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Unit 5: Rotational Motion — Online MCQ Test

Q: Two particles of mass 5 kg and 10 kg respectively are attached to the two ends of a rigid rod of length 1 m with negligible mass. The centre of mass of the system from the 5 kg particle is nearly at a distance of:

Answer: B. 67 cm — The distance of COM from mass m1 is x = (m2 * d) / (m1 + m2). x = (10 * 1) / (5 + 10) = 10/15 m = 2/3 m ≈ 0.666 m, which is 67 cm.

Q: A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v0 as shown in figure. If the string gets slack at some point P making an angle  from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v0 is:

Answer: D. sinθ / sqrt(2 + 3sinθ) — Standard circular motion problem involving string slackness conditions and energy conservation.

Q: The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.

Answer: D. 108 days — Conservation of angular momentum: I1*omega1 = I2*omega2. (2/5 MR1^2)(2pi/T1) = (2/5 MR2^2)(2pi/T2). T2 = T1 * (R2/R1)^2 = 27 * (2)^2 = 27 * 4 = 108 days.

Q: A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:

Answer: C. 7/57 — Using parallel axis theorem and mass distribution ratio. Given key indicates option 3.

Q: From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

Answer: A. 13 MR^2/32 — The mass of the removed disc is M/4. Its radius is R/2 and its center is at a distance R/2 from the center of the original disc. Using the parallel axis theorem, the moment of inertia of the removed part is I' = Icm + md^2 = (1/2)(M/4)(R/2)^2 + (M/4)(R/2)^2 = MR^2/32 + MR^2/16 = 3MR^2/32. The remaining moment of inertia is (1/2)MR^2 - 3MR^2/32 = 13MR^2/32.

Q: A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?

Answer: A. Sphere — The time to reach the bottom is proportional to √(1 + k²/R²). For a disk, k²/R² = 1/2. For a sphere, k²/R² = 2/5. Since 2/5 < 1/2, the sphere takes less time.

Q: A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s^-2. Its net acceleration in ms^-2 at the end of 2.0 s is approximately :

Answer: D. 8.0 — Angular velocity ω = αt = 2.0 * 2.0 = 4 rad/s. Centripetal acceleration a_r = Rω^2 = 0.5 * 4^2 = 8 m/s^2. Tangential acceleration a_t = Rα = 0.5 * 2.0 = 1 m/s^2. Net acceleration = sqrt(a_r^2 + a_t^2) ≈ 8 m/s^2.

Q: A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10^-4 J by the end of the second revolution after the beginning of the motion?

Answer: D. 0.1 m/s^2 — KE = 1/2 * m * v^2 = 8*10^-4 => v = 0.4 m/s. Using v^2 = u^2 + 2as where s = 2 * (2πr). Substituting values, 0.16 = 2 * a * 2 * 2 * 3.14 * 0.064, leading to a = 0.1 m/s^2.

Q: Three objects, A: (a solid sphere), B: (a thin circular disk) and C: (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed ω about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation

Answer: C. W_C > W_B > W_A — Work required = initial KE = ½Iω². Moments of inertia: I_A (solid sphere) = (2/5)MR², I_B (disk) = (1/2)MR², I_C (ring) = MR². Since MR²ω² is common factor: W ∝ I. I_C > I_B > I_A, so W_C > W_B > W_A.

Q: Two particles A and B are moving in uniform circular motion in concentric circles of radii rA and rB with speed vA and vB respectively. Their time period of rotation is the same. The ratio of angular speed of A to that of B will be :

Answer: B. 1 : 1 — Angular speed omega = 2 * pi / T. Since both particles have the same time period T, their angular speeds are identical, hence 1:1.

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Practice Unit 5: Rotational Motion with a free chapter-wise online MCQ test for NEET NEET (All) PHYSICS. This chapter covers: Centre of the mass of a two-particle system Centre of the mass of a rigid body; Basic concepts of rotational motion; moment of a force; torque angular momentum conservation of angu.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Unit 5: Rotational Motion — Important Questions & Answers (FAQ)

Frequently asked questions from NEET NEET (All) PHYSICS — Unit 5: Rotational Motion, with answers and explanations. These are sample questions; the exam has its own separate question set.

