1. Interpret the situation

A block of mass $m=\sqrt{3}\,\text{kg}$ is attached to a string fixed to a wall. A force $F$ is applied such that the string makes an angle $30^\circ$ with the vertical wall.

Since the question asks for the tension $T$, we consider the block to be in equilibrium.

2. Forces on the block

The block has:

• Weight $mg$ downward
• Tension $T$ along the string
• Applied force $F$ balancing the horizontal component

Because the string makes angle $30^\circ$ with the wall (vertical), the vertical component of tension is: $T\cos 30^\circ$

For vertical equilibrium, $T\cos 30^\circ = mg$

3. Substitute values

Mass: $m=\sqrt{3}\,\text{kg}$

Weight: $mg=\sqrt{3}\times 10=10\sqrt{3}\,\text{N}$

Also, $\cos 30^\circ = \frac{\sqrt{3}}{2}$

So, $T\cdot \frac{\sqrt{3}}{2} = 10\sqrt{3}$

4. Solve for $T$

$T = \frac{10\sqrt{3}}{\sqrt{3}/2}$

$T = 10\sqrt{3}\cdot \frac{2}{\sqrt{3}}$

$T=20\,\text{N}$

5. Check options

• A: $15\,\text{N}$
• B: $10\,\text{N}$
• C: $25\,\text{N}$
• D: $20\,\text{N}$

Hence, the correct option is D.