Two bodies of masses `m_(1) " and " m_(2)` are connected a light string which passes over a frictionless massless pulley. If the pulley is moving upward with uniform acceleration `(g)/(2)`, then tension in the string will be

Options

A. `(3m_(1)m_(2))/(m_(1)+m_(2))g`
B. `(m_(1)+m_(2))/(4m_(1)m_(2))g`
C. `(2m_(1)m_(2))/(m_(1)+m_(2))g`
D. `(m_(1)m_(2))/(m_(1)+m_(2))g`

Explanation

To solve the problem, we need to analyze the forces acting on the two masses connected by a string over a pulley that is accelerating upwards. Let's denote the masses as \( m_1 \) and \( m_2 \), and the acceleration of the pulley as \( a = \frac{g}{2} \).

### Step-by-Step Solution:

1. **Identify the Forces**:
- For mass \( m_1 \) (which is heavier), the forces acting on it are its weight \( m_1 g \) downward and the tension \( T \) upward.
- For mass \( m_2 \) (which is lighter), the forces acting on it are its weight \( m_2 g \) downward and the tension \( T \) upward.

2. **Write the Equations of Motion**:
- For mass \( m_1 \):
\[ m_1 \cdot \left( \frac{g}{2} + g \right) - T = m_1 \cdot a \]
Here, the effective acceleration acting on \( m_1 \) is \( \frac{g}{2} + g = \frac{3g}{2} \).
Thus, the equation becomes:
\[ m_1 \cdot \frac{3g}{2} - T = m_1 \cdot a \quad \text{(1)} \]

- For mass \( m_2 \):
\[ T - m_2 \cdot \left( \frac{g}{2} + g \right) = m_2 \cdot a \]
Similarly, the effective acceleration acting on \( m_2 \) is \( \frac{3g}{2} \).
Thus, the equation becomes:
\[ T - m_2 \cdot \frac{3g}{2} = m_2 \cdot a \quad \text{(2)} \]

3. **Combine the Equations**:
- Adding equations (1) and (2):
\[ \left( m_1 \cdot \frac{3g}{2} - T \right) + \left( T - m_2 \cdot \frac{3g}{2} \right) = m_1 \cdot a + m_2 \cdot a \]
This simplifies to:
\[ m_1 \cdot \frac{3g}{2} - m_2 \cdot \frac{3g}{2} = (m_1 + m_2) \cdot a \]
Rearranging gives:
\[ \frac{3g}{2} (m_1 - m_2) = (m_1 + m_2) \cdot a \]

4. **Solve for Acceleration \( a \)**:
- From the above equation, we can express \( a \):
\[ a = \frac{3g(m_1 - m_2)}{2(m_1 + m_2)} \]

5. **Substitute \( a \) back into one of the equations**:
- We can substitute this value of \( a \) back into either equation (1) or (2) to find \( T \). Using equation (2):
\[ T = m_2 \cdot \frac{3g}{2} + m_2 \cdot a \]
Substituting \( a \):
\[ T = m_2 \cdot \frac{3g}{2} + m_2 \cdot \left( \frac{3g(m_1 - m_2)}{2(m_1 + m_2)} \right) \]

6. **Simplify to find \( T \)**:
- This gives:
\[ T = \frac{3m_2g}{2} + \frac{3m_2g(m_1 - m_2)}{2(m_1 + m_2)} \]
- After simplifying, we can find:
\[ T = \frac{3m_1 m_2 g}{m_1 + m_2} \]

### Final Answer:
The tension in the string is:
\[ T = \frac{3m_1 m_2 g}{m_1 + m_2} \]