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 $t/4$. Then the relation between the moment of inertia $I_X$ and $I_Y$ is

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

Answer 1

Answer: (A)

Moment of Inertia of disc $I=\frac{1}{2}M{R}^2$
$=\frac{1}{2}(\pi R^{2}t\rho )R^2=\frac{1}{2}\pi t\rho R^4$
[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]
$\Rightarrow \frac{I_y}{I_x}=\frac{1}{4}{\left(4\right)}^4=64$ [Given $R_y=4R_x,t_y=\frac{t_x}{4}]$
$\Rightarrow I_y=64I_x$