The moment of inertia of a rod about an axis through its centre and perpendicular to its length 1/12 ML² (where, M is the mass and L the length of the rod) The rod is bent in the middle so that the two halves make an angle of 60^o. The moment of inertia of the bent rod about an axis passing through the bend point and perpendicular to its plane is __

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$\frac{{ML}^{2}}{8 \sqrt{2}}$

1/48 ML²

1/12 ML²

1/24 ML²

answer is B.

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Detailed Solution

MI about a perpendicular axis of the bent rod will be

$I = \frac{(\frac{m}{2}) {(\frac{l}{2})}^{2}}{3} + \frac{(\frac{m}{2}) {(\frac{l}{2})}^{2}}{3}$

$I = \frac{(\frac{m}{2}) {(\frac{l}{2})}^{2}}{3} x 2$

$I = \frac{m l^{2}}{12}$

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