JEE Main 2026 Jan 21 Shift 1, Physics Q14: Kinematics and Dynamics of Rotational Motion

Immediately after one string is cut, the rod begins rotating about the point where the remaining string meets the rod (call it $O$). Take torques about $O$:

$mg\cdot \frac{l}{2}=I_O\alpha =\frac{m{l}^2}{3}\alpha \Rightarrow \alpha =\frac{3g}{2l}$.

The centre of mass is at $l/2$ from $O$, so its tangential (downward) acceleration is $a_c=(l/2)\alpha =3g/4$.

Newton's second law for the COM in the vertical direction: $mg-T=m{a}_c\Rightarrow T=m(g-3g/4)=\frac{mg}{4}$.