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JEE • PHYSICS • SYSTEM OF PARTICLES

System of particles practice · Mixed difficulty

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Practice format and chapter quiz

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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241 Medium
241 of 483
Application +4 | -1
A flywheel having moment of inertia 10 kg.m210 \mathrm{~kg} . \mathrm{m}^2 is rotating at 50 rad.s−150 \mathrm{~rad.s}^{-1}. What amount of work needs to be done in order to bring this flywheel to rest in 10 seconds?
View answer and explanation
Correct option: B
12500 J
Option B (12500 J) is correct because it is the work done to bring the flywheel to rest.
242 Medium
242 of 483
Understanding +4 | -1
In free space, a shell moving with velocity 60 m/s60\,m/s along the positive x-axis. When it passes through origin, it explodes into two pieces of mass ratio 1: 2. After the explosion, the velocity of the centre of mass is:-
View answer and explanation
Correct option: B
60 m/s60\,m/s
Option B (60 m/s60\,m/s) is correct because in the absence of external forces, the velocity of the center of mass remains constant.
243 Medium
243 of 483
Understanding +4 | -1
A particle with position vector rr has a linear momentum pp. Which of the following statements is true in respect of its angular momentum L about the origin
View answer and explanation
Correct option: D
LL is maximum when pp is perpendicular to rr
Option C (They should take off spectacles) is correct because the microscope is designed to correct for normal vision, and spectacles would introduce additional lens corrections.
244 Medium
244 of 483
Application +4 | -1
A rigid body of mass mm rotates about an axis passing through its centre. If the angular velocity of the body is unity, the moment of inertia of the body will be equal to
View answer and explanation
Correct option: B
twice the kinetic energy of the body
Option B (twice the kinetic energy of the body) is correct because the moment of inertia is equal to twice the kinetic energy when the angular velocity is unity.
245 Medium
245 of 483
Understanding +4 | -1
A ball kept in a closed box moves in the box making collisions with the walls. The box is kept
on a smooth surface. The center of mass
View answer and explanation
Correct option: B
Of the box plus the ball system remains constant
Option B (Of the box plus the ball system remains constant) is correct because there is no external force acting on the box-ball system.
246 Medium
246 of 483
Remembering +4 | -1
Find the position of centre of mass from the base of a uniform solid cone of height 20 cm20 \mathrm{~cm}.
View answer and explanation
Correct option: A
5 cm5 \mathrm{~cm}
Option A (5 cm) is correct because the center of mass of a uniform solid cone is located at a height of H/4 from the base, where H is the total height.
247 Medium
247 of 483
Analyzing +4 | -1
Statement-I: A coin is placed on phonogram turn table. The motor is started, coin moves along the moving table.
Statement-II: The rotating table is providing necessary centripetal force to the coin.
View answer and explanation
Correct option: D
Both statements are false
Option D (Both statements are false) is correct because the coin initially moves with the table due to friction, but if the speed increases beyond a certain point, the friction is insufficient, and the coin will fly off tangentially.
248 Medium
248 of 483
Understanding +4 | -1
Assertion: As star collapse its angular velocity increases.
Reason: The mass of star decreases.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C (If Assertion is True but the Reason is False) is correct because as a star collapses, its moment of inertia decreases, causing its angular velocity to increase due to conservation of angular momentum, but the mass of the star does not decrease significantly during the collapse.
249 Medium
249 of 483
Application +4 | -1
An astronaut is rotated in a horizontal circle at a radius of 5 m5 \mathrm{~m}. If the astronaut can withstand a maximum acceleration of 7 g7 \mathrm{~g}, then maximum numbers of permissible revolution per second is (1 g=10 m/s2)\left(1 \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^2\right).
View answer and explanation
Correct option: A
142πrevs\frac{\sqrt{14}}{2 \pi} \frac{\mathrm{rev}}{\mathrm{s}}
Option A (142πrevs\frac{\sqrt{{14}}}{{2 \pi}} \frac{{\mathrm{{rev}}}}{{\mathrm{{s}}}}) is correct because it represents the maximum permissible revolutions per second calculated from the maximum allowable acceleration.
250 Easy
250 of 483
Application +4 | -1
A shaft rotating at 6000 rpm is transmitting a power of 2π2\pi kilowatt. Then magnitude of the driving torque is
View answer and explanation
Correct option: A
10 Nm
Option A (10 Nm) is correct because the driving torque is calculated using the formula τ=Pω\tau = \frac{P}{\omega}.
251 Medium
251 of 483
Understanding +4 | -1
When an external force is applied at the centre of mass of a rigid body at rest, then it undergoes
View answer and explanation
Correct option: A
Only translator motion
Option A (Only translator motion) is correct because a force applied at the center of mass results in translational motion without rotation.
252 Medium
252 of 483
Analyzing +4 | -1
Assertion: Location of centre of mass is independent of the reference frame.
