A bent tube is lowered into a water stream as shown in figure. The velocity of the stream is `V_(1)`. The closed upper end is located at a height `h_(0) ` and has a small orifice. To what height `h` will the liquid get spurt?

Options

A. `(v^(2))/(2g)`
B. `(v^(2))/(2g)+h_(0)`
C. `(v^(2))/(2g)-h_(0)`
D. `(v^(2))/(4g)-h_(0)`

Answer

C

Explanation

Let tube's entrance be a depth y below the surface. Take point 1 at entry, 2 at the maximum height of the fountain.
Applying Bernoulli's theorem,
`p_(1)+rho gh_(1)+(1)/(2)rhov_(1)^(2)=p_(2)+rho gh_(2)+(1)/(2)rhov_(2)^(2)`
Taking, `h_(1)=0, h_(2)=(y+h_(0)+h),v_(1)=v,v_(2)=0`
substituting, `p_(0)+rho gy+(1)/(rho)v^(2)=p_(0)+rho g(y+h_(0)+h)`
`implies (1)/(2) rho v^(2)=rho g(h_(0)+h)`
`implies h=((v^(2))/(2g)-h_(0))`