Tap Water

- Why do you think tap water has a conical shape when falling down?

- Why does water in sinks whirl in certain direction in the northern hemisphere and in the opposite direction in the southern hemisphere?

Surface tension.

Coriolis Effect.

You mean the drops ?

The are ellipsoidal. After some time, they become spherical, but they first oscilate, from the instant they were separated from bulk water.

If you refer to the tear form, forget it. It's just false. And this form isn't very good for the drag. Just another common misconception.

Hemisphere: it has no influence on the direction. Other factors, like any tiny rotation of the water, is much more important.

Maybe you could find a book that is less false?

What other shape would you like

basically any liquid (or anything if it could) forms a shape with the least surface area...

ie a sphere.. or when dropping a cone or tube

If you look careful at falling water
you will note it gains a curl... a twist as it falls.

Interesting... IMO gravity is twisted !

I think the OP means when you run water in a kitchen sink, the stream is fatter at the top than the bottom.

I would say it's because of acceleration of the water and surface tension.
The water is falling faster the farther it falls from the tap and the surface tension keeps it as a stream for a while.

I would like to agree with buttershug's interpretation and answer, and add a few calculations. Assuming that the water falls more-or-less freely (internal viscous drag is probably very small compared to the force of gravity for a stream of water trickling from a faucet), then the velocity of the water is given by the kinematic equation v^2 - (v_0)^2 = 2gy, where a is 9.81m/s^2 and y is the distance below the faucet. Now since water is incompressible, the mass flow rate m' is proportional to the volume flow rate Q. But m' is constant, since the quantity of water that lands in the sink each second is obviously the same as the quantity that leaves the tap (assuming you leave the faucet handle in the same position). Therefore, Q is constant.

Now Q = area*velocity = Av, so v = Q/A. The surface tension serves to keep the stream compact, so it doesn't immediately spray out into droplets, so the area remains circular. Thus v = Q/(pi*r^2). Putting this into the previous equation gives (Q/(pi*r^2))^2 - (Q/(pi*R^2))^2 = 2gy, where R is the radius of the stream just as it exits the faucet. Moving the Q and pi factors to the other side of the equation gives r^-4 - R^-4 = 2gy(pi/Q)^2. Just to put some numbers into this, let's suppose Q = pi cm^3/s (equivalent to 2 liters per minute, about 1/2 gallon per minute in American units), R = 0.2cm (about 1/6 inch diameter in American units), and y = 10cm (about 4 inches). Let's say g = 1000cm/s^2 approximately. Then r^-4 = 625 + 2*1000*10*(1)^2 = 20625, so r = 0.083cm, which means that the radius would be about 42% of its original size after this stream has fallen 10cm.

For these reasonable choices of faucet size, flow rate, and distance, that seems like about the amount of contraction I remember seeing.

Hope that helps!

--Stuart Anderson