Reason: Centre of mass is same as centre of gravity.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C (If Assertion is True but the Reason is False) is correct because the location of the center of mass is independent of the reference frame, but the center of mass and center of gravity are not always the same.
253 Medium
253 of 483
Analyzing +4 | -1
A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time a viscous fluid of mass ‘m’ is dropped at the centre and is allowed to spread out and finally fall. The angular velocity during this period
View answer and explanation
Correct option: B
Decreases initially and increases again
Option B (Decreases initially and increases again) is correct because the angular velocity decreases as the fluid spreads, increasing the moment of inertia, and then increases again as the fluid falls off, decreasing the moment of inertia.
254 Medium
254 of 483
Application +4 | -1
A solid cylinder of radius 4 cm4 \mathrm{~cm} and mass 250 g250 \mathrm{~g} rolls down an inclined plane (1 in 10). Calculate the acceleration and the total energy of the cylinder after 5 s5 \mathrm{~s}.
View answer and explanation
Correct option: A
0.653 ms−2,2.0 J0.653 \mathrm{~ms}^{-2}, 2.0 \mathrm{~J}
Option A (0.653 ms−2,2.0 J0.653 \mathrm{~ms}^{-2}, 2.0 \mathrm{~J}) is correct because it accurately calculates the acceleration of the rolling cylinder down the incline and its total kinetic energy after 5 seconds.
255 Medium
255 of 483
Application +4 | -1
The radius of gyration of disc of mass and radius RR about an axis which is at distance 2R2 R from its centre and in the plane of the disc will be
View answer and explanation
Correct option: C
172R\frac{\sqrt{17}}{2} R
Option C (172R\frac{\sqrt{17}}{2} R) is correct because it is derived from the parallel axis theorem and the definition of radius of gyration.
256 Medium
256 of 483
Application +4 | -1
A particle of mass 1 kg is projected with an initial velocity 10 ms−1\mathrm{m}{\mathrm{s}^{ - 1}}  at an angle of projection 450{45^0} with the horizontal. The average torque acting on the projectile, between the time at which it is projected and the time at which it strikes the ground, about the point of projection in newton-metre is
View answer and explanation
Correct option: B
50
Option B (50) is correct because the average torque is calculated using the rate of change of angular momentum, which simplifies to m(u2cos⁡2θ)m(u^2 \cos^2 \theta).
257 Medium
257 of 483
Understanding +4 | -1
Two equal and opposite forces act on a rigid body at a certain distance. Then
View answer and explanation
Correct option: B
the body will rotate about its centre of mass
Option the body will rotate about its centre of mass is correct because two equal and opposite forces acting at a distance create a torque, causing rotation about the center of mass.
258 Medium
258 of 483
Application +4 | -1
A projectile of mass 1 kg1 \mathrm{~kg} is projected with a speed of 10 m/s10 \mathrm{~m} / \mathrm{s} at an angle of 60∘60^{\circ} from the horizontal when projectile is at its highest point, its magnitude of angular momentum (about point of projection)
View answer and explanation
Correct option: B
754Nm/s\frac{75}{4} \mathrm{Nm} / \mathrm{s}
Option B (754Nm/s\frac{75}{4} \mathrm{Nm} / \mathrm{s}) is correct because it is calculated by finding the angular momentum at the highest point using the horizontal velocity and maximum height.
259 Medium
259 of 483
Application +4 | -1
A car wheel is rotated to uniform angular acceleration about its axis. Initially its angular velocity is zero. It rotates through an angle θ1{\mathrm{\theta }_1} in the first 2 s , in the next 2 s, it rotates through an additional angle θ2{\mathrm{\theta }_2}, the ratio of θ2θ1\frac{{{\mathrm{\theta }_2}}}{{{\mathrm{\theta }_1}}} is
View answer and explanation
Correct option: C
3
Option C (3) is correct because the angle rotated is proportional to the square of the time, given uniform angular acceleration.
260 Medium
260 of 483
Application +4 | -1
Assume the earth's orbit around the sun as circular and the distance between their centre as D\mathrm{D}. Mass of the earth is M\mathrm{M} and its radius is R\mathrm{R}. If earth has an angular velocity ω0\omega_0 with respect to its centre and ω\omega with respect to the centre of the sun, the total kinetic energy of earth is
View answer and explanation
Correct option: B
MR2ω025[1+52(DωRω0)2]\frac{\mathrm{MR}^2 \omega_0^2}{5}\left[1+\frac{5}{2}\left(\frac{\mathrm{D} \omega}{\mathrm{R} \omega_0}\right)^2\right]
Option B (MR2ω025[1+52(DωRω0)2]\frac{{\mathrm{{MR}}^2 \omega_0^2}}{{5}}\left[1+\frac{{5}}{{2}}\left(\frac{{\mathrm{{D}} \omega}}{{\mathrm{{R}} \omega_0}}\right)^2\right]) is correct because it is derived from the sum of the rotational kinetic energy about its center and the translational kinetic energy due to its orbit around the sun.
