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JEE Advanced Physics Rotational Motion 2021 JEE Advanced 2021 (Paper 1)

JEE Advanced Physics Question (2021) — Solution

Question

A thin rod of mass  M  and length  a  is free to rotate in horizontal plane about a fixed vertical axis passing through point  O . A thin circular disc of mass  M  and of radius  a 4  is pivoted on this rod with its centre at a distance  a 4  from the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary conserver finds the rod rotating with an angular velocity  Ω  and the disc rotating about its vertical axis with angular velocity  4 Ω . The total angular momentum of the system about the point  O  is  M a 2 Ω 48 n . The value of  n  is _________ .

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Angular momentum of the system about  O , L → = L → Rod + L → Disc ⇒ L → = L → Rod + L → C M Disc + r → C M × M D i s c V → C M ⇒ L = M a 2 Ω 3 + M a 2 8 Ω + M 3 a 4 Ω 3 a 4 ⇒ L = M a 2 3 Ω + M a 2 8 Ω + 9 M a 2 16 Ω ⇒ L = 16 M a 2 Ω + 6 M a 2 Ω + 27 M a 2 Ω 48 ⇒ L = 49 48 M a 2 Ω Compare it with  n M a 2 48 Ω , we get  n = 49 Hence,  n = 49

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