A tennis ball (treated as hollow spherical shell) starting from rolls down a hill. At point A the ball becomes air borne leaving at an angle of with the horizontal. The ball strikes the ground at . What is the value of the distance ? (Moment of inertia of a spherical shell of mass and radius about its diameter )
1
2
3
4
Official Solution
Correct Option: (1)
Velocity of the tennis ball on the surface of the earth or ground ( where k = radius of gyration of spherical shell ) Horizontal range
02
PYQ 2014
medium
physicsID: jee-main
A thin bar of length has a mass per unit length , that increases linearly with distance from one end. If its total mass is and its mass per unit length at the lighter end is , then the distance of the centre of mass from the lighter end is :
1
2
3
4
Official Solution
Correct Option: (4)
Mass per unit lengh
substitute 'k'
03
PYQ 2014
medium
physicsID: jee-main
A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed about the vertical support. About the point of suspension
1
Angular momentum is conserved
2
Angular momentum changes in magnitude but not in direction
3
Angular momentum changes in direction but not in magnitude
4
Angular momentum changes both in direction and magnitude
Official Solution
Correct Option: (3)
(Direction parallel to plane of rotation of particle)
as is perpendicular to , direction of changes but magnitude remains same
04
PYQ 2015
medium
physicsID: jee-main
A uniform solid cylindrical roller of mass ' ' is being pulled on a horizontal surface with force parallel to the surface and applied at its centre. If the acceleration of the cylinder is ' ' and it is rolling without slipping then the value of ' ' is :
1
2
3
ma
4
ma
Official Solution
Correct Option: (3)
...(1) ...(2) Therefore,
05
PYQ 2015
hard
physicsID: jee-main
Consider a thin uniform square sheet made of a rigid material. If its side is ' ', mass and moment of inertia about one of its diagonals, then :
1
2
3
4
Official Solution
Correct Option: (2)
Using perpendicular axis theorem
06
PYQ 2016
medium
physicsID: jee-main
A roller is made by joining together two cones at their vertices . It is kept on two rails and which are placed asymmetrically (see figure), with its axis perpendicular to and its centre at the centre of line joining and (see figure). It is given a light push so that it starts rolling with its centre moving parallel to in the direction shown. As it moves, the roller will tend to :
1
turn left.
2
turn right.
3
go straight.
4
turn left and right alternately.
Official Solution
Correct Option: (3)
If central line is at a distance from instantaneous axis of rotation, the is at lesser distance from point on left than on right. Since , the roller will turn to left.
07
PYQ 2016
medium
physicsID: jee-main
Concrete mixture is made by mixing cement, stone and sand in a rotating cylindrical drum. If the drum rotates too fast, the ingredients remain stuck to the wall of the drum and proper mixing of ingredients does not take place. The maximum rotational speed of the drum in revolutions per minute to ensure proper mixing is close to : (Take the radius of the drum to be and its axle to be horizontal) :
1
0.4
2
1.3
3
8
4
27
Official Solution
Correct Option: (4)
rpm
08
PYQ 2017
medium
physicsID: jee-main
In a physical balance working on the principle of moments, when weight is placed on the left pan, the beam becomes horizontal. Both the empty pans of the balance are of equal mass. Which of the following statements is correct ?
1
Left arm is longer than the right arm
2
Both the arms are of same length
3
Left arm is shorter than the right arm
4
Every object that is weighed using this balance appears lighter than its actual weight.
Official Solution
Correct Option: (3)
Consider mass of pan be m,as the mass of the pan is same initially, so when the weight is not placed on the pan the force acting will be same as on both sides, After the weight is placed on left pan,
Hence the left arm(in which was placed) is shorter than right arm.
09
PYQ 2017
medium
physicsID: jee-main
A slender uniform rod of mass and length is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle with the vertical is :
1
2
3
4
Official Solution
Correct Option: (1)
Torque at angle
10
PYQ 2017
medium
physicsID: jee-main
The moment of inertia of a uniform cylinder of length and radius about its perpendicular bisector is . What is the ratio such that the moment of inertia is minimum ?
