Question

Water from a tap emerges vertically downwards with an initial velocity $V_0$.assuming pressure is constant throughout the stream of water and the flow is steady, find the distance from the tap at which cross-sectional area of stream is half of the cross-sectional area of stream at the tap:

• A. $V_0^2/2g$
• B. ${3V}_0^2/2g$
• C. ${2V}_0^2/g$
• D. ${5V}_0^2/2g$

Answer 1

Answer: (B)

${3V}_0^2/2g$

$\begin{array}{c}\text{Applying equation of Continuity}\\ \begin{array}{cc}A,V_1& =A_2{V}_2\\ A{V}_0& =\frac{A}{2}{V}^{\prime }\\ [V^{\prime }=2V_0]& \end{array}\end{array}$

$\text{Applying Bernoul's Theorn}$

$\begin{array}{c}\frac{1}{2}\rho v^2+fg{h}_1+P_k=\text{constant}\\ \frac{1}{2}g{v}_1^2+\rho g{h}_1+p_1=\frac{1}{2}\rho v_2^2+\rho g{h}_2+P_2\end{array}$

$\begin{array}{c}\frac{1}{2}\rho (v_1^2-v_2^2)+1g(h_1-h_2)+\left(p_1/p_2\right)=0\\ \frac{1}{2}\rho \left(3v_0^2\right)+\rho gh=0\\ \therefore h=\frac{3v_0^2}{2g}\end{array}$

Hence option B is correct