Question

A bent tube is lowered into a water stream as shown in figure above. The velocity of the stream relative to the tube is equal to $V$. The closed upper end of the tube located at the height $h_0$.The height $h$ to which the water jet spurt up is______.

• A. $\frac{V^2}{2g}+h_0$
• B. $\frac{V^2}{2g}-h_0$
• C. $\frac{2V^2}{g}+h_0$
• D. $\frac{V^2}{2g}+\frac{h_0}{2}$

Answer 1

The correct option is B $\frac{V^2}{2g}-h_0$

Let the velocity of jet

$\frac{1}{2}p{v}^2=h_{o}pg+\frac{1}{2}p{v}^2$

$\Rightarrow V^1=\sqrt{V^2-2g{h}_o}\to \left(1\right)$

Now, $\frac{1}{2}p{V}^{\prime 2}=pgh$ or $h=\frac{V^{\prime 2}}{2g}$ (kinetic energy $\to P.E$)

$=\frac{V^2}{2g}-h_o=20cm$ [using (1)]