Rotational Motion Question 6

Question: A circular disc X of radius R is made from an iron plate of thickness t and another disc Y of radius 4R is made from an iron plate of thickness t/4. Then, the relation between the moment of inertia $ I _{X} $ and $ I _{Y} $ is

[AIEEE 2003]

Options:

A) $ l _{Y}=32l _{X} $

B) $ l _{Y}=16l _{X} $

C) $ l _{Y}=l _{X} $

D)$ l _{Y}=64l _{X} $

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Answer:

Correct Answer: D

Solution:

  • [d]

Mass of disc (X), $ m _{x}=\pi R^{2}t\rho $ $ (m=v\rho =At\rho =\pi R^{2}-\rho ) $

where, $ \rho = $ density of material of disc

$ \therefore $ $ l _{x}=\frac{1}{2}m _{x}R^{2}=\frac{1}{2}\pi R^{2}t\rho R^{2} $ $ l _{x}=\frac{1}{2}\pi \rho tR^{4} $ …

(i) Mass of disc (Y) $ m _{Y}=\pi {{(4R)}^{2}}\frac{t}{4}\rho =4\pi R^{2}t\rho $

and $ l _{Y}=\frac{1}{2}m _{Y}{{(4R)}^{2}}=\frac{1}{2}4R^{2}t\rho .16R^{2} $

$ \Rightarrow $ $ l _{Y}=32\pi t\rho R^{4} $ …(ii)

$ \therefore $ $ \frac{l _{Y}}{l _{X}}=\frac{32\pi t\rho R^{4}}{\frac{1}{2}\pi \rho tR^{4}}=64 $

$ \therefore $ $ l _{Y}=64l _{X} $



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