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NEET PHYSICSRotational motionEasy

Question

Consider a thin circular ring (A), a circular disc (B), a hollow cylinder (C) and a solid cylinder (D) of the same radii RR and of the same masses. If IAI_A, IBI_B, ICI_C and IDI_D are their moments of inertia about the axis shown, then choose the correct answer from the options given below:

A

IA=ICI_A = I_C and IB=IDI_B = I_D

B

IA=2IBI_A = 2I_B and 2IC=ID2I_C = I_D

C

2IA=IC2I_A = I_C and IB=2IDI_B = 2I_D

D

IA=IB=IC=2IDI_A = I_B = I_C = 2I_D

Step-by-Step Solution

The moment of inertia of various uniform bodies about their central geometric axes are given by:

  1. Thin circular ring (A): IA=MR2I_A = MR^2
  2. Circular disc (B): IB=12MR2I_B = \frac{1}{2}MR^2
  3. Hollow cylinder (C): IC=MR2I_C = MR^2
  4. Solid cylinder (D): ID=12MR2I_D = \frac{1}{2}MR^2

Comparing the above values, we can clearly see that: IA=ICI_A = I_C (both are equal to MR2MR^2) IB=IDI_B = I_D (both are equal to 12MR2\frac{1}{2}MR^2). Therefore, the correct relation is IA=ICI_A = I_C and IB=IDI_B = I_D.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Rotational motion. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRotational motionconsidercircularcircularhollowcylinder

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