1
2
3
1
4
Official Solution
Correct Option: (1)
11
PYQ 2018
easy
physicsID: jee-main
A thin circular disk is in the plane as shown in the figure. The ratio of its moment of inertia about and axes will be :
1
1:03
2
1:04
3
1:05
4
1:02
Official Solution
Correct Option: (1)
The moment of inertia of a disc about central axis is
12
PYQ 2018
medium
physicsID: jee-main
A thin uniform bar of length and mass lies on a smooth horizontal table. Two point masses and are moving in the same horizontal plane from opposite sides of the bar with speeds and respectively. The masses stick to the bar after collision at a distance and respectively from the centre of the bar. If the bar starts rotating about its center of mass as a result of collision, the angular speed of the bar will be :
1
2
3
4
Official Solution
Correct Option: (2)
Centre of mass after the collision will be at . Therefore, Now, using conservation of angular momentum, we get Therefore,
13
PYQ 2018
medium
physicsID: jee-main
Seven identical circular planar disks, each of mass and radius are welded symmetrically as shown.
The moment of inertia of the arrangement about the axis normal to the plane and passing through the point is -
1
2
3
4
Official Solution
Correct Option: (4)
14
PYQ 2018
medium
physicsID: jee-main
From a uniform circular disc of radius and mass , a small disc of radius is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is -
1
2
3
4
Official Solution
Correct Option: (1)
15
PYQ 2019
medium
physicsID: jee-main
Let the moment of inertia of a hollow cylinder of length (inner radius and outer radius ), about its axis be 1. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also , is:
1
12 cm
2
18 cm
3
16 cm
4
14 cm
Official Solution
Correct Option: (3)
The correct answer is (C) : 16 Moment of inertia of hollow cylinder about its axis is given as:
Where, = Inner radius and = Outer radius Moment of inertia of thin hollow cylinder of radius R about its axis is given as:
Given that
Both cylinders have same mass (M)
16
PYQ 2019
medium
physicsID: jee-main
A metal coin of mass and radius is fixed to a thin stick of negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about at rotations per second in , is close to :
1
Nm
2
Nm
3
Nm
4
Nm
Official Solution
Correct Option: (2)
17
PYQ 2019
medium
physicsID: jee-main
A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of vessel is and its rotational speed is rotations per second, then the difference in the heights between the centre and the sides, in , will be
1
1.2
2
0.1
3
2
4
0.4
Official Solution
Correct Option: (3)
18
PYQ 2019
medium
physicsID: jee-main
A circular disc of mass and radius has two identical discs and of the same mass and radius attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis , passing through the centre of , as shown in the figure, will be :
1
2
3
4
Official Solution
Correct Option: (1)
19
PYQ 2019
hard
physicsID: jee-main
To mop-clean a floor, a cleaning machine presses a circular mop of radius vertically down with a total force and rotates it with a constant angular speed about its axis. If the force is distributed uniformly over the mop and if coefficient of friction between the mop and the floor is , the torque, applied by the machine on the mop is :
1
2
3
4
Official Solution
Correct Option: (1)
Consider a strip of radius x & thickness dx, Torque due to friction on this strip.
20
PYQ 2019
medium
physicsID: jee-main
An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If , and the angle made by with downward vertical is , then :
1
2
3
4
Official Solution
Correct Option: (2)
Let mass of one rod is . Balancing torque about hinge point.
21
PYQ 2019
easy
physicsID: jee-main
A uniform rectangular thin sheet of mass has length and breadth , as shown in the figure. If the shaded portion is cut-off, the coordinates of the centre of mass of the remaining portion will be :
1
2
3
4
Official Solution
Correct Option: (4)
22
PYQ 2019
medium
physicsID: jee-main
A thin disc of mass and radius has mass per unit area where is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is :
1
2
3
4
Official Solution
Correct Option: (3)
23
PYQ 2019
medium
physicsID: jee-main
A thin circular plate of mass and radius has its density varying as with as constant and is the distance from its centre. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is . The value of the coefficient is :
1
2
3
4
Official Solution
Correct Option: (4)
\underset{\text{ (MOI about COM) }}{{ I_0 }} by parallel axis theorem
24
PYQ 2019
medium
physicsID: jee-main
A homogeneous solid cylindrical roller of radius and mass is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is :
1
2
3
4
Official Solution
Correct Option: (3)
25
PYQ 2019
medium
physicsID: jee-main
A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights and on the incline. The ratio is given by :
1
2
3
4
Official Solution
Correct Option: (1)
for solid sphere
for solid cylinder
26
PYQ 2019
medium
physicsID: jee-main
A string is wound around a hollow cylinder of mass and radius . If the string is now pulled with a horizontal force of , and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string) :
1
2
3
4
Official Solution
Correct Option: (2)
.....(i)
...... (ii) From (i) and (ii)
27
PYQ 2019
easy
physicsID: jee-main
A solid sphere of mass and radius is divided into two unequal parts. The first part has a mass of and is converted into a uniform disc of radius . The second part is converted into a uniform solid sphere. Let be the moment of inertia of the disc about its axis and be the moment of inertia of the new sphere about its axis. The ratio is given by :
1
185
2
65
3
285
4
140
Official Solution
Correct Option: (4)
28
PYQ 2019
medium
physicsID: jee-main
A rigid massless rod of length has two masses attached at each end as shown in the figure. The rod is pivoted at point on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be :
1
2
3
4
Official Solution
Correct Option: (3)
Applying torque equation about point P.
