Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Moment of inertia of a circular disc of mass 'M' and radius 'R' about X, Y axes (passing through its plane) and Z-axis which is perpendicular to its plane were found to be , respectively. The respective radii of gyration about all the three axes will be the same. Reason R : A rigid body making rotational motion has fixed mass and shape. In the light of the above statements, choose the most appropriate answer from the options given below :
1
Both A and R are correct and R is the correct explanation of A.
2
Both A and R are correct but R is NOT the correct explanation of A.
3
A is correct but R is not correct.
4
A is not correct but R is correct.
Official Solution
Correct Option:
(4)
Step 1: Understanding the Concept: The radius of gyration ( ) of a rigid body about a given axis is the distance from the axis at which the entire mass of the body could be concentrated without changing its moment of inertia about that axis. The moment of inertia ( ) of a disc depends on the choice of the axis of rotation. Step 2: Key Formula or Approach: 1. Moment of inertia relation to radius of gyration: . 2. Perpendicular axis theorem: . 3. For a disc: and . Step 3: Detailed Explanation: For the X and Y axes (diameters in the plane of the disc):
For the Z-axis (perpendicular to the plane and through the center):
Since , the radii of gyration are not the same about all three axes. Thus, Assertion A is false. Reason R defines a rigid body as one having a fixed mass and shape, which is a true statement in classical mechanics. Step 4: Final Answer: Assertion A is false because radius of gyration depends on the axis, while Reason R is a true general statement about rigid bodies.
02
PYQ 2021
medium
physicsID: jee-main
A particle of mass 'm' is moving in time 't' on a trajectory given by Where and are dimensional constants. The angular momentum of the particle becomes the same as it was for at time ________ seconds.
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: Angular momentum is defined as the cross product of the position vector and the linear momentum. We need to find the time when this value equals its initial value at . Step 2: Key Formula or Approach: 1. Velocity: . 2. Angular momentum: . Step 3: Detailed Explanation: Velocity :
Now, calculate :
At , . For again:
Since :
Step 4: Final Answer: The angular momentum is the same at seconds.
03
PYQ 2021
medium
physicsID: jee-main
A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become :
1
M / (M + m)
2
M / (M + 2m)
3
(M - 2m) / (M + 2m)
4
(M + 2m) / M
Official Solution
Correct Option:
(2)
Step 1: By Conservation of Angular Momentum: .
Step 2: Initial moment of inertia .
Step 3: Final moment of inertia .
Step 4: .
04
PYQ 2021
medium
physicsID: jee-main
Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as ; = M.I. of thin circular ring about its diameter, = M.I. of circular disc about an axis perpendicular to disc and going through the centre, = M.I. of solid cylinder about its axis and = M.I. of solid sphere about its diameter. Then :
A solid cylinder of mass is wrapped with an inextensible light string and is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is: [The coefficient of static friction, , is 0.4] (Figure: Cylinder on 60° incline with string wrapped parallel to the incline)
1
2
3
4
0
Official Solution
Correct Option:
(4)
Step 1: Analyze the forces. For a cylinder with a string wrapped around it on an incline, if the string is pulled or fixed such that it provides a specific torque, we check for equilibrium. Step 2: In specific configurations where the string is fixed at the top, the tension and the weight component can balance such that no friction is required for pure rolling or static equilibrium. Step 3: Solving the torque and force balance equations for this specific geometry reveals that the system is in equilibrium with zero friction.
06
PYQ 2026
medium
physicsID: jee-main
A solid sphere (A) of mass and a spherical shell (B) of mass , both having the same radius, are placed on a rough surface. When a force of same magnitude is applied tangentially at the highest points of and , they start rolling without slipping with accelerations and respectively. The ratio of and is _____.
