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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 moment of inertia I_x and I_y is

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a

$I_{y} = 32 I_{x}$

b

$I_{y} = I_{x}$

c

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

d

$I_{y} = 16 I_{x}$

answer is D.

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Detailed Solution

$I = \frac{m R^{2}}{2} \Rightarrow I α R^{2} \times t \times R^{2} \Rightarrow I α R^{4} t \frac{I_{x}}{I_{y}} = {[\frac{R}{4 R}]}^{4} \times [\frac{t}{t / 4}] = \frac{1}{4 \times 4 \times 4 \times 4} \times 4 = \frac{1}{64} \Rightarrow I_{y} = 64 I_{x}$

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