2023 A/L Physics Past Paper · Question 27 (English Medium)

Answer: (3) $50.02\,\text{m}$

Given information

What we know:

The steel tape is right at $\theta_0 = 20\,^\circ\text{C}$.

It is used at $\theta = 40\,^\circ\text{C}$.

The reading on the tape is $L_{\text{read}} = 50.00\,\text{m}$.

The linear expansivity of steel is $\alpha = 2 \times 10^{-5}\,^\circ\text{C}^{-1}$.

What the question asks

We want: the actual length $L_{\text{true}}$ that was measured.

What each letter means:

$L_{\text{read}}$ ($\text{m}$): the value read off the tape.

$L_{\text{true}}$ ($\text{m}$): the actual length.

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

$\Delta\theta$ ($^\circ\text{C}$): the rise of temperature of the tape.

$f$: the number that tells how many times longer each mark of the tape has become.

Relevant theory

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

The amount of increase of length of an object is directly proportional to; 1. The original length (l0) 2. The rise of temperature

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.

The first quote says a rod grows longer when it gets warmer. The second quote says the growth is in proportion to the original length and to the rise of temperature. The third quote says the linear expansivity is the fraction of its length that a body gains for each degree of rise.

So a length $l$ grows by $\alpha\,l\,\Delta\theta$.

For example, think of a shopkeeper who measures cloth with a metal metre rule, and the rule has grown a little in the hot sun. Each "metre" on that rule is now a bit longer than a true metre.

The shopkeeper counts the marks and says the cloth is so many metres. But the cloth is a bit longer than the count, because each mark covers more than a metre.

Required equations

$\Delta l = \alpha\,l\,\Delta\theta$

$l' = l\,(1 + \alpha\,\Delta\theta)$

$f = 1 + \alpha\,\Delta\theta$

$L_{\text{true}} = L_{\text{read}} \times f$

$\alpha$ is in $^\circ\text{C}^{-1}$ and $\Delta\theta$ is in $^\circ\text{C}$, so $\alpha\,\Delta\theta$ has no unit. The answer is in $\text{m}$.

Applying it to this question

$20$

$\Delta\theta = \theta - \theta_0 = 40 - 20$

$4 \times 10^{-4}$

$\alpha\,\Delta\theta = 2 \times 10^{-5} \times 20$

$1.0004$

$f = 1 + 4 \times 10^{-4}$

$50.02$

$L_{\text{true}} = 50.00 \times 1.0004$

$r * (1 + a * t)$

Reasoning and calculation

Now match it with the five options.

(1) $49.96\,\text{m}$ is less than the reading. It does not match.

(2) $49.98\,\text{m}$ is less than the reading. It does not match.

(3) $50.02\,\text{m}$ is what we found. This matches.

(4) $50.04\,\text{m}$ is twice the growth we found. It does not match.

(5) $50.06\,\text{m}$ is three times the growth we found. It does not match.

Checks

Do it a second way. Work out how much the whole length gains.

$$\Delta l = \alpha\,l\,\Delta\theta = 2 \times 10^{-5} \times 50.00 \times 20 = 0.02\,\text{m}$$

Add it to the reading.

$$50.00 + 0.02 = 50.02\,\text{m}$$

This is the same. The change is small, $0.02\,\text{m}$ in $50\,\text{m}$, as it should be for steel in a $20\,^\circ\text{C}$ rise. And the actual length is more than the reading, as a tape that has grown would give.

Conclusion

$L_{\text{true}} = 50.02\,\text{m}$

Conclusion

The quotes give the growth of a length with a rise of temperature. That each mark of a tape grows by the same fraction, so that the reading must be multiplied by $1 + \alpha\,\Delta\theta$, is our reasoning from them.

Teacher's note

A steel tape that is warmer than when it was marked is longer. So it reads too small, and the actual length is more than the reading.

Use the rise of temperature, $\theta - \theta_0$, and not the temperature itself.

Exam tip