A lift is moving up with an acceleration of 3.675 `m//sec_(2)`. The weight of a man-

Options

A. increases by `37.5%`
B. decreases by`37.5%`
C. increases by `137.5%`
D. remains the same

Answer

To solve the problem of how the weight of a man changes when a lift accelerates upwards, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Forces Acting on the Man**:
- When the lift is stationary or moving at constant velocity, the only forces acting on the man are his weight (downward force) and the normal force (upward force) exerted by the lift floor.
- The weight of the man is given by \( W = mg \), where \( m \) is the mass of the man and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

2. **Identify the Situation with Upward Acceleration**:
- When the lift accelerates upwards with an acceleration \( a = 3.675 \, \text{m/s}^2 \), the effective weight of the man increases.
- The normal force \( N \) acting on the man can be expressed as:
\[ N = mg + ma \]
- Here, \( mg \) is the weight of the man, and \( ma \) is the additional force due to the lift's upward acceleration.

3. **Calculate the Normal Force**:
- Substitute the values into the equation for normal force:
\[ N = m(g + a) = m(9.8 + 3.675) \]
- This simplifies to:
\[ N = m(13.475) \, \text{N} \]

4. **Calculate the Percentage Increase in Weight**:
- The original weight of the man is \( mg = m \times 9.8 \).
- The increase in weight due to the lift's acceleration is:
\[ \Delta W = N - mg = m(13.475) - m(9.8) = m(3.675) \]
- The percentage increase in weight is given by:
\[ \text{Percentage Increase} = \left( \frac{\Delta W}{mg} \right) \times 100 = \left( \frac{m(3.675)}{m(9.8)} \right) \times 100 \]
- The mass \( m \) cancels out:
\[ \text{Percentage Increase} = \left( \frac{3.675}{9.8} \right) \times 100 \]

5. **Perform the Calculation**:
- Calculate the fraction:
\[ \frac{3.675}{9.8} \approx 0.3745 \]
- Convert to percentage:
\[ 0.3745 \times 100 \approx 37.45\% \]

6. **Conclusion**:
- The weight of the man increases by approximately \( 37.5\% \) when the lift accelerates upwards at \( 3.675 \, \text{m/s}^2 \).

### Final Answer:
The weight of the man increases by approximately **37.5%**.