Thank you for reporting, we will resolve it shortly
Q.
The moment of inertia of a rod about an axis through its centre and perpendicular to it
is I. The rod is bent in the middle, so that the two halves make an angle $60^{\circ}$. Now the
new M.I about the same axis is
Solution:
$I =\frac{1}{12} ML ^{2}$
$I ^{'}=2 \times \frac{1}{3}\left(\frac{ M }{2}\right)\left(\frac{ L }{2}\right)^{2}$
$=\frac{1}{2} ML ^{2}= I$