A boy whose mass is 50 kg stands on a spring balance inside a lift. The lift starts to ascent with an acceleration of `2ms^(-2)`. The reading of the machine or balance `(g=10ms^(-2))` is

Options

A. 50 kg
B. Zero
C. 49 kg
D. 60 kg

Answer

To solve the problem, we need to calculate the reading of the spring balance when the lift is accelerating upwards. The reading on the spring balance will be equal to the apparent weight of the boy, which is affected by the acceleration of the lift.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the boy, \( m = 50 \, \text{kg} \)
- Acceleration of the lift, \( a = 2 \, \text{m/s}^2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Weight of the Boy:**
- The weight \( W \) of the boy is given by the formula:
\[ W = m \cdot g \]
- Substituting the values:
\[ W = 50 \, \text{kg} \times 10 \, \text{m/s}^2 = 500 \, \text{N} \]

3. **Calculate the Apparent Weight in the Accelerating Lift:**
- When the lift accelerates upwards, the apparent weight \( W' \) can be calculated using the formula:
\[ W' = m \cdot (g + a) \]
- Substituting the values:
\[ W' = 50 \, \text{kg} \times (10 \, \text{m/s}^2 + 2 \, \text{m/s}^2) = 50 \, \text{kg} \times 12 \, \text{m/s}^2 \]
- Calculating the result:
\[ W' = 50 \times 12 = 600 \, \text{N} \]

4. **Conclusion:**
- The reading of the spring balance, which represents the apparent weight of the boy when the lift is accelerating upwards, is \( 600 \, \text{N} \).

### Final Answer:
The reading of the spring balance is **600 N**.