A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of 45$$^\circ$$ with the vertical. Then F equals: (Take g = 10 m s$$^{-2}$$ and the rope to be massless)

Let us denote the ceiling point by $$O$$, the mid-point of the rope where the horizontal force is applied by $$P$$, and the point where the 10 kg mass is attached by $$M$$. The rope is massless and its total length is $$4\ \text{m}$$, so each half measures $$2\ \text{m}$$.

The upper half $$OP$$ of the rope is given to make an angle of $$45^{\circ}$$ with the vertical. Writing the length of this segment and resolving it into vertical and horizontal projections, we have

$$OP = 2\ \text{m},\qquad \angle(OP,\text{vertical}) = 45^{\circ}.$$

Therefore

vertical projection $$= OP\cos 45^{\circ}=2\cos45^{\circ}=2\dfrac{\sqrt2}{2}=\sqrt2\ \text{m},$$
horizontal projection $$= OP\sin45^{\circ}=2\sin45^{\circ}=2\dfrac{\sqrt2}{2}=\sqrt2\ \text{m}.$$

Because the horizontal force is applied only at $$P$$ and the rope is light, the lower half $$PM$$ adjusts until the 10 kg mass remains in equilibrium. For the mass to be at rest it must experience no horizontal force at all; hence the tension in the lower half must act vertically. This immediately shows that the segment $$PM$$ is vertical.

On the mass $$M$$ two forces act: its weight downward $$mg$$ and the tension $$T_2$$ of the lower half upward along the same vertical line. Writing the condition of translational equilibrium,

$$T_2 - mg = 0.$$

Substituting the data $$m = 10\ \text{kg},\ g = 10\ \text{m s}^{-2},$$ we obtain

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

Now we inspect the force balance at the junction point $$P$$. Three forces meet there:

1. The tension $$T_1$$ in the upper half, directed from $$P$$ toward the ceiling along $$PO$$.
2. The tension $$T_2 = 100\ \text{N}$$ in the lower half, directed vertically downward toward the mass.
3. The horizontal force $$F$$ applied to the right (say) at $$P$$.

Because the rope makes $$45^{\circ}$$ with the vertical in the upper half, we resolve $$T_1$$ into its components using

$$T_1^{(\text{vertical})}=T_1\cos45^{\circ},\qquad T_1^{(\text{horizontal})}=T_1\sin45^{\circ}.$$

The equilibrium conditions at $$P$$ are

Vertical: $$T_1\cos45^{\circ}-T_2=0,$$

Horizontal: $$F-T_1\sin45^{\circ}=0.$$

From the vertical balance we solve for $$T_1$$:

$$T_1\cos45^{\circ}=T_2\quad\Longrightarrow\quad T_1=\dfrac{T_2}{\cos45^{\circ}} =\dfrac{100}{\frac{\sqrt2}{2}} =100\cdot\dfrac{2}{\sqrt2} =100\sqrt2\ \text{N}.$$

Substituting this value of $$T_1$$ into the horizontal balance gives

$$F = T_1\sin45^{\circ} =(100\sqrt2)\left(\dfrac{\sqrt2}{2}\right) =100\cdot\dfrac{2}{2} =100\ \text{N}.$$

Hence, the correct answer is Option A.