JEE Physics System Of Particles and Rotational Motion MCQs Set 10

Multiple Choice Questions (MCQs) for JEE Physics: System Of Particles and Rotational Motion

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Multiple Choice Questions

Question 1: A solid sphere rolls without slipping. The ratio of its rotational kinetic energy to its total kinetic energy is:
(a) 1/2
(b) 5/7
(c) 2/5
(d) 2/7
Show Answer & Explanation

Answer: (d) 2/7

Explanation:
1. \( K_{rot} = \tfrac12\cdot\tfrac25MR^2\omega^2 = \tfrac15Mv^2 \), \( K_{tot} = \tfrac{7}{10}Mv^2 \).
2. Ratio = 2/7.

Teacher's Note:
1. Total = translational + rotational.

Question 2: A ring rolls without slipping. The fraction of total kinetic energy that is translational is:
(a) 2/3
(b) 1/3
(c) 3/4
(d) 1/2
Show Answer & Explanation

Answer: (d) 1/2

Explanation:
1. For a ring, \( K_{rot} = \tfrac12Mv^2 = K_{trans} \).

Teacher's Note:
1. I = MR² makes the two equal.

Question 3: A solid cylinder of mass 2 kg rolls without slipping at 4 m/s. Its total kinetic energy is:
(a) 32 J
(b) 24 J
(c) 16 J
(d) 8 J
Show Answer & Explanation

Answer: (b) 24 J

Explanation:
1. \( K = \tfrac12Mv^2(1+\tfrac12) = \tfrac34(2)(16) = 24 \) J.

Teacher's Note:
1. Factor (1 + k²/R²) = 3/2 for a solid cylinder.

Question 4: A disc rolls down an inclined plane of angle θ without slipping. Its acceleration is:
(a) \( \tfrac35g\sin\theta \)
(b) \( \tfrac12g\sin\theta \)
(c) \( g\sin\theta \)
(d) \( \tfrac23g\sin\theta \)
Show Answer & Explanation

Answer: (d) \( \tfrac23g\sin\theta \)

Explanation:
1. \( a = \dfrac{g\sin\theta}{1 + k^2/R^2} = \dfrac{g\sin\theta}{3/2} \).

Teacher's Note:
1. Use the general rolling formula.

Question 5: A solid sphere rolls down an incline of 30 degrees. Its acceleration (g = 10 m/s²) is:
(a) 5 m/s²
(b) 7 m/s²
(c) 3.57 m/s²
(d) 3.33 m/s²
Show Answer & Explanation

Answer: (c) 3.57 m/s²

Explanation:
1. \( a = \dfrac{5}{7}g\sin30^\circ = \dfrac57\times5 = 3.57 \) m/s².

Teacher's Note:
1. k²/R² = 2/5 for a solid sphere.

Question 6: A ring, a disc and a solid sphere of the same mass and radius are released from rest at the top of the same incline and roll without slipping. The order in which they reach the bottom (first to last) is:
(a) all together
(b) ring, disc, sphere
(c) sphere, disc, ring
(d) disc, sphere, ring
Show Answer & Explanation

Answer: (c) sphere, disc, ring

Explanation:
1. Acceleration is \( g\sin\theta/(1+k^2/R^2) \): sphere (5/7), disc (2/3), ring (1/2) times g sin θ.

Teacher's Note:
1. Smaller k²/R² means greater acceleration.

Question 7: A solid sphere rolls down from height h without slipping. Its speed at the bottom is:
(a) \( \sqrt{\dfrac{10gh}{7}} \)
(b) \( \sqrt{gh} \)
(c) \( \sqrt{\dfrac{4gh}{3}} \)
(d) \( \sqrt{2gh} \)
Show Answer & Explanation

Answer: (a) \( \sqrt{\dfrac{10gh}{7}} \)

Explanation:
1. \( mgh = \tfrac12mv^2(1+\tfrac25) \Rightarrow v^2 = \tfrac{10}{7}gh \).

Teacher's Note:
1. Energy conservation with rolling.

Question 8: A ring rolls down from height h. Its speed at the bottom is:
(a) \( \sqrt{gh} \)
(b) \( \sqrt{\dfrac{4gh}{3}} \)
(c) \( \sqrt{2gh} \)
(d) \( \sqrt{\dfrac{gh}{2}} \)
Show Answer & Explanation

Answer: (a) \( \sqrt{gh} \)

Explanation:
1. \( mgh = \tfrac12mv^2(1+1) \Rightarrow v = \sqrt{gh} \).

Teacher's Note:
1. Half the energy goes into rotation.

Question 9: For a body rolling down an incline without slipping, the minimum coefficient of friction needed for a solid cylinder is:
(a) \( \tfrac27\tan\theta \)
(b) \( \tfrac13\tan\theta \)
(c) \( \tfrac12\tan\theta \)
(d) \( \tan\theta \)
Show Answer & Explanation

Answer: (b) \( \tfrac13\tan\theta \)

Explanation:
1. \( f = \dfrac{Mg\sin\theta}{1 + R^2/k^2} = \tfrac13Mg\sin\theta \le \mu Mg\cos\theta \).

