A body of mass `32 kg` is suspended by a spring balance from the roof of vertically operating lift and going downwards from rest. At the instant the lift has covered 20 m and 50 m, the spiring balance showed 30 kg and 36 kg respectively. The velocity of the lift is

Options

A. decreasing at 20 m and increasing at 50 m
B. increasing at 20 m and decreasing at 50 m
C. continuously increasing at constant rate throughtout the journey
D. remaing constant throught the journey

Explanation

To solve the problem, we will analyze the forces acting on the body suspended in the lift at two different positions (20 m and 50 m) and use Newton's second law of motion to find the acceleration of the lift.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The weight of the body (W) acting downward is given by \( W = mg = 32 \, \text{kg} \times g \).
- The tension (T) in the spring balance acts upward.

2. **Write the Force Equation for the Lift**:
- According to Newton's second law, the net force acting on the body is equal to the mass times the acceleration of the body:
\[ W - T = ma \]
where \( W = 32g \) and \( m = 32 \, \text{kg} \).

3. **Case 1: Lift covers 20 m**:
- At this point, the spring balance shows 30 kg, which means:
\[ T = 30g \]
- Substitute into the force equation:
\[ 32g - 30g = 32a \]
- Simplifying gives:
\[ 2g = 32a \implies a = \frac{2g}{32} = \frac{g}{16} \]

4. **Case 2: Lift covers 50 m**:
- At this point, the spring balance shows 36 kg, which means:
\[ T = 36g \]
- Substitute into the force equation:
\[ 32g - 36g = 32a \]
- Simplifying gives:
\[ -4g = 32a \implies a = -\frac{4g}{32} = -\frac{g}{8} \]

5. **Determine the Velocity of the Lift**:
- The lift starts from rest and has covered 20 m and 50 m. We can use the kinematic equation to find the velocity at these points.
- Using the equation \( v^2 = u^2 + 2as \) where \( u = 0 \):
- For 20 m:
\[ v^2 = 0 + 2 \left(\frac{g}{16}\right)(20) = \frac{20g}{8} = \frac{5g}{2} \]
- For 50 m:
\[ v^2 = 0 + 2 \left(-\frac{g}{8}\right)(50) = -\frac{100g}{8} = -12.5g \quad \text{(This indicates a change in direction)} \]

6. **Conclusion**:
- The velocity of the lift at 20 m is \( v = \sqrt{\frac{5g}{2}} \) and at 50 m, it is negative indicating that the lift is decelerating.

### Final Answer:
The velocity of the lift is changing as it moves downwards, with specific values depending on the acceleration calculated at each distance.