A 50 kg man stuck in flood is being lifted vertically by an army helicopter with the help of light rope which can bear a maximum tension of 70 kg-wt. The maximum acceleration with which helicopter can rise so that rope does not breaks is `(g = 9.8 ms^-2)`

Options

A. `4.00 ms^-2`
B. `3.92 ms^-2`
C. `3.52 ms^-2`
D. `3.00 ms^-2`

Answer

To solve the problem step by step, we will follow the physics principles related to forces, tension, and acceleration.

### Step 1: Identify the Forces Acting on the Man
The forces acting on the man are:
- The weight of the man (downward force) given by \( W = mg \)
- The tension in the rope (upward force) which we denote as \( T \)

### Step 2: Calculate the Weight of the Man
Given:
- Mass of the man, \( m = 50 \, \text{kg} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

The weight of the man can be calculated as:
\[ W = mg = 50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 490 \, \text{N} \]

### Step 3: Determine the Maximum Tension in the Rope
The maximum tension that the rope can bear is given as 70 kg-wt. We need to convert this to Newtons:
\[ T_{\text{max}} = 70 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 686 \, \text{N} \]

### Step 4: Apply Newton's Second Law
According to Newton's second law, the net force acting on the man can be expressed as:
\[ T - W = ma \]
Where:
- \( T \) is the tension in the rope
- \( W \) is the weight of the man
- \( a \) is the acceleration of the man

### Step 5: Rearrange the Equation to Solve for Acceleration
We can rearrange the equation to find the acceleration \( a \):
\[ a = \frac{T - W}{m} \]

### Step 6: Substitute the Known Values
Now we substitute the known values into the equation:
\[ a = \frac{T_{\text{max}} - W}{m} = \frac{686 \, \text{N} - 490 \, \text{N}}{50 \, \text{kg}} \]

### Step 7: Calculate the Acceleration
Calculating the numerator:
\[ 686 \, \text{N} - 490 \, \text{N} = 196 \, \text{N} \]
Now substituting back into the equation:
\[ a = \frac{196 \, \text{N}}{50 \, \text{kg}} = 3.92 \, \text{m/s}^2 \]

### Conclusion
The maximum acceleration with which the helicopter can rise, so that the rope does not break, is:
\[ \boxed{3.92 \, \text{m/s}^2} \]