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JEE • PHYSICS • SYSTEM OF PARTICLES

System of particles practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 181-210 of 483 questions

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181 Medium
181 of 483
Application +4 | -1
The velocity of a solid sphere on reaching the bottom of an inclined plane is v′v^{\prime} when it slides down. The value of v/v′v / v^{\prime} will be
View answer and explanation
Correct option: B
75\sqrt{\frac{7}{5}}
Option B (75\sqrt{\frac{7}{5}}) is correct because it is the ratio of the velocities when the sphere slides and rolls down the inclined plane, derived from energy conservation.
182 Medium
182 of 483
Application +4 | -1
A man of 50 kg mass in standing in a gravity free space at a height of 10m above the floor. He throws a stone of 0.5 kg mass downwards with a speed of 2m/s. When the stone reaches the floor, the distance of the man above the floor will be
View answer and explanation
Correct option: C
10.1 m
Option C (10.1 m) is correct because the center of mass of the man-stone system remains constant. The man moves upward slightly as the stone falls.
183 Medium
183 of 483
Application +4 | -1
A smooth sphere AA is moving on a frictionless horizontal plane with angular speed ω\omega and centre of mass velocity vv. It collides elastically and head on with an identical sphere BB at rest. Neglect friction everywhere. After the collision, their angular speeds are ωA\omega_A and ωB\omega_B, respectively. Then
View answer and explanation
Correct option: C
ωA=ω\omega_{\mathrm{A}}=\omega
Option C (ωA=ω\omega_{\mathrm{A}}=\omega) is correct because in an elastic head-on collision with an identical sphere at rest, sphere A transfers its linear velocity to sphere B, but its angular velocity remains unchanged due to the absence of friction.
184 Medium
184 of 483
Analyzing +4 | -1
The density of a rod is increasing from end A to B it is easier to set in rotation by
View answer and explanation
Correct option: D
Clamping the rod at B and applying a force F at A perpendicular to rod
Option D (Clamping the rod at B and applying a force F at A perpendicular to rod) is correct because it minimizes the moment of inertia and maximizes the torque for easier rotation.
185 Medium
185 of 483
Application +4 | -1
A flywheel is making 3000/π3000 / \pi revolutions per minute about its axis. If the moment of inertia of the flywheel about that axis is 400 kgm2400 \,\mathrm{kgm}^2, its rotational kinetic energy is
View answer and explanation
Correct option: A
2×106 J2 \times 10^6 \mathrm{~J}
Option A (2×106 J2 \times 10^6 \mathrm{~J}) is correct because it is calculated using the rotational kinetic energy formula with correct angular velocity.
186 Medium
186 of 483
Analyzing +4 | -1
Assertion: There is a line passing through the tangent of a disc; the disc's moment of inertia about that line is 74MR2\frac{7}{4} M R^2

Reason: If the disc's moment of inertia about its diameter is known then to know the moment of inertia about the tangent line, we need to use the parallel axis theorem.
View answer and explanation
Correct option: D
Assertion is false, but Reason is true
Assertion is false, but Reason is true. The moment of inertia of a disc about a tangent is 54MR2\frac{5}{4}MR^2, not 74MR2\frac{7}{4}MR^2. The parallel axis theorem is indeed needed.
187 Medium
187 of 483
Understanding +4 | -1
Assertion: If a person jumps from a height with his legs rigid, then his ankle is more injured.
Reason: The force will be maximum on its ankle during the collision with ground.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because a rigid landing results in a shorter impact time, leading to a greater force on the ankle.
188 Medium
188 of 483
Application +4 | -1
A child is standing with folded hands at the centre of a platform rotating about its central axis. The kinetic energy of the system is K. The child stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is
View answer and explanation
Correct option: B
K2\frac{K}{2}
Option B (K2\frac{K}{2}) is correct because when the moment of inertia doubles, the angular velocity halves due to conservation of angular momentum, resulting in the kinetic energy being halved.
189 Medium
189 of 483
Understanding +4 | -1
What is moment of inertia in terms of angular momentum (L) and kinetic energy (K)?
View answer and explanation
Correct option: B
L22K\frac{{{L^2}}}{{2K}}
Option B (L22K\frac{{{L^2}}}{{2K}}) is correct because it is derived directly from the definitions of angular momentum (L) and kinetic energy (K) for rotational motion.
190 Medium
190 of 483
Understanding +4 | -1
Statement I: The centre of mass of a circular disc lies always at the centre of the disc.

