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The radius of a solid sphere is R and its density D. When it is made to rotate about an axis passing through any diameter of sphere, expression for its moment of inertia is

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$\frac{28}{5} π {DR}^{5}$

$\frac{8}{15} π {DR}^{5}$

$\frac{8}{7} π {DR}^{5}$

$\frac{28}{15} π {DR}^{5}$

answer is B.

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Detailed Solution

$I = \frac{2}{5} {MR}^{2} = \frac{2}{5} [\frac{4}{3} {πR}^{3}] {DR}^{2} = \frac{8}{15} {πDR}^{5}$

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