A bent tube is lowered into a water stream as shown in figure. 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. To what height `h` will the water jet spurt?

Explanation

Let the velocity of the water jet, near the orifice be `v^'`, then applying Bernoulli's theorem,
`1/2rhov^2=h_0rhog+1/2rhov^2`
or, `v^'=sqrt(v^2-2gh_0)` (1)
Here the pressure term on both sides is the same and equal to atmospheric pressure. (In the problem book figure should be more clear.)
Now, if it rised upto a height h, then at this height, whole of its kinetic-energy will be converted into potential energy. So,
`1/2rhov^('^2)=rhogh` or `h=(v^('^2))/(2g)`
`=(v^2)/(2g)-h_0=20cm`, [using Eq. (1)]