2024 A/L Physics Past Paper · Question 17 (English Medium)

Answer: (5) $5\ ^\circ\text{C}$

Given information

This question is about a steel ruler that grows a little when it warms up.

We read the ruler in millimetres. If the ruler warms, every millimetre mark gets a tiny bit longer, and the reading is no longer exact.

What we must find: the largest change in temperature that still keeps each millimetre reading accurate to $5\times10^{-5}\ \text{mm}$.

How we find it: take one millimetre mark as the length that grows. Its growth may be at most $5\times10^{-5}\ \text{mm}$.

One millimetre mark: $l_0=1\ \text{mm}$. The largest growth allowed: $\Delta l=5\times10^{-5}\ \text{mm}$. For steel, $\alpha=1\times10^{-5}\ ^\circ\text{C}^{-1}$.

What the question asks

$l_0$ ($\text{mm}$): original length of one millimetre mark.

$\Delta l$ ($\text{mm}$): increase of that length.

$\alpha$ ($^\circ\text{C}^{-1}$): linear expansivity of steel.

$\Delta\theta$ ($^\circ\text{C}$): change in temperature.

Relevant theory

If we consider only the increase of length of an object with rise of temperature, then it is called linear expansion.

This constant is known as the linear expansivity of the material of the object and is defined as the fractional increase of length per unit rise of temperature.

What is happening: when the ruler warms, every part of it grows. A long part grows more, and a short part grows less.

We read the ruler one millimetre at a time. So the length that matters is one millimetre, $l_0=1\ \text{mm}$, not the whole metre.

A bigger temperature change makes a bigger growth. The largest change allowed is the one that makes the growth exactly $5\times10^{-5}\ \text{mm}$.

For example, look at a railway track on a hot afternoon.

The workers leave a little space between the steel rails. In the morning each space is wide.

By noon the sun has heated the rails. Each rail grows a little, and each space gets narrower.

In words: the hotter the steel gets, the more it grows. That space is the largest growth the track can take.

Required equations

$\Delta l=\alpha l_0\Delta\theta$

$\Delta\theta=\dfrac{\Delta l}{\alpha l_0}$

Applying it to this question

$1\times10^{-5}$

$\alpha l_0=1\times10^{-5}\times1$

$5\times10^{-5}$

$\Delta l$

$\Delta\theta=\dfrac{5\times10^{-5}}{1\times10^{-5}}$

$5$

$\Delta\theta$

Reasoning and calculation

Any change up to $5\ ^\circ\text{C}$ keeps the growth of each millimetre within $5\times10^{-5}\ \text{mm}$. A larger change would make it too big. So the largest change allowed is $5\ ^\circ\text{C}$, option (5).

At $0.1\ ^\circ\text{C}$ each millimetre grows by only $1\times10^{-6}\ \text{mm}$. That is safe, but the question asks for the largest change allowed, and $5\times10^{-5}\ \text{mm}$ is fifty times more.

At $0.2\ ^\circ\text{C}$ each millimetre grows by $2\times10^{-6}\ \text{mm}$, still twenty-five times less than the $5\times10^{-5}\ \text{mm}$ allowed. The temperature may change more.

At $1\ ^\circ\text{C}$ each millimetre grows by $1\times10^{-5}\ \text{mm}$, only a fifth of the $5\times10^{-5}\ \text{mm}$ allowed. This is the change for one degree, not the largest change.

At $2\ ^\circ\text{C}$ each millimetre grows by $2\times10^{-5}\ \text{mm}$. That is less than the $5\times10^{-5}\ \text{mm}$ allowed, so a bigger change is still safe.

Checks

$dl/(a*l0)$

$\Delta l=1\times10^{-5}\times1\times5$

$\Delta l=5\times10^{-5}\ \text{mm}$

$\Delta l/(\alpha l_0)$ is $\text{mm}$ divided by $^\circ\text{C}^{-1}\times\text{mm}$, which leaves $^\circ\text{C}$, a temperature change.

If we used the whole metre, $l_0=1000\ \text{mm}$, we would get $0.005\ ^\circ\text{C}$. That is not an option, and it is not what the question measures: each reading is one millimetre.

Conclusion

$\Delta\theta=5\ ^\circ\text{C}$

The largest temperature change allowed is $5\ ^\circ\text{C}$. The answer is (5).