What a complete plan contains, and how to set it out

Choosing apparatus by range and resolution, and judging it with percentage uncertainty

Reading vernier calipers (0.1 mm) and micrometer screw gauges (0.01 mm), including zero errors

Measuring techniques: avoiding parallax, set squares, fiducial marks, timing many oscillations, repeating and averaging

Variables, repeats, hazards and precautions

Choosing the graph that tests a relationship and gives the quantity you want

Reducing uncertainty and commenting on the physics

1. A plan names every instrument with its resolution, says what is varied, what is measured and what is kept constant, and ends with the graph and how the result comes from it.

2. Choose the instrument whose resolution makes the percentage uncertainty small: uncertainty ≈ half the resolution.

3. Vernier caliper: main scale (mm) + the vernier line that lines up × 0.1 mm. Micrometer: sleeve (to 0.5 mm) + thimble × 0.01 mm.

4. Corrected reading = reading − zero error. A zero error is systematic: repeating does not remove it.

5. Linearise the relationship so the graph is a straight line; the constant you want is in the gradient or intercept.

A planning question describes an aim, such as "investigate how the deflection of a strip depends on its length", and asks how you would do it. Every good plan answers the same short list of questions.

Range is the largest (and smallest) value the instrument can measure; it must cover every reading you expect. Resolution is the smallest change the instrument can show: one scale division, or the last digit on a digital display. At IAS level, take the uncertainty in a single reading as half the resolution, and judge an instrument by the percentage uncertainty it gives for the size of quantity you are measuring.

WORKED EXAMPLE · QUESTION

A student needs the diameter of a copper wire, which is about 0.38 mm. She has a metre rule, a vernier caliper and a micrometer screw gauge. Use percentage uncertainties to justify the choice of instrument.

SOLUTION

Half the resolution: metre rule ±0.5 mm, vernier ±0.05 mm, micrometer ±0.005 mm. Percentage uncertainty = uncertainty ÷ reading × 100%:

0.5/0.38 ×100 = 132%, 0.05/0.38 ×100 = 13%, 0.005/0.38 ×100 = 1.3%.

Micrometer screw gauge: the only instrument giving a percentage uncertainty of a few per cent or less.

Vernier caliper (0.1 mm)

The main scale is marked in millimetres. The sliding vernier scale has 10 divisions that together span 9 mm, so each vernier division is 0.9 mm long, 0.1 mm shorter than a main-scale division.

1. Read the main scale at the vernier zero: the whole number of millimetres to its left.

2. Find the vernier line that lines up best with any main-scale line. Its number gives the tenths of a millimetre.

3. Reading = main-scale millimetres + (number of the coinciding vernier line × 0.1 mm). Record it to 0.1 mm, for example 23.4 mm, never as 23.40 mm.

Close the jaws first: if the vernier zero does not line up with the main zero, there is a zero error. Outside jaws measure external diameters, inside jaws internal diameters, the tail depths. Close gently: squeezing a soft object makes it read small.

WORKED EXAMPLE · QUESTION

The scales above show a vernier caliper with its jaws closed (a) and measuring the outside diameter of a plastic pipe (b). Determine the diameter of the pipe and its percentage uncertainty.

SOLUTION

Read (b): main scale 23 mm, vernier line 4 coincides, so reading = 23 + 4 ×0.1 = 23.4 mm. Zero error from (a): +0.2 mm, so diameter = 23.4 - 0.2 = 23.2 mm. Uncertainty = half the resolution = 0.05 mm, so 0.05/23.2 ×100 = 0.22%.

Micrometer screw gauge (0.01 mm)

The sleeve has a horizontal datum line with millimetre marks above it and half-millimetre marks below it. The thimble is a screw with a pitch of 0.50 mm: one full turn moves the spindle 0.50 mm. Its edge carries 50 divisions, so each division is 0.50/50 = 0.01 mm.

Reading = the last mark visible on the sleeve (to 0.5 mm) + (thimble division on the datum line × 0.01 mm). Record to 0.01 mm.

Turn the ratchet, not the thimble, to close on the object: the ratchet slips at the same gentle force every time, so soft wires are not squashed and readings are consistent.

Check the zero with the anvil and spindle closed (and clean) before measuring.

WORKED EXAMPLE · QUESTION

The micrometer above is measuring the diameter of a steel ball bearing. With the anvil and spindle closed, the datum line lies 2 divisions below the zero on the thimble, so the instrument reads −0.02 mm. (a) State the reading shown. (b) Determine the diameter of the ball bearing. (c) Calculate the percentage uncertainty in the diameter.

