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

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 :

\[{{A}_{1}}{{\text{v}}_{1}}={{A}_{2}}{{\text{v}}_{2}}\] \[{{10}^{-4}}\times 1={{A}_{2}}{{\text{v}}_{2}}\] \[{{A}_{2}}{{\text{v}}_{2}}={{10}^{-4}}\]                                           ?.(1) \[P+\frac{1}{2}\rho (v_{1}^{2}-v_{2}^{2})+\rho gh=P\] \[v_{2}^{2}=v_{1}^{2}+2gh\] \[v_{2}^{{}}=\sqrt{\text{v}_{1}^{2}+2gh}\] \[=\sqrt{1+2\times 10\times 0.15}\] \[\frac{{{10}^{-4}}}{{{A}_{2}}}=2\] \[{{A}_{2}}=5\times {{10}^{-5}}{{m}^{2}}\]