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 $I_Y$ is

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

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 $(Y),m_Y=\pi (4R{)}^2\frac{t}{4}\rho =4\pi R^{2}t\rho$
$\therefore I_Y=\frac{1}{2}{m}_Y(4R{)}^2$
$=\frac{1}{2}4\pi R^{2}t\rho \left(16R^2\right)=32\pi t\rho R^4$
$\therefore \frac{I_Y}{I_X}=\frac{32\pi t\rho R^4}{\frac{1}{2}\pi \rho t{R}^4}=64$ or $I_Y=64I_X$