1.3 Measurements and uncertainties — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. State the uncertainty rule you use at each step.

Warm-up: remember

Write $0.800\,\text{mm}$ in metres.
Find the mean of $12.4$ , $12.6$ and $12.5$ .
Work out the percentage $\dfrac{0.2}{8.0} \times 100\%$ .

Answers

$8.00 \times 10^{-4}\,\text{m}$ .
$12.5$ .
$2.5\%$ .

If question 1 goes wrong, use the prefix table in Section 1.1: milli means $10^{-3}$ . If question 2 goes wrong, add the three values and divide by 3. If question 3 goes wrong, divide first and then multiply by 100.

Practise with help

1. A student measures the diameter of a wire four times and records $0.542\,\text{mm}$ , $0.540\,\text{mm}$ , $0.544\,\text{mm}$ and $0.538\,\text{mm}$ . Find the mean and the uncertainty using the half-range convention.

Hint

Find the range first: the largest reading minus the smallest reading.

Next step

The range is $0.544 - 0.538 = 0.006\,\text{mm}$ . The uncertainty is half of that.

Full working

\begin{aligned} \text{mean} &= \frac{0.542 + 0.540 + 0.544 + 0.538}{4} = 0.541\,\text{mm} \\ \text{uncertainty} &= \frac{0.006}{2} = 0.003\,\text{mm} \end{aligned}

The diameter is $0.541 \pm 0.003\,\text{mm}$ .

2. A single length is read from a metre rule as $0.542\,\text{m}$ . The rule has a smallest division of 1 mm. Find the absolute uncertainty and the percentage uncertainty in the length.

Hint

Start with the reading uncertainty of one reading, not the half-range of repeats.

Next step

The absolute uncertainty is half the smallest division, $0.0005\,\text{m}$ . Now divide that by the value and multiply by 100.

Full working

\begin{aligned} \Delta L &= 0.0005\,\text{m} \\ \text{percentage uncertainty} &= \frac{0.0005}{0.542} \times 100\% \\ &= 0.09\% \quad (2\ \text{s.f.}) \end{aligned}

The length is $0.542 \pm 0.0005\,\text{m}$ , and the percentage uncertainty is 0.09%.

3. A cylinder has radius $r = 1.25 \pm 0.05\,\text{cm}$ and height $h = 8.00 \pm 0.05\,\text{cm}$ . Find the volume and the percentage uncertainty in the volume.

Hint

The volume is $V = \pi r^2 h$ . The radius is squared, so its percentage uncertainty is multiplied by 2.

Next step

Find the percentage uncertainty in $r$ and in $h$ , double the radius value, then add the two.

Full working

\begin{aligned} \%r &= \frac{0.05}{1.25} \times 100\% = 4.0\% \\ \%r^2 &= 2 \times 4.0\% = 8.0\% \\ \%h &= \frac{0.05}{8.00} \times 100\% = 0.63\% \\ \%V &= 8.0\% + 0.63\% = 8.6\% \end{aligned}

The volume is

V = \pi \times 1.25^2 \times 8.00 = 39.3\,\text{cm}^3

and its absolute uncertainty is $39.3 \times 0.086 = 3.4\,\text{cm}^3$ . The volume is $39 \pm 3\,\text{cm}^3$ to one significant figure in the uncertainty.

4. Two lengths of a rod are measured as $12.4 \pm 0.1\,\text{cm}$ and $8.6 \pm 0.1\,\text{cm}$ . Find the total length and its absolute uncertainty.

Hint

The two lengths are added, so their uncertainties are added too.

Next step

Write the total as $12.4 + 8.6$ and the uncertainty as $0.1 + 0.1$ .

Full working

\begin{aligned} L &= 12.4 + 8.6 = 21.0\,\text{cm} \\ \Delta L &= 0.1 + 0.1 = 0.2\,\text{cm} \end{aligned}

The total length is $21.0 \pm 0.2\,\text{cm}$ . Adding absolute uncertainties is the rule for a sum and for a difference.

5. A timer gives a value of $4.567\,\text{s}$ with an uncertainty of $0.1\,\text{s}$ . Write the result in the correct form.

Hint

The uncertainty is 0.1 s, which is in the first decimal place. The value must be quoted to the same decimal place.

Next step

Round $4.567$ to one decimal place and keep the uncertainty at 0.1 s.

Full working

t = 4.6 \pm 0.1\,\text{s}

Quoting $4.567\,\text{s}$ would suggest that the timer knows the value to the third decimal place, which the uncertainty of 0.1 s does not support.

6. A student wants the percentage uncertainty in $g$ to be below 0.5% when using a pendulum. The timing uncertainty for one measurement is about 0.2 s. Decide how many oscillations the student should time. Use $T = 1.795\,\text{s}$ .

Hint

Write the percentage uncertainty in $g$ in terms of the number of oscillations $N$ , then make it less than 0.5.

Next step

The percentage uncertainty in $T$ is $\dfrac{0.2}{1.795N} \times 100$ . Doubling it gives the percentage uncertainty in $g$ . Set the result below 0.5.

Full working

\begin{aligned} \%g &= 2 \times \frac{0.2}{1.795N} \times 100 \\ &= \frac{40}{1.795N} \\ \frac{40}{1.795N} &< 0.5 \\ N &> \frac{40}{1.795 \times 0.5} \\ N &> 44.6 \end{aligned}

The student should time at least 45 oscillations. Timing 20 gives about 0.56% from the reaction time alone, a little above the target. Repeating the timing also helps, because the half-range of several repeats is usually smaller than the reaction-time estimate for a single timing.