SOLUTION

(a) 6.5 + 23 ×0.01 = 6.73 mm.

(b) Corrected reading = reading − zero error: 6.73 - (-0.02) = 6.75 mm. A negative zero error makes every reading too small, so it is added back.

(c) Uncertainty = 0.005 mm, so 0.005/6.75 ×100 = 0.074%.

A zero error adds (or removes) the same amount from every reading: it is a systematic error. Repeating and averaging cannot remove it, so you must correct for it: corrected reading = reading − zero error.

Checking zero before use: close the jaws of a caliper or micrometer; switch on a meter with no current or p.d. and check it reads zero; tare (zero) a balance with the empty container on it; check the end of a metre rule is not worn, or measure from the 10.0 cm mark and subtract.

Calibration means checking an instrument against known values and correcting for any difference: for example hanging known masses on a newton meter, or placing a thermometer in melting ice (0 °C) and in steam above boiling water (about 100 °C).

Avoiding parallax

Parallax error occurs when a pointer (or meniscus) is not touching the scale and the eye is not level with it. Put the eye at the same level as the pointer, perpendicular to the scale; move the scale as close to the pointer as possible; for meters, use one with a mirror behind the pointer and line the pointer up with its image.

WORKED EXAMPLE · QUESTION

A spring hangs beside a vertical metre rule, with readings increasing downwards. A pointer on the bottom of the spring is 20 mm in front of the rule. A student's eye is 200 mm from the rule and 50 mm above the pointer when she reads the stretched position. The unstretched reading was 350.0 mm and the true stretched reading is 412.0 mm. (a) Calculate the reading she obtains. (b) Calculate the percentage error this causes in the extension. (c) Suggest two ways to avoid it.

SOLUTION

(a) Similar triangles: the line of sight drops 50 mm over 180 mm, so over a further 20 mm it drops 50 ×20/180 = 5.6 mm. Reading = 412.0 + 5.6 = 417.6 mm.

(b) True extension = 412.0 - 350.0 = 62 mm, so 5.6/62 ×100 = 9%, far larger than the 0.5 mm reading uncertainty.

(c) Eye level with the pointer, looking perpendicular to the rule; move the rule (or pointer) so they almost touch.

Other techniques worth naming

Set square: stand it on the bench against a metre rule to check the rule is vertical, or use it to line up the bottom of an object with the scale. A spirit level does the same job for horizontal surfaces.

Fiducial mark: a fixed reference line or pin at the centre of an oscillation (where the object moves fastest), so each pass is timed at the same, sharply defined point.

Time many cycles: time 10 or 20 oscillations and divide. The reaction-time error happens once at the start and once at the end, so it is shared across all the cycles.

Several places and directions: measure a wire's diameter at several points, turning the micrometer through 90° each time, and take the mean.

Electronic timing (light gates) removes reaction time for short events.

WORKED EXAMPLE · QUESTION

A hacksaw blade is clamped vertically with a small mass on its top and set vibrating. A student times single oscillations three times: 0.40 s, 0.47 s, 0.37 s. He then times 20 oscillations three times: 8.46 s, 8.31 s, 8.55 s. (a) Calculate the period and its percentage uncertainty from each set. (b) Explain why timing 20 oscillations is better. (c) Suggest one further improvement.

SOLUTION

(a) Single oscillations: mean = 0.41 s; uncertainty = half the range = (0.47 - 0.37)/2 = 0.05 s; percentage = 0.05/0.413 ×100 = 12%. Twenty oscillations: mean = 8.44 s; uncertainty = (8.55 - 8.31)/2 = 0.12 s; percentage = 0.12/8.44 ×100 = 1.4%; period = 8.44/20 = 0.422 s (1.4%).

(b) The reaction-time error at start and stop is about the same, but it is now spread over a time 20 times longer, so it is a much smaller fraction of the total and the percentage uncertainty falls sharply (1.4% instead of 12%).

(c) A fiducial mark at the centre of the swing; start the count at "zero" as the blade passes it.

For each control variable say how it is kept constant ("use the same 0.200 kg mass throughout"), not just that it is: "fair test" alone is not an answer.

Range and number of readings: at least five or six values of the independent variable, spread across the widest range the apparatus allows, so the shape of the graph is clear.

Repeats are useful when readings scatter randomly: timing by hand, rebound heights, readings that drift. Repeat, check for an anomaly, and take the mean.

