1. Set up the tensions

Let:

• $T_1$ = tension in the upper half of the rope
• $T_2$ = tension in the lower half of the rope

The mass is $10\,\text{kg}$, so its weight is $mg = 10 \times 10 = 100\,\text{N}.$

Since the lower end supports the mass and the lower half of the rope is vertical, for the mass in equilibrium: $T_2 = 100\,\text{N}.$

2. Analyse the midpoint where horizontal force $F$ is applied

At the midpoint, three forces act:

• upper tension $T_1$ along the upper rope, making $45^\circ$ with vertical,
• lower tension $T_2 = 100\,\text{N}$ vertically downward,
• applied force $F$ horizontally.

The system is in equilibrium, so resolve $T_1$ into components.

Since the upper rope makes $45^\circ$ with the vertical,

• 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 at midpoint

Upward vertical component balances downward tension $T_2$: $T_1\cos 45^\circ = 100.$

Using $\cos 45^\circ = \frac{1}{\sqrt{2}}$, $T_1 \cdot \frac{1}{\sqrt{2}} = 100$ $T_1 = 100\sqrt{2}\,\text{N}.$

4. Horizontal equilibrium at midpoint

Horizontal component of $T_1$ balances the applied force $F$: $F = T_1\sin 45^\circ.$

Using $\sin 45^\circ = \frac{1}{\sqrt{2}}$, $F = 100\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 100\,\text{N}.$

5. Final answer

$\boxed{F = 100\,\text{N}}$

So the correct option is A.