A body of mass `sqrt(3)` kg is suspended by a string to a rigid support. The body is pulled horizontally by a force F until the string makes an angle of 30° with the vertical. The value of F and tension in the string are.

Options

A. 9.8 JN, 9.8 N
B. 9.8 N, 19.6 N
C. 19.6 N, 19.6 N
D. 19.6 N, 9.8 N

Answer

To solve the problem, we need to analyze the forces acting on the body suspended by the string when it is pulled horizontally, causing the string to make an angle of 30° with the vertical.

### Step-by-step Solution:

1. **Identify the Forces Acting on the Body:**
- The weight of the body (W) acting downwards:
\[ W = mg = \sqrt{3} \, \text{kg} \times g \]
where \( g = 9.8 \, \text{m/s}^2 \).
- The tension (T) in the string acting at an angle of 30° with the vertical.
- The horizontal force (F) pulling the body.

2. **Calculate the Weight of the Body:**
\[ W = \sqrt{3} \times 9.8 = 9.8\sqrt{3} \, \text{N} \]

3. **Resolve the Tension into Components:**
- The vertical component of the tension (T) balances the weight:
\[ T \cos(30°) = W \]
- The horizontal component of the tension (T) balances the applied force (F):
\[ T \sin(30°) = F \]

4. **Substituting the Values:**
- We know that \( \cos(30°) = \frac{\sqrt{3}}{2} \) and \( \sin(30°) = \frac{1}{2} \).
- Thus, the equations become:
\[ T \cdot \frac{\sqrt{3}}{2} = 9.8\sqrt{3} \]
\[ T \cdot \frac{1}{2} = F \]

5. **Solving for Tension (T):**
- From the first equation:
\[ T = \frac{9.8\sqrt{3}}{\frac{\sqrt{3}}{2}} = 9.8 \times 2 = 19.6 \, \text{N} \]

6. **Finding the Applied Force (F):**
- Substitute the value of T into the second equation:
\[ F = T \cdot \frac{1}{2} = 19.6 \cdot \frac{1}{2} = 9.8 \, \text{N} \]

### Final Results:
- The applied force \( F \) is \( 9.8 \, \text{N} \).
- The tension \( T \) in the string is \( 19.6 \, \text{N} \).

### Summary:
- The values of \( F \) and \( T \) are \( 9.8 \, \text{N} \) and \( 19.6 \, \text{N} \), respectively.

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