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NEET • PHYSICS • SYSTEM OF PARTICLES

System of particles practice · Mixed difficulty

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121 Medium
121 of 239
Application +4 | -1
A body of mass 1 kg thrown with a velocity of 10  ms−110\;\mathrm{m}{\mathrm{s}^{ - 1}} at an angle of 600{60^0} with the horizontal. Its momentum at the highest point is
View answer and explanation
Correct option: D
5  kg  ms−15\mathrm{\;kg\;m}{\mathrm{s}^{ - 1}}
Option D (5  kg  ms−15\mathrm{\;kg\;m}{\mathrm{s}^{ - 1}}) is correct because the momentum at the highest point is only due to the horizontal component of the initial velocity.
122 Medium
122 of 239
Application +4 | -1
Two bodies of mass 10 kg and 2 kg are moving with velocities (2i^−7j^+3k^) m/s\left( {2\hat i - 7\hat j + 3\hat k} \right)\,m/s and (−10i^+35j^−3k^) m/s\left( { - 10\hat i + 35\hat j - 3\hat k} \right)\,m/s respectively. The velocity of their centre of mass is
View answer and explanation
Correct option: B
2k^ m/s2\hat k{\mkern 1mu} m/s
Option B (2k^ m/s2\hat k{\mkern 1mu} m/s) is correct because the velocity of the center of mass is calculated using the formula vcm=(m1v1+m2v2)/(m1+m2)v_{cm} = (m_1v_1 + m_2v_2) / (m_1 + m_2).
123 Medium
123 of 239
Analyzing +4 | -1
A particle of mass m moving with a velocity (3i^+2j^)ms−1\left( {3\hat i + 2\hat j} \right)m{s^{ - 1}} collides with a stationary body mass M and finally moves with a velocity (−2i^+j^)ms−1. If mM=113,\left( { - 2\hat i + \hat j} \right)\mathrm{m}{\mathrm{s}^{ - 1}}.\,If\,\frac{m}{M} = \frac{1}{{13}}, then
View answer and explanation
Correct option: D
All the above are correct
Option D (All the above are correct) is correct because the impulse, velocity of M, and coefficient of restitution are all consistent with the conservation of momentum and the given conditions.
124 Medium
124 of 239
Analyzing +4 | -1
A tank filled with water is connected to another empty tank lying to the right side of the first tank. Until the level of water in the tanks become same
a) Centre of mass of water shifts down at constant rate
b) Centre of mass of water shifts down and rate of change of shift in centre of mass decreases gradually and becomes zero
c) Centre of mass shifts towards right as well
View answer and explanation
Correct option: B
Only b and c are correct
Option B (Only b and c are correct) is correct because as water flows to the right tank, the rate of change of the shift in the center of mass decreases and the center of mass shifts towards the right.
125 Easy
125 of 239
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 torque is calculated using the formula P=τωP = \tau \omega, where P is power and ω\omega is angular velocity.
126 Medium
126 of 239
Remembering +4 | -1
The moment of inertia of a straight thin rod of mass M and length L about an axis perpendicular to its length and passing through one end is
View answer and explanation
Correct option: B
ML23\frac{{M{L^2}}}{{3}}
Option B (ML²/3) is correct because the moment of inertia of a rod about one end is ML²/3.
127 Hard
127 of 239
Application +4 | -1
A ring is rolling down an inclined plane which is inclined at an angle of 30∘30^{\circ} to the horizontal. Then the speed of ring after travelling a distance of 10 m10 \mathrm{~m} is :-
View answer and explanation
Correct option: A
10 ms−110 \mathrm{~ms}^{-1}
Option A (10 ms−110 \mathrm{~ms}^{-1}) is correct because the speed is calculated using the conservation of energy, considering both translational and rotational kinetic energy of the ring.
