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^(@)` at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g`=10 ms^(-2)`)

Options

A. 70 N
B. 75 N
C. 100 N
D. 90 N

Answer

To solve the problem, we need to analyze the forces acting on the mass and the rope when a horizontal force is applied, causing the rope to deviate at an angle of 45 degrees.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The weight of the mass (W) acting downward:
\[ W = mg = 10 \, \text{kg} \times 10 \, \text{m/s}^2 = 100 \, \text{N} \]
- The tension (T) in the rope acting at an angle of 45 degrees from the vertical.
- The horizontal force (F) applied to the rope.

2. **Draw a Free Body Diagram**:
- At the point where the rope is attached to the roof, the tension T can be resolved into two components:
- Vertical component: \( T \cos(45^\circ) \)
- Horizontal component: \( T \sin(45^\circ) \)

3. **Set Up the Equilibrium Conditions**:
- Since the mass is in equilibrium, the sum of the vertical forces must equal zero:
\[ T \cos(45^\circ) = mg \]
- The sum of the horizontal forces must also equal zero:
\[ T \sin(45^\circ) = F \]

4. **Substituting Values**:
- From the vertical equilibrium condition:
\[ T \cos(45^\circ) = 100 \, \text{N} \]
Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ T \cdot \frac{1}{\sqrt{2}} = 100 \implies T = 100\sqrt{2} \, \text{N} \]

5. **Finding the Horizontal Force**:
- Now substitute T into the horizontal equilibrium condition:
\[ T \sin(45^\circ) = F \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ F = (100\sqrt{2}) \cdot \frac{1}{\sqrt{2}} = 100 \, \text{N} \]

6. **Conclusion**:
- The magnitude of the horizontal force applied is:
\[ F = 100 \, \text{N} \]

### Final Answer:
The magnitude of the force applied is **100 N**.