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)/(mM) g`
D. `(Mm)/(M + m) g`

Explanation

To find the tension in the string connecting a block of mass \( M \) on a frictionless table and a block of mass \( m \) hanging vertically, we can follow these steps:

### Step 1: Analyze the Forces on Each Block
- For the block of mass \( M \) on the table:
- The only horizontal force acting on it is the tension \( T \) in the string.
- According to Newton's second law, we can write the equation:
\[ T = M \cdot a \quad \text{(1)} \]

- For the hanging block of mass \( m \):
- The forces acting on it are the gravitational force \( mg \) downward and the tension \( T \) upward.
- The net force acting on this block can be expressed as:
\[ mg - T = m \cdot a \quad \text{(2)} \]

### Step 2: Set Up the Equations
From equation (1), we have:
\[ T = M \cdot a \]

From equation (2), we can rearrange it to express \( T \):
\[ T = mg - m \cdot a \]

### Step 3: Solve for Acceleration
Now, we can set the two expressions for \( T \) equal to each other:
\[ M \cdot a = mg - m \cdot a \]

Rearranging gives:
\[ M \cdot a + m \cdot a = mg \]
\[ a(M + m) = mg \]
\[ a = \frac{mg}{M + m} \]

### Step 4: Substitute Back to Find Tension
Now that we have \( a \), we can substitute it back into equation (1) to find the tension \( T \):
\[ T = M \cdot a = M \cdot \left(\frac{mg}{M + m}\right) \]
\[ T = \frac{M \cdot m \cdot g}{M + m} \]

### Final Result
Thus, the tension in the string is given by:
\[ T = \frac{M \cdot m \cdot g}{M + m} \]