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, we need to find the maximum acceleration with which the helicopter can rise without breaking the rope. We will use Newton's second law of motion to derive the solution step by step.

### Step 1: Convert the maximum tension from kg-wt to Newtons
The maximum tension that the rope can bear is given as 70 kg-wt. To convert this to Newtons, we use the relation:
\[ \text{Tension (T)} = \text{mass (m)} \times g \]
where \( g = 9.8 \, \text{m/s}^2 \).

Calculating the tension:
\[ T = 70 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 686 \, \text{N} \]

### Step 2: Calculate the weight of the man
The weight of the man (W) can be calculated using the formula:
\[ W = m \times g \]
where \( m = 50 \, \text{kg} \).

Calculating the weight:
\[ W = 50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 490 \, \text{N} \]

### Step 3: Apply Newton's second law of motion
According to Newton's second law, the net force acting on the man can be expressed as:
\[ F_{\text{net}} = T - W \]
where \( T \) is the tension in the rope and \( W \) is the weight of the man.

### Step 4: Set up the equation for maximum acceleration
The net force can also be expressed in terms of mass and acceleration:
\[ F_{\text{net}} = m \times a \]
Substituting the expressions for net force:
\[ T - W = m \times a \]

### Step 5: Substitute known values into the equation
Substituting the values of \( T \) and \( W \):
\[ 686 \, \text{N} - 490 \, \text{N} = 50 \, \text{kg} \times a \]
This simplifies to:
\[ 196 \, \text{N} = 50 \, \text{kg} \times a \]

### Step 6: Solve for acceleration (a)
Now, we can solve for \( a \):
\[ a = \frac{196 \, \text{N}}{50 \, \text{kg}} = 3.92 \, \text{m/s}^2 \]

### Final Answer
The maximum acceleration with which the helicopter can rise without breaking the rope is:
\[ \boxed{3.92 \, \text{m/s}^2} \]