128 Medium
128 of 239
Understanding +4 | -1
Two particles move towards each other from rest under a mutual force of attraction. At the instant, when the speed of A is V, the speed of B is V/2. The speed of centre of mass of the system is
View answer and explanation
Correct option: D
zero
Option D is correct. Since the particles move under mutual attraction and start from rest, there are no external forces. Therefore, the velocity of the center of mass of the system is zero.
129 Easy
129 of 239
Application +4 | -1
The torque of a force 5i^+3j^−7k^5 \hat{i}+3 \hat{j}-7 \hat{k} about the origin is τ\tau. If the force acts on a particle whose position vector is 2i^+2j^+k^2 \hat{i}+2 \hat{j}+\hat{k}, then the value of τ\tau will be
View answer and explanation
Correct option: C
−17i^+19j^−4k^-17 \hat{i}+19 \hat{j}-4 \hat{k}
Option C (−17i^+19j^−4k^-17 \hat{i}+19 \hat{j}-4 \hat{k}) is correct because it is the result of the cross product of the position vector and the force vector.
130 Hard
130 of 239
Application +4 | -1
As observed in the laboratory system, a 6 MeV6 \mathrm{~MeV} proton is incident on a stationary 12 C12 \mathrm{~C} target. The velocity of center of mass of the system is (Take mass of proton to be 1amu1 \mathrm{amu})
View answer and explanation
Correct option: A
2.6×106 ms−12.6 \times 10^6 \mathrm{~ms}^{-1}
Option A (2.6×106 ms−12.6 \times 10^6 \mathrm{~ms}^{-1}) is correct because the velocity of the center of mass is calculated using the formula vcm=m1v1+m2v2m1+m2v_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}.
131 Easy
131 of 239
Understanding +4 | -1
The centre of mass of system of particles does not depend on
View answer and explanation
Correct option: D
Force acting on the particle
Option D (Force acting on the particle) is correct because the position of the center of mass depends only on the masses and positions of the particles, not on the forces acting on them.
132 Medium
132 of 239
Analyzing +4 | -1
A solid sphere rolls down two different inclined planes of same length, but of different inclinations. In both cases
View answer and explanation
Correct option: B
Speed will be same, but time of descent will be different
Option B (Speed will be same, but time of descent will be different) is correct because the final speed depends only on the height, while the time of descent depends on the inclination of the plane.
133 Medium
133 of 239
Application +4 | -1
Two points of a rod move with velocities 3v and v perpendicular to the rod and in the same direction separated by a distance 'r'. The angular velocity of the rod is
View answer and explanation
Correct option: D
2Vr\frac{{2V}}{r}
Option D (2V/r) is correct because the angular velocity is the difference in velocities divided by the distance between the points.
134 Easy
134 of 239
Application +4 | -1
Three particle of masses 8 kg, 4 kg, 4kg are at (4,1) (-2,2) and (1,-3) respectively. If they are acted upon by forces 16j,−6i16j, -6i and 14i14i newton’s respectively, acceleration of the centre of mass of the system is
View answer and explanation
Correct option: D
None
None of the options are correct. The acceleration of the center of mass is calculated by dividing the net force acting on the system by the total mass. The correct acceleration is (2/4)i + (16/4)j = i/2 + 4j m/s2, which is not among the given options.
135 Medium
135 of 239
Application +4 | -1
A flywheel of mass 50 kg and radius of gyration about its axis of rotation of 0.5m is acted upon by a constant torque of 12.5 Nm. Its angular velocity at t=5s is
View answer and explanation
Correct option: B
5 rad/s
Option B (5 rad/s) is correct because the angular velocity is calculated using the formula ω=αt\omega = \alpha t, where α\alpha is the angular acceleration and t is the time.
136 Medium
136 of 239
Remembering +4 | -1
The moment of inertia (I)(I) of a sphere of radius RR and mass MM is given by
View answer and explanation
Correct option: D
I=(2/5)MR2I=(2 / 5) M R^2
Option D (I=(2/5)MR2I=(2 / 5) M R^2) is correct because it represents the moment of inertia of a solid sphere about an axis through its center.
