Q 11-09-142JEE MainJEE Main 2019 (10 Apr, Shift 2)Easy
Water from a tap emerges vertically downwards with an initial speed of $1.0\ \text{m s}^{-1}$. The cross-sectional area of the tap is $10^{-4}\ \text{m}^2$. Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross-sectional area of the stream, $0.15\ \text{m}$ below the tap would be: (Take $g = 10\ \text{m s}^{-2}$)
Answer: (D) $5\times10^{-5}\ \text{m}^2$
With constant pressure the water falls freely: $v^2 = 1^2 + 2\times10\times0.15 = 4$, so $v = 2\ \text{m/s}$.
Continuity, $A_1v_1 = A_2v_2$:
$$A_2 = \frac{10^{-4}\times1}{2} = 5\times10^{-5}\ \text{m}^2$$
Solution by Sreeraj P, M.Sc Physics