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 top:

• A. $V_0^2/2g$
• B. $3V_0^2/2g$
• C. $2V_0^2/g$
• D. $5V_0^2/2g$

Answer 1

Answer: (B)

$3V_0^2/2g$

velocity equation $V_2^2=V_0^2+2gh$ and continuity equation $A_1{V}_0=A_2{V}_2$ .....(i)
Solving, we get $\frac{A_2}{A_1}=\frac{V_0}{\sqrt{V_0^2+2gh}}$
$A_2/A_1=\frac{1}{2}=\frac{V_0}{\sqrt{V_0^2+2gh}}$
$4V_0^2=V_0^2+2gh$
$h=\frac{3V_0^2}{2g}$