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The moment of inertia of a rod about an axis through its centre and perpendicular to it is $\frac{1}{12} M L^2$ (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 $60^{\circ}$. The moment of inertia of the bent rod about the same axis would be
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Verified Answer
The correct answer is:
$\frac{1}{12} M L^2$
Moment of inertia of uniform rod about one end $=\frac{M L^2}{3}$. $\therefore$ Moment of inertia of a rod about an axis passing through
$O$ and perpendicular to the plane of the rod will be
$$
=2\left(\frac{M}{2}\right) \frac{(L / 2)^2}{3}=\frac{M L^2}{4 \times 3}=\frac{M L^2}{12}
$$
$O$ and perpendicular to the plane of the rod will be
$$
=2\left(\frac{M}{2}\right) \frac{(L / 2)^2}{3}=\frac{M L^2}{4 \times 3}=\frac{M L^2}{12}
$$
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