A body stands on a weighing machine inside a lift. When the lift is going down wit acceleration `g//4`, the machine shows a reading 30 kg. When the lift goes upwards with acceleration `g//4`, the reading would be

Options

A. 18 k
B. 37.5 kg
C. 50 kg
D. 67.5 kg

Explanation

To solve the problem, we need to analyze the forces acting on the body when the lift is accelerating downwards and upwards.

### Step-by-Step Solution:

1. **Understanding the Situation**:
- The body is standing on a weighing machine inside a lift.
- When the lift accelerates downwards with an acceleration of \( \frac{g}{4} \), the weighing machine shows a reading of 30 kg.

2. **Calculating the Weight Reading**:
- The reading on the weighing machine (normal force \( N \)) when the lift is going down can be expressed as:
\[ N = mg - ma \]
- Here, \( a = \frac{g}{4} \). Therefore, substituting \( a \):
\[ N = mg - m\left(\frac{g}{4}\right) = mg - \frac{mg}{4} = mg\left(1 - \frac{1}{4}\right) = mg\left(\frac{3}{4}\right) \]
- Given that the reading is 30 kg, we convert this to Newtons (using \( g \approx 10 \, \text{m/s}^2 \)):
\[ N = 30 \times 10 = 300 \, \text{N} \]
- Thus, we have:
\[ mg\left(\frac{3}{4}\right) = 300 \quad \Rightarrow \quad mg = 300 \times \frac{4}{3} = 400 \, \text{N} \]

3. **Finding the Mass**:
- The mass \( m \) can be calculated as:
\[ m = \frac{mg}{g} = \frac{400}{10} = 40 \, \text{kg} \]

4. **Calculating the Reading When the Lift Goes Up**:
- Now, when the lift accelerates upwards with the same acceleration \( \frac{g}{4} \), the normal force \( N' \) can be expressed as:
\[ N' = mg + ma \]
- Substituting \( a = \frac{g}{4} \):
\[ N' = mg + m\left(\frac{g}{4}\right) = mg + \frac{mg}{4} = mg\left(1 + \frac{1}{4}\right) = mg\left(\frac{5}{4}\right) \]
- We already found \( mg = 400 \, \text{N} \). Therefore:
\[ N' = 400 \times \frac{5}{4} = 500 \, \text{N} \]

5. **Converting the Normal Force Back to kg**:
- Finally, to find the reading on the weighing machine in kg:
\[ \text{Reading in kg} = \frac{N'}{g} = \frac{500}{10} = 50 \, \text{kg} \]

### Final Answer:
The reading on the weighing machine when the lift goes upwards with an acceleration of \( \frac{g}{4} \) will be **50 kg**.