A mass is hanging on a spring balance which is kept in a lift. The lift ascends. The spring balance will show in its reading

Options

A. Increase
B. Decrease
C. No change
D. Change depending upon velocity

Answer

To solve the problem of how a spring balance reading changes when a mass is hanging in a lift that is ascending, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Mass:**
- When the mass is hanging in the lift, two main forces are acting on it:
- The gravitational force (weight) acting downwards, which is given by \( mg \) (where \( m \) is the mass and \( g \) is the acceleration due to gravity).
- The spring force acting upwards, which is the force exerted by the spring balance, denoted as \( F_s \).

2. **Consider the Lift's Acceleration:**
- Since the lift is ascending, it will have an upward acceleration, denoted as \( a \).
- According to Newton's second law, the net force acting on the mass can be expressed as:
\[ F_{\text{net}} = F_{\text{up}} - F_{\text{down}} = F_s - mg \]

3. **Apply Newton's Second Law:**
- The net force is also equal to the mass times its acceleration:
\[ F_{\text{net}} = ma \]
- Therefore, we can set the two expressions for net force equal to each other:
\[ F_s - mg = ma \]

4. **Rearranging the Equation:**
- Rearranging the equation gives us the spring force:
\[ F_s = mg + ma \]

5. **Factor Out the Mass:**
- We can factor out the mass \( m \) from the right side:
\[ F_s = m(g + a) \]

6. **Interpret the Spring Balance Reading:**
- The reading on the spring balance \( F_s \) represents the apparent weight of the mass when the lift is accelerating upwards. Thus, the reading increases compared to the normal weight \( mg \) when the lift is stationary or moving at constant velocity.

### Conclusion:
- The spring balance will show a reading of \( m(g + a) \) when the lift is ascending, indicating that the reading increases due to the upward acceleration of the lift.