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 $v=2.5m/s$. The closed upper end of the tube located at the height $h_0=12cm$ has a small orifice. The height $h$ the water jet spurt?

Answer 1

Let the velocity of the water jet, near the orifice be $v^{\prime }$, then applying Bernoulli's theorem,

$\frac{1}{2}\rho v^2=h_0\rho g+\frac{1}{2}\rho v^2$

or, $v^{\prime }=\sqrt{v^2-2g{h}_0}$ (1)

Here the pressure term on both sides is the same and equal to atmospheric pressure.

Now, if it rises upto a height $h$, then at this height, whole of its kinetic energy will be converted into potential energy. So,

$\frac{1}{2}\rho {v^{\prime }}^2=\rho gh$ or $h=\frac{{v^{\prime }}^2}{2g}$

$=\frac{v^2}{2g}-h_0=20cm$, [using equation (1)]