A string tied on a roof can bear a maximum tension of 50 kg wt. The minimum acceleration that can be acquired by a man of 98 kg to descend will be [take g = ` 9*8 m/s^(2)`]

Options

A. `9*8 m/s^(2)`
B. `4*9m/s^(2)`
C. `4*8 m/s^(2)`
D. `5 m/s^(2)`

Answer

To solve the problem, we need to find the minimum acceleration that a man of mass 98 kg can acquire while descending on a string that can bear a maximum tension of 50 kg weight.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Maximum tension \( T_{max} = 50 \, \text{kg wt} \)
- Mass of the man \( m = 98 \, \text{kg} \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)

2. **Convert Maximum Tension to Newtons:**
\[ T_{max} = 50 \, \text{kg} \times g = 50 \times 9.8 = 490 \, \text{N} \]

3. **Write the Equation of Motion:**
When the man is descending, the forces acting on him are:
- Weight \( W = mg \) acting downwards
- Tension \( T \) acting upwards

According to Newton's second law:
\[ mg - T = ma \]
where \( a \) is the acceleration of the man.

4. **Substituting Known Values:**
Substitute \( T = T_{max} = 490 \, \text{N} \) and \( m = 98 \, \text{kg} \):
\[ 98 \times 9.8 - 490 = 98 \times a \]

5. **Calculate the Weight of the Man:**
\[ mg = 98 \times 9.8 = 960.4 \, \text{N} \]

6. **Plugging in the Values:**
\[ 960.4 - 490 = 98a \]
\[ 470.4 = 98a \]

7. **Solving for Acceleration \( a \):**
\[ a = \frac{470.4}{98} \approx 4.8 \, \text{m/s}^2 \]

8. **Conclusion:**
The minimum acceleration that can be acquired by the man to descend is approximately \( 4.8 \, \text{m/s}^2 \).

### Final Answer:
The minimum acceleration that can be acquired by the man is **4.8 m/s²**.

---