When a thin tube is put into water, the water will climb up some distance in the

Question

When a thin tube is put into water, the water will climb up

some distance in the tube as shown in the figure at the right. This is because

the water sticks to the glass and the surface tension of the water stuck to the

glass pulls the water upward until the surface tension force balances the

force of gravity pulling the extra water down. We therefore expect that the

height, h, that the water rises in the tube depends on the surface tension, ,

the radius of the tube, R, the mass density of the water, , and the

graviational field, g. [The surface tension parameter, , is defined so that

along a line of length L on a liquid surface, the force by which each part of

the surface pulls on the surface on the other side of the line is Fst= L.]

If we replaced water with a denser liquid but everything else was the same, would you

expect that the height of the liquid in the tube would be greater, the same, or less?

If we did the experiment using a liquid with a greater surface tension, but everything else

was the same, would you expect that the height of the liquid in the tube would be greater, the same, or less?

Use dimensional analysis and your reasoning for the previous parts of the question to

propose an equation for h in terms of , R, , and g. (There are dimensionless geometric factors, but we will ignore them for now.) Show your reasoning.

A dolphin can adjust its volume so that its buoyant force exactly balances its weight (neutral buoyancy). It can then easily swim up and down without either having to fight gravity or buoyancy with its muscles. Calculate the volume a 275 kg dolphin must adjust itself to in order to be neutrally buoyant in seawater at 5 oC (density 1.03 g/cm3). If you need g, take it to be 9.8 N/kg.

Explanation / Answer

Due to surface tension, the liquid rises in the tube in such a way so that the net force due to surface tension will balance the weight of risen liquid so if

* Density of the liquid is increased then it's weight for the same volume will increase and hence net force in download direction will increase so height of liquid in the tube will decrease and hence we can say height of liquid risen will be inversely proportional to the density. h1/

* Surface tension of the liquid is increased then net upwards force will increase and to balance this force liquid will rise more in the tube so we can say height of liquid rises in the tube will be directly proportional to the surface tension
h

Now we can say h aRbcgd
(Since this relation is having 4 parameters on RHS so relation can't be developed using dimensional analysis, while solving it will have 4 unknown a,b,c and d and three equations)
The actual relation is
h=k(/Rg)
Where k is a constant.

Mass of dolphin (m) = 275 kg
Let volume of dolphin be V m³
Density of sea water () = 1.03g/cm³.
So weight of dolphin = buoyant force
mg=Vg
m = V
V= m/
V= 275/(1.03*103)
V= 0.26699 m³

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