A liquid flowing from a vertical pipe has a very definite shape as it flows from the pipe. To get the equation for this shapes assume that the liquid is falling under gravity once it leaves th pipe. Just as it leaves the pipe, the liquid has speed `V_(0)` and the radius of the stream of liquid is `r_(0)`
Combining the above result with the equation of continuity the expression for the radius of the stream as a function of ..y..is

Options

A. `r=(r_(0)sqrt(V_(0)))/((V_(0)^(2)+2gy)^(1//2))`
B. `r=(r_(0)sqrt(V_(0)))/((V_(0)^(2)+2gy)^(1//4))`
C. `r=(V_(0)^(2)+2gy)^(1//2)`
D. `r=(V_(0)^(2)+2gy)^(1//4)`

Answer

B