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NEET • PHYSICS • SYSTEM OF PARTICLES

System of particles practice · Mixed difficulty

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Showing 181-210 of 239 questions
181 Easy
181 of 239
Understanding +4 | -1
The total linear momentum of a system of particle about the centre of mass is
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because the total linear momentum of a system of particles about the center of mass is always zero.
182 Medium
182 of 239
Analyzing +4 | -1
S-I: In rolling, all points of a rigid body have the same linear speed.
S-II: The rotational motion does not affect the linear velocity of rigid body.
View answer and explanation
Correct option: D
Both S-I and S-II are incorrect
Option D is correct because in rolling motion, points on a rigid body have different linear speeds and rotational motion affects the linear velocity.
183 Medium
183 of 239
Understanding +4 | -1
A body falling vertically downwards under gravity breaks in two parts of unequal masses. The centre of mass of the two parts taken together shifts horizontally towards
View answer and explanation
Correct option: C
does not shift horizontally
Option C (does not shift horizontally) is correct because the explosion is an internal force, and internal forces do not affect the motion of the center of mass.
184 Medium
184 of 239
Understanding +4 | -1
A loded spring gun of mass M fires a shot of mass m with a velocity v at an angle of elevation θ\theta. The gun was initially at rest on a horizontal frictionless surface. After firing, the centre of mass of gun-shot system
View answer and explanation
Correct option: C
Remain at rest
Option C (Remain at rest) is correct because there are no external forces acting on the gun-shot system in the horizontal direction, so the center of mass remains at rest.
185 Medium
185 of 239
Understanding +4 | -1
A mass is revolving in a circle which is in the plane of paper. The direction of angular acceleration is
View answer and explanation
Correct option: A
Upward the radius
Option D (At right angle to angular velocity) is correct because angular acceleration is perpendicular to angular velocity when the magnitude of angular velocity changes.
186 Medium
186 of 239
Remembering +4 | -1
What is the moment of inertia of solid sphere of density ρ\rho and radius R about its diameter?
View answer and explanation
Correct option: C
176105R5ρ\frac{{176}}{{105}}{R^5}\mathrm{\rho }
Option C (176105R5ρ\frac{{176}}{{105}}{R^5}\mathrm{\rho }) is correct because it represents the moment of inertia of a solid sphere in terms of its density and radius, derived using the formula for moment of inertia and the relationship between mass, density, and volume.
187 Hard
187 of 239
Application +4 | -1
A solid cylinder starts rolling from a height h{h} on an inclined plane. At some instant tt, the ratio of its rotational K.E. and the total K.E. would be:
View answer and explanation
Correct option: B
1:31: 3
Option B (1:31: 3) is correct because the rotational kinetic energy is one-third of the total kinetic energy for a solid cylinder rolling down an inclined plane.
188 Medium
188 of 239
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 is correct. The velocity of the center of mass remains constant unless acted upon by an external force. Since the shell explodes due to internal forces, the velocity of the center of mass remains the same, which is 60 m/s.
189 Easy
189 of 239
Application +4 | -1
A sphere of mass 10kg and radius 1m rotates about a tangent. The moment of inertia of the solid sphere is
View answer and explanation
Correct option: A
14 kg−m214\,kg - {m^2}
Option A (14 kg−m214\,kg - {m^2}) is correct because the moment of inertia about a tangent is I=ICM+Md2=25MR2+MR2I = I_{CM} + Md^2 = \frac{2}{5}MR^2 + MR^2, where d=Rd = R.
190 Hard
190 of 239
Understanding +4 | -1
Which of the following statements are correct:

(a) Centre of mass of a body always coincides with the centre of gravity of the body.

(b) A couple on a body produce pure rotational motion in a body.

(c) A couple on a body produce both translational and rotational motion in a body.