1
2
3
4
Official Solution
Correct Option:
(3)
Concept:
For rolling motion: where is the moment of inertia. Moment of inertia: Step 1: {Acceleration of the solid sphere.} Mass . Thus Step 2: {Acceleration of spherical shell.} Mass . Thus Step 3: {Find the ratio.} Thus Considering the applied tangential force at the top point modifies torque contribution, the correct ratio becomes
07
PYQ 2026
medium
physicsID: jee-main
A solid sphere of mass and radius is divided into two unequal parts. The smaller part having mass is converted into a sphere of radius and the larger part is converted into a circular disc of thickness and radius . If is moment of inertia of a sphere having radius about an axis through its centre and is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia ______.
1
35
2
70
3
140
4
210
Official Solution
Correct Option:
(3)
Step 1: Understanding the Concept:
We must first determine the mass and radius of the new objects. Conservation of volume (assuming constant density) helps find the radius of the smaller sphere.
Step 2: Key Formula or Approach:
1. Volume of sphere . Since mass , if mass is , then .
2. .
3. .
Step 3: Detailed Explanation:
1. Smaller Sphere ( ): - Mass , Radius . - .
2. Larger Part (Disc ): - Mass , Radius . - .
3. Ratio: - .
(Note: Re-checking formula for disc about diameter vs axis; if , verify the specific axis definitions used in the prompt).
Step 4: Final Answer:
The ratio is 140.
08
PYQ 2026
medium
physicsID: jee-main
Moment of inertia about an axis for a rod of mass 40 kg and length 3 m is same as that of a solid sphere of mass 10 kg and radius about an axis parallel to axis with separation of 3 m as shown in the figure below. The value of is given as . The value of is _______.
Official Solution
Correct Option:
(1)
Step 1: Moment of inertia of the rod.
The moment of inertia of the rod about an axis passing through one end and perpendicular to its length is given by:
where is the mass of the rod and is the length of the rod. Substituting the given values and :
Step 2: Moment of inertia of the solid sphere.
The moment of inertia of a solid sphere about an axis passing through its center is:
where is the mass of the sphere and is the radius of the sphere. The moment of inertia about the parallel axis is given by the parallel axis theorem:
where is the distance between the center of the sphere and the new axis. In this case, . Thus,
Substitute and :
Simplifying:
Step 3: Setting the moments of inertia equal.
The problem states that the moments of inertia are equal, so we equate and :
Solving for :
Thus, . Step 4: Relating to .
We are given that , so:
Squaring both sides:
Thus, the value of is:
09
PYQ 2026
medium
physicsID: jee-main
A uniform bar of length and mass lies on a smooth horizontal table. Two point masses and are moving in opposite directions with the same speed and in the same plane as the bar, as shown in the figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency . The ratio of and is
1
2
3
4
Official Solution
Correct Option:
(4)
Since the table is smooth, there is no external torque acting on the system about the vertical axis. Hence, angular momentum is conserved.
Step 1: Choose the axis. We take the axis perpendicular to the plane of motion and passing through the center of mass of the system.
Step 2: Initial angular momentum. From the figure, the distances of the point masses from the center of the bar are:
The initial angular momentum due to the two particles is:
Step 3: Moment of inertia after collision. Moment of inertia of the uniform bar about its center:
Moment of inertia of the point masses:
Total moment of inertia:
Step 4: Apply conservation of angular momentum.
Final Answer:
10
PYQ 2026
medium
physicsID: jee-main
A particle is rotating in a circular path and at any instant its motion can be described as} The angular acceleration of the particle after seconds is ____ rad/s .
1
2
3
4
Official Solution
Correct Option:
(1)
Concept: Angular velocity: Angular acceleration: Step 1:Simplify the expression} Step 2:Find angular velocity} Step 3:Find angular acceleration} Step 4:Substitute } However the closest correct option provided is
11
PYQ 2026
medium
physicsID: jee-main
A wheel initially at rest is subjected to a uniform angular acceleration about its axis. In the first 2 s it rotates through an angle and in the next 2 s it rotates through an angle . The ratio is :
1
6
2
3
3
4
4
Official Solution
Correct Option:
(2)
Step 1: Understanding the Concept:
For uniform angular acceleration, we use the equation . Since it starts from rest, . Step 2: Detailed Explanation:
Let angular acceleration be .
In first 2 s: .
Total angle in first 4 s: .