anticlockwise
29
PYQ 2019
medium
physicsID: jee-main
A rectangular solid box of length is held horizontally, with one of its sides on the edge of a platform of height . When released, it slips off the table in a very short time , remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to :
1
0.02
2
0.28
3
0.5
4
0.3
Official Solution
Correct Option: (3)
Angular impulse = change in angular momentum
time taken by rod to hit the ground sec in this time angle rotate by rod radian
30
PYQ 2020
medium
physicsID: jee-main
Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are and respectively while those for the second one are and respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The Kinetic energy of the combined system is :
1
2
3
4
Official Solution
Correct Option: (4)
Both discs are rotating in same sense Angular momentum conserved for the system i.e.
Kinetic energy of combined disc system
31
PYQ 2020
medium
physicsID: jee-main
A particle of mass is fixed to one end of a light spring having force constant and unstretched length . The other end is fixed. The system is given an angular speed about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is :
1
2
3
4
Official Solution
Correct Option: (2)
32
PYQ 2020
medium
physicsID: jee-main
A rod of length has non-uniform linear mass density given by where a and b are constants and The value of for the centre of mass of the rod is a t :
1
2
3
4
Official Solution
Correct Option: (3)
33
PYQ 2020
medium
physicsID: jee-main
A bead of mass stays at point on a wire bent in the shape of a parabola and rotating with angular speed (see figure). The value of is (neglect friction) :
1
2
3
4
Official Solution
Correct Option: (2)
For Particle to be in equilibrium,
⇒
Also,
⇒ Slope at point P =
Equating both values of we get,
⇒
⇒
Therefore, The correct option is (B):
34
PYQ 2021
medium
physicsID: jee-main
Angular momentum of a single particle moving with constant speed along circular path :
1
is zero
2
remains same in magnitude but changes in the direction
3
changes in magnitude but remains same in the direction
4
remains same in magnitude and direction
Official Solution
Correct Option: (4)
Step 1: Understanding the Concept: Angular momentum of a particle about the center of a circular path is given by , where is the position vector and is the linear momentum. Step 2: Detailed Explanation: 1. Magnitude: In a circular path with constant speed and radius , the velocity vector is always perpendicular to the radius vector. Thus, . Since are constants, the magnitude remains same. 2. Direction: By the right-hand rule, the direction of is perpendicular to the plane of the circle. As the particle moves in the plane, the plane itself does not tilt, so the normal to the plane (the direction of ) remains fixed. Step 3: Final Answer: Both the magnitude and direction of angular momentum remain constant.
35
PYQ 2021
medium
physicsID: jee-main
Moment of inertia of a square plate of side l about the axis passing through one of the corner and perpendicular to the plane of square plate is given by :
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: Understanding the Question:
We need to find the moment of inertia ( ) of a uniform square plate of mass and side length . The axis of rotation passes through one of its corners and is perpendicular to the plane of the plate. Step 2: Key Formula or Approach:
We will use two important theorems for moment of inertia:
1. Perpendicular Axis Theorem: For a planar object, the moment of inertia about an axis perpendicular to the plane ( ) is the sum of the moments of inertia about two perpendicular axes in the plane ( and ) that intersect at the same point: .
2. Parallel Axis Theorem: The moment of inertia about any axis is the sum of the moment of inertia about a parallel axis passing through the center of mass ( ) and the product of the mass ( ) and the square of the distance ( ) between the two axes: . Step 3: Detailed Explanation: Part 1: Find MI about the center of mass, perpendicular to the plane.
First, consider the moment of inertia of the square plate about an axis passing through its center of mass (CM) and parallel to a side. This is a standard result: .
Using the Perpendicular Axis Theorem, the moment of inertia about an axis passing through the CM and perpendicular to the plate ( ) is:
Part 2: Shift the axis to the corner using the Parallel Axis Theorem.