Two particles of mass 5 kg and 10 kg respectively are attached to the two ends of a rigid rod of length 1 m with negligible mass. The centre of mass of the system from the 5 kg particle is nearly at a distance of:

A. 50 cm
B. 67 cm ✓
C. 80 cm
D. 33 cm

Answer: B. 67 cm
The distance of COM from mass m1 is x = (m2 * d) / (m1 + m2). x = (10 * 1) / (5 + 10) = 10/15 m = 2/3 m ≈ 0.666 m, which is 67 cm.

A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v0 as shown in figure. If the string gets slack at some point P making an angle  from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v0 is:

A. 1 / 2sinθ
B. 1 / sqrt(2 + 3sinθ)
C. cosθ / sqrt(2 + 3sinθ)
D. sinθ / sqrt(2 + 3sinθ) ✓

Answer: D. sinθ / sqrt(2 + 3sinθ)
Standard circular motion problem involving string slackness conditions and energy conservation.

The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.

A. 100 days
B. 105 days
C. 115 days
D. 108 days ✓

Answer: D. 108 days
Conservation of angular momentum: I1*omega1 = I2*omega2. (2/5 MR1^2)(2pi/T1) = (2/5 MR2^2)(2pi/T2). T2 = T1 * (R2/R1)^2 = 27 * (2)^2 = 27 * 4 = 108 days.

A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:

A. 7/8
B. 7/40
C. 7/57 ✓
D. 7/64

Answer: C. 7/57
Using parallel axis theorem and mass distribution ratio. Given key indicates option 3.

From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

A. 13 MR^2/32 ✓
B. 11 MR^2/32
C. 9 MR^2/32
D. 15 MR^2/32

Answer: A. 13 MR^2/32
The mass of the removed disc is M/4. Its radius is R/2 and its center is at a distance R/2 from the center of the original disc. Using the parallel axis theorem, the moment of inertia of the removed part is I' = Icm + md^2 = (1/2)(M/4)(R/2)^2 + (M/4)(R/2)^2 = MR^2/32 + MR^2/16 = 3MR^2/32. The remaining moment of inertia is (1/2)MR^2 - 3MR^2/32 = 13MR^2/32.

A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?

A. Sphere ✓
B. Both reach at the same time
C. Depends on their masses
D. Disk

Answer: A. Sphere
The time to reach the bottom is proportional to √(1 + k²/R²). For a disk, k²/R² = 1/2. For a sphere, k²/R² = 2/5. Since 2/5 < 1/2, the sphere takes less time.

A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s^-2. Its net acceleration in ms^-2 at the end of 2.0 s is approximately :

A. 7.0
B. 6.0
C. 3.0
D. 8.0 ✓

Answer: D. 8.0
Angular velocity ω = αt = 2.0 * 2.0 = 4 rad/s. Centripetal acceleration a_r = Rω^2 = 0.5 * 4^2 = 8 m/s^2. Tangential acceleration a_t = Rα = 0.5 * 2.0 = 1 m/s^2. Net acceleration = sqrt(a_r^2 + a_t^2) ≈ 8 m/s^2.

A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10^-4 J by the end of the second revolution after the beginning of the motion?

A. 0.15 m/s^2
B. 0.18 m/s^2
C. 0.2 m/s^2
D. 0.1 m/s^2 ✓

Answer: D. 0.1 m/s^2
KE = 1/2 * m * v^2 = 8*10^-4 => v = 0.4 m/s. Using v^2 = u^2 + 2as where s = 2 * (2πr). Substituting values, 0.16 = 2 * a * 2 * 2 * 3.14 * 0.064, leading to a = 0.1 m/s^2.

Three objects, A: (a solid sphere), B: (a thin circular disk) and C: (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed ω about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation

A. W_A > W_B > W_C
B. W_B > W_A > W_C
C. W_C > W_B > W_A ✓
D. W_A > W_C > W_B

Answer: C. W_C > W_B > W_A
Work required = initial KE = ½Iω². Moments of inertia: I_A (solid sphere) = (2/5)MR², I_B (disk) = (1/2)MR², I_C (ring) = MR². Since MR²ω² is common factor: W ∝ I. I_C > I_B > I_A, so W_C > W_B > W_A.

Two particles A and B are moving in uniform circular motion in concentric circles of radii rA and rB with speed vA and vB respectively. Their time period of rotation is the same. The ratio of angular speed of A to that of B will be :

A. rB : rA
B. 1 : 1 ✓
C. rA : rB
D. vA : vB

Answer: B. 1 : 1
Angular speed omega = 2 * pi / T. Since both particles have the same time period T, their angular speeds are identical, hence 1:1.

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