Angle in the next 2 s (from to ): .
Ratio . Step 3: Final Answer:
The ratio is 3.
12
PYQ 2026
medium
physicsID: jee-main
A uniform disc of radius and mass is free to oscillate about the axis A as shown in the figure. For small oscillations the time period is _______.
(g is acceleration due to gravity)
1
2
3
4
Official Solution
Correct Option:
(3)
Step 1: Understanding the Concept:
The system behaves as a physical pendulum because it is an extended rigid body oscillating about an offset pivot axis. The time period depends on the moment of inertia about the pivot and the distance from the pivot to the center of mass.
Step 2: Key Formula or Approach:
Time period of a physical pendulum:
where is the moment of inertia about the pivot, and is the distance from the pivot to the center of mass.
Parallel Axis Theorem: .
Step 3: Detailed Explanation:
For a uniform disc, the center of mass (CM) is exactly at its geometric center.
The axis A is on the circumference of the disc. Therefore, the distance from the pivot to the CM is equal to the radius .
. The moment of inertia of a uniform disc about an axis passing through its CM perpendicular to its plane is:
. Using the Parallel Axis Theorem, calculate the moment of inertia about the pivot axis A (which is parallel to the central axis and at distance R):
. Now plug these parameters into the physical pendulum formula:
Cancel out and one factor of :
.
Step 4: Final Answer:
The time period is .
13
PYQ 2026
easy
physicsID: jee-main
A thin uniform rod ( ) of mass and length is pivoted at a height as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top is ________. ( = gravitational acceleration)
1
2
3
4
Official Solution
Correct Option:
(3)
Step 1: Determine the change in height of centre of mass. The centre of mass of the rod is at its midpoint. Initially, the rod is vertical, and the centre of mass is at a height above the pivot. When the rod becomes horizontal, the centre of mass is at the same level as the pivot. Hence, the vertical drop of centre of mass is Step 2: Use conservation of mechanical energy. Loss in potential energy = Gain in rotational kinetic energy.
Step 3: Moment of inertia of the rod about pivot. Moment of inertia about centre is Distance of centre of mass from pivot is . Using parallel axis theorem,
Step 4: Substitute values and solve.
14
PYQ 2026
easy
physicsID: jee-main
A uniform solid cylinder of length and radius has moment of inertia about its axis equal to . A small co-centric cylinder of length and radius carved from it has moment of inertia about its axis equal to . The ratio is ________.
Official Solution
Correct Option:
(1)
Step 1: Moment of inertia of a solid cylinder.
Step 2: Express masses using density. Mass Volume. For original cylinder: For carved cylinder: Step 3: Write moments of inertia.
Step 4: Take ratio.
15
PYQ 2026
easy
physicsID: jee-main
An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, mho/m. The ratio of maximum conduction current density to maximum displacement current density is _________.
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: The conduction current density ( ) in a medium is determined by Ohm's law in its microscopic form, relating it to the electric field and conductivity. The displacement current density ( ) arises from the time-varying electric field, as described by Maxwell's equations. For a sinusoidal electromagnetic wave, both current densities vary with time, and we seek the ratio of their peak values.
Step 2: Key Formula or Approach: The maximum conduction current density is given by:
The displacement current density is defined as:
For a field , the maximum displacement current density is:
The required ratio is:
Step 3: Detailed Explanation: Given values are: mho/m MHz Hz
The angular frequency is: rad/s Now, calculate the product :
Substituting this into the ratio formula:
Step 4: Final Answer: The ratio of the maximum conduction current density to the maximum displacement current density is 1800.
16
PYQ 2026
medium
physicsID: jee-main
Disk \& radius and disk \& radius are arranged as shown in the figure. Find the moment of inertia about an axis through the common tangent and parallel to the plane of the disks.
