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.
• B.
• C.
• D.

Answer 1

The correct option is A

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