Explanation

(d) When one string is cut off, the rod will rotate about the other point $A$. Let $a$ be the linear acceleration of centre of mass of the rod and $\alpha$ be the linear acceleration of centre of mass of the rod and $\alpha$ be the angular acceleration of the rod about $A$. As is clear from figure,
[image:7_image165.emf]
$mg - T = ma$ …(i)
$\alpha = \frac{\tau}{I} = \frac{mg(l/2)}{ml^{2}/3} = \frac{3g}{4}$ …(ii)
$a = r\alpha = \frac{l}{2}\alpha = \frac{l}{2}\frac{3g}{2l} = \frac{3g}{4}$
From Eq. (i), $T = mg - ma = mg - \frac{3mg}{4} = \frac{mg}{4}$