261 Medium
261 of 483
Analyzing +4 | -1
STATEMENT-1: If there is no external torque on a body about its center of mass, then the velocity of the center of mass remains constant.

STATEMENT-2: The linear momentum of an isolated system remains constant.
View answer and explanation
Correct option: D
Statement- 1 is False, Statement-2 is True
Option D (Both A and R are correct and R is the correct explanation of A) is correct because the change in phase difference is directly caused by the difference in wavelengths in different media.
262 Medium
262 of 483
Application +4 | -1
Two uniform rods AA and BB of lengths 5 m5 {~m} and 3 m3{~m} are placed end to end. If their linear densities are 3 kg/m3 {~kg} / {m} and 2 kg/m2 {~kg} / {m}, the position of their centre of mass from their interface is
View answer and explanation
Correct option: C
19/14 m19 / 14 {~m} on the side of heavier rod
Option C (19/14 m19 / 14 {~m} on the side of heavier rod) is correct because it's the calculated center of mass position relative to the interface.
263 Medium
263 of 483
Application +4 | -1
A flywheel rotates with uniform angular acceleration of 2πrad/s22 \pi {rad} / {s}^2. Its initial angular velocity is 20πrad/s20 \pi {rad} / {s}. What are the total number of rotations made in 20 s20 {~s} ?
View answer and explanation
Correct option: C
400
Option C (400) is correct because the total number of rotations is calculated using the kinematic equation for angular displacement and then dividing by 2π2\pi.
264 Medium
264 of 483
Analyzing +4 | -1
Statement-1: If the centre of mass of a system is the origin then the total mass to the right of origin is same as total mass to left of origin.
Statement-2: xcm=ΣmixiΣmix_{\mathrm{cm}}=\frac{\Sigma_{m_i x_i}}{\Sigma_{m_i}}
View answer and explanation
Correct option: D
Statement-1 is false, Statement-2 is true
Option D is correct because Statement-1 is false (it's about the sum of m_i*x_i, not total mass), and Statement-2 is true (it's the correct formula for the center of mass).
265 Medium
265 of 483
Understanding +4 | -1
(A): If body or a system has neither translational nor rotational motion then it is in complete mechanical equilibrium.
(R): A body or system is said to be in mechanical equilibrium when net force and net torque acting on the body are zero.
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 both the assertion and reason are true, and the reason correctly explains why a body is in complete mechanical equilibrium when it has neither translational nor rotational motion.
266 Medium
266 of 483
Application +4 | -1
Given that force (5i^+7j^−3k^)N(5 \hat{i}+7 \hat{j}-3 \hat{k}) \mathrm{N} acts on a particle at position (i^+j^−k^)m(\hat{i}+\hat{j}-\hat{k}) \mathrm{m}. Find torque of this force on the particle about origin.
View answer and explanation
Correct option: A
4i^−2j^+2k^4 \hat{i}-2 \hat{j}+2 \hat{k}
Option A (4i^−2j^+2k^4 \hat{i}-2 \hat{j}+2 \hat{k}) is correct because it results from the correct cross product calculation of the position vector and the force vector.
267 Easy
267 of 483
Application +4 | -1
Radius of gyration of a thin rod of mass 1 kg about an axis perpendicular to its length and passing through its centre is 3 m\sqrt 3 \,m . Then length of the rod is
View answer and explanation
Correct option: B
6 m
Option B (20∘C20^{\circ} \mathrm{C}) is correct because it is the temperature at which the pendulum clock neither gains nor loses time.
268 Medium
268 of 483
Understanding +4 | -1
A non-zero external force acts on a systems of particles. The velocity and acceleration of the centre of mass are found to be v0 and a0{v_0}\,and\,{a_0} at an instant t. It is possible that
View answer and explanation
Correct option: B
v0=0,a0≠0{v_0} = 0,{a_0} \ne 0
Option B (v0=0,a0≠0{v_0} = 0,{a_0} \ne 0) is correct because a non-zero external force implies a non-zero acceleration of the center of mass, but the velocity at a particular instant can be zero.
269 Medium
269 of 483
Application +4 | -1
Two particles of equal mass have velocities v1=4i^ and v2=4j^ ms−1{v_1} = 4\hat i\,and\,{v_2} = 4\hat j\,m{s^{ - 1}}. First particle has an acceleration a1=(5i^+5j^)ms−2{a_1} = \left( {5\hat i + 5\hat j} \right)m{s^{ - 2}} , while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of
View answer and explanation
Correct option: A
Straight line
Option A (Straight line) is correct because the center of mass has a constant acceleration, resulting in a straight-line path.
270 Medium
270 of 483
Understanding +4 | -1
When no external torque acts on a rotating system,
View answer and explanation
Correct option: D
Its rotational kinetic energy is inversely proportional to moment of inertia
Option D (Its rotational kinetic energy is inversely proportional to moment of inertia) is correct because when no external torque acts, angular momentum is conserved, and kinetic energy is inversely proportional to the moment of inertia.
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