1. Understand the configuration

A mass of $1\,\text{kg}$ hangs at the lower end of a rope. A horizontal force $F$ is applied at the midpoint of the rope, and the upper half of the rope makes an angle $45^\circ$ with the vertical.

Since the force is applied at the midpoint, the rope is effectively divided into two segments:

• upper segment: between ceiling and midpoint
• lower segment: between midpoint and mass

The system is in equilibrium.

2. Tension in the lower half of the rope

Consider the mass alone.

For the mass to be in equilibrium, the upward tension in the lower segment must balance its weight:

$T_2 = mg = 1 \times 10 = 10\,\text{N}$

So, the tension in the lower half is

$T_2 = 10\,\text{N}$

3. Direction of the lower segment

The mass is acted upon only by:

• its weight downward
• tension in the rope upward along the rope

For these two forces alone to balance, they must be along the same vertical line. Hence the lower half of the rope must be vertical.

So at the midpoint:

• the lower segment pulls downward with magnitude $10\,\text{N}$
• the upper segment is inclined at $45^\circ$ to the vertical
• the applied force $F$ is horizontal

4. Equilibrium of the midpoint

Let $T_1$ be the tension in the upper half.

At the midpoint, three forces act:

• $T_1$ along the upper rope, making $45^\circ$ with vertical
• $T_2 = 10\,\text{N}$ downward
• horizontal force $F$

Resolve $T_1$ into components.

Since it makes angle $45^\circ$ with vertical:

• vertical component = $T_1 \cos 45^\circ$
• horizontal component = $T_1 \sin 45^\circ$

Vertical equilibrium:

$T_1 \cos 45^\circ = 10$

$T_1 \cdot \frac{1}{\sqrt{2}} = 10$

$T_1 = 10\sqrt{2}\,\text{N}$

Horizontal equilibrium:

The horizontal component of $T_1$ is balanced by $F$:

$F = T_1 \sin 45^\circ$

$F = 10\sqrt{2} \cdot \frac{1}{\sqrt{2}}$

$F = 10\,\text{N}$

5. Final answer

$\boxed{F = 10\,\text{N}}$

So the correct option is B.