(d) Mechanical advantage greater than one means that small effort can be used to lift a large load.
View answer and explanation
Correct option: B
b and d
Option B (b and d) is correct. A couple produces pure rotational motion, and a mechanical advantage greater than one allows a small effort to lift a large load.
191 Medium
191 of 239
Analyzing +4 | -1
What happens to the centripetal acceleration of a particle, when its speed is doubled and angular velocity is halved?
View answer and explanation
Correct option: C
Remains unchanged
Option C (Remains unchanged) is correct because centripetal acceleration is given by ac=vωa_c = v\omega, and doubling the speed and halving the angular velocity cancels out, leaving the acceleration unchanged.
192 Easy
192 of 239
Application +4 | -1
A point mass (m)(m) is located at (0,a)(0, a). Where should the 2m2 m mass be kept, so that centre of mass lies at origin?
View answer and explanation
Correct option: A
(0,−a2)\left(0, \frac{-a}{2}\right)
Option A (0,−a2)\left(0, \frac{-a}{2}\right) is correct because it places the 2m mass such that the weighted average position is at the origin.
193 Easy
193 of 239
Application +4 | -1
A boat of mass 20 kg is at rest. A dog of mass 1 kg in the boat with a velocity of 5 ms−15\,m{s^{ - 1}}. What is the velocity of boat (nearly)?
View answer and explanation
Correct option: D
0.24 ms−10.24\,m{s^{ - 1}}
Option D is correct. Using conservation of momentum, the velocity of the boat is calculated to be approximately 0.24 m/s in the opposite direction of the dog.
194 Medium
194 of 239
Application +4 | -1
A particle is under uniform circular motion with angular momentum L. If its kinetic energy is tripled and frequency of motion is halved then new angular momentum becomes
View answer and explanation
Correct option: D
6 L
Option D (6 L) is correct. Angular momentum is related to kinetic energy and frequency, and changing these parameters affects the angular momentum.
195 Easy
195 of 239
Remembering +4 | -1
Centre of mass for a uniform semi-circular ring will be at (from centre)
View answer and explanation
Correct option: A
2Rπ\frac{{2R}}{\pi }
Option A (2Rπ\frac{{2R}}{\pi }) is correct because the center of mass of a uniform semi-circular ring is located at a distance of 2Rπ\frac{{2R}}{\pi } from the center.
196 Medium
196 of 239
Application +4 | -1
From a circular disc of radius R and mass 4m, a small disc of radius R2\frac{R}{2} is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is
View answer and explanation
Correct option: B
158mR2\frac{{15}}{8}m{R^2}
Option B (15/8 mR²) is correct because it is the result of subtracting the moment of inertia of the removed disc from the original disc.
197 Hard
197 of 239
Application +4 | -1
A uniform rod of length L=60 cm\mathrm{L}=60 \mathrm{~cm} is suspended through an end and is set into oscillation with small amplitude under gravity. Find the time period of oscillation. (g=10 m/s2)\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)
View answer and explanation
Correct option: A
0.4πs0.4 \pi \mathrm{s}
Option A (0.4πs0.4 \pi \mathrm{s}) is correct because it is the time period of a physical pendulum oscillating with small amplitude.
198 Medium
198 of 239
Understanding +4 | -1
Moment of inertia of an object does not depend upon
View answer and explanation
Correct option: C
Angular velocity
Option C (Angular velocity) is correct because the moment of inertia depends on the mass, mass distribution, and axis of rotation, but not on the angular velocity.
199 Medium
199 of 239
Application +4 | -1
If linear density of a rod of length 3m varies as λ=2+x\lambda = 2 + x, then the position of the centre of gravity of the rod is
View answer and explanation
Correct option: C
107m\frac{{10}}{7}m
Option C (107m\frac{{10}}{7}m) is correct because it is the calculated position of the center of gravity using the integral formula for a varying linear density.
200 Medium
200 of 239
Application +4 | -1
A body of mass ‘M’ is moving on a circular track of radius ‘r’ in such a way that its kinetic energy ‘k’ depends on the distance travelled by the body ‘s’ according to relation k=βsk = \beta s, where β\beta is a constant. The angular acceleration of the particle is
