1. Understand the setup

A mass $m=10\,\text{kg}$ hangs from the roof by a rope of total length $5\,\text{m}$. A horizontal force of $30\,\text{N}$ is applied at the midpoint of the rope.

So the rope is effectively divided into two halves:

• Upper half: makes angle $\theta$ with the vertical.
• Lower half: remains vertical because the mass hangs directly below the midpoint.

The weight of the mass is $mg = 10 \times 10 = 100\,\text{N}.$

2. Forces on the mass

The mass is in equilibrium.

For the mass, only two forces act:

• weight $100\,\text{N}$ downward,
• tension in the lower half of rope, say $T_2$, upward.

Thus, $T_2 = 100\,\text{N}.$

3. Forces on the midpoint of the rope

At the midpoint, three forces act:

• tension in upper half, $T_1$, along the upper rope,
• tension in lower half, $T_2=100\,\text{N}$ downward,
• applied horizontal force $30\,\text{N}$.

Since the midpoint is in equilibrium, resolve $T_1$ into components.

• Vertical balance: $T_1 \cos\theta = 100$

• Horizontal balance: $T_1 \sin\theta = 30$

4. Find $\tan\theta$

Divide the horizontal equation by the vertical equation: $\tan\theta = \frac{30}{100} = \frac{3}{10}.$

So, $\theta = \tan^{-1}\left(\frac{3}{10}\right).$

Given in the question: $\theta = \tan^{-1}(x \times 10^{-1}).$

Thus, $x \times 10^{-1} = \frac{3}{10}.$

Hence, $x=3.$

5. Comparison with stored answer

Derived answer: $3$

Stored correct answer: $3$

They match.