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 $I_X\text{\hspace{0.17em}and\hspace{0.17em}}{I}_Y\text{\hspace{0.17em}is}$

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

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

[As $M=V\times \rho =\pi R^{2}t\rho$ where $t=\text{thickness},\rho =\text{density}$] 

$\begin{array}{l}\therefore \frac{I_y}{I_x}=\frac{t_y}{t_x}{\left(\frac{R_y}{R_x}\right)}^4 \left[\text{If\hspace{0.17em}}\rho =\text{constant}\right]\\ \Rightarrow \frac{I_y}{I_x}=\frac{1}{4}{\left(4\right)}^4=64 [\text{Given\hspace{0.17em}}{R}_y=4R_x,t_y=\frac{t_x}{4}]\\ \Rightarrow I_y=64I_x\end{array}$