137 Medium
137 of 239
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 (Statement-1 is false, Statement-2 is true) is correct because Statement-1 is only true if the origin is the center of mass and the masses are distributed symmetrically, and Statement-2 is the correct formula for the center of mass.
138 Easy
138 of 239
Application +4 | -1
The angular speed of fly wheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in rad/s2\mathrm{rad/}{\mathrm{s}^2} is:
View answer and explanation
Correct option: C
4 π4\,\pi
Option C (4π4\pi) is correct because the angular acceleration is calculated using the change in angular speed over time, converting rpm to rad/s.
139 Medium
139 of 239
Understanding +4 | -1
A particle performing uniform circular motion has
View answer and explanation
Correct option: C
Transverse velocity and radial acceleration
Option C (Transverse velocity and radial acceleration) is correct because in uniform circular motion, the velocity is tangential (transverse) and the acceleration is directed towards the center (radial).
140 Medium
140 of 239
Remembering +4 | -1
Which of the following relation is correct? (Where symbol have their usual meanings)
View answer and explanation
Correct option: B
τ⃗=r⃗×F⃗\vec{\tau}=\vec{r} \times \vec{F}
Option B (τ⃗=r⃗×F⃗\vec{\tau}=\vec{r} \times \vec{F}) is correct because torque is defined as the cross product of the position vector and the force vector.
141 Medium
141 of 239
Understanding +4 | -1
The two persons of masses 55 kg and 65 kg respectively, are at at opposite ends of a boat. The length of the boat is 30 m and weight 100 kg. The 55 kg man walkup to the 65 kg man and its with him. If the boat is in still water the centre of mass of the system shifts by
View answer and explanation
Correct option: C
Zero
Option C is correct. Since the boat is in still water and there are no external forces acting on the system (boat + two persons), the center of mass of the system does not shift.
142 Easy
142 of 239
Application +4 | -1
A ring of radius 1 m and mass 20 kg is rotating about is diameter with angular velocity of 40 rad/sec. Its kinetic energy is
View answer and explanation
Correct option: C
8000 J
Option C (8000 J) is correct because the kinetic energy of a rotating ring is given by KE=12Iω2KE = \frac{1}{2}I\omega^2, where I is the moment of inertia about the diameter and ω\omega is the angular velocity.
143 Easy
143 of 239
Application +4 | -1
The speed of a motor decreases from 1200 rpm to 600 rpm in 20s. The total number of rotations it makes before coming to rest.
View answer and explanation
Correct option: A
400
Option A (400) is correct because it is calculated by finding the angular deceleration, time to stop, and then using the angular displacement formula.
144 Easy
144 of 239
Application +4 | -1
An automobile engine develops 50 kW when rotating at a speed of 900 rev/min. The torque it delivers (in N-m)
View answer and explanation
Correct option: B
530.7
Option B (530.7) is the closest to the correct answer because the torque is calculated using the formula P=τωP = \tau \omega, where P is power and ω\omega is angular speed in rad/s.
145 Easy
145 of 239
Remembering +4 | -1
1∘1^{\circ} is equal to
View answer and explanation
Correct option: D
Both (1) and (2)
Option D (Both (1) and (2)) is correct because 1 degree is equal to π180\frac{\pi}{180} radians and also equal to 60 arcminutes.
146 Hard
146 of 239
Application +4 | -1
A solid cylinder begins to rolls down on 300{30^0} incline of length 9 m without slipping. The speed of centre of mass at the bottom of plane is
View answer and explanation
Correct option: D
6g\sqrt {6g}
Option D (6g\sqrt {6g}) is correct because it is the final velocity calculated using conservation of energy for rolling without slipping.
