2024 A/L Physics Past Paper · Question 21 (English Medium)

Answer: (4) $\dfrac{1}{2\pi}\ \text{N m}^{-1}$

Given information

This question is about pulling a thin ring up out of a liquid.

The ring lies just under the surface. When we lift it, the liquid surface clings to the ring and pulls it back down. Just before the film breaks, this pull is as big as it can be.

What we must find: the surface tension of the liquid.

How we find it: surface tension is the pull on each metre of edge. Find the total edge the film pulls along, then divide the force by it.

Radius of the ring $r=2\ \text{cm}=0.02\ \text{m}$. The ring is light, so its weight is ignored. The force just before the film breaks is $F=0.04\ \text{N}$.

What the question asks

$r$ ($\text{m}$): radius of the ring.

$l$ ($\text{m}$): total length of edge the film pulls along.

$F$ ($\text{N}$): force needed to pull the ring out.

$T$ ($\text{N m}^{-1}$): surface tension of the liquid.

Relevant theory

The surface tension of a liquid is defined as the force acting along the surface of the liquid

on a unit length of an imaginary line drawn on the surface of the liquid, normal to and

towards one side of the line.

What is happening: as the ring rises, it lifts a thin film of liquid with it.

The film has two faces, one inside the ring and one outside. Each face pulls down along the whole ring.

So the film pulls along twice the circumference: $l=2\times2\pi r$.

For example, dip a bubble wand into soapy water and lift it out slowly.

A shiny film stretches across the ring and tugs on it. You can feel the wand pulled back for a moment.

Then the film snaps and the pull is gone.

In words: the liquid skin holds on all the way round the edge, and the longer the edge, the bigger the pull.

Required equations

$T=\dfrac{F}{l}$

$l=2\times2\pi r$

$T=\dfrac{F}{4\pi r}$

Applying it to this question

$0.04\pi$

$2\pi r=2\pi\times0.02$

$0.08\pi$

$l=2\times0.04\pi$

$T=\dfrac{0.04}{0.08\pi}$

$\dfrac{1}{2\pi}$

$T$

Reasoning and calculation

The film pulls along $0.08\pi\ \text{m}$ of edge with $0.04\ \text{N}$, so each metre carries $\dfrac{1}{2\pi}\ \text{N}$. This is option (4).

$4\ \text{N m}^{-1}$ comes from $0.04/0.01$. That divides the force by a number that is not the length of the edge. The film pulls along the whole edge, $0.08\pi\ \text{m}$.

$2\ \text{N m}^{-1}$ comes from $0.04/0.02$, the force divided by the radius. Surface tension is force per length of edge, and the edge is the circumference, not the radius.

$\dfrac{1}{\pi}\ \text{N m}^{-1}$ uses only one side: $0.04/(2\pi\times0.02)$. The film touches the ring on the inside and on the outside, so the length is doubled.

$\dfrac{1}{4\pi}\ \text{N m}^{-1}$ uses the diameter, $4\ \text{cm}$, in place of the radius. The radius is $2\ \text{cm}$.

Checks

$F/(4*pi*r)$

$F=T\times4\pi r=\dfrac{1}{2\pi}\times4\pi\times0.02$

$F=0.04\ \text{N}$

$F/l$ is $\text{N}$ divided by $\text{m}$, so $T$ comes out in $\text{N m}^{-1}$, the unit of surface tension.

$\dfrac{1}{2\pi}\approx0.16\ \text{N m}^{-1}$. That is a little larger than for water, about $0.07\ \text{N m}^{-1}$, which is a fair value for a liquid.

Conclusion

$T=\dfrac{1}{2\pi}\ \text{N m}^{-1}$

The surface tension of the liquid is $\dfrac{1}{2\pi}\ \text{N m}^{-1}$. The answer is (4).