A thin rod of mass $6 M$ and length $6 l$ is bent into regular hexagon. The MI of the hexagon about a normal axis to its place and through centre of system is $x M l^{2}$ , then the value of $x$ is ___

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$3 M l^{2}$

$5 M l^{2}$

$7 M l^{2}$

$8 M l^{2}$

answer is B.

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Detailed Solution

Applying parallel axes theorem $I = I_{c} + m d^{2}$
$\frac{m l^{2}}{12} + m {[\frac{\sqrt{3} l}{2}]}^{2}; I = \frac{5 m l^{2}}{6}$
For 6 rods total moment of inertia $I = 5 m l^{2}$

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