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 system 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 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: Analyze the forces acting on the system
Since the table is frictionless, the only forces acting on the system are the tension in the string (T) and the applied force (F) at body B. The entire system (A + B) will accelerate together at 2 m/s².

### Step 3: Calculate the total mass of the system
Total mass (m_total) = m_A + m_B = 30 kg + 50 kg = 80 kg

### Step 4: Use Newton's second law for the entire system
According to Newton's second law, the net force (F_net) acting on the system is equal to the total mass multiplied by the acceleration:
\[ F_{net} = m_{total} \cdot a \]
\[ F_{net} = 80 \, \text{kg} \cdot 2 \, \text{m/s}^2 = 160 \, \text{N} \]

### Step 5: Analyze the forces acting on body A
For body A, the only force acting on it is the tension (T) in the string, which causes it to accelerate. According to Newton's second law:
\[ T = m_A \cdot a \]
Substituting the values:
\[ T = 30 \, \text{kg} \cdot 2 \, \text{m/s}^2 = 60 \, \text{N} \]

### Conclusion
The tension in the string is **60 N**.

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