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NEET • PHYSICS • SYSTEM OF PARTICLES

System of particles practice · Hard

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1 Hard
1 of 24
Application +4 | -1
A sphere rolls without slipping on an inclined plane of inclination θ\theta . Then the minimum value of coefficient of friction so that the sphere rolls with out slipping is
View answer and explanation
Correct option: A
27tan⁡θ\frac{2}{7}\tan \theta
Option A (27tan⁡θ\frac{2}{7}\tan \theta) is correct because it represents the minimum coefficient of friction required for a sphere to roll without slipping on an inclined plane.
2 Hard
2 of 24
Application +4 | -1
Two discs of same moment of inertia rotating about
their regular axis passing through centre and
perpendicular to the plane of disc with angular
velocity 2 rad/s and 4 rad/s. They are brought into
contact face to face coinciding the axis of rotation.
The expression for loss of energy during this procss
is:
View answer and explanation
Correct option: A
II
Option D (All of the above) is correct because pulling the weights inwards decreases the moment of inertia, increasing angular velocity and kinetic energy while conserving angular momentum.
3 Hard
3 of 24
Application +4 | -1
The surface density of a circular disc of radius a depends on the distance as ρ(r)=A+Br\rho(r)=A+B r. The moment of inertia about the line perpendicular to the plane of the disc is
View answer and explanation
Correct option: A
πa4(A2+2a5B)\pi a^4\left(\frac{A}{2}+\frac{2 a}{5} B\right)
Option A (πa4(A2+2a5B)\pi a^4(\frac{A}{2}+\frac{2 a}{5} B)) is correct because it is the result of integrating the moment of inertia over the area of the disc with the given density function.
4 Hard
4 of 24
Application +4 | -1
A thin rod of mass 6 m and length 6l6l is bent into regular hexagon. The moment of inertia of the hexagon about an axis normal to its plane and through the centre of mass of the system
View answer and explanation
Correct option: C
5 ml25\,m{l^2}
Option C (5 ml25\,m{l^2}) is correct because it is the moment of inertia calculated by summing the moments of inertia of each rod segment about the center.
5 Hard
5 of 24
Application +4 | -1
A solid sphere of mass 2 kg2 \mathrm{~kg} is rolling on a frictional horizontal surface with velocity 6 m/s6 \mathrm{~m} / \mathrm{s}. It collides on the free end of an ideal spring whose other end in fixed. The maximum compression produced in the spring will be x m\sqrt{x} \mathrm{~m}. Find the value of xx (Force constant of the spring is 36 N/m36 \mathrm{~N} / \mathrm{m} ).
View answer and explanation
Correct option: A
2.8
Option A (2.8) is correct because the maximum compression is found by equating the total kinetic energy (translational and rotational) to the potential energy stored in the spring.
6 Hard
6 of 24
Application +4 | -1
A solid cylinder of mass M{M} and radius R{R} rolls down an inclined plane of height h{h}. The angular velocity of the cylinder when it reaches the bottom of the plane is :
View answer and explanation
Correct option: B
2Rgh3\frac{2}{R} \sqrt{\frac{g h}{3}}
Option B (2Rgh3\frac{2}{R} \sqrt{\frac{g h}{3}}) is correct because it is the correct angular velocity when the cylinder reaches the bottom.
7 Hard
7 of 24
Application +4 | -1
 A{~A} wheel has mass of the rim 1 kg1 {~kg}, having 50 spokes each of mass 5 g5 {~g}. The radius of the wheel is 40 cm40 {~cm}. The moment of inertia
View answer and explanation
Correct option: C
0.173 kg−m20.173 {~kg}-{m}^2
Option C (0.173 kg−m2{kg}-{m}^2) is correct because the moment of inertia is calculated by summing the moment of inertia of the rim and the spokes.
8 Hard
8 of 24
Application +4 | -1
A sphere of mass M{M} and radius R{R} is released from the top of an inclined plane of inclination θ\theta. The minimum coefficient of friction between the plane and the sphere so that it rolls down the plane without sliding is given by
View answer and explanation
Correct option: D
μ=27tan⁡θ\mu=\frac{2}{7} \tan \theta
Option D (μ=27tan⁡θ\mu=\frac{2}{7} \tan \theta) is correct because it represents the minimum coefficient of friction required for a sphere to roll down an inclined plane without sliding.
9 Hard
9 of 24
Analyzing +4 | -1
A uniform cubical box of side aa is placed on a rough floor and it is to be moved by applying minimum possible force FF at a point which is at a distance bb above its centre of mass. If the coeflicient of friction is μ=0.4\mu=0.4, then what is the maximum possible value of 100×ba100 \times \frac{b}{a} for which the box will not topple before moving?
View answer and explanation
Correct option: A
50
Option A is correct. The box will not topple if the torque due to the applied force is less than or equal to the torque due to the weight of the box. This condition leads to the maximum possible value of b/a being 0.5, hence 100 * (b/a) = 50.
10 Hard
10 of 24
Application +4 | -1
A solid sphere of mass 1 kg rolls down an inclined plane from rest from a height of 7m. On reaching the bottom rotational kinetic energy K(g=10 ms−2)K\left( {g = 10\,m{s^{ - 2}}} \right)
View answer and explanation
Correct option: B
20 J
Option B (20 J) is correct because the rotational kinetic energy is 2/7 of the total potential energy converted to kinetic energy.
11 Hard
11 of 24
Understanding +4 | -1
Moment of inertia of a uniform annular disc of internal radius rr and external radis RR and mass MM about an axis through its centre and perpendicular to its plane is
View answer and explanation
Correct option: B
12M(R2+r2)\frac{1}{2} M\left(R^2+r^2\right)
Option B (12M(R2+r2)\frac{1}{2} M(R^2+r^2)) is correct because it's the standard formula for the moment of inertia of an annular disc.
12 Hard
12 of 24
Application +4 | -1
A spherical shell and a solid cylinder of same radius rolls down an inclined plane. The ratio of their accelerations will be:
View answer and explanation
Correct option: B
9:109: 10
Option B (9:109: 10) is correct because it represents the ratio of the accelerations of a spherical shell and a solid cylinder rolling down an inclined plane.
13 Hard
13 of 24
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.
14 Hard
14 of 24
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}.
15 Hard
15 of 24
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.