Statement II: Circular disc is a symmetrical body.
View answer and explanation
Correct option: A
If both Statement - I and Statement - II are true, and Statement - II is the correct explanation of Statement - I
Option A is correct because the center of mass of a symmetrical body like a circular disc lies at its geometric center, and symmetry is the reason for this.
191 Medium
191 of 483
Understanding +4 | -1
If ' II ' is the moment of inertia and ' LL ' is angular momentum of a rotating body, then L22I\frac{L^2}{2 I} is its
View answer and explanation
Correct option: B
rotational kinetic energy
Option B (rotational kinetic energy) is correct because the formula for rotational kinetic energy is 12Iω2\frac{{1}}{{2}}I\omega^2, and L=IωL = I\omega, so L22I=12Iω2\frac{{L^2}}{{2I}} = \frac{{1}}{{2}}I\omega^2.
192 Medium
192 of 483
Analyzing +4 | -1
If a force 10i^+15j^+25k^10\hat i + 15\hat j + 25\hat k acts on a system and gives an acceleration 2i^+3j^−5k^2\hat i + 3\hat j - 5\hat k to the centre of mass of the system, the mass of the system is
View answer and explanation
Correct option: D
Given data is not correct
Option D is correct because the force and acceleration vectors are not in the same direction, implying the given data is inconsistent with Newton's Second Law.
193 Medium
193 of 483
Application +4 | -1
A thin uniform circular disc of mass MM and radius RR is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity ω\omega. Another disc of same thickness and radius but of mass 18M\frac{1}{8} \mathrm{M} is placed gently on the first disc co-axially. The angular velocity of the system is now
View answer and explanation
Correct option: A
89ω\frac{8}{9} \omega
Option A (89ω\frac{8}{9} \omega) is correct because it is derived from the conservation of angular momentum when the second disc is placed on the first.
194 Medium
194 of 483
Application +4 | -1
A uniform disc of radius aa and mass mm is rotating freely with angular speed ω\omega in a horizontal plane about a smooth fixed vertical axis through its centre. A particle of mass mm is suddenly attached to the rim of the disc and rotates with it. The new angular speed is
View answer and explanation
Correct option: B
ω3\frac{\omega}{3}
Option B (ω3\frac{\omega}{3}) is correct because the angular momentum is conserved when the particle is attached to the disc.
195 Medium
195 of 483
Understanding +4 | -1
A dancer on ice spins faster when she folds her arms. This is due to
View answer and explanation
Correct option: C
Constant angular momentum and increase in kinetic energy
The correct answer is C (Constant angular momentum and increase in kinetic energy) because folding her arms decreases her moment of inertia, and angular momentum is conserved.
196 Medium
196 of 483
Application +4 | -1
If \mathop \mathrm{F}\limits^ \to = 2\,\widehat \mathrm{i} - 3\widehat \mathrm{j}\,\,\mathrm{N}\mathop \mathrm{F}\limits^ \to = 2\,\widehat \mathrm{i} - 3\widehat \mathrm{j}\,\,\mathrm{N} and \mathop r\limits^ \to = 3\,\widehat \mathrm{i} + 2\widehat \mathrm{j}\,\,m\mathop r\limits^ \to = 3\,\widehat \mathrm{i} + 2\widehat \mathrm{j}\,\,m then torque τ→\overrightarrow \tau is
View answer and explanation
Correct option: D
- 13\widehat \mathrm{k}- 13\widehat \mathrm{k}
Option D (- 13\widehat \mathrm{k}- 13\widehat \mathrm{k}) is correct because it is the torque calculated as the cross product of the position vector and the force vector.
197 Medium
197 of 483
Application +4 | -1
A particle is moving with uniform speed 0.5 m/s0.5\,m/s along a circle of radius 1m1m, then the angular velocity of particle is ( in rads−1rad{s^{ - 1}})
View answer and explanation
Correct option: D
0.5
Option D (0.5) is correct because the angular velocity is calculated using ω = V/r.
198 Medium
198 of 483
Analyzing +4 | -1
A car is accelerating on a rough horizontal track. The wheels are rolling purely.
Statement-1: Friction does positive work on car.
Statement-2: The friction force is responsible for the acceleration of the car on a rough horizontal track.
