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) $ I _{Y}=32I _{X} $
B) $ I _{Y}=16I _{X} $
C) $ I _{Y}=I _{X} $
D)$ I _{Y}=64I _{X} $
Show Answer
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 $ $ I _{x}=\frac{1}{2}m _{x}R^{2}=\frac{1}{2}\pi R^{2}t\rho R^{2} $ $ I _{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 $ I _{Y}=\frac{1}{2}m _{Y}{{(4R)}^{2}}=\frac{1}{2}4R^{2}t\rho .16R^{2} $
$ \Rightarrow $ $ I _{Y}=32\pi t\rho R^{4} $ …(ii)
$ \therefore $ $ \frac{I _{Y}}{I _{X}}=\frac{32\pi t\rho R^{4}}{\frac{1}{2}\pi \rho tR^{4}}=64 $
$ \therefore $ $ I _{Y}=64I _{X} $
BETA
