A block of mass 10kg is suspended from string of length 4m. When pulled by a force F along horizontal from midpoint. Upper half of string makes `45^@` with vertical, value of F is

Options

A. 100N
B. 75N
C. 90N
D. 85N

Explanation

To solve the problem, we need to analyze the forces acting on the block when it is pulled horizontally, causing the string to make an angle of \(45^\circ\) with the vertical.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The weight of the block \(W = mg\) acts downward, where \(m = 10 \, \text{kg}\) and \(g = 9.8 \, \text{m/s}^2\).
- The tension \(T\) in the string acts along the string at an angle of \(45^\circ\) with the vertical.
- The horizontal force \(F\) is applied at the midpoint of the string.

2. **Calculate the Weight of the Block**:
\[ W = mg = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98 \, \text{N} \]

3. **Resolve the Tension into Components**:
- The vertical component of the tension \(T\) is given by:
\[ T_y = T \cos(45^\circ) = T \cdot \frac{1}{\sqrt{2}} = \frac{T}{\sqrt{2}} \]
- The horizontal component of the tension \(T\) is given by:
\[ T_x = T \sin(45^\circ) = T \cdot \frac{1}{\sqrt{2}} = \frac{T}{\sqrt{2}} \]

4. **Apply the Equilibrium Condition in the Vertical Direction**:
Since the block is in equilibrium vertically, the upward tension must balance the weight:
\[ T_y = W \implies \frac{T}{\sqrt{2}} = 98 \, \text{N} \]
Solving for \(T\):
\[ T = 98 \sqrt{2} \, \text{N} \]

5. **Apply the Equilibrium Condition in the Horizontal Direction**:
The horizontal component of the tension must equal the applied force \(F\):
\[ T_x = F \implies \frac{T}{\sqrt{2}} = F \]
Substituting \(T\) from the previous step:
\[ F = \frac{98 \sqrt{2}}{\sqrt{2}} = 98 \, \text{N} \]

### Final Answer:
The value of the force \(F\) is \(98 \, \text{N}\).