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 $\frac{t}{4}$. Then the relation between the moment of inertia $I_X$ and $I_Y$ about their natural axis is -

• A. $I_Y=64I_X$
• B. $I_Y=32I_X$
• C. $I_Y=16I_X$
• D. $I_X=I_Y$

Answer 1

The correct option is D $I_X=I_Y$

$I=\frac{1}{2}m{R}^2$
$M\propto t\propto R^2$
For disc $X,I_X=\frac{1}{2}\left(m\right)(R{)}^2$
For disc $Y,I_Y=\frac{1}{2}\left[\pi \right(4R{)}^2.t/4\left]\right[4R{]}^2$
$\Rightarrow \frac{I_X}{I_Y}=\frac{1}{(4{)}^3}\Rightarrow I_Y=64I_Y$