The apparent weight of a man in a lift is `W_(1)` when lift moves upwards with some acceleration and is `W_(2)` when it is accerating down with same acceleration. Find the true weight of the man and acceleration of lift .

Options

A. `( w_(1) + w_(2))/( 2)`
B. `( w_(1) - w_(2))/( 2)`
C. `2w_(1)`
D. `2w_(2)`

Answer

To solve the problem, we need to find the true weight of the man (W) and the acceleration of the lift (a) based on the apparent weights (W1 and W2) when the lift is accelerating upwards and downwards, respectively.

### Step-by-Step Solution:

1. **Understanding Apparent Weight**:
- When the lift accelerates upwards, the apparent weight (W1) of the man is given by:
\[ W_1 = m(g + a) \]
- When the lift accelerates downwards, the apparent weight (W2) is given by:
\[ W_2 = m(g - a) \]

2. **Setting Up the Equations**:
- From the first equation, we can express the true weight (W) in terms of W1:
\[ W = W_1 - ma \]
- From the second equation, we can express the true weight (W) in terms of W2:
\[ W = W_2 + ma \]

3. **Equating the Two Expressions for True Weight**:
- Since both expressions equal the true weight (W), we can set them equal to each other:
\[ W_1 - ma = W_2 + ma \]

4. **Rearranging the Equation**:
- Rearranging the equation gives:
\[ W_1 - W_2 = 2ma \]

5. **Finding the Acceleration (a)**:
- We can solve for the acceleration (a):
\[ a = \frac{W_1 - W_2}{2m} \]

6. **Finding the True Weight (W)**:
- Now, we can substitute the value of a back into either equation for W. Let's use the first equation:
\[ W = W_1 - m\left(\frac{W_1 - W_2}{2m}\right) \]
- Simplifying this gives:
\[ W = W_1 - \frac{W_1 - W_2}{2} = \frac{W_1 + W_2}{2} \]

### Final Answers:
- The true weight of the man (W) is:
\[ W = \frac{W_1 + W_2}{2} \]
- The acceleration of the lift (a) is:
\[ a = \frac{W_1 - W_2}{2m} \]