1. Free body diagram and equilibrium conditions

A mass $m$ is suspended by two strings with tensions $T_1$ and $T_2$ making angles $\theta_1$ and $\theta_2$ with the horizontal.

Given: $T_1=\sqrt{3}\,T_2$

Since the body is in equilibrium:

• Horizontal components must cancel.
• Vertical components must balance the weight $mg$.

So, $T_1\cos\theta_1=T_2\cos\theta_2 \quad \text{(1)}$ $T_1\sin\theta_1+T_2\sin\theta_2=mg \quad \text{(2)}$

2. Use the given relation $T_1=\sqrt{3}T_2$ in the horizontal balance

From (1): $\sqrt{3}T_2\cos\theta_1=T_2\cos\theta_2$ $\sqrt{3}\cos\theta_1=\cos\theta_2$

Now test the angle pairs given in the options.

3. Check option A / C: $\theta_1=30^\circ,\ \theta_2=60^\circ$

Then, $\sqrt{3}\cos30^\circ=\sqrt{3}\cdot \frac{\sqrt{3}}{2}=\frac{3}{2}$ while $\cos60^\circ=\frac{1}{2}$

These are not equal, so this angle combination is impossible.

Hence A and C are rejected.

4. Check option B: $\theta_1=45^\circ,\ \theta_2=45^\circ$

Then (1) gives: $T_1\cos45^\circ=T_2\cos45^\circ \Rightarrow T_1=T_2$ But given: $T_1=\sqrt{3}T_2$ This is not possible.

Hence B is rejected.

5. Check option D: $\theta_1=60^\circ,\ \theta_2=30^\circ$

From horizontal equilibrium: $T_1\cos60^\circ=T_2\cos30^\circ$ $T_1\cdot \frac12=T_2\cdot \frac{\sqrt3}{2}$ $T_1=\sqrt3\,T_2$

This exactly matches the given condition, so the angle pair is correct.

Now use vertical equilibrium: $T_1\sin60^\circ+T_2\sin30^\circ=mg$ Substitute $T_1=\sqrt3 T_2$: $\sqrt3 T_2\cdot \frac{\sqrt3}{2}+T_2\cdot \frac12=mg$ $\frac{3T_2}{2}+\frac{T_2}{2}=mg$ $2T_2=mg$ $T_2=\frac{mg}{2}$

This matches option D.

6. Final answer

The correct option is: $\boxed{D}$ with $\theta_1=60^\circ,\quad \theta_2=30^\circ,\quad T_2=\frac{mg}{2}$