Unit 5: Rotational Motion — Online MCQ Test
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Practice Unit 5: Rotational Motion with a free chapter-wise online MCQ test.
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....
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Unit 5: Rotational Motion — Important Questions & Answers
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.
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.
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.
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.
Using parallel axis theorem and mass distribution ratio. Given key indicates option 3.
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.
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.