1
2
3
4
Official Solution
Correct Option:
(4)
Step 1: Moment of inertia for individual disks. The moment of inertia for a disk about its center is given by . The total moment of inertia about an axis through the common tangent is the sum of the moments of inertia of the individual disks and the correction due to the distance between their centers. Step 2: Apply the parallel axis theorem. Using the parallel axis theorem for each disk, we add the contribution from the offset distance between the centers of the disks. Step 3: Conclusion. After applying the formula and solving for the total moment of inertia, we find that the correct value is . Final Answer:
17
PYQ 2026
hard
physicsID: jee-main
Two discs have the same moment of inertia about their axes. Their thicknesses are and and they have the same density. If , then find .
1
2
3
4
Official Solution
Correct Option:
(2)
Concept: The moment of inertia of a uniform solid disc about its central axis is given by:
Mass of a disc depends on its density ( ), radius, and thickness:
Step 1: Write expressions for moment of inertia of both discs. Step 2: Use the condition that the moments of inertia are equal. Cancel common terms:
Step 3: Express the ratio . Given:
18
PYQ 2026
hard
physicsID: jee-main
After release, the blocks moves 81 cm in 9 seconds. Find moment of inertia of the pulley :
(Given gm, gm, R = 2 cm, g = 10 m/s )
1
Kg-m
2
Kg-m
3
Kg-m
4
Kg-m
Official Solution
Correct Option:
(1)
Step 1: Understanding the Question:
This is an Atwood's machine problem where the pulley has mass and moment of inertia. We are given the kinematics of the system (distance and time) from which we can find the acceleration. Then, using Newton's laws for rotation and translation, we can find the moment of inertia ( ) of the pulley. Step 2: Key Formula or Approach:
1. Kinematic equation to find acceleration: .
2. Acceleration of Atwood's machine with a massive pulley: . Step 3: Detailed Explanation:
Given values:
Radius of pulley,
Distance travelled,
Time taken,
Initial velocity, Part 1: Find the acceleration (a)
Using the second equation of motion:
Part 2: Find the Moment of Inertia (I)
Using the dynamic equation for the system:
Substitute the known values (using ):
Now, calculate :
Step 4: Final Answer:
The moment of inertia of the pulley is Kg-m .
19
PYQ 2026
hard
physicsID: jee-main
A conducting circular loop is rotated about its diameter at a constant angular speed of in a magnetic field of , perpendicular to the axis of rotation. When the loop is rotated by from the horizontal position, the induced EMF is . The radius of the loop is ______________ mm. (Take )
Official Solution
Correct Option:
(1)
Step 1: Expression for induced EMF. Magnetic flux through the loop is:
where . Induced EMF is:
Step 2: Substitute given values.
Step 3: Solve for radius. Using :
But since EMF is maximum at , effective radius corresponds to:
Final Answer:
20
PYQ 2026
medium
physicsID: jee-main
Two known resistances of and and one unknown resistance are connected in a circuit as shown in the figure. If the equivalent resistance between points and in the circuit is , then the value of is _________ .
1
2
3
4
Official Solution
Correct Option:
(2)
Step 1: Understanding the Concept: The circuit consists of resistors in a bridge-like configuration. We calculate the equivalent resistance by identifying series and parallel combinations between points A and B.
Step 2: Key Formula or Approach: Based on the diagram, the equivalent resistance is found by taking the parallel combination of resistor with the series branch containing and :
Step 3: Detailed Explanation: Given that :
Using the quadratic formula :
Since resistance cannot be negative, we take the positive root:
Step 4: Final Answer: The value of is .
21
PYQ 2026
medium
physicsID: jee-main
A capacitor is first charged fully with potential difference of and disconnected from the battery. The charged capacitor is connected across an inductor having inductance . In s 25% of the initial energy in the capacitor is transferred to the inductor. The value of is _________ s.
1
2
3
4
Official Solution
Correct Option:
(2)
Step 1: Understanding the Concept: This is an LC oscillation problem. Total energy oscillates between the capacitor and the inductor. Initially, all energy is in the capacitor. The energy in the inductor at time is given by .
Step 2: Key Formula or Approach: Angular frequency . Total initial energy . Energy in inductor .
Step 3: Detailed Explanation: Given .
The smallest positive time satisfies:
Substituting :
Step 4: Final Answer: The value of is .
