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For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is

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answer is B.

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Detailed Solution

$I_{0} = \frac{M}{12} ((80)^{2} + (60)^{2})$

$I_{O^{'}} = I_{0} + M (50)^{2}$

$\Rightarrow Ratio = \frac{I_{O}}{I_{O^{'}}} = \frac{\frac{M}{12} (10000)}{\frac{M}{12} (10000) + (M) (2500)} = \frac{1}{4}$

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