Let the two thin rods be $AB$ and $BC$, each of length $L$ and same linear mass density $\lambda$. Hence each rod has equal mass.

When the L-shaped object is suspended from point $A$, it comes to rest such that its centre of mass lies vertically below $A$. So we only need the direction of the position vector of the centre of mass from $A$.

1. Choose axes along the rods

Take point $A$ as origin.

• Let rod $AB$ lie along the $x$-axis.
• Let rod $BC$ be perpendicular to $AB$, starting from $B$.

Then:

• Midpoint of rod $AB$ is at $\left(\frac{L}{2},0\right)$
• Since $BC$ is perpendicular to $AB$ and starts at $B(L,0)$, its midpoint is at $\left(L,\frac{L}{2}\right)$

Because the rods have equal mass, the centre of mass of the whole system is the average of these two midpoint positions.

2. Compute the centre of mass

Thus, $x_{cm}=\frac{\frac{L}{2}+L}{2}=\frac{3L}{4}$ $y_{cm}=\frac{0+\frac{L}{2}}{2}=\frac{L}{4}$

So relative to point $A$, the centre of mass is at $\left(\frac{3L}{4},\frac{L}{4}\right)$

3. Relate this to the suspended position

In equilibrium, the line from $A$ to the centre of mass must be vertically downward.

The rod $AB$ makes angle $\theta$ with the downward vertical. Therefore $\theta$ is also the angle between rod $AB$ and the line joining $A$ to the centre of mass.

From the coordinates above, $\tan\theta=\frac{\text{perpendicular component}}{\text{parallel component}}=\frac{y_{cm}}{x_{cm}}=\frac{\frac{L}{4}}{\frac{3L}{4}}=\frac{1}{3}$

4. Final answer

$\boxed{\tan\theta=\frac{1}{3}}$ So the correct option is D.

5. Comparison with stored answer

Stored correct answer: D

My derived answer also gives D, so they agree.