A person standing on a spring balance inside a stationary lift measures 60 kg. The weight of that person if the lift descends with uniform downward acceleration of `1.8 m//s^(2)` will be_N.
`[g=10m//s^2]`

Answer

To solve the problem, we need to determine the weight of a person standing on a spring balance inside a lift that is descending with a uniform downward acceleration of \(1.8 \, \text{m/s}^2\). The mass of the person is given as \(60 \, \text{kg}\) and the acceleration due to gravity \(g\) is \(10 \, \text{m/s}^2\).

### Step-by-Step Solution:

1. **Identify Forces Acting on the Person**:
- The gravitational force acting on the person (weight) is given by:
\[ F_g = m \cdot g \]
where \(m = 60 \, \text{kg}\) and \(g = 10 \, \text{m/s}^2\).

2. **Calculate the Gravitational Force**:
\[ F_g = 60 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 600 \, \text{N} \]

3. **Consider the Lift's Acceleration**:
- The lift is accelerating downward with \(a = 1.8 \, \text{m/s}^2\). In the frame of the lift, a pseudo force acts upward on the person due to this acceleration.

4. **Calculate the Pseudo Force**:
- The pseudo force \(F_{ps}\) can be calculated as:
\[ F_{ps} = m \cdot a = 60 \, \text{kg} \cdot 1.8 \, \text{m/s}^2 = 108 \, \text{N} \]

5. **Determine the Net Force**:
- In the lift's frame, the normal force \(N\) (the reading on the spring balance) can be found using the equation:
\[ N + F_{ps} = F_g \]
- Rearranging gives:
\[ N = F_g - F_{ps} \]

6. **Substitute Values**:
\[ N = 600 \, \text{N} - 108 \, \text{N} = 492 \, \text{N} \]

7. **Conclusion**:
- The weight of the person as measured by the spring balance when the lift descends with a uniform acceleration of \(1.8 \, \text{m/s}^2\) is \(492 \, \text{N}\).

### Final Answer:
The weight of the person when the lift descends with a uniform downward acceleration of \(1.8 \, \text{m/s}^2\) will be \(492 \, \text{N}\).