2024 A/L Physics Past Paper · Question 20 (English Medium)

Answer: (3) Wire $C$ has the largest extension.

Given information

This question is about four wires that are pulled with the same force.

All four wires have the same length. They are made of two different materials, and they come in two thicknesses.

What we must find: which wire stretches the most.

How we find it: write the extension using Young’s modulus. Then see which wire has the smallest area times Young’s modulus.

The same tension $F$ and the same length $l$ for all. $A$: $Y=2.0$, $d=1.0$. $B$: $Y=2.0$, $d=2.0$. $C$: $Y=1.0$, $d=1.0$. $D$: $Y=1.0$, $d=2.0$. Here $Y$ is in units of $10^{11}\ \text{N m}^{-2}$ and $d$ in $\text{mm}$.

What the question asks

$F$ ($\text{N}$): tension in each wire.

$l$ ($\text{m}$): length of each wire.

$d$ ($\text{mm}$): diameter.

$A$ ($\text{m}^2$): area of cross-section, $A=\pi d^2/4$.

$Y$ ($\text{N m}^{-2}$): Young’s modulus of the material.

$e$ ($\text{m}$): extension.

Relevant theory

The ratio of tensile stress to tensile strain is called the Young's modulus of the material

of the wire.

The extension per unit length of the wire is known as the tensile strain.

What is happening: each wire is pulled with the same force and has the same length.

A thick wire shares the pull over a bigger area, so it stretches less. A stiff material, with a big $Y$, also stretches less.

So the wire that stretches most is the one that is both thin and made of the less stiff material.

For example, hang the same water bottle from a thin rubber band and from a thick one.

The thin band stretches a long way. The thick band hardly stretches.

Now try a band made of stiffer rubber. Even a thin one stretches less.

In words: thinner and softer means more stretch for the same pull.

Required equations

$Y=\dfrac{F/A}{e/l}$

$e=\dfrac{Fl}{AY}$

$e\propto\dfrac{1}{d^2Y}$

Applying it to this question

$2$

$A:\ 1.0^2\times2.0$

$8$

$B:\ 2.0^2\times2.0$

$1$

$C:\ 1.0^2\times1.0$

$4$

$D:\ 2.0^2\times1.0$

$e_A:e_B:e_C:e_D=\dfrac{1}{2}:\dfrac{1}{8}:1:\dfrac{1}{4}$

$4:1:8:2$

$e_A:e_B:e_C:e_D$

$e_C=2e_A=8e_B=4e_D$

$C$

$\text{largest}$

Reasoning and calculation

Wire $C$ has the smallest $d^2Y$, so it has the largest extension, twice that of $A$ and eight times that of $B$. This is option (3).

Wire $A$ is as thin as $C$, but its material is twice as stiff ($Y=2.0$ against $1.0$). So it stretches only half as much as $C$.

Wire $B$ is the thickest and is made of the stiffer material. Both make the extension smaller, so $B$ stretches the least, one eighth of $C$.

Wire $D$ has the softer material like $C$, but twice the diameter, so four times the area. It stretches only a quarter as much as $C$.

The same extension needs the same area and the same Young’s modulus as well as the same tension and length. Here both the diameter and $Y$ differ between the wires.

Checks

$(dA*dA*YA)/(dC*dC*YC)$

$e=Fl/(AY)$ is $\text{N}\times\text{m}$ divided by $\text{m}^2\times\text{N m}^{-2}$, which leaves $\text{m}$, a length, as an extension should be.

$C$ is the thin wire made of the less stiff material, so it should stretch the most. $B$, thick and stiff, should stretch the least. The numbers agree.

Conclusion

$e_C\ \text{largest}$

Wire $C$ has the largest extension. The answer is (3).