A block of mass M placed on a frictionless horizontal table is pulled by another block of mass m hanging vertically by a massless string passing over a frictionless pulley. The tension in the string is :

Options

A. `(m)/(M + m )g`
B. `(M)/(M + m) g`
C. `(M + m)/(mn) g`
D. `(Mn)/(M + n) g`

Explanation

To find the tension in the string connecting the two blocks, we can follow these steps:

### Step 1: Identify the forces acting on each block
- For the hanging block of mass \( m \):
- The downward force due to gravity is \( mg \) (where \( g \) is the acceleration due to gravity).
- The upward force due to tension in the string is \( T \).

- For the block of mass \( M \) on the table:
- The only horizontal force acting on it is the tension \( T \).

### Step 2: Write the equations of motion
- For the hanging block (mass \( m \)):
\[ mg - T = ma \quad \text{(1)} \]
Here, \( a \) is the acceleration of the system.

- For the block on the table (mass \( M \)):
\[ T = Ma \quad \text{(2)} \]

### Step 3: Solve the equations simultaneously
- From equation (2), we can express \( a \):
\[ a = \frac{T}{M} \quad \text{(3)} \]

- Substitute equation (3) into equation (1):
\[ mg - T = m \left(\frac{T}{M}\right) \]

### Step 4: Rearrange the equation
- Rearranging gives:
\[ mg - T = \frac{mT}{M} \]
\[ mg = T + \frac{mT}{M} \]
\[ mg = T \left(1 + \frac{m}{M}\right) \]

### Step 5: Solve for tension \( T \)
- Now, isolate \( T \):
\[ T = \frac{mg}{1 + \frac{m}{M}} \]
\[ T = \frac{mg}{\frac{M + m}{M}} = \frac{mgM}{M + m} \]

### Final Result
The tension in the string is:
\[ T = \frac{mgM}{M + m} \]

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