A man of mass 70 kg stands on a weighing scale in a lift which is moving downward with the uniform acceleration of `5 m/s^2` . The reading on the weighing scale is

Options

A. 70 kg
B. 35 kg
C. 105 kg
D. 700 kg

Answer

To solve the problem, we need to calculate the reading on the weighing scale when a man of mass 70 kg is standing in a lift that is moving downward with a uniform acceleration of 5 m/s².

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Man**:
- The gravitational force acting on the man (weight) is given by:
\[ W = mg \]
where \( m = 70 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \) (approximately).

2. **Calculate the Gravitational Force**:
- Substituting the values:
\[ W = 70 \, \text{kg} \times 10 \, \text{m/s}^2 = 700 \, \text{N} \]

3. **Consider the Acceleration of the Lift**:
- The lift is accelerating downward at \( a = 5 \, \text{m/s}^2 \). This means there is a pseudo-force acting on the man in the upward direction due to the downward acceleration of the lift.

4. **Calculate the Net Force on the Man**:
- The net force acting on the man can be expressed as:
\[ F_{\text{net}} = W - F_{\text{pseudo}} = mg - ma \]
- Here, \( F_{\text{pseudo}} = ma = 70 \, \text{kg} \times 5 \, \text{m/s}^2 = 350 \, \text{N} \).

5. **Calculate the Effective Weight**:
- The effective weight (apparent weight) that the scale reads is:
\[ F_{\text{scale}} = mg - ma = 700 \, \text{N} - 350 \, \text{N} = 350 \, \text{N} \]

6. **Convert the Force Reading to Mass**:
- The reading on the scale in terms of mass can be calculated using the formula:
\[ m' = \frac{F_{\text{scale}}}{g} \]
- Substituting the values:
\[ m' = \frac{350 \, \text{N}}{10 \, \text{m/s}^2} = 35 \, \text{kg} \]

7. **Final Answer**:
- The reading on the weighing scale is \( 35 \, \text{kg} \).

### Summary:
The reading on the weighing scale when the lift is moving downward with an acceleration of \( 5 \, \text{m/s}^2 \) is \( 35 \, \text{kg} \).