Q 11-02-115JEE MainJEE Main 2021 (27 Aug, Shift 2)Easy
Water drops are falling from a nozzle of a shower onto the floor from a height of $9.8$ m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.
Answer: (D) $7.35$ m
Time for a drop to fall $9.8$ m: $t = \sqrt{\dfrac{2\times9.8}{9.8}} = \sqrt{2}$ s.
The third drop starts as the first lands, so there are two intervals in $\sqrt{2}$ s: interval $= \dfrac{\sqrt{2}}{2}$ s. The second drop has fallen for this time:
$$s = \frac{1}{2}\times9.8\times\frac{1}{2} = 2.45\ \text{m}$$
Height above the floor $= 9.8 - 2.45 = 7.35$ m.
Solution by Sreeraj P, M.Sc Physics