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
Let the interval between drops be $\tau$. When the $4^{\text{th}}$ drop starts, the $1^{\text{st}}$ has fallen for $3\tau$ and just reached the floor:
$$4.5 = \frac{1}{2}g(3\tau)^2$$
The $2^{\text{nd}}$ drop has fallen for $2\tau$:
$$h_2 = \frac{1}{2}g(2\tau)^2 = \frac{4}{9} \times 4.5 = 2\ \text{m below the tap}$$
Height above the floor $= 4.5 - 2 = 2.5$ m.
Solution by Sreeraj P, M.Sc Physics