Teacher's Note:
1. Friction supplies the torque for rotation.

Question 10: In pure rolling, the velocity of the point of contact with the ground is:
(a) zero
(b) v
(c) v/2
(d) 2v
Show Answer & Explanation

Answer: (a) zero

Explanation:
1. The contact point is instantaneously at rest: v_cm = ωR cancels the rotational velocity there.

Teacher's Note:
1. Rolling without slipping.

Question 11: In pure rolling with centre speed v, the speed of the topmost point is:
(a) zero
(b) 2v
(c) v/2
(d) v
Show Answer & Explanation

Answer: (b) 2v

Explanation:
1. Top point: v + ωR = 2v.

Teacher's Note:
1. Velocity of centre plus rotational velocity.

Question 12: A wheel of radius R rolls on a horizontal surface with centre speed v. A point on the rim at the level of the centre has speed:
(a) \( \sqrt2\,v \)
(b) \( v \)
(c) \( 2v \)
(d) \( v/\sqrt2 \)
Show Answer & Explanation

Answer: (a) \( \sqrt2\,v \)

Explanation:
1. Velocity = v horizontally plus v vertically (rotation), magnitude \( \sqrt2v \).

Teacher's Note:
1. Vector sum of two perpendicular components.

Question 13: A cylinder rolls up an incline with initial speed v and comes to rest at height h. For a solid cylinder, h is:
(a) \( \dfrac{v^2}{2g} \)
(b) \( \dfrac{2v^2}{3g} \)
(c) \( \dfrac{v^2}{g} \)
(d) \( \dfrac{3v^2}{4g} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{3v^2}{4g} \)

Explanation:
1. \( \tfrac34mv^2 = mgh \Rightarrow h = \dfrac{3v^2}{4g} \).

Teacher's Note:
1. Total KE (translation + rotation) is converted to potential energy.

Question 14: A disc rolls without slipping on a horizontal plane at 6 m/s. Its total kinetic energy equals what fraction of the translational energy?
(a) 5/4
(b) 2
(c) 3/2
(d) 1
Show Answer & Explanation

Answer: (c) 3/2

Explanation:
1. \( K = \tfrac12mv^2(1+\tfrac12) \).

Teacher's Note:
1. Disc: k²/R² = 1/2.

Question 15: A solid sphere of mass 1 kg and radius 0.1 m spins at 10 rad/s about a diameter. Its rotational KE is:
(a) 0.2 J
(b) 0.4 J
(c) 1 J
(d) 2 J
Show Answer & Explanation

Answer: (a) 0.2 J

Explanation:
1. \( I = \tfrac25(1)(0.01) = 0.004 \); \( K = \tfrac12(0.004)(100) = 0.2 \) J.

Teacher's Note:
1. K = ½Iω².

Question 16: Work done by a torque of 10 N m in rotating a body through 2 radians is:
(a) 20 J
(b) 12 J
(c) 8 J
(d) 5 J
Show Answer & Explanation

Answer: (a) 20 J

Explanation:
1. \( W = \tau\theta = 20 \) J.

Teacher's Note:
1. Rotational work formula.

Question 17: The power delivered by a torque of 50 N m acting on a shaft rotating at 20 rad/s is:
(a) 70 W
(b) 1000 W
(c) 2.5 W
(d) 30 W
Show Answer & Explanation

Answer: (b) 1000 W

Explanation:
1. \( P = \tau\omega = 1000 \) W.

Teacher's Note:
1. Power equals torque times angular velocity.

Question 18: A motor develops a torque of 100 N m at 3000 rpm. Its power is closest to:
(a) 3.14 kW
(b) 31.4 kW
(c) 10 kW
(d) 314 kW
Show Answer & Explanation

Answer: (b) 31.4 kW

Explanation:
1. \( \omega = 100\pi \) rad/s, \( P = 100\times100\pi = 31.4 \) kW.

Teacher's Note:
1. Convert rpm to rad/s.

Question 19: A uniform disc of mass M and radius R rotates freely about a vertical axis at ω. A small mass m is placed at the rim. The new angular speed is:
(a) \( \dfrac{2M\omega}{M+m} \)
(b) \( \dfrac{M\omega}{M+2m} \)
(c) \( \omega \)
(d) \( \dfrac{M\omega}{M+m} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{M\omega}{M+2m} \)

Explanation:
1. \( \tfrac12MR^2\omega = (\tfrac12M+m)R^2\omega' \Rightarrow \omega' = \dfrac{M\omega}{M+2m} \).

Teacher's Note:
1. Angular momentum is conserved.

Question 20: A solid cylinder and a hollow cylinder of the same mass and radius are released together on an incline and roll. Which reaches the bottom first?
(a) both together
(b) hollow cylinder
(c) depends on the mass
(d) solid cylinder
Show Answer & Explanation

Answer: (d) solid cylinder

Explanation:
1. Solid cylinder k²/R² = 1/2 is smaller than 1, so its acceleration is greater.

Teacher's Note:
1. The result is independent of mass and radius.

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Multiple Choice Questions (MCQs) for JEE Physics System Of Particles and Rotational Motion

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