question_answer

Water from a tap emerges vertically downwards with an initial speed of \[1.0m{{s}^{-1}}.\]The cross-sectional area of the tap is \[{{10}^{-4}}{{m}^{2}}.\] 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 m below the tap would be: (Take \[g=10m{{s}^{-2}}\]) [JEE Main 10-4-2019 Afternoon]

A) s  \[1\times {{10}^{5}}{{m}^{2}}\]      

B)                                                          \[5\times {{10}^{5}}{{m}^{2}}\]

C)  \[2\times {{10}^{5}}{{m}^{2}}\]      

D)       \[5\times {{10}^{4}}{{m}^{2}}\]

Correct Answer: B

Solution :

[b]

\[{{A}_{1}}{{\text{v}}_{1}}={{A}_{2}}{{\text{v}}_{2}}\]

\[{{10}^{-4}}\times 1={{A}_{2}}{{\text{v}}_{2}}\]

\[{{A}_{2}}{{\text{v}}_{2}}={{10}^{-4}}\]                                        .(1)

\[P+\frac{1}{2}\rho (v_{1}^{2}-v_{2}^{2})+\rho gh=P\]

\[v_{2}^{2}=v_{1}^{2}+2gh\]

\[v_{2}^{{}}=\sqrt{\text{v}_{1}^{2}+2gh}\]

\[=\sqrt{1+2\times 10\times 0.15}\]

\[\frac{{{10}^{-4}}}{{{A}_{2}}}=2\]

\[{{A}_{2}}=5\times {{10}^{-5}}{{m}^{2}}\]