A man pulls a bucket of water from a depth of h from a well. If the mass of the rope and that of bucket full of water are m and M respectively, the work done by the man is :

Options

A. `(M+m)gh`
B. `(M+m/2)gh`
C. `((M+m)/2)gh`
D. `(M/2+m)gh`

Answer

To solve the problem of calculating the work done by a man pulling a bucket of water from a depth \( h \) in a well, we can break it down into a series of steps:

### Step-by-Step Solution:

1. **Identify the Forces Involved**:
- The man pulls a bucket of mass \( M \) and a rope of mass \( m \).
- The gravitational force acting on the bucket is \( Mg \) (where \( g \) is the acceleration due to gravity).
- The gravitational force acting on the rope is \( mg \).

2. **Determine the Displacement**:
- The bucket is pulled from a depth \( h \) to the top of the well, which means it is displaced by a distance \( h \).
- The center of mass of the rope, which has a uniform mass distribution, is located at a height of \( \frac{h}{2} \) when the entire rope is pulled up.

3. **Calculate Work Done on the Bucket**:
- The work done \( W_b \) on the bucket can be calculated using the formula:
\[ W_b = \text{Force} \times \text{Displacement} = Mg \times h \]

4. **Calculate Work Done on the Rope**:
- The work done \( W_r \) on the rope is calculated by considering the center of mass of the rope being lifted by \( \frac{h}{2} \):
\[ W_r = \text{Force} \times \text{Displacement} = mg \times \frac{h}{2} \]

5. **Total Work Done by the Man**:
- The total work done \( W \) by the man is the sum of the work done on the bucket and the work done on the rope:
\[ W = W_b + W_r = Mg \cdot h + mg \cdot \frac{h}{2} \]

6. **Factor Out Common Terms**:
- We can factor out \( gh \) from the total work done:
\[ W = gh \left( M + \frac{m}{2} \right) \]

### Final Answer:
The work done by the man in pulling the bucket of water from a depth \( h \) is:
\[ W = gh \left( M + \frac{m}{2} \right) \]