2021 A/L Physics Past Paper · Question 07 (English Medium)

Answer: (4) $2.5 \times 10^{-5}\ \text{m}^2$

Given information

What we know:

The rod is aluminium. It breaks when it is stretched by $0.2\%$ of its length. The force on it is $3.5 \times 10^3\ \text{N}$. Young's modulus of aluminium is $7.0 \times 10^{10}\ \text{N m}^{-2}$.

What each letter means:

$F$ ($\text{N}$): tensile force, the pull on the rod, $3.5 \times 10^3\ \text{N}$.

$A$ ($\text{m}^2$): area of cross-section, how thick the rod is.

$\sigma$ ("sigma", $\text{N m}^{-2}$): tensile stress, the pull on each square metre of the cut face.

$\varepsilon$ ("epsilon"): tensile strain, the stretch for each metre of length. It has no unit.

$Y$ ($\text{N m}^{-2}$): Young's modulus, how stiff the material is, $7.0 \times 10^{10}\ \text{N m}^{-2}$.

What the question asks

We want: the smallest area $A$ for which the rod does not break.

The plan: first find the biggest stress the rod can take before it breaks. Then find the area that keeps the stress at that value.

Relevant theory

The tensile force acting on a unit area of cross section of the wire is known as the tensile stress.

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

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

This point is referred to as the breaking load and there after the wire gets thinner and breaks at a point such as D. The point D is called the breaking point.

Required equations

$\sigma = \dfrac{F}{A}$

$Y = \dfrac{\sigma}{\varepsilon}$

$\sigma = Y \varepsilon$

Applying it to this question

Think of a rope made of many threads. One thin thread snaps under a heavy bag. A thick rope with many threads holds it, because each thread carries only a small part of the pull.

A rod is the same. A thicker rod shares the pull over a bigger area, so the stress is smaller.

The rod breaks when its strain reaches $0.2\%$. So the strain must not go above $0.2\%$.

$0.002$

$\varepsilon = \dfrac{0.2}{100}$

Now find the biggest stress the rod can take.

$$\sigma = Y \varepsilon$$

This says: a stiffer material, or a bigger strain, needs a bigger stress.

$1.4 \times 10^{8}$

$\sigma = 7.0 \times 10^{10} \times 0.002$

The stress is the force shared over the area. So the area is the force divided by the stress.

$$A = \dfrac{F}{\sigma}$$

This says: a bigger force, or a smaller stress allowed, needs a bigger area.

$2.5 \times 10^{-5}$

$A = \dfrac{3.5 \times 10^{3}}{1.4 \times 10^{8}}$

Reasoning and calculation

Why is this the smallest area?

If the rod is thinner, the same force is shared over less area. The stress goes up, and the strain goes above $0.2\%$. The rod breaks.

If the rod is thicker, it holds easily. But that is not the smallest.

So the smallest area is $2.5 \times 10^{-5}\ \text{m}^2$. It is option (4).

Checks

Does it make sense?

Put the answer back. The stress is $\dfrac{3.5 \times 10^3}{2.5 \times 10^{-5}} = 1.4 \times 10^8\ \text{N m}^{-2}$. The strain is $\dfrac{1.4 \times 10^8}{7.0 \times 10^{10}} = 0.002$, which is exactly $0.2\%$.

The area is about $25\ \text{mm}^2$, a rod about $6\ \text{mm}$ across. That is a sensible size for holding a $350\ \text{kg}$ load.

The unit: $\dfrac{\text{N}}{\text{N m}^{-2}} = \text{m}^2$. An area, as it should be.

$f / (y * s)$

Conclusion

$A = 2.5 \times 10^{-5}\ \text{m}^2$

The rod breaks at a strain of $0.002$. So the biggest stress is $Y\varepsilon = 1.4 \times 10^8\ \text{N m}^{-2}$.

The smallest area is $A = \dfrac{F}{\sigma} = \dfrac{3.5 \times 10^3}{1.4 \times 10^8} = 2.5 \times 10^{-5}\ \text{m}^2$. The answer is (4).

Turn the percentage into a number first: $0.2\% = 0.002$.

Then the biggest stress is $Y \times$ breaking strain, and the smallest area is the force divided by that stress.

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