In the question number 91, if a force Fis applied to 20 kg block, then the tension in the string is

Options

A. 100N
B. 200N
C. 300N
D. 400N

Explanation

To solve the problem of finding the tension in the string when a force \( F \) is applied to a 20 kg block, we can follow these steps:

### Step 1: Identify the masses and forces
Let:
- \( m_1 = 10 \, \text{kg} \) (mass of block A)
- \( m_2 = 20 \, \text{kg} \) (mass of block B)
- \( F \) = applied force on block B
- \( T \) = tension in the string
- \( a \) = acceleration of the system

### Step 2: Draw Free Body Diagrams
For block B (20 kg):
- The applied force \( F \) acts in the positive direction.
- The tension \( T \) acts in the opposite direction.

For block A (10 kg):
- The tension \( T \) acts in the positive direction.
- There are no other forces acting on it in the horizontal direction.

### Step 3: Write the equations of motion
For block B:
\[ F - T = m_2 a \quad \text{(1)} \]

For block A:
\[ T = m_1 a \quad \text{(2)} \]

### Step 4: Substitute and solve for acceleration
From equation (2), we can express \( a \) in terms of \( T \):
\[ a = \frac{T}{m_1} \]

Substituting this into equation (1):
\[ F - T = m_2 \left(\frac{T}{m_1}\right) \]

Rearranging gives:
\[ F = T + \frac{m_2 T}{m_1} \]

Factoring out \( T \):
\[ F = T \left(1 + \frac{m_2}{m_1}\right) \]

### Step 5: Solve for tension \( T \)
Now, we can solve for \( T \):
\[ T = \frac{F}{1 + \frac{m_2}{m_1}} = \frac{F}{\frac{m_1 + m_2}{m_1}} = \frac{F \cdot m_1}{m_1 + m_2} \]

### Step 6: Substitute known values
Given \( F = 600 \, \text{N} \), \( m_1 = 10 \, \text{kg} \), and \( m_2 = 20 \, \text{kg} \):
\[ T = \frac{600 \cdot 10}{10 + 20} = \frac{6000}{30} = 200 \, \text{N} \]

### Final Answer:
The tension in the string is \( T = 200 \, \text{N} \).

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