Repeats are not useful when the reading cannot change (a steady digital display, a fixed length), or when the measurement changes the object (a wire stretched beyond its elastic limit, a sample that is destroyed). Then say why, and take readings at more values instead.

For safety, give a hazard that belongs to this experiment and a precaution that deals with it; "wear goggles" only when something could hit the eyes.

Rearrange the equation linking the variables into the form y = mx + c and plot the y-quantity against the x-quantity. State the expected shape ("a straight line through the origin" confirms a proportional relationship) and what the gradient or intercept equals.

WORKED EXAMPLE · QUESTION

A wooden strip 25.0 mm wide and 6.0 mm thick is clamped to a bench, as in the first figure. A load of weight W = 0.200 ×9.81 = 1.96 N hangs from its free end. Theory predicts y = 4WL³/Ebd³, where L is the length beyond the bench, b the width, d the thickness and E the Young modulus of the wood. (a) Write a plan to test the relationship and find E. (b) Use the graph above to determine E.

SOLUTION

(a) Variables: vary L; measure y; keep W, the strip and the load position fixed. Measurements: L with a metre rule (1 mm) from the bench edge to the hanger; y as the change in the pin's position on a vertical metre rule, unloaded then loaded; b with a vernier (0.1 mm) and d with a micrometer (0.01 mm), each at several places, averaged. Technique and safety: rule checked vertical with a set square; eye level with the pin; six lengths from 0.300 m to 0.800 m; box of sand under the load. Analysis: plot y against L³: a straight line through the origin confirms the relationship; gradient = 4W/(Ebd³).

(b) Gradient = (60.7 - 6.0) mm ÷ (0.50 - 0.05) m³ = 122 mm m⁻³ = 0.122 m m⁻³.

Then E = (4 ×1.96)/(0.0250 ×0.0060³ ×0.122), so E = 1.2 ×10¹⁰ Pa.

Note the unit conversion: y was in mm, so divide the gradient by 1000 before using it in SI units.

Find the biggest source of uncertainty: the measurement with the largest percentage uncertainty. In the cantilever data the smallest deflection, 3 mm, read to ±0.5 mm, is uncertain by 17%, far more than any length.

Reduce it by making the quantity bigger (longer lengths or heavier loads, within the elastic limit), by choosing an instrument with a finer resolution (a travelling microscope or a vernier for small deflections), or by measuring many units at once (20 oscillations; the thickness of 10 sheets).

Systematic errors (zero errors, a rule that is not vertical, a thermometer that reads high) shift every reading the same way. Remove them by checking zero, calibrating and using good technique; repeats do not help.

Random errors scatter readings either side of the true value. Reduce their effect by repeating and averaging and by plotting a graph.

Commenting on the physics: link the result to its context. Is the value sensible for the material? Would the relationship still hold for very large deflections (no: the theory assumes small ones)?

What does the result mean for a real use, such as a shelf or a diving board?

2 Uncertainties and graphs; 3.3 Core practicals

Page 3.2 covers recording, plotting and evaluating the readings (significant figures, half-range uncertainties, best-fit lines, gradients). Every core practical uses the instruments and techniques on this page.

Materials (Unit 1)

The Young modulus and Hooke's law supply many planning contexts; measuring a wire's diameter with a micrometer is the key step in finding stress.

1 Planning at A2

The same planning skills return with A2 physics, together with combined uncertainties, precision, accuracy and sensitivity.

1. On a vernier caliper the vernier zero is just past the 41 mm mark and vernier line 6 lines up with a main-scale line. What is the reading?

2. A micrometer sleeve shows the 3 mm mark and the half-millimetre mark after it; thimble division 12 is on the datum line. What is the reading?

3. An ammeter reads 0.02 A before the circuit is switched on. It then reads 0.46 A. What is the current?

4. Why should the diameter of a wire be measured at several points and in two perpendicular directions?

5. A trolley is released from rest on a ramp and a light gate measures its speed v after it has rolled a distance s. Which graph gives the acceleration, and how?

1. 41 + 6 ×0.1 = 41.6 mm. (Vernier line 6 lines up with the 47 mm line: 41.6 + 6 ×0.9 = 47.0.)

2. 3.5 + 12 ×0.01 = 3.62 mm.

3. The zero error is +0.02 A, so the current is 0.46 - 0.02 = 0.44 A. Better still, zero the meter before use.

4. The wire may not be uniform along its length or perfectly circular; the mean of several readings is a better estimate, and the spread shows how variable it is.

5. From rest, v² = 2as. Vary s and plot v² against s: a straight line through the origin with gradient 2a, so a = gradient ÷ 2.