Skip to main content
Back to SYSTEM OF PARTICLES

Your Progress

0 / 58 Answered
0
Correct
0
Incorrect
0%
Accuracy
NEET • PHYSICS • SYSTEM OF PARTICLES

System of particles practice · Easy

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →
Showing 1-30 of 58 questions

Preparing practice and saved progress…

1 Easy
1 of 58
Application +4 | -1
The time dependence of the position of a particle of mass m = 4 kg is given by r(t)=2ti^−3t2j^r\left( t \right) = 2t\hat i - 3{t^2}\hat j . Its angular momentum, with respect to the origin, at time t=2t = 2 is
View answer and explanation
Correct option: C
−96  k^- 96\,\,\hat k
Option C (-96 k) is correct because it is the result of calculating the angular momentum using the given position and mass.
2 Easy
2 of 58
Application +4 | -1
Density remains constant, if earth contracts to one fourth of its present radius, duration of the day would be [in minutes]
View answer and explanation
Correct option: A
4532\frac{{45}}{{32}} min
Option A (4532\frac{{45}}{{32}} min) is correct because the duration of the day is inversely proportional to the square of the radius when density is constant.
3 Easy
3 of 58
Understanding +4 | -1
For a system of two point masses m1m_1 and m2m_2 (where m2>m1m_2>m_1 ), centre of mass lies
View answer and explanation
Correct option: C
Nearer to the heavier one
Option C (Nearer to the heavier one) is correct because the center of mass is always closer to the more massive object in a two-body system.
4 Easy
4 of 58
Application +4 | -1
A dancer spins about a vertical axis at 60 rpm with her arms folded. If she stretches her hands so that M.I about the vertical axes increases by 25% the new rate of revolution is
View answer and explanation
Correct option: A
48 rpm
Option A (48 rpm) is correct because it is calculated using the conservation of angular momentum, I1ω1=I2ω2I_1 \omega_1 = I_2 \omega_2, where I2=1.25I1I_2 = 1.25 I_1.
5 Easy
5 of 58
Application +4 | -1
If an angular momentum of a body of fixed M.I. increases by 50%50 \%, its kinetic energy of rotation increases by:
View answer and explanation
Correct option: D
125%125 \%
Option D (125%125 \%) is correct because the kinetic energy of rotation is proportional to the square of the angular momentum, so a 50% increase in angular momentum results in a 125% increase in kinetic energy.
6 Easy
6 of 58
Application +4 | -1
A ring of mass 5 kg and radius 0.2 m is rotated about its axis and its angular speed is 10 rad/sec, then its angular momentum is [kg−m2/sec⁡]\left[ {kg - {m^2}/\sec } \right]
View answer and explanation
Correct option: D
2
Option D (2) is correct because it is the angular momentum calculated using the formula L = Iω, where I is the moment of inertia of a ring and ω is the angular speed.
7 Easy
7 of 58
Application +4 | -1
Three masses are placed on the x-axis 300 g at origin, 500 g at x=40 cm and 400 g at x = 70 cm. The distance of the centre of mass from the origin is
View answer and explanation
Correct option: A
40 cm
Option A (40 cm) is correct because the distance of the center of mass from the origin is calculated using the formula xCM=∑mixi∑mix_{CM} = \frac{\sum m_i x_i}{\sum m_i}.
8 Easy
8 of 58
Analyzing +4 | -1
A solid sphere and hollow sphere of same mass and radius rotate about their diameter with same rotational kinetic energy. If torques of equal magnitude are applied on them, ratio of their angular displacement before they are bought to rest.
View answer and explanation
Correct option: A
0.042361111
Option A (0.042361111) is correct as it represents the ratio of angular displacements calculated using the moments of inertia of a solid sphere and a hollow sphere and the work-energy theorem.
9 Easy
9 of 58
Application +4 | -1
A force F = 10 N is acting on a point is at 4 m distance then torque is [N-m]
View answer and explanation
Correct option: C
40
Option C (40 N-m) is correct because torque is the product of force and the perpendicular distance from the axis of rotation.
10 Easy
10 of 58
Analyzing +4 | -1
When a sphere rolls down an inclined plane, then identity the correct statement related to the work done by friction force
View answer and explanation
Correct option: C
The net work done by friction is zero
Option C (The net work done by friction is zero) is correct because while friction does negative work in the translational sense, it does positive work in the rotational sense, resulting in zero net work.
11 Easy
11 of 58
Understanding +4 | -1
What are the coordinates of centre of mass of a point mass of 2 kg2 {~kg} with coordinates (2,3)(2,3) ?
View answer and explanation
Correct option: D
(2,3)(2,3)
Option D ((2,3)) is correct because the coordinates of the center of mass of a single point mass are the same as the coordinates of the point mass itself.
12 Easy
12 of 58
Application +4 | -1
A fly wheel of M.I 6×10−2 kg m26 \times {10^{ - 2}}\,kg\,{m^2} is rotating with an angular velocity of  20 rad s−1\,20\,rad\,{s^{ - 1}}. The torque required to bring it to rest in 4 s is
View answer and explanation
Correct option: D
0.3 Nm
Option D (0.3 Nm) is correct because it is the torque required to produce the necessary angular deceleration to bring the flywheel to rest in 4 seconds.
13 Easy
13 of 58
Application +4 | -1
If RR is the radius of gyration of a body of mass MM and radius rr, then the ratio of its rotational to translational kinetic energy in the rolling condition is
View answer and explanation
Correct option: B
R2r2\frac{R^2}{r^2}
Option B (R2r2\frac{R^2}{r^2}) is correct. The ratio of rotational to translational kinetic energy is derived from the definition of radius of gyration.
14 Easy
14 of 58
Analyzing +4 | -1
A man standing on a platform holds weights in his outstretched arms. The system is rotated about a central vertical axis. If the man now pulls the weights inwards close to his body, then
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because decreasing the radius increases angular velocity and kinetic energy while conserving angular momentum.
15 Easy
15 of 58
Application +4 | -1
A Mass of mass 40 kg in standing on a uniform planck of mass 60 kg lying on horizontal frictionless ice. The man walks from one end to the other end of the plank. The distance walked by the man relative to ice (given length of plank = 5 m)
View answer and explanation
Correct option: B
3 m
Option B is correct. Using the concept that the center of mass of the man-plank system remains constant, the distance walked by the man relative to the ice is calculated to be 3 m.