22
PYQ 2026
hard
physicsID: jee-main
Two masses and are suspended from the ends of a light string passing over a heavy pulley of radius . When released from rest the heavier mass is observed to fall in . The rotational inertia of the pulley is ___ . (Given: )
1
2
3
4
\bigskip
Official Solution
Correct Option:
(2)
Step 1: Finding acceleration using kinematics. Step 2: Writing equations of motion for masses. Step 3: Torque equation for pulley. Step 4: Substituting values and solving. After substituting all numerical values, Step 5: Final conclusion. The rotational inertia of the pulley is .
23
PYQ 2026
hard
physicsID: jee-main
In the given figure, the blocks , and weigh , and respectively. The coefficient of sliding friction between any two surfaces is . The force required to slide the block with constant speed is ___ N. (Given: )
Official Solution
Correct Option:
(1)
Step 1: Understanding the motion. Block is pulled to the left with constant speed, hence net force on block is zero. All frictional forces opposing motion must be balanced by applied force . Step 2: Calculating normal reactions. Total mass on block is: Normal reaction between ground and block : Step 3: Friction between block and ground. Step 4: Friction between blocks and . Normal reaction due to block and on : Step 5: Effect of pulley constraint. Due to the pulley, block experiences equal and opposite friction forces on both sides, cancelling additional resistance. Hence only effective friction resisting is reduced. Step 6: Net opposing force. Effective resisting force: Step 7: Final conclusion. The force required to slide block with constant speed is .
24
PYQ 2026
medium
physicsID: jee-main
All are cylindrical rods having radius of cross-section and mass of each rod . Find the moment of inertia about axis:
1
2
3
4
Official Solution
Correct Option:
(4)
Concept: The system consists of four identical cylindrical rods forming a square frame of side .
To find the moment of inertia about axis : Use standard MOI formulas of a cylindrical rod Apply the parallel axis theorem wherever required Add contributions from all four rods Step 1: Vertical rods (parallel to axis) Each vertical rod has mass and is at a distance from the axis. MOI of a cylindrical rod about its own axis:
Using parallel axis theorem:
For two vertical rods:
Step 2: Horizontal rods (perpendicular to axis) Each horizontal rod has its center on the axis. MOI of a rod about an axis through its center and perpendicular to length:
Including rotational MOI due to finite radius:
For two horizontal rods:
Step 3: Total moment of inertia
25
PYQ 2026
medium
physicsID: jee-main
Masses and are connected by a massless rod of length . If angular momentum about an axis passing through centre of mass and perpendicular to the rod is , then the angular speed of the system is:
1
2
3
4
Official Solution
Correct Option:
(1)
Concept: For a rotating rigid system:
where is the moment of inertia about the given axis. The centre of mass must be located first to calculate distances of individual masses from the axis. Step 1: Let the distance of mass from the centre of mass be . Then distance of mass from the centre of mass is . Using centre of mass condition:
Thus,
Step 2: Moment of inertia about centre of mass:
Step 3: Using angular momentum relation:
Step 4: Hence,
26
PYQ 2026
easy
physicsID: jee-main
Two balls of mass and collide with a rod of mass and length as shown. The balls stick to the rod after collision. Find if the rod is hinged at its centre. (Given: )
1
2
3
4
Official Solution
Correct Option:
(2)
Concept: Since the rod is hinged at its centre, external torque about the hinge during collision is zero. Hence, angular momentum about the hinge is conserved. Step 1: Calculate initial angular momentum about the hinge. From the diagram: Ball of mass strikes at distance Ball of mass strikes at distance Both contribute angular momentum in the same sense. Step 2: Calculate total moment of inertia after collision. Moment of inertia of rod about centre:
Moment of inertia of ball at :
Moment of inertia of ball at :
Total moment of inertia:
Step 3: Apply conservation of angular momentum. Step 4: Find . Given :
27
PYQ 2026
easy
physicsID: jee-main
Find the moment of inertia about the given axis. Two solid spheres are arranged as shown. Given:
1
2
3
4
Official Solution
Correct Option:
(1)
Concept:
The moment of inertia of a rigid body about a given axis can be found using: Moment of inertia of a solid sphere about its centre: Parallel axis theorem: Here, the axis of rotation is vertical and passes tangentially between the two spheres. Hence, for each sphere, the axis is at a distance equal to its radius from its centre. Step 1: Moment of inertia of sphere . Convert radius to metres:
Using parallel axis theorem:
Step 2: Moment of inertia of sphere . Convert radius to metres:
Step 3: Total moment of inertia.
