The radius of gyration of solid hemisphere of mass M and radius R about an axis parallel to the diameter at a distance 3R / 4 from center and lying in the same plane is given by (center of mass of the hemisphere of the hemisphere lies at a height 3R / 8 from the base)

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$\frac{3 R}{\sqrt{10}}$

$\frac{5 R}{4}$

$\sqrt{\frac{2}{5}} R$

$\frac{5 R}{8}$

answer is D.

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Detailed Solution

$I_{XX'} = \frac{2}{5} {MR}^{2}$

The axes PQ and XX' are at an equal distance from center of mass of the hemisphere

$\Rightarrow I_{PQ} = I_{XX'}$

${MK}^{2} = \frac{2}{5} {MR}^{2} \Rightarrow K = \sqrt{\frac{2}{5}} R$

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