Physics - NLM Question with Solution | TestHub

At equilibrium, the centre of mass will be below the point of suspension.

Taking torque about point O,

$mg\frac{l}{2}\mathrm{s}\mathrm{i}\mathrm{n}\theta +mg\left(l\mathrm{s}\mathrm{i}\mathrm{n}\theta -\frac{l}{2}\mathrm{c}\mathrm{o}\mathrm{s}\theta \right)=0$

$\Rightarrow \frac{3l}{2}\mathrm{s}\mathrm{i}\mathrm{n}\theta =\frac{l}{2}\mathrm{c}\mathrm{o}\mathrm{s}\theta$

$\Rightarrow \mathrm{t}\mathrm{a}\mathrm{n}\theta =\frac{1}{3}$

$\Rightarrow \theta ={\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}\frac{1}{3}$