A person stands on a weighing machine kept on the floor of an elevator .When the elevator is at rest then the apparent weight of the person is

Options

A. equal to his true weight
B. less than his true weight
C. more than his true weight
D. more or less than his true weight

Explanation

To solve the problem of determining the apparent weight of a person standing on a weighing machine in a stationary elevator, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Weight**: The true weight of a person is given by the formula:
\[ W = mg \]
where \( m \) is the mass of the person and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

2. **Situation in the Elevator**: When the elevator is at rest, it is not accelerating. Therefore, the forces acting on the person are balanced. The gravitational force acting downwards is equal to the normal force exerted by the weighing machine acting upwards.

3. **Force Analysis**: In this case:
- The downward force (weight of the person) is \( mg \).
- The upward force (normal force from the weighing machine) is also \( N \).

4. **Equilibrium Condition**: Since the elevator is at rest, we can apply Newton's first law of motion. The sum of the forces acting on the person is zero:
\[ N - mg = 0 \]
This implies:
\[ N = mg \]

5. **Conclusion**: The normal force \( N \) is what the weighing machine measures as the apparent weight of the person. Since \( N = mg \), the apparent weight of the person when the elevator is at rest is equal to their true weight.

### Final Answer:
The apparent weight of the person when the elevator is at rest is equal to their true weight.