Question

Consider a bent tube lowered slightly into a water stream as shown in figure. The velocity of stream relative to the tube is equal to $v=5{\text{ms}}^{-1}$. The closed upper end of the tube located at the height $h_0=10\text{cm}$ has a small orifice. The height ${}^{\prime }{h}^{\prime }$ to which water jet spurts (in $m$) is

Answer 1

Let $v_1$ be the velocity of water jet coming out of orifice.
Applying conservation of energy principle,

$(\text{K.E}{)}_{\text{lower end}}=\text{Pressure energy}+K.E_{orifice}$
$\Rightarrow \frac{1}{2}\rho v^2=h_0\rho g+\frac{1}{2}\rho v_1^2$
We get, $v_1=\sqrt{v^2-2g{h}_0}$
The kinetic energy at the opening end is converted into potential energy at topmost point. Thus,
$\frac{1}{2}\rho v_1^2=\rho gh\Rightarrow v_1^2=2gh$
From the equations above, we get,
$v^2-2g{h}_0=2gh$
$\Rightarrow h=\frac{v^2}{2g}-h_0=\frac{(5{)}^2}{2\left(10\right)}-0.1$
$\Rightarrow h=\frac{5}{4}-0.1=1.25-0.1=1.15\text{m}$