(c)

Let first stone mass $m_1$ is dropped at instant $t=0$.

Then at time $t$, its velocity and displacement respectively, are

$v_1=-g t \text { and } s_1=-\frac{1}{2} g t^2$

As, second stone mass $m_2$ is dropped $\Delta t$ time after, so its velocity and displacement at instant $t$ respectively, are

$\quad \quad v_2 =-g(t-\Delta t)$

$\text { and }$ $s_2 =-\frac{1}{2}(g)(t-\Delta t)^2$

Difference in speeds of stones is

$\Delta v=v_1-v_2$

$=(-g t)-(-g(t-\Delta t))$

$=-g t+g t-g \Delta t=-g \Delta t$

As both $g$ and $\Delta t$ are constants.

$\therefore \Delta v$ is constant and its value does not changes with time $t$.

The mutual separation $\Delta s$ of the stones is

$\Delta s=s_1-s_2$

$=-\frac{1}{2} g t^2-\left(-\frac{1}{2} g(t-\Delta t)^2\right)$

$=\frac{1}{2} g\left((t-\Delta t)^2-t^2\right)$

$=\frac{1}{2} g\left(t^2+\Delta t^2-2 t \Delta t-t^2\right)$

$=\frac{1}{2} g\left(-2 t \Delta t+\Delta t^2\right)$

$\Rightarrow \quad \Delta s=\frac{1}{2} g\left(\Delta t^2-2 t \Delta t\right)$

Clearly, $\Delta s$ decreases with time and becomes zero when $2 t=\Delta t$.