View answer and explanation
Correct option: D
βMr\frac{\beta }{{Mr}}
Option D (βMr\frac{\beta}{Mr}) is correct because the angular acceleration is derived from the relationship between kinetic energy, distance traveled, and the radius of the circular track.
201 Easy
201 of 239
Application +4 | -1
A wheel of mass 16 kg has a moment of inertia of 160 kg-m2 about its own axis, the radius of gyration will be
View answer and explanation
Correct option: D
3.2 m
Option D (3.2 m) is correct because the radius of gyration is calculated using the formula k=IMk = \sqrt{\frac{I}{M}}, where I is the moment of inertia and M is the mass.
202 Easy
202 of 239
Analyzing +4 | -1
For the same total mass, which of the following will have the largest moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the body
View answer and explanation
Correct option: D
Four rods forming a square of side 2R2 R
Option D (Four rods forming a square of side 2R2 R) is correct because it has the mass furthest from the axis of rotation.
203 Medium
203 of 239
Remembering +4 | -1
Position vector of centre of mass of two particles system is given by:
View answer and explanation
Correct option: D
R⃗=m1r⃗1+m2r⃗2m1+m2\vec{R}=\frac{m_1 \vec{r}_1+m_2 \vec{r}_2}{m_1+m_2}
Option D (R⃗=m1r⃗1+m2r⃗2m1+m2\vec{{R}}=\frac{{m_1 \vec{{r}}_1+m_2 \vec{{r}}_2}}{{m_1+m_2}}) is correct because it represents the standard formula for the position vector of the center of mass of a two-particle system.
204 Medium
204 of 239
Application +4 | -1
A thin uniform 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 angular velocity ω\omega. Another disc of same dimension but mass M2\frac{M}{2} is placed gently on the first disc co-axially. The angular velocity of the system now is
View answer and explanation
Correct option: D
2ω3\frac{{2\omega }}{3}
Option D (2ω3\frac{{2\omega }}{{3}}) is correct because the angular momentum is conserved when the second disc is placed on the first, leading to a decrease in angular velocity.
205 Medium
205 of 239
Application +4 | -1
In the equation of angular displacement of a particle moving on a circular path is given as θ=2t3+0.5,θ\theta = 2{t^3} + 0.5,\theta is in radian and t in second. The angular velocity of the particle at t=2s is
View answer and explanation
Correct option: C
24 rad/s
Option C (24 rad/s) is correct because angular velocity is the derivative of angular displacement with respect to time, evaluated at t=2s.
206 Medium
206 of 239
Analyzing +4 | -1
A metre stick is placed vertically at the origin on a frictionless surface. A gentle push in +x+x-direction is given to the top most point of the rod. When it has fallen completely, the xx-coordinate of centre of rod is at
View answer and explanation
Correct option: A
Origin
The correct answer is Origin because there are no external horizontal forces acting on the system, so the center of mass remains at the origin.
207 Medium
207 of 239
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 is zero, as the center of mass is at rest.
208 Medium
208 of 239
Analyzing +4 | -1
A large tray of mass M{M} has in it a cubical block of ice of mass m{m} and edge length L{L}. The ice starts melting. The shift in the height of centre of mass of system (ice + tray) is
View answer and explanation
Correct option: C
12mL(m+M)\frac{1}{2} \frac{{mL}}{({m}+{M})}; downward
Option C (12mL(m+M)\frac{1}{2} \frac{{mL}}{({m}+{M})}; downward) is correct because when the ice melts, the center of mass of the ice lowers by L/2, and the overall shift in the center of mass is calculated considering both masses.
209 Medium
209 of 239
Remembering +4 | -1
The correct relation between moment of inertial II, radius of gyration KK and mass MM of the body is:
View answer and explanation
Correct option: D
K=IMK=\sqrt{\frac{I}{M}}
Option D (K=IMK=\sqrt{\frac{{I}}{{M}}}) is correct because the radius of gyration is defined as K=IMK = \sqrt{\frac{{I}}{{M}}}.
210 Medium
210 of 239
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 correctly relates moment of inertia to angular momentum and kinetic energy.
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