147 Hard
147 of 239
Analyzing +4 | -1
The position vectors of two 1 kg particles, (A) and (B), are given by r→A=(α1t2i^+α2tj^+α3tk^)m\overrightarrow{\mathrm{r}}_{\mathrm{A}}=\left(\alpha_1 \mathrm{t}^2 \hat{\mathrm{i}}+\alpha_2 \text{t}\hat{\mathrm{j}}+\alpha_3 \text{t} \hat{\mathrm{k}}\right) \mathrm{m} and r→B=(β1ti^+β2t2j^+β3tk^)m,\overrightarrow{\mathrm{r}}_{\mathrm{B}}=\left(\beta_1 \mathrm{t} \hat{\mathrm{i}}+\beta_2 \mathrm{t}^2 \hat{\mathrm{j}}+\beta_3 \mathrm{t} \hat{\mathrm{k}}\right) \mathrm{m}, respectively; (α1=1 m/s2,α2=3 nm/s,α3=2 m/s,β1=2 m/s,β2=−1 m/s2,β3=4pm/s),\left(\alpha_1=1 \mathrm{~m} / \mathrm{s}^2, \alpha_2=3 \mathrm{~nm} / \mathrm{s}, \alpha_3=2 \mathrm{~m} / \mathrm{s}, \beta_1=2 \mathrm{~m} / \mathrm{s},\beta_2=-1 \mathrm{~m} / \mathrm{s}^2, \beta_3=4 \mathrm{p} \mathrm{m} / \mathrm{s}\right), where t is time, n and p are constants, At t=1 s,\text{t}=1 \mathrm{~s}, ∣V→A∣=∣V→B∣\left|\overrightarrow{\mathrm{V}}_{\mathrm{A}}\right|=\left|\overrightarrow{\mathrm{V}}_{\mathrm{B}}\right| and velocities V⃗A\mathrm{\vec{V}_A} and V⃗B\mathrm{\vec{V}_B} of the particles are orthogonal to each other. At t=1 s,\text{t}=1 \mathrm{~s}, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is L kgm2 s−1.\sqrt{\mathrm{L}} \mathrm{~kgm}^2 \mathrm{~s}^{-1}. The value of 'L' is
View answer and explanation
Correct option: A
90
Option A (90) is correct because after calculating the velocities and ensuring orthogonality, the angular momentum magnitude squared (L) is found to be 90 kgm²/s².
148 Medium
148 of 239
Application +4 | -1
An automobile engine develops 105 W{10^5}\,\mathrm{W} power when rotating at an angular speed of 900 rad/minute. The instantaneous torque delivered by the engine is
View answer and explanation
Correct option: C
23×104N−m\frac{2}{3} \times {10^4}N - m
Option C (23×104N−m\frac{2}{3} \times {10^4}N - m) is correct because the torque is calculated using the formula P=τωP = \tau \omega, where P is power and ω\omega is angular speed.
149 Medium
149 of 239
Analyzing +4 | -1
A disc of radius 'rr' is removed from the disc of radius 'RR' then
a) The minimum shift in centre of mass is zero
b) The maximum shift in centre of mass cannot be greater than r2/(R+r)r^2 /(R+r)
c) Centre of mass must lie where mass exists
d) The shift in centre of mass is r2/(R+r)r^2 /(R+r)
View answer and explanation
Correct option: A
Only a and b are correct
Option A (Only a and b are correct) is correct because the minimum shift in the center of mass is zero, and the maximum shift cannot be greater than r2/(R+r)r^2/(R+r).
150 Medium
150 of 239
Analyzing +4 | -1
A solid sphere of radius R{R} is placed on a smooth horizontal surface. A horizontal force FF is applied at height hh from the lowest point. For the maximum acceleration of centre of mass, which is correct?
View answer and explanation
Correct option: D
No relation between hh and R{R}
Option D (No relation between h and R) is correct because the acceleration of the center of mass depends only on the applied force and the mass of the sphere, not the height at which the force is applied.
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