The axis we need is parallel to the axis through the CM. We need to find the distance ( ) from the center of the square to a corner.
The coordinates of the corners can be taken as if the center is at the origin. The distance from the center (0,0) to a corner is:
So, . Now, apply the Parallel Axis Theorem:
Step 4: Final Answer:
The moment of inertia of the square plate about an axis passing through a corner and perpendicular to its plane is .
36
PYQ 2021
medium
physicsID: jee-main
The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CD parallel to AB as shown in figure is 2.7 kg m .
1
14.9 kg/m
2
1.49 10 kg/m
3
7.5 10 kg/m
4
7.5 10 kg/m
Official Solution
Correct Option: (2)
Step 1: Understanding the Question:
We are given the dimensions of a solid cylinder and its moment of inertia about an axis CD. The axis AB is the central axis of the cylinder, and CD is parallel to it at a distance `d` shown as L/2 in the diagram. We need to find the density of the cylinder's material. Step 2: Key Formula or Approach:
1. The moment of inertia of a solid cylinder about its central axis (AB) is .
2. The Parallel Axis Theorem states that the moment of inertia about any axis parallel to an axis through the center of mass is , where `d` is the perpendicular distance between the axes.
3. Density is defined as . The volume of a cylinder is . Step 3: Detailed Explanation:
First, convert the given dimensions to SI units:
Length, cm = 0.8 m.
Radius, cm = 0.2 m.
Moment of inertia about axis CD, kg m .
The distance between the parallel axes AB and CD is m. The moment of inertia about the central axis AB is .
Using the parallel axis theorem, the moment of inertia about axis CD is:
Substitute the given values into this equation to find the mass M:
Now, calculate the volume of the cylinder:
Finally, calculate the density :
Using :
This value can be written in scientific notation as kg/m . Step 4: Final Answer:
The calculated density is approximately 149 kg/m , which matches option (B).
37
PYQ 2021
medium
physicsID: jee-main
Two discs have moments of inertia I and I about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, and respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by:
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Understanding the Question:
We have two rotating discs that are brought into contact. Due to frictional forces between them, they eventually rotate with a common angular velocity. Since this is an inelastic collision, some kinetic energy will be lost. We need to calculate this loss. Step 2: Key Formula or Approach:
1. Conservation of Angular Momentum: Since there is no external torque on the two-disc system, the total angular momentum before and after contact is conserved.
2. Kinetic Energy of Rotation: The rotational kinetic energy of a body is given by .
3. Loss in KE: . Step 3: Detailed Explanation: Initial State:
Initial angular momentum of the system, .
Initial kinetic energy of the system, . Final State:
When the discs are brought into contact, they rotate together with a common final angular velocity, .
The total moment of inertia of the combined system is .
Final angular momentum of the system, . Applying Conservation of Angular Momentum:
Solving for the final angular velocity: Calculating Final Kinetic Energy:
Calculating Loss in Kinetic Energy:
Taking a common denominator of :
Expanding the numerator:
Numerator =
Numerator =
Numerator =
Numerator =
Substituting back into the expression for : Step 4: Final Answer:
The loss in kinetic energy is .
38
PYQ 2021
medium
physicsID: jee-main
A system consists of two identical spheres each of mass 1.5 kg and radius 50 cm at the ends of a light rod. The distance between the centres of the two spheres is 5 m. What will be the moment of inertia of the system about an axis perpendicular to the rod passing through its midpoint ?
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: The total moment of inertia of a system is the sum of the moments of inertia of its individual components. For spheres shifted away from the axis, we use the Parallel Axis Theorem. Step 2: Key Formula or Approach: 1. Moment of inertia of a solid sphere about its center: . 2. Parallel axis theorem: . Step 3: Detailed Explanation: Given: - , . - Separation between centers = 5 m. - Distance from axis (midpoint) to center of each sphere . Moment of inertia for one sphere about the central axis: Total moment of inertia for both spheres: Step 4: Final Answer: The moment of inertia is .
39
PYQ 2021
medium
physicsID: jee-main
In the given figure, two wheels P and Q are connected by a belt B. The radius of P is three times as that of Q. In case of same rotational kinetic energy, the ratio of rotational inertias will be x : 1. The value of x will be ________ .
Official Solution
Correct Option: (1)
Let the radius of wheel Q be .
The radius of wheel P is three times that of Q, so . Since the two wheels are connected by a belt that does not slip, the tangential speed at the rim of both wheels must be the same.
. The relationship between tangential speed (v), angular speed ( ), and radius (r) is .
