Two stones of mass `m_1 and m_2` (such that `m_1 gt m_2`) are dropped `Deltat` time apart from the same height towards the ground. At a later time t the difference in their speed is `DeltaV` and their mutual separation is `DeltaS`. While both stones are in flight

Options

A. `DeltaV` decreases with time and `DeltaS` increases with time
B. Both `DeltaV` and `DeltaS` increase with time
C. `DeltaV` remains constant with time and `DeltaS` decreases with time
D. `DeltaV` remains constant with time and `DeltaS` increases with time

Explanation

To solve the problem, we need to find the expressions for the difference in speed (\(\Delta V\)) and the mutual separation (\(\Delta S\)) of the two stones dropped from the same height at different times.

### Step-by-step Solution:

1. **Understanding the Setup**:
- Let the mass of the first stone be \(m_1\) and the second stone be \(m_2\) such that \(m_1 > m_2\).
- The first stone is dropped at time \(t = 0\).
- The second stone is dropped after a time interval \(\Delta t\).
- We are interested in the time \(t\) after the second stone is dropped.

2. **Time Calculation**:
- The time \(t_1\) for which the first stone has been falling when we start observing is:
\[ t_1 = t + \Delta t \]
- The time \(t_2\) for which the second stone has been falling is:
\[ t_2 = t \]

3. **Velocity Calculation**:
- The velocity of the first stone (\(v_1\)) after time \(t_1\) is given by:
\[ v_1 = g t_1 = g(t + \Delta t) \]
- The velocity of the second stone (\(v_2\)) after time \(t_2\) is:
\[ v_2 = g t_2 = g t \]
- The difference in their speeds (\(\Delta V\)) is:
\[ \Delta V = v_1 - v_2 = g(t + \Delta t) - gt = g \Delta t \]

4. **Separation Calculation**:
- The distance fallen by the first stone (\(S_1\)) is:
\[ S_1 = \frac{1}{2} g t_1^2 = \frac{1}{2} g (t + \Delta t)^2 \]
- The distance fallen by the second stone (\(S_2\)) is:
\[ S_2 = \frac{1}{2} g t_2^2 = \frac{1}{2} g t^2 \]
- The mutual separation (\(\Delta S\)) between the two stones is:
\[ \Delta S = S_1 - S_2 = \frac{1}{2} g \left( (t + \Delta t)^2 - t^2 \right) \]
- Expanding the square:
\[ (t + \Delta t)^2 = t^2 + 2t \Delta t + (\Delta t)^2 \]
- Therefore,
\[ \Delta S = \frac{1}{2} g \left( 2t \Delta t + (\Delta t)^2 \right) \]

5. **Final Expressions**:
- The difference in speed is:
\[ \Delta V = g \Delta t \]
- The mutual separation is:
\[ \Delta S = \frac{1}{2} g (2t \Delta t + (\Delta t)^2) \]