The three identical rods each of mass M and length L are joined as shown in figure. Find the moment of inertia about $3 - 3$

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${ML}^{2} / 3$

$4 {ML}^{2} / 3$

$2 {ML}^{2} / 3$

$5 {ML}^{2} / 3$

answer is C.

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Detailed Solution

$\begin{array}{l} I_{3 - 3} = \frac{{ML}^{2}}{3} + [I_{a} + {ML}^{2}] + 0 \\ = \frac{{ML}^{2}}{3} + [0 + {ML}^{2}] \\ = \frac{4 {ML}^{2}}{3} \end{array}$

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