View answer and explanation
Correct option: D
Statement-1 is false, Statement-2 is true
Option D (Statement-1 is false, Statement-2 is true) is correct because friction acts as the driving force for the car's acceleration, but the work done by static friction on the point of contact is zero.
199 Medium
199 of 483
Understanding +4 | -1
(A): The size and shape of the rigid body remains unaffected under the effect of extemal forces.
(R): The distance between two particles remains constant in a rigid body.
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A)
Option A is correct because the definition of a rigid body implies that its size and shape remain unchanged under external forces, and this is because the distance between any two particles within the body remains constant.
200 Medium
200 of 483
Understanding +4 | -1
Select incorrect statement
View answer and explanation
Correct option: B
Centre of mass of regular bodies is always at its geometrical centre.
Option B is the correct answer because the center of mass of regular bodies is at its geometrical center only if the mass distribution is uniform.
201 Medium
201 of 483
Understanding +4 | -1
A uniform disc of mass 2 kg and radius 15 cm is revolving around the central axis by 4 rads−1\mathrm{rad}{\mathrm{s}^{ - 1}}. The linear momentum of disc will be
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the linear momentum of a rotating disc about its center of mass is zero.
202 Medium
202 of 483
Application +4 | -1
A coin is of mass 4.8 kg and radius one metre rolling on a horizontal surface without sliding with angular velocity 600 rotation/min. What is total kinetic energy of the coin ?
View answer and explanation
Correct option: B
1440 π2  J{\mathrm{\pi }^2}\mathrm{\;J}
Option B (1440 π2  J{\mathrm{\pi }^2}\mathrm{\;J}) is correct because it correctly calculates the total kinetic energy considering both translational and rotational kinetic energy.
203 Medium
203 of 483
Understanding +4 | -1
The motion of the centre of mass is the result of
View answer and explanation
Correct option: B
External forces
Option B (External forces) is correct because the motion of the center of mass is determined solely by the net external force acting on the system.
204 Medium
204 of 483
Analyzing +4 | -1
If we hollow the ball of pendulum and fill it with sand, what would be the effect of change on the center of mass?
View answer and explanation
Correct option: D
Its value will first decrease and then increase
Option D (Its value will first decrease and then increase) is correct because the center of mass shifts as sand is added, initially lowering and then rising as the cavity fills.
205 Medium
205 of 483
Application +4 | -1
A solid sphere rolls down without slipping from rest on an 300\mathrm{3}{\mathrm{0}^0} incline. Its linear acceleration is
View answer and explanation
Correct option: B
5g14\frac{{5g}}{{14}}
Option B (5g14\frac{{5g}}{{14}}) is correct because the linear acceleration is calculated using the formula for rolling without slipping on an incline, considering the moment of inertia of a solid sphere.
206 Medium
206 of 483
Application +4 | -1
One quarter of the disc of mass m is removed. If rr be the radius of the disc, the new moment of inertia is
View answer and explanation
Correct option: C
38mr2\frac{3}{8}m{r^2}
Option C (38mr2\frac{3}{8}m{r^2}) is correct because the moment of inertia is proportional to the mass, and removing one-quarter of the disc reduces the mass to three-quarters.
207 Medium
207 of 483
Understanding +4 | -1
Assertion: The centre of mass of a body may lie outside the body.
Reason: The centre of mass of a body depends on the distribution of mass in the body.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of the assertion
Option A is correct because the center of mass depends on mass distribution, and it can indeed lie outside the body, such as in a ring.
208 Medium
208 of 483
Understanding +4 | -1
A ballet dancer suddenly folds her outstretched
arms. Her angular velocity
View answer and explanation
Correct option: A
Increases
Option A (Increases) is correct because when a ballet dancer folds her arms, her moment of inertia decreases, and to conserve angular momentum, her angular velocity must increase.
209 Medium
209 of 483
Application +4 | -1
If the earth were to contract such that its
radius becomes one quarter, without change in
its mass, the duration of one full day would be
View answer and explanation
Correct option: B
1.5 h
Option B (1.5 h) is correct because it is calculated using the conservation of angular momentum, considering the change in the Earth's radius.
210 Medium
210 of 483
Application +4 | -1
A flywheel is in the form of solid circular wheel of mass 72 kg and radius of 0.5m and it takes 70 r.p.m, then the energy of revolution is
View answer and explanation
Correct option: B
240 J
The correct answer is 240 J. The energy of revolution is calculated using the formula for rotational kinetic energy, considering the moment of inertia of a solid circular wheel.
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