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A circular disc X of radius R is made from an iron plate of thickness t and another circular disc Y of radius 4R is made from an iron plate of thickness t/4. Then the relation between moments of inertia about the normal axes through centre I_x and I_y is

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I_y = 32 L_x

I_y = 16 I_x

I_y = I_x

I_y = 64 I_x

answer is D.

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Detailed Solution

$I = \frac{{MR}^{2}}{2} = \frac{{ρVR}^{2}}{2} = \frac{{πρR}^{4} t}{2}$

$\frac{I_{x}}{I_{y}} = \frac{R^{4} t}{4^{4} R^{4} t} \times 4 = \frac{1}{64}$

$I_{y} = 64 I_{x}$

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