An $L$-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in the figure. If $AB = BC$, and the angle made by $AB$ with the downward vertical is $\theta$, then
Answer: (B) $\tan\theta = \dfrac13$
Let each rod have mass $m$ and length $a$. Take unit vectors $\hat u$ along $AB$ (from $A$ to $B$) and $\hat w$ along $BC$, with $\hat u \perp \hat w$.
Centre of mass of $AB$: $\frac a2\hat u$ from $A$. Centre of mass of $BC$: $a\hat u + \frac a2\hat w$ from $A$. So the overall centre of mass is
$$\vec r_{cm} = \frac12\left(\frac{3a}{2}\hat u + \frac a2\hat w\right)$$
In equilibrium the centre of mass lies vertically below the point of suspension $A$. The angle between $\vec r_{cm}$ (the vertical) and $AB$ satisfies
$$\tan\theta = \frac{a/2}{3a/2} = \frac13$$
Solution by Sreeraj P, M.Sc Physics