which of the following graph represents correctly the variation of potential difference across it's plates with respect to the charge on the condenser?
1
2
3
4
Official Solution
Correct Option: (1)
The correct option is (A)
24
PYQ 2024
medium
physicsID: jee-main
Two particles each of mass kg are places as shown in plane. If the distance of centre of mass from origin is find :
Official Solution
Correct Option: (1)
The Correct Answer is : 2
25
PYQ 2025
hard
physicsID: jee-main
The magnitude of heat exchanged by a system for the given cyclic process ABC (as shown in the figure) is (in SI units):
1
2
3
4
zero
Official Solution
Correct Option: (3)
In thermodynamics, the heat exchanged by a system in a cyclic process is equal to the area enclosed by the process curve on a diagram. In the given problem, the process involves a rectangle on the diagram (since the pressure-volume graph forms a closed loop between points A, B, and C). The area of this rectangle can be calculated as: Therefore, the magnitude of heat exchanged is units. Final Answer: .
26
PYQ 2025
hard
physicsID: jee-main
A particle of mass and charge is fastened to one end of a massless string having equilibrium length , whose other end is fixed at point . The whole system is placed on a frictionless horizontal plane and is initially at rest. If a uniform electric field is switched on along the direction as shown in the figure, then the speed of the particle when it crosses the x-axis is:
1
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3
4
Official Solution
Correct Option: (2)
The particle moves under the influence of an electric field. We will use the work-energy principle to find its speed when it crosses the x-axis. Step 1: The electric force acting on the particle is given by: where is the electric field. Step 2: The work done by this force in moving the particle a distance along the x-axis is: Step 3: The kinetic energy gained by the particle is equal to the work done: So, equating the work and kinetic energy: Step 4: Solve for : Final Conclusion: The speed of the particle when it crosses the x-axis is , which is Option (2).
27
PYQ 2025
hard
physicsID: jee-main
The magnitude of heat exchanged by a system for the given cyclic process ABC (as shown in the figure) is (in SI units):
1
2
3
4
zero
Official Solution
Correct Option: (3)
In thermodynamics, the heat exchanged by a system in a cyclic process is equal to the area enclosed by the process curve on a diagram. In the given problem, the process involves a rectangle on the diagram (since the pressure-volume graph forms a closed loop between points A, B, and C). The area of this rectangle can be calculated as: Therefore, the magnitude of heat exchanged is units. Final Answer: .
28
PYQ 2025
medium
physicsID: jee-main
A particle of mass and charge is fastened to one end of a massless string having equilibrium length , whose other end is fixed at point . The whole system is placed on a frictionless horizontal plane and is initially at rest. If a uniform electric field is switched on along the direction as shown in the figure, then the speed of the particle when it crosses the x-axis is:
1
2
3
4
Official Solution
Correct Option: (2)
The particle moves under the influence of an electric field. We will use the work-energy principle to find its speed when it crosses the x-axis.
Step 1: The electric force acting on the particle is given by:
where is the electric field.
Step 2: The work done by this force in moving the particle a distance along the x-axis is:
Step 3: The kinetic energy gained by the particle is equal to the work done:
So, equating the work and kinetic energy:
Step 4: Solve for :
Final Conclusion: The speed of the particle when it crosses the x-axis is , which is Option (2).
29
PYQ 2025
medium
physicsID: jee-main
As shown below, bob A of a pendulum having massless string of length is released from to the vertical. It hits another bob B of half the mass that is at rest on a frictionless table in the center. Assuming elastic collision, the magnitude of the velocity of bob A after the collision will be (take as acceleration due to gravity):
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3
4
Official Solution
Correct Option: (3)
Using the principle of conservation of mechanical energy, the velocity of bob A just before the collision is given by:
Since the collision is elastic and the mass of bob B is half that of bob A, we can use the conservation of momentum and energy to solve for the final velocity of bob A after the collision. The final velocity of bob A, after the elastic collision, can be derived using the relative velocities and momentum conservation. The final velocity is given by:
Thus, the correct answer is .
