Two mass m and 2m are attached with each other by a rope passing over a frictionless and massless pulley. If the pulley is accelerated upwards with an acceleration 'a', what is the value of tension?

Options

A. `(g+a)/(3)`
B. `(g-a)/(3)`
C. `(4m(g+a))/(3)`
D. `(m(g-a))/(3)`

Explanation

To solve the problem, we need to analyze the forces acting on the two masses and the effect of the upward acceleration of the pulley. Here’s a step-by-step solution:

### Step 1: Identify the forces acting on each mass
- For mass \( m \):
- Weight \( W_m = mg \) (downward)
- Tension \( T \) (upward)

- For mass \( 2m \):
- Weight \( W_{2m} = 2mg \) (downward)
- Tension \( T \) (upward)

### Step 2: Consider the upward acceleration of the pulley
Since the pulley is accelerating upwards with an acceleration \( a \), we need to consider the effect of this acceleration on the forces acting on the masses. The effective gravitational acceleration acting on the masses will be \( g + a \).

### Step 3: Write the equations of motion for each mass
- For mass \( m \):
\[ T - mg = ma \quad \text{(1)} \]

- For mass \( 2m \):
\[ 2mg - T = 2ma \quad \text{(2)} \]

### Step 4: Solve the equations simultaneously
From equation (1):
\[ T = mg + ma \quad \text{(3)} \]

Substituting equation (3) into equation (2):
\[ 2mg - (mg + ma) = 2ma \]
\[ 2mg - mg - ma = 2ma \]
\[ mg = 3ma \]
\[ g = 3a \quad \text{(4)} \]

### Step 5: Substitute back to find the tension
Now substituting \( g = 3a \) back into equation (3):
\[ T = mg + ma \]
\[ T = m(3a) + ma \]
\[ T = 3ma + ma \]
\[ T = 4ma \]

### Final Result
Thus, the tension \( T \) in the rope is:
\[ T = 4ma \]