A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45$$^{\circ}$$ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is ($$g = 10$$ m s$$^{-2}$$)

Let the mass be $$m = 10 \text{ kg}$$. Its weight acts vertically downward and is given by the well-known relation $$W = mg$$.

Substituting $$g = 10 \text{ m s}^{-2}$$, we get

$$W = m g = 10 \times 10 = 100 \text{ N}.$$

The horizontal pull applied somewhere on the rope makes the upper part of the rope tilt so that, at the roof, the rope now forms an angle $$\theta = 45^{\circ}$$ with the vertical. The whole system is stated to be in equilibrium; therefore the net force on the suspended mass is zero.

The only forces acting on the mass itself are

(i) its weight $$W = 100 \text{ N}$$ downward, and

(ii) the tension in the rope, $$T$$, directed along the rope, i.e. at $$45^{\circ}$$ to the vertical.

We resolve the tension into mutually perpendicular components, remembering the standard trigonometric relations

$$T_{\text{vertical}} = T \cos\theta, \qquad T_{\text{horizontal}} = T \sin\theta.$$

Because the mass neither rises nor falls, the vertical components must balance:

$$T\cos\theta = W.$$

Substituting $$\theta = 45^{\circ}$$ and $$W = 100 \text{ N}$$, we have

$$T \cos45^{\circ} = 100.$$

Using $$\cos45^{\circ} = \dfrac{\sqrt2}{2},$$ we obtain

$$T \left(\dfrac{\sqrt2}{2}\right) = 100 \;\;\Longrightarrow\;\; T = 100 \times \dfrac{2}{\sqrt2} = 100\sqrt2 \text{ N}.$$

Now we look at the horizontal balance. In equilibrium the horizontal component of the tension must be cancelled by the externally applied horizontal force, which we call $$F$$. Hence

$$F = T\sin\theta.$$

Again inserting $$\theta = 45^{\circ}$$ and $$T = 100\sqrt2 \text{ N},$$ we write

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

Simplifying the product of the radicals,

$$F = 100 \left(\dfrac{\sqrt2 \times \sqrt2}{2}\right) = 100 \left(\dfrac{2}{2}\right) = 100 \text{ N}.$$

Hence, the correct answer is Option A.