A 60kg man stands on a spring scales in a lift. At some instant. He finds that the scale reading has changed from 60kg to 50 kg for a while and then comes back to original mark. What should be concluded?

Options

A. The lift was in constant motion upward.
B. The lift was in constant motion downward.
C. The lift while in motion downward suddenly stopped.
D. The lift while in motion upward suddenly stopped.

Explanation

To analyze the situation where a 60 kg man stands on a spring scale in a lift, we need to consider the forces acting on him and how the lift's motion affects the scale reading.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- The man has a mass of 60 kg. Under normal circumstances (when at rest or moving with constant velocity), the scale would read his weight, which is given by the formula:
\[ W = mg \]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)). Thus, the scale reads:
\[ W = 60 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 588 \, \text{N} \]
- The scale reading in kg is simply the mass, which is 60 kg.

2. **Analyzing the Change in Scale Reading**:
- The scale reading drops to 50 kg. This indicates that the effective weight of the man has decreased. The effective weight is influenced by the acceleration of the lift.
- The scale reading in kg can be interpreted as the normal force \( N \) acting on the man:
\[ N = m \cdot g_{\text{effective}} \]
- When the scale reads 50 kg, we can express this as:
\[ N = 50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 490 \, \text{N} \]

3. **Relating the Forces**:
- The effective weight can be expressed as:
\[ g_{\text{effective}} = g - a \]
where \( a \) is the acceleration of the lift.
- Setting up the equation:
\[ 490 \, \text{N} = 60 \, \text{kg} \times (9.8 \, \text{m/s}^2 - a) \]
- Solving for \( a \):
\[ 490 = 588 - 60a \]
\[ 60a = 588 - 490 \]
\[ 60a = 98 \]
\[ a = \frac{98}{60} \approx 1.63 \, \text{m/s}^2 \]

4. **Conclusion About the Lift's Motion**:
- The negative acceleration indicates that the lift is accelerating downwards. This is why the scale reading decreased temporarily.
- After this, the lift returns to its original state, and the scale reading goes back to 60 kg, indicating that the lift either moves with uniform velocity again or comes to rest.

### Final Conclusion:
The conclusion is that the lift was initially moving upwards with a uniform velocity, then it experienced a downward acceleration (deceleration) which caused the scale reading to drop to 50 kg, and finally, it returned to its original state.