2018 A/L Physics Past Paper · Question 44 (English Medium)

Capillary tube — the height of the water column with an oil layer

When a long capillary tube of uniform bore radius is dipped vertically in a beaker of water of density $d_w$ , the water column in the capillary tube rises to a height $h_0$ as shown in figure (a). Now, an oil of density $d_0\,(< d_w)$ is poured onto the surface of the water in the beaker slowly without disturbing the water as shown in figure (b). Assume that the oil and water are immiscible liquids. The variation of height $H$ of the water column inside the capillary tube, measured from the water surface, with the height $h$ of the oil layer is best represented in

(1) Figure (1)
(2) Figure (2)
(3) Figure (3)
(4) Figure (4)
(5) Figure (5)

Answer and explanation

Answer: (4) Figure (4)

What the question asks

We want: how the height $H$ of the water column in the tube changes as the oil layer $h$ gets thicker.

Relevant theory

When a clean glass capillary tube is placed in a liquid of which the angle of contact is an acute angle, a liquid column rises in the glass tube and stands ata level higher than that of the outside liquid level.

Consider spherical air bubbles in a liquid and the equilibrium of one half of it.

In the tube, just below the water meniscus, the pressure is $\frac{2T}{r}$ lower than atmospheric pressure.

That is why the water rises to $h_0$ .

$h_0 d_w g = \frac{2T}{r}$

Required equations

$h_0 d_w g = \frac{2T}{r}$

$P_0 + h d_0 g = P_0 - \frac{2T}{r} + H d_w g$

$H = h_0 + \frac{d_0}{d_w}\,h$

Reasoning and calculation

The axes.

Across: the height $h$ of the oil layer.

Up: the height $H$ of the water column above the water surface.

At the start there is no oil. So $h = 0$ .

Then the water stands at its capillary rise: $H = h_0$ .

As long as the meniscus is a water–air surface above the oil, the behaviour is the same.

So there is only one part.

At the level of the water surface, the pressure outside is $P_0 + h d_0 g$ .

To match it inside the tube, the water column must rise more.

It rises by a further $\frac{d_0}{d_w}h$ .

$H d_w g = \frac{2T}{r} + h d_0 g$

$H d_w g = h_0 d_w g + h d_0 g$

$H = h_0 + \frac{d_0}{d_w}h$

The shape: a straight line.

It starts at $h_0$ .

Its slope is $\frac{d_0}{d_w}$ . It is positive and less than 1.

Reading the options

The graph that rises in a straight line from $h_0$ is figure (4).

Why the other options fail

This option says: figure (1). $H$ stays the same.

But the oil presses on the water surface. So $H$ grows.

So it is wrong.

This option says: figure (2). The curve rises and then levels off.

But each mm of oil adds the same rise. So it is a straight line.

This option says: figure (3). The curve rises faster and faster.

But the slope $\frac{d_0}{d_w}$ is constant. So it is a straight line.

This option says: figure (5). $H$ falls later.

But more oil means more pressure. So $H$ never falls.

Conclusion

$H = h_0 + \frac{d_0}{d_w}\,h$

$H = h_0 + \frac{d_0}{d_w}h$

It grows in a straight line from $h_0$ .

The answer is option (4).

This result holds while the meniscus is above the oil.

$H$ grows more slowly than $h$ , because $\frac{d_0}{d_w} < 1$ .

With a very thick oil layer the meniscus could go inside the oil. The question does not look at that case.

A note on the limit

Explore

The water rises $h_0$ up the capillary tube.

Oil is poured slowly onto the water. It does not mix with the water.

The oil presses on the water surface. The water in the tube rises higher.

Each mm of oil adds the same rise. So $H$ grows in a straight line from $h_0$ . That is figure (4).

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