JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Rotational Mechanics

At equilibrium the string must pass through the system's centre of mass, so set the net torque about the suspension point to zero. Take each arm to have mass $m$ and length $a$, with the centres of mass of $AB$ and $BC$ at their midpoints. Writing the moment arms of the two weights in terms of $\theta$ and balancing the clockwise torque of one arm against the anticlockwise torque of the other gives $\frac{3a}{2}\sin \theta =\frac{a}{2}\cos \theta$, which collapses neatly to $\tan \theta =\frac{1}{3}$.