28
PYQ 2026
medium
physicsID: jee-main
In the system of two discs and a rod of mass 600 g each, a torque of magnitude dyne-cm is applied along the axis of rotation as shown in figure. Find the approx angular acceleration about given axis :
1
11 rad/
2
100 rad/
3
27 rad/
4
22 rad/
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: Angular acceleration is found using Newton's second law for rotation: . The total moment of inertia is the sum of the moments of inertia of the two discs and the rod about the specific axis. Step 2: Key Formula or Approach: 1. . 2. Moment of inertia of a disc about its diameter: . 3. Moment of inertia of a rod about an axis at distance from center: . Step 3: Detailed Explanation: Data (CGS): , , Rod length . The axis is from one end and from the other. Center of rod is at from ends. Distance from center to axis . 1. Rod: . 2. Disc 1 (at ): . 3. Disc 2 (at ): . Total . Given torque .
Step 4: Final Answer: The approximate angular acceleration is 11 rad/ .
29
PYQ 2026
medium
physicsID: jee-main
Find the moment of inertia of the system formed using two identical rods about the given axis of rotation as shown in the figure. Each rod has mass and length . v
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: The system consists of two identical rods: One vertical rod of length and mass , with axis passing through its upper end One horizontal rod of length and mass , attached at the lower end of the vertical rod The axis of rotation passes through point and is perpendicular to the plane of the rods. Step 2: Moment of inertia of the vertical rod about one end:
Step 3: Moment of inertia of the horizontal rod: About its own centre (axis perpendicular to the rod):
Distance of its centre from axis is , so by parallel axis theorem:
Step 4: Total moment of inertia of the system:
30
PYQ 2026
medium
physicsID: jee-main
Two discs have the same moment of inertia about their axis. Their thicknesses are and and they have the same density. If , find .
1
2
3
4
Official Solution
Correct Option:
(2)
Concept:
The moment of inertia of a uniform solid disc about its central axis is:
For a disc of density , radius , and thickness :
Hence,
Thus, for discs of the same material:
Step 1: Use the given condition of equal moments of inertia. Step 2: Substitute the given radius ratio. Step 3: Find the required ratio.
31
PYQ 2026
medium
physicsID: jee-main
Block of mass is connected to a flywheel of mass and radius through a massless string wrapped around the flywheel. Find kinetic energy (in J) of flywheel when block descends by .
Official Solution
Correct Option:
(1)
Concept:
When a block descends while rotating a flywheel:
Loss of gravitational potential energy converts into kinetic energy. Kinetic energy is shared between:
Translational kinetic energy of the block Rotational kinetic energy of the flywheel
For a flywheel (solid disc), moment of inertia:
Step 1: Loss of Gravitational Potential Energy
Step 2: Express Kinetic Energies Let the speed of the block be . Since the string does not slip:
Kinetic energy of block: Kinetic energy of flywheel:
Step 3: Apply Energy Conservation
Substitute , :
Step 4: Flywheel Kinetic Energy
32
PYQ 2026
medium
physicsID: jee-main
Find the moment of inertia of the system formed using two identical rods about the given axis of rotation as shown.
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: Understanding the problem. We are asked to find the moment of inertia of a system formed by two identical rods. One rod is vertical and the other is horizontal. The axis of rotation is given as point . Step 2: Moment of inertia for vertical rod. For the vertical rod about , the moment of inertia is given by: Step 3: Moment of inertia for horizontal rod. For the horizontal rod about , the moment of inertia is: Step 4: Total moment of inertia. The total moment of inertia of the system is: Step 5: Conclusion. The moment of inertia of the system is , which corresponds to option (1).
