The masses `(M+m)` and `(M-m)` are attached to the ends of a light inextensible string and the string is made to pass over the surface of a smooth fixed pulley. When the masses are relased from rest the acceleraion of the system is .

Options

A. `gm//M`
B. `2gM//m`
C. `gm//2M`
D. `g(M^(2)-m^(2))//2M`

Explanation

To solve the problem of finding the acceleration of the system with masses \( (M+m) \) and \( (M-m) \) attached to a light inextensible string over a smooth fixed pulley, we will follow these steps:

### Step 1: Identify the masses
Let:
- \( M_1 = M + m \) (the mass on one side of the pulley)
- \( M_2 = M - m \) (the mass on the other side of the pulley)

### Step 2: Write the equations of motion
For mass \( M_1 \) (which is \( M + m \)):
- The force acting downwards is \( M_1 g \) (weight).
- The force acting upwards is the tension \( T \).
- According to Newton's second law:
\[ M_1 g - T = M_1 a \quad \text{(1)} \]

For mass \( M_2 \) (which is \( M - m \)):
- The force acting downwards is \( M_2 g \) (weight).
- The force acting upwards is the tension \( T \).
- According to Newton's second law:
\[ T - M_2 g = M_2 a \quad \text{(2)} \]

### Step 3: Substitute the masses into the equations
Substituting \( M_1 \) and \( M_2 \) into equations (1) and (2):
1. For \( M_1 \):
\[ (M + m) g - T = (M + m) a \quad \text{(3)} \]
2. For \( M_2 \):
\[ T - (M - m) g = (M - m) a \quad \text{(4)} \]

### Step 4: Add equations (3) and (4)
Adding equations (3) and (4) to eliminate \( T \):
\[ (M + m) g - T + T - (M - m) g = (M + m) a + (M - m) a \]
This simplifies to:
\[ (M + m) g - (M - m) g = (M + m + M - m) a \]
\[ (2m) g = (2M) a \]

### Step 5: Solve for acceleration \( a \)
Now, we can isolate \( a \):
\[ a = \frac{(2m) g}{(2M)} = \frac{mg}{M} \]

### Final Answer
Thus, the acceleration of the system is:
\[ a = \frac{mg}{M} \]