1.3 Measurements and uncertainties — Practise

Work through these on your own. For each calculation, write the uncertainty rule you are using before you substitute any numbers.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A student measures the diameter of a wire four times and records $2.14\,\text{mm}$ , $2.16\,\text{mm}$ , $2.15\,\text{mm}$ and $2.12\,\text{mm}$ . Find the mean and the uncertainty using the half-range convention.
A length is measured as $0.542\,\text{m}$ with an uncertainty of $0.001\,\text{m}$ . Calculate the percentage uncertainty.
The length of a pendulum is found from two metre-rule readings: $0.000\,\text{m}$ at the pivot and $0.800\,\text{m}$ at the centre of the bob. Each reading has an uncertainty of $0.0005\,\text{m}$ . Find the length and the uncertainty in the length.
A cylinder has radius $r = 2.50\,\text{cm}$ with a percentage uncertainty of $1.2\%$ , and height $h = 6.00\,\text{cm}$ with a percentage uncertainty of $0.8\%$ . Find the percentage uncertainty in the volume $V = \pi r^2 h$ .
The density of a liquid is found from $\rho = \dfrac{m}{V}$ . The mass is $25.6 \pm 0.1\,\text{g}$ and the volume is $10.0 \pm 0.2\,\text{cm}^3$ . Find the density and its absolute uncertainty.
A measurement gives $E = 12.345\,\text{J}$ with an uncertainty of $0.2\,\text{J}$ . Write the result in the correct form.
A student measures the acceleration due to gravity as $9.80 \pm 0.04\,\text{m}\,\text{s}^{-2}$ . The accepted value is $9.81\,\text{m}\,\text{s}^{-2}$ . Does the measurement agree with the accepted value? Give a reason.
A pendulum has length $0.400\,\text{m}$ . Its period is measured as $1.27\,\text{s}$ . Use $g = \dfrac{4\pi^2 L}{T^2}$ to find the acceleration due to gravity.
You must find the density of a small irregular stone. State which measurements you would take, which instrument you would use for each, and how you would obtain the uncertainty in the density.
A student times 20 oscillations three times and records $35.90\,\text{s}$ , $35.85\,\text{s}$ and $35.95\,\text{s}$ . Find the mean time, its uncertainty, and the percentage uncertainty in the mean.
A student says: "A smaller percentage uncertainty always means a more accurate result." Explain why this is not always true.
The mass of an empty beaker is $52.4 \pm 0.1\,\text{g}$ . The mass of the beaker and the water is $128.7 \pm 0.1\,\text{g}$ . Find the mass of the water and its uncertainty.

Answers and concise solutions

1. The mean and its uncertainty

Answer: The mean is $\dfrac{2.14 + 2.16 + 2.15 + 2.12}{4} = 2.1425\,\text{mm}$ . The range is $2.16 - 2.12 = 0.04\,\text{mm}$ , so the uncertainty is half the range, $0.02\,\text{mm}$ . The uncertainty is quoted to one significant figure and the value to the same decimal place, so the diameter is $2.14 \pm 0.02\,\text{mm}$ .

2. The percentage uncertainty

Answer: The percentage uncertainty is $\dfrac{0.001}{0.542} \times 100\% = 0.18\%$ to two significant figures.

3. The length and its uncertainty

Answer: The length is $0.800 - 0.000 = 0.800\,\text{m}$ . A difference adds the absolute uncertainties, so the uncertainty is $0.0005 + 0.0005 = 0.001\,\text{m}$ . The length is $0.800 \pm 0.001\,\text{m}$ .

4. The percentage uncertainty in the volume

Answer: The constant $\pi$ has no uncertainty, so it does not add any. The radius appears squared, so its percentage uncertainty is doubled: $2 \times 1.2\% = 2.4\%$ . Add the height: $2.4\% + 0.8\% = 3.2\%$ .

5. The density and its absolute uncertainty

Answer: The percentage uncertainty in the mass is $\dfrac{0.1}{25.6} \times 100\% = 0.39\%$ . In the volume it is $\dfrac{0.2}{10.0} \times 100\% = 2.0\%$ . For a quotient, add the percentages: $0.39\% + 2.0\% = 2.4\%$ . The density is $\dfrac{25.6}{10.0} = 2.56\,\text{g}\,\text{cm}^{-3}$ , so its absolute uncertainty is $2.56 \times 0.024 = 0.06\,\text{g}\,\text{cm}^{-3}$ . The density is $2.56 \pm 0.06\,\text{g}\,\text{cm}^{-3}$ .

6. The result in the correct form

Answer: The uncertainty is $0.2\,\text{J}$ to one significant figure. The value is quoted to the same decimal place, so $E = 12.3 \pm 0.2\,\text{J}$ .

7. Whether the results agree

The measured value could lie between 9.76 and 9.84 m s^-2.
The accepted value 9.81 m s^-2 lies inside that range, so the measurement agrees with it within the stated uncertainty.

8. The value of g

Answer: $g = \dfrac{4\pi^2 \times 0.400}{1.27^2} = 9.79\,\text{m}\,\text{s}^{-2}$ to three significant figures.

9. The measurements and the uncertainty

Measure the mass on a top-pan balance, and use the balance resolution to find the uncertainty in the mass.
Find the volume from the volume of water displaced in a measuring cylinder, reading the level before and after the stone is added. The two readings each carry half the smallest division, and the uncertainties add.
The density is mass divided by volume, so find the percentage uncertainty in each and add them to get the percentage uncertainty in the density.

10. The mean time and its percentage uncertainty

Answer: The mean is $\dfrac{35.90 + 35.85 + 35.95}{3} = 35.90\,\text{s}$ . The range is $0.10\,\text{s}$ , so the uncertainty is half the range, $0.05\,\text{s}$ . The percentage uncertainty is $\dfrac{0.05}{35.90} \times 100\% = 0.14\%$ to two significant figures.

11. Why a small uncertainty is not the same as accuracy

Percentage uncertainty measures the random scatter in the readings, so repeating the measurement can make it smaller.
A systematic error shifts every reading in the same direction and is not reduced by repeating. A precise set of readings can therefore still be a long way from the true value.

12. The mass of water and its uncertainty

Answer: The mass of water is $128.7 - 52.4 = 76.3\,\text{g}$ . A difference adds the absolute uncertainties, so the uncertainty is $0.1 + 0.1 = 0.2\,\text{g}$ . The mass is $76.3 \pm 0.2\,\text{g}$ .