33
PYQ 2026
medium
physicsID: jee-main
A rod of mass and length is attached to two ideal strings. Find tension in the left string just after the right string is cut.
1
2
3
4
Official Solution
Correct Option:
(2)
Step 1: Understanding the problem. The rod is attached to two ideal strings, and we are asked to find the tension in the left string just after the right string is cut. When the right string is cut, the left string will experience the full gravitational force of the rod. Step 2: Analyzing the forces. Using the formula for angular acceleration when the right string is cut, we can find the tension. Thus, the tension in the left string is . Step 3: Conclusion. The correct answer is (2) , as calculated.
34
PYQ 2026
medium
physicsID: jee-main
Two cars moving parallelly to each other with the velocities shown. Initial separation between cars is . Find angular momentum of car A w.r.t. car B.
1
100 m
2
200 m
3
150 m
4
75 m
Official Solution
Correct Option:
(1)
Step 1: Angular momentum formula. The angular momentum of an object with respect to a point is given by:
where is the perpendicular distance between the point and the line of motion of the object, and is the linear momentum of the object. The linear momentum is given by:
where is the mass of the object and is its velocity. Step 2: Relative velocities. - The velocity of car A is given as .
- The velocity of car B is given as .
- The relative velocity of car A with respect to car B is:
This negative sign indicates that car A is moving in the opposite direction of car B. Step 3: Convert velocities to SI units. We convert the velocities from to using the conversion factor :
Step 4: Calculate the angular momentum of car A w.r.t. car B. The mass of car A is , and the mass of car B is . Now, the angular momentum of car A w.r.t. car B is:
The magnitude of angular momentum is:
Thus, the angular momentum of car A w.r.t. car B is .
35
PYQ 2026
medium
physicsID: jee-main
In the diagram given below, pulley is a ring of mass and radius fitted with two rods each of mass and length along the diameter such that if the pulley rotates, the rods also rotate with the same angular velocity. Find the magnitude of the acceleration of when the system is released.
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: Understand the system. The system consists of a pulley (a ring) with two rods attached. When the system is released, the pulley and rods rotate with the same angular velocity. The tension in the rods leads to the acceleration of the mass hanging from the pulley. Step 2: Apply the rotational dynamics of the pulley. The pulley is a ring, and its moment of inertia is given by:
For each rod, since they rotate about the center of the pulley, the moment of inertia is:
There are two rods, so the total moment of inertia for the rods is:
Step 3: Use Newton’s second law for the hanging mass. For the mass , the force due to gravity is , and the tension in the rod provides the upward force. The net force on gives the acceleration :
where is the tension in the rope and is the acceleration of . Step 4: Apply rotational dynamics to the pulley. The torque on the pulley due to the tension in the rods is:
This causes the angular acceleration of the pulley, and using , we get:
Since the rods rotate with the pulley, the angular acceleration is related to the linear acceleration of mass by:
Substituting this into the equation for torque:
Simplifying:
Solving for :
Step 5: Combine the equations. Substitute into the equation :
Rearranging to solve for :
Substituting the appropriate values, we get:
Thus, the magnitude of the acceleration is .
36
PYQ 2026
hard
physicsID: jee-main
Two discs having the same moment of inertia about their axis. Thickness is and , and they have the same density. If , then find .
1
2
16
3
4
4
Official Solution
Correct Option:
(2)
Step 1: Moment of inertia for a disc. The moment of inertia of a disc is given by the formula , where is the mass and is the radius. Since both discs have the same moment of inertia, we set their moments equal to each other. Step 2: Relating the thickness and radius. Since the discs have the same density, the mass of each disc is proportional to its volume. Therefore, . Given , we can solve for the ratio based on the equality of moments of inertia. Step 3: Conclusion. The ratio simplifies to 16. Final Answer:
About Rotational Mechanics - JEE-MAIN
Rotational Mechanics is a vital chapter for JEE-MAIN aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.
By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.
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