1. Understand the setup

A mass of $10\,\text{kg}$ hangs at rest from a rope. A horizontal force is applied at some point on the rope, causing the upper part of the rope (near the roof) to make an angle $45^\circ$ with the vertical.

Since the mass is in equilibrium, the tension in the lower vertical part of the rope must balance its weight.

$T_2 = mg = 10 \times 10 = 100\,\text{N}$

2. Force balance at the point where horizontal force is applied

At the point where the horizontal force is applied, three forces act:

• Tension $T_1$ in the upper inclined part of the rope,
• Tension $T_2 = 100\,\text{N}$ in the lower vertical part,
• Horizontal applied force $F$.

Because the point is in equilibrium, resolve $T_1$ into components.

The upper rope makes $45^\circ$ with the vertical, so:

• Vertical component of $T_1$ is $T_1 \cos 45^\circ$
• Horizontal component of $T_1$ is $T_1 \sin 45^\circ$

3. Vertical equilibrium

The vertical component of $T_1$ balances the downward pull of the lower rope:

$T_1 \cos 45^\circ = 100$

$T_1 \cdot \frac{1}{\sqrt{2}} = 100$

$T_1 = 100\sqrt{2}$

4. Horizontal equilibrium

The applied force $F$ balances the horizontal component of $T_1$:

$F = T_1 \sin 45^\circ$

$F = 100\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 100\,\text{N}$

5. Check options

• A: $200\,\text{N}$
• B: $140\,\text{N}$
• C: $70\,\text{N}$
• D: $100\,\text{N}$

So the correct option is D.

Derived Answer: $100\,\text{N}$