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 l_X and I_Y is

Moment of Inertia of disc $I=\frac{1}{2}M{R}^2$

                                         $=\frac{1}{2}\left(\pi R^{2}t\rho \right)R^2=\frac{1}{2}\pi t\rho R^4$

$\begin{array}{l}\text{ [As }\left.M=V\times \rho =\pi R^{2}t\rho \text{ where }t=\text{ thickness, }\rho =\text{ density }\right]\\ \therefore \frac{I_y}{I_x}=\frac{t_y}{t_x}{\left(\frac{R_y}{R_x}\right)}^4 [\text{ If }\rho =\text{ constant }]\\ \Rightarrow \frac{I_y}{I_x}=\frac{1}{4}(4{)}^4=64 \left[\text{ Given }{R}_y=4R_x,t_y=\frac{t_x}{4}\right]\\ \Rightarrow I_y=64I_x\end{array}$