Moment of inertia of a uniform quarter disc of radius R and mass M about an axis through its centre of mass and perpendicular to its plane is

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$\frac{M R^{2}}{2} - M {[\sqrt{2} \frac{4 R}{3 π}]}^{2}$

$\frac{M R^{2}}{2} + M {[\sqrt{2} \frac{4 R}{3 π}]}^{2}$

$\frac{M R^{2}}{2} + M {[\frac{4 R}{3 π}]}^{2}$

$\frac{M R^{2}}{2} - M {[\frac{4 R}{3 π}]}^{2}$

answer is B.

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Detailed Solution

$I_{2} = \frac{M R^{2}}{2}; d = (\frac{4 R}{3 π}) \sqrt{2} I_{c} = I_{c m} + m d^{2} \Rightarrow I_{c m} = \frac{M R^{2}}{2} - M {[\sqrt{2} \frac{4 R}{3 π}]}^{2}$

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