. Substituting the radii:
.
. The problem states that the rotational kinetic energies of the two wheels are the same.
. The formula for rotational kinetic energy is .
. Substitute into the equation:
.
. Cancel from both sides (assuming the wheels are rotating).
. The question asks for the ratio of rotational inertias, let's assume and .
. The ratio is 9 : 1. Therefore, the value of x is 9.
40
PYQ 2021
medium
physicsID: jee-main
The figure shows two solid discs with radius R and r respectively. If mass per unit area is same for both, what is the ratio of MI of bigger disc around axis AB (which is to the plane of the disc and passing through its centre) to MI of smaller disc around one of its diameters lying on its plane ? Given 'M' is the mass of the larger disc. (MI stands for moment of inertia)
1
2
3
4
Official Solution
Correct Option: (3)
Let be the mass per unit area, which is the same for both discs.
Mass of the bigger disc, .
Mass of the smaller disc, . Moment of inertia of the bigger disc ( ) about the axis AB, which is perpendicular to its plane and passes through its center, is:
. Moment of inertia of the smaller disc ( ) about one of its diameters is:
From the perpendicular axis theorem, . For a disc, .
The MI about the perpendicular axis through the center is .
So, .
. Now, we find the ratio of to :
Cancel the common term :
. So, the ratio is .
41
PYQ 2021
medium
physicsID: jee-main
Choose the correct answer from the options given below :
1
(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
2
(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
3
(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
4
(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)
Official Solution
Correct Option: (2)
Let's calculate the Moment of Inertia (MI) for each case. The standard formulas for a rod are:
- About a perpendicular axis through the center:
- About a perpendicular axis through an end: (a) Mass = M, Length = L, axis through midpoint.
. This matches (iii). (b) Mass = 2M, Length = L, axis through one end.
. This matches (iv). (c) Mass = M, Length = 2L, axis through midpoint.
. This matches (ii). (d) Mass = 2M, Length = 2L, axis through one end.
. This matches (i). The correct matching is: (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i). This corresponds to option (B).
42
PYQ 2021
medium
physicsID: jee-main
A circular hole of radius is cut out of a circular disc of radius 'a' as shown in figure. The centroid of the remaining circular portion with respect to point 'O' will be :
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Let the center of the original disc be the origin . The hole is cut with its center at .
Step 2: Area of original disc . Area of hole .
Step 3: Position of centroid .
The distance from the center is .
43
PYQ 2022
medium
physicsID: jee-main
The distance of centre of mass from end of a one dimensional rod ( ) having mass density and length L. (in meter) is . The value of is......... (where is the distance form end )
Official Solution
Correct Option: (1)
⇒
44
PYQ 2023
easy
physicsID: jee-main
A solid cylinder of radius R and length L have moment of inertia I1 and a second solid cylinder of radius R2 and length L2 cut from it have moment of inertia I2. Find 11/I2.
A solid sphere and a ring have equal masses and equal radius of gyration. If the sphere is rotating about its diameter and ring about an axis passing through and perpendicular to its plane, then the ratio of radius is then find the value of x.
Official Solution
Correct Option: (1)
Therefore, the value of x is 5.
46
PYQ 2023
medium
physicsID: jee-main
A body is rotating with kinetic energy E. If angular velocity of body is increased to three times of initial angular velocity then kinetic energy become nE. Find n.
Official Solution
Correct Option: (1)
The kinetic energy of a rotating body can be expressed as: E = where, E is the initial kinetic energy, I is the moment of inertia, w is the initial angular velocity. If the angular velocity is increased by a factor of three, the new angular velocity is 3w. The kinetic energy of the body can be expressed as: nE = = = 9E Therefore, n = 9. So, the kinetic energy becomes nine times the initial kinetic energy when the angular velocity is increased to three times the initial angular velocity. Answer. 9
47
PYQ 2024
medium
physicsID: jee-main
A particle is moving in circular path of radius r speed v such that speed is proportional to radius as . Then how does time period of revolution depends on r i.e Trn then is.
1
2
3
4
Official Solution
Correct Option: (2)
The correct option is (B):
48
PYQ 2024
medium
physicsID: jee-main
A particle moves in a straight line so that its displacement at any time is given by . Its acceleration at any time is where ____.
Official Solution
Correct Option: (1)
The problem provides the relationship between the displacement of a particle and time as . We are asked to find the value of if the particle's acceleration is given by the expression .
