A fireman is sliding down with the help of a rope. If the breaking strength of the rope is `( 2mg)/( 3)` where m is the mass of the man, the minimum acceleration with which the fireman can slide, so that the rope does not break, is `:`

Options

A. `( g)/( 2)`
B. `(g)/( 3)`
C. `( g)/( 4)`
D. `( 2g)/( 3)`

Answer

To solve the problem, we need to analyze the forces acting on the fireman as he slides down the rope. The key is to apply Newton's second law of motion and understand the relationship between tension, weight, and acceleration.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The weight of the fireman acting downwards: \( W = mg \)
- The tension in the rope acting upwards: \( T \)

2. **Write the Equation of Motion**:
- According to Newton's second law, the net force acting on the fireman is equal to the mass times the acceleration (\( a \)):
\[ mg - T = ma \]
Rearranging gives:
\[ T = mg - ma \]

3. **Determine Maximum Tension**:
- The breaking strength of the rope is given as:
\[ T_{max} = \frac{2mg}{3} \]

4. **Set Up the Inequality**:
- To ensure the rope does not break, the tension must not exceed the maximum tension:
\[ T \leq T_{max} \]
Substituting for \( T \) from the equation of motion:
\[ mg - ma \leq \frac{2mg}{3} \]

5. **Solve for Acceleration**:
- Rearranging the inequality:
\[ mg - \frac{2mg}{3} \leq ma \]
Simplifying the left side:
\[ \frac{mg}{3} \leq ma \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{g}{3} \leq a \]
This implies:
\[ a \geq \frac{g}{3} \]

6. **Conclusion**:
- The minimum acceleration with which the fireman can slide down without breaking the rope is:
\[ a_{min} = \frac{g}{3} \]

### Final Answer:
The minimum acceleration with which the fireman can slide, so that the rope does not break, is \( \frac{g}{3} \).