A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R,as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is $I_{O}$ and $I_{P}$ respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $I_{P} / I_{O}$ to the nearest integer is

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answer is 3.

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Detailed Solution

Let $σ$ be the surface mass density.
Moment of inertia of the lamina about axes passing through ‘O’
$I_{O} = \frac{1}{2} σ [π {(2 R)}^{2}] \times {(2 R)}^{2} - [\frac{1}{2} {(σ π R^{2})}^{2} + σ (π R^{2}) \times R^{2}]$
$= \frac{13}{2} π σ R^{4}$
Moment of inertia of the lamina about axes passing through ‘P’
$I_{P} = 8 π σ R^{4} + π σ {(2 R)}^{2} \times {(2 R)}^{2} [\frac{1}{2} σ (π R^{2}) R^{2} + σ (π R^{2}) (\sqrt{{(2 R)}^{2} + R^{2}})]$
$= 24 π σ R^{4} - 5.5 σ π R^{4} = 18.5 π σ R^{4}$
$∴ \frac{I_{P}}{I_{O}} = \frac{18.5 π σ R^{4}}{\frac{13}{2} π σ R^{4}} = \frac{37}{13} \approx 3$