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Four parties each of mass $m$ are lying symmetrically on the rim of a disc of mass M and radius $R$ . MI of this system about an axis passing through one of the particles and $⊥$ to plane of disc is

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$(3 m + 12 M) \frac{R^{2}}{2}$

$(3 M + 16 m) \frac{R^{2}}{2}$

Zero

$16 m R^{2}$

answer is B.

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Detailed Solution

According to the theorem of $∥$ axes, MI of disc about an axis passing through K and $⊥$ to plane of disc is $= \frac{1}{2} {MR}^{2} + {MR}^{2} = \frac{3}{2} {MR}^{2}$

Total MI of the system $= \frac{3}{2} {MR}^{2} + m {(2 R)}^{2} + m {(\sqrt{2} R)}^{2} + m {(\sqrt{2} R)}^{2}$

$= (3 M + 16 m) \frac{R^{2}}{2}$ .

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