Q 11-02-033NEETJEE MainMedium
Water drops fall from a tap at regular intervals onto the floor $4.5$ m below. The first drop strikes the floor just as the fourth drop begins to fall. At that instant, the height of the second drop above the floor is ($g = 10\ \text{m/s}^2$)
Answer: (B) $2.5$ m
Concept. With interval $\tau$, the first drop has fallen for $3\tau$ and the second for $2\tau$; distance fallen $\propto t^2$.
Step 1. First drop. $4.5 = \dfrac12g(3\tau)^2$.
Step 2. Second drop.
$$\frac12g(2\tau)^2 = \frac49 \times 4.5 = 2\ \text{m below the tap}$$
Step 3. Height above the floor. $4.5 - 2 = 2.5$ m.
Answer. $2.5$ m (Option B).
Solution by Sreeraj P, M.Sc Physics