Concept Used:
The solution involves finding the acceleration of the particle, which is the second derivative of displacement with respect to time. The key steps are:
Velocity ( ): The velocity is the first derivative of displacement with respect to time, .
Acceleration ( ): The acceleration is the second derivative of displacement, or the first derivative of velocity, with respect to time, .
Implicit Differentiation: Since the displacement is not given explicitly as a function of , we use implicit differentiation to find the derivatives.
Quotient Rule: The quotient rule for differentiation, , is used to find the derivative of the velocity.
Step-by-Step Solution:
Step 1: Find the velocity ( ) by differentiating the given equation with respect to time .
The given relation is:
Differentiating both sides with respect to :
Using the chain rule on the left side:
Since , we have:
Solving for :
Step 2: Find the acceleration ( ) by differentiating the velocity with respect to time .
Acceleration is . We use the quotient rule to differentiate :
We know that and . So,
Step 3: Substitute the expression for velocity into the acceleration equation.
Substitute into the equation for acceleration:
To simplify the complex fraction, multiply the numerator and denominator by :
Step 4: Express the acceleration solely in terms of .
From the original equation, , we can express as . Substitute this into the expression for acceleration:
Step 5: Compare the result with the given form to find .
The calculated acceleration is , which can be written as:
We are given that the acceleration is of the form . By comparing the two expressions,
we can conclude that .
The value of is 3.
49
PYQ 2025
medium
physicsID: jee-main
The center of mass of a thin rectangular plate (fig - x) with sides of length and , whose mass per unit area ( ) varies as (where is a constant), would be
1
2
3
4
Official Solution
Correct Option: (1)
Using the variable density equation and integrating over the entire area, the -coordinate of the center of mass is given by . After integration and applying the mass distribution, the resulting coordinates for and are and respectively.
50
PYQ 2025
medium
physicsID: jee-main
The temperature of a body in air falls from to in 4 minutes. The temperature of the air is . The temperature of the body in the next 4 minutes will be:
1
2
3
4
Official Solution
Correct Option: (1)
This problem involves Newton's Law of Cooling, which can be expressed as:
where is the temperature of the body, is the ambient temperature (temperature of the air), and is a constant. The temperature changes from to in 4 minutes. Using Newton's Law of Cooling, we can compute the constant and then apply it to determine the temperature change in the next 4 minutes. Based on the given information and applying the necessary calculations, the temperature after 4 more minutes is:
Final Answer: .
51
PYQ 2025
hard
physicsID: jee-main
The temperature of a body in air falls from to in 4 minutes. The temperature of the air is . The temperature of the body in the next 4 minutes will be:
1
28/3 °C
2
14/3 °C
3
56/3 °C
4
42/3 °C
Official Solution
Correct Option: (1)
This problem involves Newton's Law of Cooling, which can be expressed as: where is the temperature of the body, is the ambient temperature (temperature of the air), and is a constant. The temperature changes from to in 4 minutes. Using Newton's Law of Cooling, we can compute the constant and then apply it to determine the temperature change in the next 4 minutes. Based on the given information and applying the necessary calculations, the temperature after 4 more minutes is:
Final Answer:
52
PYQ 2025
medium
physicsID: jee-main
The increase in pressure required to decrease the volume of a water sample by 0.2percentage is . Bulk modulus of water is . The value of is ________________.
Official Solution
Correct Option: (1)
Step 1: Understand the Given Information
The problem provides the following information:
The percentage change in volume of water is 0.2% (i.e., the volume decreases by 0.2%).
The bulk modulus of water is given as .
The pressure change required to decrease the volume by this percentage is given by , and we are asked to find the value of .
Step 2: Formula for Bulk Modulus
The bulk modulus is related to the change in pressure and the corresponding fractional change in volume by the formula:
where: - is the bulk modulus, - is the change in pressure, - is the fractional change in volume.
Step 3: Substitute the Given Values
We are given the fractional change in volume as 0.2%, which is equal to . The bulk modulus is given as , so we can write the equation as: Solving for , we get:
Step 4: Relating the Pressure Change to
We are given that the pressure change is , so: Solving for , we get:
Conclusion
The value of is 43.
53
PYQ 2025
medium
physicsID: jee-main
A string of length is fixed at one end and carries a mass of at the other end. The mass makes rotations per second about the vertical axis passing through the end of the string as shown. The tension in the string is ________________ ML.
Official Solution
Correct Option: (1)
Step 1: Understand the Given Information
The problem involves a rotating mass attached to a string of length , making rotations per second. The question asks for the tension in the string. The mass describes a circular path with a radius , and the string makes an angle with the vertical axis.
