Two masses `m_(1) and m_(2) ` are connected to the end of a string passing over a smooth pulley . The magnitude of tension in the string is T and the masses are moving with magnitude of acceleration a . If the masses are interchanged , then

Options

A. Both a and T will change
B. a will change but T will remain unchanged
C. T will change but a will remain unchanged
D. None of these will change

Answer

To solve the problem of two masses \( m_1 \) and \( m_2 \) connected by a string over a smooth pulley, we need to analyze the forces acting on each mass and derive expressions for tension and acceleration. Let’s go through the solution step by step.

### Step 1: Identify the Forces
- For mass \( m_1 \) (which is moving upwards), the forces acting on it are:
- Weight: \( m_1 g \) (downward)
- Tension: \( T \) (upward)

- For mass \( m_2 \) (which is moving downwards), the forces acting on it are:
- Weight: \( m_2 g \) (downward)
- Tension: \( T \) (upward)

### Step 2: Write the Equations of Motion
For mass \( m_1 \):
\[ T - m_1 g = m_1 a \quad \text{(1)} \]

For mass \( m_2 \):
\[ m_2 g - T = m_2 a \quad \text{(2)} \]

### Step 3: Solve the Equations
Now, we can add equations (1) and (2) to eliminate \( T \):
\[ (T - m_1 g) + (m_2 g - T) = m_1 a + m_2 a \]
This simplifies to:
\[ m_2 g - m_1 g = (m_1 + m_2) a \]
Rearranging gives:
\[ (m_2 - m_1) g = (m_1 + m_2) a \]
Thus, the acceleration \( a \) can be expressed as:
\[ a = \frac{(m_2 - m_1) g}{(m_1 + m_2)} \quad \text{(3)} \]

### Step 4: Find the Tension
Now, we can substitute \( a \) back into either equation (1) or (2) to find the tension \( T \). Using equation (1):
\[ T = m_1 g + m_1 a \]
Substituting \( a \) from equation (3):
\[ T = m_1 g + m_1 \left(\frac{(m_2 - m_1) g}{(m_1 + m_2)}\right) \]
This simplifies to:
\[ T = m_1 g \left(1 + \frac{(m_2 - m_1)}{(m_1 + m_2)}\right) \]
Combining terms:
\[ T = \frac{m_1 g (m_1 + m_2 + m_2 - m_1)}{(m_1 + m_2)} = \frac{m_1 m_2 g}{(m_1 + m_2)} \quad \text{(4)} \]

### Step 5: Interchanging the Masses
When the masses are interchanged, \( m_1 \) becomes \( m_2 \) and \( m_2 \) becomes \( m_1 \). The new acceleration \( a' \) will be:
\[ a' = \frac{(m_1 - m_2) g}{(m_1 + m_2)} \quad \text{(5)} \]
The tension \( T' \) will also be:
\[ T' = \frac{m_2 m_1 g}{(m_1 + m_2)} \quad \text{(6)} \]

### Conclusion
From equations (4) and (6), we see that the tension \( T \) remains the same regardless of the interchange of masses. The magnitude of acceleration changes direction but not magnitude.

### Final Answer
- The magnitude of tension \( T \) does not change.
- The magnitude of acceleration \( a \) remains the same, but its direction reverses.