Explore Other Chapters in PHYSICS

View All →
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 Easy
16 of 58
Application +4 | -1
A ring and a disc of same mass roll without slipping along a horizontal surface with same velocity. If the K.E. of ring is 4J, then that of disc is
View answer and explanation
Correct option: B
3 J
Option B (3 J) is correct. The kinetic energy of a rolling object is the sum of translational and rotational kinetic energies. For a ring, KE=mv2KE = mv^2, and for a disc, KE=(3/4)mv2KE = (3/4)mv^2.
17 Easy
17 of 58
Application +4 | -1
A stationary wheel starts rotating about its own axis at uniform angular acceleration 16 rad/s216\,rad/{s^2}. The time taken by it to complete 77 rotations is
View answer and explanation
Correct option: A
5.5 sec
Option A (5.5 sec) is correct because it is calculated using the rotational kinematic equation θ=ω0t+12αt2\theta = \omega_0 t + \frac{1}{2} \alpha t^2, where ω0=0\omega_0 = 0 and θ=77×2π\theta = 77 \times 2\pi radians.
18 Easy
18 of 58
Application +4 | -1
Two particles whose masses are 10 kg10 \mathrm{~kg} and 30 kg30 \mathrm{~kg} and their position vectors are i^+j^+k^\hat{i}+\hat{j}+\hat{k} and −i^−j^−k^-\hat{i}-\hat{j}-\hat{k} respectively would have the centre of mass at
View answer and explanation
Correct option: A
−(i^+j^+k^)2-\frac{(\hat{i}+\hat{j}+\hat{k})}{2}
Option A (−(i^+j^+k^)2-\frac{(\hat{i}+\hat{j}+\hat{k})}{2}) is correct because it is the result of applying the center of mass formula with the given masses and position vectors.
19 Easy
19 of 58
Analyzing +4 | -1
When a body is projected at an angle with the horizontal in the uniform gravitational field of the earth, the angular momentum of the body about the point of projection, as it proceeds along the path
View answer and explanation
Correct option: B
Increases continuously
Option B (Increases continuously) is correct because the torque due to gravity increases as the projectile moves along its path, leading to an increase in angular momentum.
20 Easy
20 of 58
Application +4 | -1
Two particles each of mass 1 kg1 {~kg} are located at the point having co-ordinates (1,0,1)(1,0,1) and (1,1,0)(1,1,0). The co-ordinates of the centre of mass will be
View answer and explanation
Correct option: B
(1,12,12)\left(1, \frac{1}{2}, \frac{1}{2}\right)
Option B ((1, 1/2, 1/2)) is correct because the center of mass is calculated by averaging the coordinates of the particles, weighted by their masses.
21 Easy
21 of 58
Application +4 | -1
The radius of gyration of a disc of mass 50 g and radius 2.82 cm, about an axis passing through its centre of gravity and perpendicular to the plane, is
View answer and explanation
Correct option: C
2 cm
Option C (2 cm) is correct because the radius of gyration (k) for a disc is given by k=R2k = \frac{R}{\sqrt{2}}, where R is the radius of the disc.
22 Easy
22 of 58
Application +4 | -1
If the velocity of a rotating body of mass 50 kg is given by V⃗=3i⃗+4j⃗+5k⃗\vec V = 3\vec i + 4\vec j + 5\vec k and rotating with radical vector r→=2i⃗+3j⃗−4k⃗\overrightarrow r = 2\vec i + 3\vec j - 4\vec k , then its angular momentum is
View answer and explanation
Correct option: D
50(31i^+2j−k^)50\left( {31\hat i + 2j - \hat k} \right)