Step 2: Angular Velocity
The number of rotations per second (frequency) is given as rotations per second. The angular velocity is related to the frequency by:
Step 3: Centripetal Force
The mass is undergoing circular motion with a radius . The centripetal force required to keep the mass in its circular path is given by:
Substituting the value of (which is 6 rad/s), we get:
Step 4: Tension and Vertical Component
The tension in the string has both vertical and horizontal components. The vertical component of the tension balances the gravitational force acting on the mass:
The horizontal component provides the centripetal force:
Step 5: Finding Tension
Using the relationship between the vertical and horizontal components of tension:
This simplifies to:
Step 6: Final Expression for Tension
Finally, the tension in the string can be expressed as:
Conclusion
The tension in the string is .
54
PYQ 2025
hard
physicsID: jee-main
The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in a similar way. The moment of inertia of a solid sphere which has the same radius as the disc and rotating in similar way, is times higher than the moment of inertia of the given ring. Here, :
Official Solution
Correct Option: (1)
Step 1: Moment of Inertia of the Ring
For a ring rotating about its diameter:
Step 2: Moment of Inertia of the Disc
For a solid disc rotating about its diameter:
According to the problem, the moment of inertia of the disc is 2.5 times the moment of inertia of the ring:
Substituting the values:
Solving this will confirm the relationship.
Step 3: Moment of Inertia of the Solid Sphere
For a solid sphere rotating about its diameter:
Step 4: Finding the Ratio
The moment of inertia of the solid sphere is times the moment of inertia of the ring.
Substituting the values:
Therefore,
55
PYQ 2025
medium
physicsID: jee-main
The center of mass of a thin rectangular plate (fig - x) with sides of length and , whose mass per unit area ( ) varies as (where is a constant), would be
1
2
3
4
Official Solution
Correct Option: (1)
Let the rectangular plate be defined by and .
The mass per unit area is given as:
$ (x_{cm}, y_{cm}) M M x_{cm} y_{cm} \left(\frac{2}{3}a, \frac{b}{2}\right) $
56
PYQ 2025
medium
physicsID: jee-main
Two iron solid discs of negligible thickness have radii and and moment of inertia and , respectively. For , the ratio of and would be , where is:
Official Solution
Correct Option: (1)
Step 1: Formula for Moment of Inertia of a Solid Disc
The moment of inertia of a solid disc about its central axis (perpendicular to its plane) is:
Step 2: Moment of Inertia of Discs and
For the first disc: For the second disc:
Step 3: Given Condition
It is given that:
Since the discs are made of iron (same material), the mass is proportional to the area (assuming thickness is negligible). So:
Step 4: Finding the Ratio of and
Substituting the values:
Ratio :
Therefore,
57
PYQ 2025
hard
physicsID: jee-main
Two iron solid discs of negligible thickness have radii and and moment of inertia and , respectively. For , the ratio of and would be , where is:
Official Solution
Correct Option: (1)
The moment of inertia of a disc is given by . Step 1: For the two discs, we have:
Step 2: For , we substitute into the equation for :
Step 3: Thus, the ratio of and is , meaning . Final Conclusion:
The value of is 4, which corresponds to Option (2).
58
PYQ 2025
medium
physicsID: jee-main
The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in a similar way. The moment of inertia of a solid sphere which has the same radius as the disc and rotating in similar way, is times higher than the moment of inertia of the given ring. Here, :
Official Solution
Correct Option: (1)
For a disc, the moment of inertia about its diameter is , and for a ring, the moment of inertia is .
Step 1: The moment of inertia of a solid sphere about its center is .
Step 2: The relationship between the sphere and the ring's moment of inertia is:
Step 3: Substituting values, we find that the ratio is 1, so .
Final Conclusion: The value of is 1, which corresponds to Option (1).
59
PYQ 2025
medium
physicsID: jee-main
Match List - I with List - II:List - I:
(A) Electric field inside (distance from center) of a uniformly charged spherical shell with surface charge density , and radius .
(B) Electric field at distance from a uniformly charged infinite plane sheet with surface charge density .
(C) Electric field outside (distance from center) of a uniformly charged spherical shell with surface charge density , and radius .
(D) Electric field between two oppositely charged infinite plane parallel sheets with uniform surface charge density .
List - II:
(I)
(II)
(III) 0
(IV)
Choose the correct answer from the options given below:
1
2
3
4
Official Solution
Correct Option: (1)
- The electric field inside a uniformly charged spherical shell is 0 (Coulomb’s Law), hence . - The electric field due to a uniformly charged infinite plane sheet is , hence . - The electric field outside a uniformly charged spherical shell behaves like that of a point charge and is , hence . - The electric field between two oppositely charged infinite plane sheets is , hence . Thus, the correct answer is .
60
PYQ 2025
medium
physicsID: jee-main
The increase in pressure required to decrease the volume of a water sample by 0.2percentage is . Bulk modulus of water is . The value of is ………..
Official Solution
Correct Option: (1)
The bulk modulus is defined as the ratio of the increase in pressure to the relative decrease in volume :
Rearranging the equation to solve for , we get:
The negative sign indicates a decrease in volume, but we are concerned only with the magnitude. Given that and the volume change is , we can calculate as:
Thus, the value of is . Final Answer: 4.6.
61
PYQ 2025
medium
physicsID: jee-main
A string of length is fixed at one end and carries a mass of at the other end. The mass makes rotations per second about the vertical axis passing through the end of the string as shown. The tension in the string is \_\_\_\_ ML.
Official Solution
Correct Option: (1)
In this problem, the system consists of a rotating mass at the end of a string. The mass is undergoing circular motion, so we can apply the formula for centripetal force to find the tension in the string. The tension in the string provides the centripetal force required for the circular motion. The centripetal force is given by:
where:
- is the mass,
- is the angular velocity,
- is the radius of the circular path. The angular velocity can be related to the number of rotations per second (which is rotations per second) by:
Substituting the given value for , we get:
Thus, the tension in the string is:
Final Answer: .
62
PYQ 2025
easy
physicsID: jee-main
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank is _________ cm. (Take ).
Official Solution
Correct Option: (1)
The observer on the river bank sees the ball’s velocity as a combination of the velocity of the boat and the upward velocity of the ball. The horizontal velocity of the ball is the same as the velocity of the boat, which is . The time for the ball to reach its maximum height is given by:
The horizontal range is the horizontal velocity multiplied by the time of flight:
Thus, the correct answer is .
63
PYQ 2025
medium
physicsID: jee-main
A container of fixed volume contains a gas at 27°C. To double the pressure of the gas, the temperature of the gas should be raised to …. C.
Official Solution
Correct Option: (1)
For a gas in a fixed volume, the pressure is directly proportional to the temperature (in Kelvin) according to Gay-Lussac’s Law:
Let the initial temperature be . To double the pressure, the final temperature should be , which is equivalent to . Thus, the correct answer is .
64
PYQ 2026
medium
physicsID: jee-main
A cylindrical tube of length , closed at both ends, contains an ideal gas of mol having molecular weight . The tube is rotated in a horizontal plane with constant angular velocity about an axis perpendicular to and passing through the edge at end , as shown in the figure. If and are the pressures at and respectively, then (consider the temperature to be same at all points in the tube)
1
2
3
4
Official Solution
Correct Option: (1)
The tube is rotating with angular velocity about point . A gas molecule at a distance from experiences a centrifugal force given by
where is the mass of a molecule. Step 1: Pressure gradient due to centrifugal force. For equilibrium of a thin gas element of thickness ,
where is the mass density of the gas. Using the ideal gas relation,
we get
Step 2: Integrate from to . At , and pressure is . At , and pressure is . Step 3: Final expression. Taking exponential on both sides,
Final Answer:
65
PYQ 2026
medium
physicsID: jee-main
As shown in the figure, radius of gyration about the axis shown in cm for a solid sphere. Find 'n'.
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: The radius of gyration is defined by the relation . For an axis not passing through the center of mass, we use the parallel axis theorem. Step 2: Key Formula or Approach: 1. Moment of inertia of a solid sphere about its center: . 2. Parallel axis theorem: . 3. Radius of gyration: . Step 3: Detailed Explanation: Given: Radius and distance to the axis . The total moment of inertia about the given axis is:
Since , we can write:
Substitute the given values into the equation:
Therefore, . Comparing this with , we find . Step 4: Final Answer: The value of is 265.
66
PYQ 2026
medium
physicsID: jee-main
When rod becomes horizontal find its angular velocity. It is pivoted at point A as shown.
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Identifying the system. A uniform rod of length is pivoted at point A. When released, it rotates under gravity. Step 2: Change in potential energy. The center of mass falls by a vertical distance of .
Step 3: Rotational kinetic energy. Moment of inertia of rod about point A is:
