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 $\frac{t}{4}$. Then the relation between the moment of inertia $I_X$ and $I_Y$ is

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

Answer 1

Answer: (D)

Mass of disc $(X), m_X=\pi R^2 t \rho$
where $\rho=$ density of material of disc.
$\therefore I_X=\frac{1}{2} m_X R^2=\frac{1}{2} \pi R^2 t \rho R^2$
or $I_X=\frac{1}{2} \pi \rho t R^4$
Mass of disc $(Y), m_Y=\pi(4 R)^2 \frac{t}{4} \rho=4 \pi R^2 t \rho$
$\therefore I_Y=\frac{1}{2} m_Y(4 R)^2$
$=\frac{1}{2} 4 \pi R^2 t \rho\left(16 R^2\right)=32 \pi t \rho R^4 $
$\therefore \frac{I_Y}{I_X}=\frac{32 \pi t \rho R^4}{\frac{1}{2} \pi \rho t R^4}=64$ or $I_Y=64 I_X$