Two bodies A (30 kg) and B (50 kg) tied with a light string are placed on a friction less table. A force F acting at B pulls this ystem with an acceleration of `2ms^(-2)`. The tension in the string is:

Options

A. 60N,60N,60N
B. 100N
C. 35N
D. 140N

Explanation

To find the tension in the string connecting two bodies A and B, we can follow these steps:

### Step 1: Identify the masses and the acceleration
- Mass of body A (m_A) = 30 kg
- Mass of body B (m_B) = 50 kg
- Acceleration of the system (a) = 2 m/s²

### Step 2: Calculate the total mass of the system
The total mass (m_total) of the system is the sum of the masses of both bodies:
\[ m_{total} = m_A + m_B = 30\, \text{kg} + 50\, \text{kg} = 80\, \text{kg} \]

### Step 3: Calculate the total force acting on the system
Using Newton's second law, the total force (F) acting on the system can be calculated as:
\[ F = m_{total} \cdot a = 80\, \text{kg} \cdot 2\, \text{m/s}^2 = 160\, \text{N} \]

### Step 4: Analyze the forces acting on body A
For body A, the only force acting on it is the tension (T) in the string. According to Newton's second law, we can write:
\[ T = m_A \cdot a \]
Substituting the values:
\[ T = 30\, \text{kg} \cdot 2\, \text{m/s}^2 = 60\, \text{N} \]

### Step 5: Conclusion
The tension in the string is:
\[ T = 60\, \text{N} \]