STATEMENT-1: Two balls are dropped one after the other form a tall tower. The distance between them increases linearly with time (elapsed after the second ball is dropped and before the first hits ground).
STATEMENT-2: Relative acceleration is zero, whereas relative velocity non-zero in the above situation.

Options

A. Statement -1 is True, Statement-2 is Ture, Statement-2 is a correct explanantion for Statement-1.
B. Statement-1 is Ture, Statement-2 is Ture, Statement-2 is Not a correct explanantion for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Answer

To solve the problem, we need to analyze the two statements provided and determine their validity based on the principles of kinematics.

### Step-by-Step Solution:

1. **Understanding the Scenario**:
- Two balls are dropped from a tower: the first ball is dropped at time \( t = 0 \) and the second ball is dropped at time \( t = t_0 \) (where \( t_0 \) is some positive time after the first ball).
- We need to analyze the motion of both balls and the distance between them after the second ball is dropped until the first ball hits the ground.

2. **Motion of the First Ball**:
- The first ball is in free fall with an initial velocity \( u_1 = 0 \) and an acceleration \( a_1 = g \) (acceleration due to gravity).
- The distance fallen by the first ball after \( t_1 \) seconds (where \( t_1 \) is the time since it was dropped) is given by:
\[ y_1 = \frac{1}{2} g t_1^2 \]

3. **Motion of the Second Ball**:
- The second ball is also in free fall with an initial velocity \( u_2 = 0 \) and an acceleration \( a_2 = g \).
- The time of fall for the second ball after it is dropped is \( t_2 = t - t_0 \) (where \( t \) is the total time since the first ball was dropped).
- The distance fallen by the second ball is:
\[ y_2 = \frac{1}{2} g (t - t_0)^2 \]

4. **Distance Between the Two Balls**:
- The distance \( D \) between the two balls after the second ball is dropped can be expressed as:
\[ D = y_1 - y_2 = \frac{1}{2} g t_1^2 - \frac{1}{2} g (t - t_0)^2 \]
- Since \( t_1 = t - t_0 \), we can substitute this into the equation:
\[ D = \frac{1}{2} g (t - t_0)^2 - \frac{1}{2} g (t - t_0)^2 = \frac{1}{2} g (t_0)(2(t - t_0)) = g(t_0)(t - t_0) \]
- This shows that the distance \( D \) increases linearly with time \( t \).

5. **Relative Acceleration and Velocity**:
- Both balls experience the same acceleration \( g \) downwards. Therefore, their relative acceleration is:
\[ a_{rel} = a_1 - a_2 = g - g = 0 \]
- The relative velocity \( v_{rel} \) is non-zero because the first ball has a velocity that increases with time, while the second ball starts from rest:
\[ v_{rel} = v_1 - v_2 = g t_1 - 0 = g t_1 \]

6. **Conclusion**:
- **Statement 1** is true: The distance between the two balls increases linearly with time.
- **Statement 2** is also true: The relative acceleration is zero, while the relative velocity is non-zero.

### Final Answer:
Both statements are correct, and Statement 2 is the correct explanation of Statement 1. Thus, the answer is **A**.

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