A circular disc $$X$$  of radius $$R$$  is made froth 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-

Solution :
We know that density   $$\left( d \right) = \frac{{{\text{mass}}\left( M \right)}}{{{\text{volume}}\left( V \right)}}$$
$$\therefore M = d \times V = d \times \left( {\pi {R^2} \times t} \right)$$
The moment of inertia of a disc is given by $$I = \frac{1}{2}M{R^2}$$

$$\eqalign{ & \therefore \,\,I = \frac{1}{2}\left( {d \times \pi {R^2} \times t} \right){R^2} = \frac{{\pi d}}{2}t \times {R^4} \cr & \therefore \,\,\frac{{{I_X}}}{{{I_Y}}} = \frac{{{t_X}R_X^4}}{{{t_Y}R_Y^4}} \cr & = \frac{{t \times {R^4}}}{{\frac{t}{4} \times {{\left( {4R} \right)}^4}}} \cr & = \frac{1}{{64}} \cr & {\text{So, }}{I_Y} = 64{I_X} \cr} $$