Question 885774
Disclaimer/apology:
I apologize to you in advance for possible mistakes. I try to keep all my units straight, and calculate correctly, but I am prone to punching in the wrong numbers in the computer and calculator/
I also apologize to professor Walter Lewin.
My grasp of physics was never strong enough for my taste, even after paying attention at some (though not all) of his lectures.
If you do not like my answer ask again.
If this website will not produce a suitable answer, maybe the forum at artofproblemsolving.com will (they have Physics sections).

The flow rate in {{{cm^3/second}}} is
{{{125/16.3}}} .
Assuming a perfectly circular cross section, with {{{diameter=0.96cm}}}-->{{{radius=0.48cm}}} ,
the cross section of the stream at the faucet (in {{{cm^2}}} ) is
{{{pi*0.48^2}}}
The linear velocity of the water at the faucet (in cm/second) is
{{{((125/16.3))/(pi*0.48^2)=about 10.6}}} .
Under no other force but gravity, that linear velocity would increase with time.
Let {{{t}}}= time in seconds from the moment the water leaves the faucet.
{{{v(t)}}}= linear velocity of the water (in cm/second) at {{{t}}} seconds.
{{{d(t)}}}= distance between the water and the faucet (in cm) at {{{t}}} seconds.
Let's take the acceleration of gravity as {{{g=9.8}}}{{{m/second^2=980}}}{{{cm/second^2}}} .
Then, {{{v(t)=10.6+980t}}} and {{{d(t)=10.6t+980t^2/2=10.6t+490t^2}}}
When {{{d(t)=13}}} {{{10.6t+490t^2=13}}}<--->{{{10.6t+490t^2-13=0}}}
That quadratic equation would have two solutions, but we are only interested in the positive one.
So we calculate {{{t=(-10.6 +- sqrt(10.6^2-4*490*(-13)))/(2*490)}}}= approximately{{{149.4/980=about0.152}}} seconds.
Then, at that time, water has traveled 13 cm from the faucet and it is moving at a velocity (in cm/s) of
{{{v(0.1542)=10.6+980t=10.6+149.4=160}}}
The flow rate {{{cm^3/second}}} at that point (13 cm from the faucet) is
the same as before, and is equal to
the the linear velocity times the cross section of the stream at that point.
From the flow rate, we could calculate the cross section radius and diameter.
However, it is easier to calculate the diameter knowing that the linear velocity is inversely proportional to the square of the diameter.
The linear velocity increased by a factor of {{{160/10.6=about15.09}}} ,
so the diameter must have decreased by a factor of {{{sqrt(10.09)=about3.9}}}.
So the diameter would be {{{0.96cm/3.9=0.25cm}}} .
The calculation may not quite agree with reality because other physical influences and phenomena may be at work.