Q 11-02-049JEE MainJEE Main 2026 (28 Jan, Shift 1)Medium
Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from ground, at the instant when the first drop strikes the ground is ______ m.
$(g = 10\ \text{m/s}^2)$
Answer: (C) 4.2
Time for a drop to fall 5 m: $5 = \frac12(10)t^2 \Rightarrow t = 1$ s.
When drop 1 lands, drop 6 is just starting, so there are 5 equal gaps in 1 s: interval $= 0.2$ s.
At that instant drop 4 has been falling for $2\times0.2 = 0.4$ s:
$$h_{\text{fallen}} = \frac12(10)(0.4)^2 = 0.8\ \text{m}$$
Height above the ground $= 5 - 0.8 = 4.2$ m.
Solution by Sreeraj P, M.Sc Physics