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 find the tension in the string connecting two masses \( m_1 \) and \( m_2 \) over a frictionless, massless pulley that is accelerating upwards with an acceleration of \( \frac{g}{2} \), we can follow these steps:

### Step 1: Identify the Forces Acting on Each Mass
1. **For mass \( m_1 \)** (which is lighter):
- Weight acting downwards: \( m_1 g \)
- Tension \( T \) acting upwards.
- Since the pulley is accelerating upwards, the effective acceleration acting on \( m_1 \) is \( \frac{g}{2} \) upwards.

The net force equation for \( m_1 \) can be written as:
\[ T - m_1 g = -m_1 \left(\frac{g}{2}\right) \]
(The negative sign indicates that the acceleration is upwards against the weight.)

2. **For mass \( m_2 \)** (which is heavier):
- Weight acting downwards: \( m_2 g \)
- Tension \( T \) acting upwards.
- The effective acceleration acting on \( m_2 \) is also \( \frac{g}{2} \) upwards.

The net force equation for \( m_2 \) can be written as:
\[ m_2 g - T = m_2 \left(\frac{g}{2}\right) \]

### Step 2: Set Up the Equations
From the equations derived from the forces acting on \( m_1 \) and \( m_2 \):

1. For \( m_1 \):
\[ T - m_1 g = -\frac{m_1 g}{2} \quad \text{(1)} \]

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

### Step 3: Solve the Equations
From equation (1):
\[ T = m_1 g - \frac{m_1 g}{2} = \frac{m_1 g}{2} \quad \text{(3)} \]

From equation (2):
\[ T = m_2 g - \frac{m_2 g}{2} = \frac{m_2 g}{2} \quad \text{(4)} \]

### Step 4: Equate the Two Expressions for Tension
From equations (3) and (4):
\[ \frac{m_1 g}{2} = m_2 g - \frac{m_2 g}{2} \]

### Step 5: Simplify and Solve for Tension
Combine and simplify:
\[ \frac{m_1 g}{2} + \frac{m_2 g}{2} = m_2 g \]
\[ \frac{(m_1 + m_2) g}{2} = m_2 g \]
This leads to:
\[ T = \frac{m_1 g + m_2 g}{2} = \frac{(m_1 + m_2) g}{2} \]

### Final Expression for Tension
Thus, the tension \( T \) in the string is given by:
\[ T = \frac{(m_1 + m_2) g}{2} \]