Rotational Motion Question 130
Question: A circular disc x of radius $R$ is made from an iron plate of thickness $t$ , and another disc $Y$ of radius 4$R$ is made from an iron plate of thickness $t$ /4. Then the relation between the moment of inertia ${{I} _{x}}andI _{Y}$ is
Options:
A) $I _{Y}=64I _{X}$
B) $I _{Y}=32I _{X}$
C) $I _{Y}=16I _{X}$
D) $I _{Y}=I _{X}$
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Answer:
Correct Answer: A
Solution:
[a]Moment of Inertia of disc $
$I=\frac{1}{2}MR^{2}$
$=\frac{1}{2}(\pi R^{2}t\rho )R^{2}=\frac{1}{2}\pi t\rho R^{4}$
As M=$V\times \rho =\pi R^{2}t\rho $
$ where t=thickness,\rho =density$
$\therefore \frac{I _{y}}{I _{x}}=\frac{t _{y}}{t _{x}}{{\left( \frac{R _{y}}{R _{x}} \right)}^{4}}$
$if\rho =constant$
$\Rightarrow \frac{I _{y}}{I _{x}}=\frac{1}{4}{{(4)}^{4}}=64$
Given $R _{y}=4R _{x},t _{y}=\frac{t _{x}}{4}$
$\Rightarrow I _{y}=64I _{x}$
BETA
