The weight of a man in a lift moving upwards with an acceleration 'a' is 600 N. When the lift moves downwards with the same acceleration his weight is found to be 360 N. The real weight of the man is

Options

A. 380 N
B. 600 N
C. 480 N
D. 700 N

Explanation

To solve the problem, we need to analyze the situation using the concepts of forces acting on the man in the lift.

### Step-by-Step Solution:

1. **Understand the Forces**:
- When the lift is moving upwards with acceleration \( a \), the apparent weight \( W_1 \) of the man is given by:
\[ W_1 = m(g + a) \]
- When the lift is moving downwards with the same acceleration \( a \), the apparent weight \( W_2 \) of the man is:
\[ W_2 = m(g - a) \]

2. **Set Up the Equations**:
- From the problem, we know:
\[ W_1 = 600 \, \text{N} \quad \text{(upward)} \]
\[ W_2 = 360 \, \text{N} \quad \text{(downward)} \]
- Thus, we can write two equations:
\[ m(g + a) = 600 \quad \text{(1)} \]
\[ m(g - a) = 360 \quad \text{(2)} \]

3. **Add the Two Equations**:
- Adding equations (1) and (2):
\[ m(g + a) + m(g - a) = 600 + 360 \]
- This simplifies to:
\[ 2mg = 960 \]

4. **Solve for \( mg \)**:
- Dividing both sides by 2:
\[ mg = 480 \, \text{N} \]

5. **Conclusion**:
- The real weight of the man, which is \( mg \), is 480 N.

### Final Answer:
The real weight of the man is **480 N**.