Water emerges straight down from a faucet with a 2.51-cm diameter at a speed of 3.04 m/s. (because of the construction of the faucet, there is no variation in speed across the stream.) what is the diameter of the stream 0.200 m below the faucet? neglect any effects due to surface tension.

This is a question on conservation of energy. That is,

mgh + KE1 = KE2

mgh + 1/2mv1^2 = 1/2mv2^2

gh + 1/2v1^2 = 1/2v2^2

Where, h = 0.2 m, v1 = 3.04 m/s

Therefore,

v2 = Sqrt [2 (gh+1/2v1^2) ] = Sqrt [2 (9.81*0.2 + 1/2*3.04^2) ] = 7.26 m/s

Now, Volumetric flow rate, V/time, t = Surface area, A*velocity, v

Where,

V = Av = πD^2/4*3.04 = π * (2.51/100) ^2*1/4*3.04 = 1.504*10^-3 m^3/s

At 0.2 m below,

V = 1.504*10^-3 m^3/s = A*7.26

A = (1.504*10^-3) / 7.26 = 2.072*10^-4 m^2

But, A = πr^2

Then,

r = Sqrt (A/π) = Sqrt (2.072*10^-4/π) = 0.121*10^-3 m

Diameter = 2r = 0.0162 m = 1.62 cm