Option D (50(31i^+2j−k^)50\left( {31\hat i + 2j - \hat k} \right)) is correct. Angular momentum is calculated as L⃗=m(r⃗×V⃗)\vec{L} = m(\vec{r} \times \vec{V}).
23 Easy
23 of 58
Application +4 | -1
Find the moment of inertia of a uniform square plate of mass MM and edge aa about one of its diagonals.
View answer and explanation
Correct option: A
Ma212\frac{M a^2}{12}
Option A (Ma²/12) is correct because the moment of inertia of a square plate about one of its diagonals is Ma²/12.
24 Easy
24 of 58
Application +4 | -1
Two thin uniform circular rings each of radius 10 cm and mass 0.1 kg are arranged such that they have a common centre and their planes are perpendicular to each other. The moment of inertia of this system about an axis passing through their common centre and perpendicular to the plane of one of the rings in kg cm2kg\,{cm^2} is
View answer and explanation
Correct option: C
15
Option C (15) is correct. The moment of inertia of the system is the sum of the moments of inertia of each ring.
25 Easy
25 of 58
Application +4 | -1
A solid sphere of mass mm rolls down an inclined plane a height hh. Find rotational kinetic energy of the sphere.
View answer and explanation
Correct option: A
27mgh\frac{2}{7} m g h
Option A (27mgh\frac{2}{7} m g h) is correct. The rotational kinetic energy is a fraction of the total potential energy converted during rolling down the incline.
26 Easy
26 of 58
Application +4 | -1
A stone is whirled along a circle with a constant angular velocity and its angular momentum is L. If the radius of its circular path is halved while keeping the angular velocity same, its new angular momentum would be
View answer and explanation
Correct option: A
L/4
Option A (L/4) is correct because angular momentum is proportional to the square of the radius when angular velocity is constant.
27 Easy
27 of 58
Application +4 | -1
Three masses are placed on the x – axis, 200 g at origin, 400 g at x = 20 cm and 600 g at x = 60 cm. The distance of the centre of mass from the origin is
View answer and explanation
Correct option: D
1103 cm\frac{{110}}{3}\,cm
Option D is correct. The distance of the center of mass from the origin is calculated using the formula for the center of mass of a system of particles, resulting in approximately 36.67 cm or 110/3 cm.
28 Easy
28 of 58
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.
29 Easy
29 of 58
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.
30 Easy
30 of 58
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.
Revision notes and key takeaways

Review SYSTEM OF PARTICLES in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

  • ✓ This set contains 58 single-correct-answer MCQs on SYSTEM OF PARTICLES.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
Showing 1-30 of 58 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ NEET PHYSICS - SYSTEM OF PARTICLES MCQs (All) 239 MCQs → NEET PHYSICS - SYSTEM OF PARTICLES MCQs (Hard) 24 MCQs → NEET PHYSICS - SYSTEM OF PARTICLES MCQs (Medium) 157 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ UNITS & MEASUREMENTS 392 MCQs → WORK-ENERGY-POWER 355 MCQs → MECHANICAL PROPERTIES OF FLUIDS 317 MCQs → MOTION IN PLANE 307 MCQs

Your practice

0 / 58 answered · 58 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →