A vertical water jet flows out of a round hole. One of the horizontal sections of the jet has the diameter `d = 2.0 mm` while the other section section located `l = 20 mm` lower has the diameter which is `n = 1.5` times less. Find the volume of the water flowing from the hole each second.

Explanation

From the equation of continuity
`(pi)/(4) d^2. v = (pi)/(4)((d)/(n))^2 . V` or `V = n^2 v`
We then apply Bernoulli's theorem
`(p)/(rho) + (1)/(2) v^2 + Phi =` constant
The pressure `p` differs from the atmospheric pressure by capillary effects. At the upper section
`p = p_0 + (2 alpha)/(d)`
neglecting the curvature in the vertical plane. Thus,
`(p_0 + (2 alpha)/(d))/(rho) + (1)/(2) v^2 +g l = = (p_0 + (2 n alpha)/(d))/(rho) +(1)/(2) n^4 v^2`
or `v = sqrt((2 gl - (4 alpha)/(pd)(n -1))/(n^4 - 1))`
Finally, the liquid coming out per second is,
`V = (1)/(4) pi d^2 sqrt((2 gl - (4 alpha)/(pd)(n -1))/(n^4 - 1))`.