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  • Water from a tap emerges vertically down with an initial speed of 1.0m/s. The cross-sectional area of tap is <img alt="10^{-4}" src="https://entrancecorner.oncodecogs.com/gif.latex?10%5E%7B-4%

Water from a tap emerges vertically down with an initial speed of 1.0m/s. The cross-sectional area of tap is 10^{-4}m^2. Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-sectional area of the stream 0.15m below the tap is-

Option: 1

5.0\times10^{-4}m^2


Option: 2

1.0\times10^{-5}m^2


Option: 3

5.0\times10^{-5}m^2


Option: 4

2.0\times10^{-5}m^2


Answers (1)

best_answer

Decrease in P.E = Increase in K.E

\begin{gathered} \rho g h=\frac{1}{2} \rho\left(v_f^2-v_i^2\right) \\\\ 2(10)(0.15)=v_f^2-(1.0)^2 \\\\ v_f=2 \mathrm{~m} / \mathrm{s} \end{gathered}

Now from continuity equation, 

A_1 V_1=A_2 V_2 \text { or } A \propto \frac{1}{V}

velocity has become two times. Hence, area of cross section will remain half.

Posted by

SANGALDEEP SINGH

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