Explore Other Chapters in PHYSICS

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UNITS & MEASUREMENTS 392 MCQs WORK-ENERGY-POWER 355 MCQs MECHANICAL PROPERTIES OF FLUIDS 317 MCQs MOTION IN PLANE 307 MCQs MAGNETIC EFFECT OF CURRENT 282 MCQs THERMAL PROPERTIES OF MATTER 269 MCQs
16 Hard
16 of 24
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.
17 Hard
17 of 24
Application +4 | -1
A uniform metal rod is bent into L shape with the bent arms at 90∘90^\circ to each other. The distance of the centre of mass from the bent point is
View answer and explanation
Correct option: A
142\frac{1}{{4\sqrt 2 }}
Option A (142\frac{1}{{4\sqrt 2 }}) is correct because the center of mass of each arm is at L/2, and then the center of mass of the L shape is calculated considering the symmetry and equal masses.
18 Hard
18 of 24
Application +4 | -1
The moment of interia of a rod about its perpendicular bisector is II. When the temperature of the rod is increased by ΔT\Delta T, the increase in the moment of inertia of the rod about same axis is (Here, α\alpha is the coefficient of linear expansion o the rod) :
View answer and explanation
Correct option: B
2αIΔT2 \alpha {I} \Delta {T}
Option B (2αIΔT2 \alpha {I} \Delta {T}) is correct because the increase in moment of inertia is proportional to the original moment of inertia, the coefficient of linear expansion, and the change in temperature.
19 Hard
19 of 24
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.
20 Hard
20 of 24
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.
21 Hard
21 of 24
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.
22 Hard
22 of 24
Application +4 | -1
A rope is wound around a hollow cylinder of mass
4 kg and radius 20 cm. What is the angular
acceleration of the cylinder if the rope is pulled
with a force of 15 N:
View answer and explanation
Correct option: A
37.5 rad/s237.5 \mathrm{~rad} / \mathrm{s}^2
Option A (37.5 rad/s237.5 \mathrm{~rad} / \mathrm{s}^2) is correct because it is the angular acceleration calculated using the formula torque = I*alpha, where I is the moment of inertia of the hollow cylinder.
23 Hard
23 of 24
Application +4 | -1
A thin hollow sphere of mass mm is completely filled with non viscous liquid of mass mm. When the sphere roll-on horizontal ground such that centre moves with velocity vv, kinetic energy of the system is equal to
View answer and explanation
Correct option: B
43mv2\frac{4}{3} m v^2
Option B (43mv2\frac{4}{3} m v^2) is correct because it is the total kinetic energy of the system, considering both the sphere's rolling motion and the liquid's translational motion.
24 Hard
24 of 24
Analyzing +4 | -1
Assertion (A): A disc is rolling on a rough horizontal surface without slipping. The velocity of centre of mass is v.v . Then all the other points on the disc lying on a circular arc with point of contact as the center and this arc passing through center of mass of disc will have the same speed v.v.
Reason (R): When a disc is rotating without sliding on a rough horizontal surface the magnitude of velocities of all the points at a distance rr from point of contact is same.
View answer and explanation
Correct option: A
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Option A is correct because both the assertion and reason are true, and the reason correctly explains the assertion regarding rolling motion.
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