1. Moment of inertia of a disc about its central axis

For a uniform solid disc, $I = \frac{1}{2}Mr^2$ So we need to compare the masses first.

2. Mass of each disc

Since both are made of iron, density $\rho$ is same.

Mass $M =$ density $\times$ volume $M = \rho \cdot \pi r^2 \cdot h$ where $h$ is thickness.

Disc $X$

Radius $= R$, thickness $= t$ $M_X = \rho \pi R^2 t$ Hence, $I_X = \frac{1}{2}M_X R^2 = \frac{1}{2}(\rho \pi R^2 t)R^2 = \frac{1}{2}\rho \pi t R^4$

Disc $Y$

Radius $= 4R$, thickness $= \frac{t}{4}$ $M_Y = \rho \pi (4R)^2 \cdot \frac{t}{4}$ $M_Y = \rho \pi \cdot 16R^2 \cdot \frac{t}{4} = 4\rho \pi R^2 t$ So, $M_Y = 4M_X$

Now, $I_Y = \frac{1}{2}M_Y(4R)^2$ $I_Y = \frac{1}{2}(4M_X)(16R^2)$ $I_Y = 32 M_X R^2$

But since $I_X = \frac{1}{2}M_XR^2,$ we get $\frac{I_Y}{I_X} = \frac{32M_XR^2}{\frac{1}{2}M_XR^2} = 64$

Therefore, $I_Y = 64I_X$

3. Option check

• A: $I_Y = 32I_X$ ❌
• B: $I_Y = 16I_X$ ❌
• C: $I_Y = I_X$ ❌
• D: $I_Y = 64I_X$ ✅

So the correct option is D.