A 60 kg person is weighed by a balance as 54 kg in a lift which is accelerated downwards. The acceleration of the lift is

Options

A. `1.26 ms^(-2)`
B. `1.76 ms^(-2)`
C. `1.98 ms^(-2)`
D. `0.98 ms^(-2)`

Explanation

To solve the problem, we need to determine the acceleration of the lift when a 60 kg person is weighed as 54 kg in the lift that is accelerating downwards.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the person, \( m = 60 \, \text{kg} \)
- Weight measured in the lift, \( m_{\text{effective}} = 54 \, \text{kg} \)
- Acceleration due to gravity, \( g \approx 9.81 \, \text{m/s}^2 \)

2. **Understand the Forces Acting on the Person:**
- When the lift accelerates downwards, the effective weight (apparent weight) of the person decreases. The effective weight can be expressed as:
\[ m_{\text{effective}} = m - F \]
where \( F \) is the force due to the downward acceleration of the lift.

3. **Write the Equation for Effective Weight:**
- The effective weight of the person in the lift can be given by:
\[ m_{\text{effective}} = m(g - a) \]
where \( a \) is the acceleration of the lift.

4. **Set Up the Equation:**
- Substitute the known values into the equation:
\[ 54 = 60(g - a) \]

5. **Substitute the Value of \( g \):**
- Using \( g \approx 9.81 \, \text{m/s}^2 \):
\[ 54 = 60(9.81 - a) \]

6. **Solve for \( a \):**
- Rearranging the equation gives:
\[ 54 = 588.6 - 60a \]
- Rearranging further:
\[ 60a = 588.6 - 54 \]
\[ 60a = 534.6 \]
\[ a = \frac{534.6}{60} \approx 8.91 \, \text{m/s}^2 \]

7. **Calculate the Acceleration:**
- Therefore, the acceleration of the lift is:
\[ a \approx 8.91 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of the lift is approximately \( 8.91 \, \text{m/s}^2 \) downwards.