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 $R _{y}=4R _{x},t _{y}=\frac{t _{x}}{4}$

$\Rightarrow I _{y}=64I _{x}$



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