Errors and uncertainties — Paper 2 · A Level Physics 9702

1 (a) Distinguish between systematic errors and random errors. systematic errors … … random errors … … [2] (b) A cylinder of length L has a circular cross-section of radius R, as shown in Fig. 1.1. R L Fig. 1.1 The volume V of the cylinder is given by the expression V = πR2L . The volume and length of the cylinder are measured as V = 15.0 ± 0.5 cm3 L = 20.0 ± 0.1 cm. Calculate the radius of the cylinder, with its uncertainty. radius = … ± … cm [5]

7 marks

Mark scheme: 1 (a) systematic: e.g. constant error (in all readings) cannot be eliminated by averaging error in measuring instrument B1 random: e.g. readings scattered (equally) about true value error due to observer can be eliminated by averaging (only if averaging not included for systematic) B1 [2] (b) 15 = π × R2 × 20 R = 0.4886 cm (accept any number of s.f.) C1 % uncertainty in V = 3.3 % (or 0.5/15) C1 % uncertainty in L = 0.5 % (or 0.1/20) C1 % uncertainty in R = 1.9 % (i.e. one half of the sum) C1 R = 0.489 ± 0.009 cm A1 [5]

This question in 9702/21 Oct/Nov 2007

1 A simple pendulum may be used to determine a value for the acceleration of free fall g. Measurements are made of the length L of the pendulum and the period T of oscillation. The values obtained, with their uncertainties, are as shown. T = (1.93 ± 0.03) s L = (92 ± 1) cm (a) Calculate the percentage uncertainty in the measurement of (i) the period T, uncertainty = … % [1] (ii) the length L. uncertainty = … % [1] (b) The relationship between T, L and g is given by For Examiner’s 4 2L Use g = 2 . T Using your answers in (a), calculate the percentage uncertainty in the value of g. uncertainty = … % [1] (c) The values of L and T are used to calculate a value of g as 9.751 m s–2. (i) By reference to the measurements of L and T, suggest why it would not be correct to quote the value of g as 9.751 m s–2. … … [1] (ii) Use your answer in (b) to determine the absolute uncertainty in g. Hence state the value of g, with its uncertainty, to an appropriate number of significant figures. g = … ± … m s–2 [2]

6 marks

Mark scheme: 1 (a) (i) either 1.55% or 1.6% …(not 1.5 or 2) … A1 [1] (ii) either 1.09% or 1.1% …(not 1.0 or 1) … A1 [1] (b) answer of {(ii) + 2 × (i)} to any number of sig. fig. either 4.2% or 4.3% … A1 [1] (c) (i) either the value has more significant figures than the data or uncertainty of ±0.4 renders more than 2 s.f. meaningless) … B1 [1] (ii) uncertainty in g = ±0.41 / ±0.42 to any number of s.f. … C1 g = (9.8 ± 0.4) m s-2 … A1 [2] [Total: 6]

This question in 9702/22 Oct/Nov 2009

1 A metal wire has a cross-section of diameter approximately 0.8 mm. (a) State what instrument should be used to measure the diameter of the wire. … [1] (b) State how the instrument in (a) is (i) checked so as to avoid a systematic error in the measurements, … … [1] (ii) used so as to reduce random errors. … … … [2]

4 marks

Mark scheme: 1 (a) micrometer/screw gauge/digital callipers ………………………………………. B1 [1] (b) (i) look/check for zero error ……………………………………………………. B1 [1] (ii) take several readings ……………………………………………………….. M1 around the circumference/along the wire …………………………………. A1 [2]

This question in 9702/22 May/June 2010

1 A digital voltmeter with a three-digit display is used to measure the potential difference across a resistor. The manufacturers of the meter state that its accuracy is ±1% and ±1 digit. The reading on the voltmeter is 2.05 V. (a) For this reading, calculate, to the nearest digit, (i) a change of 1% in the voltmeter reading, change = … V [1] (ii) the maximum possible value of the potential difference across the resistor. maximum value = … V [1] (b) The reading on the voltmeter has high precision. State and explain why the reading may not be accurate. … … … [2]

4 marks

Mark scheme: 1 (a) (i) 1% of ±2.05 is ±0.02 A1 [1] (ii) max. value is 2.08 V A1 [1] (b) there may be a zero error/calibration error/systematic error M1 which makes all readings either higher or lower than true value A1 [2]

This question in 9702/23 May/June 2010

1 (a) For each of the following, tick [✓] one box to indicate whether the experimental technique For would reduce random error, systematic error or neither. The first row has been completed Examiner’s as an example. Use random error systematic error neither keeping your eye in line with the scale and the ✓ liquid level for a single reading of a thermometer averaging many readings of the time taken for a ball to roll down a slope using a linear scale on an ammeter correcting for a non-zero reading when a micrometer screw gauge is closed [2] (b) The measurement of a particular time interval is repeated many times. The readings are found to vary. The results are shown in Fig. 1.1. number 8 of readings 6 4 2 0 10.0 10.2 10.4 10.6 10.8 reading of time interval / s Fig. 1.1 The true value of the time interval is 10.1 s. (i) State how the readings on Fig. 1.1 show the presence of For Examiner’s 1. a systematic error, Use … … [1] 2. a random error. … … [1] (ii) State the expected changes to Fig. 1.1 for experimental measurements that are 1. more accurate, … … [1] 2. more precise. … … [1]

6 marks

Mark scheme: 1 (a) 2nd row random, 3rd row neither, 4th row systematic all correct B2 two correct scores 1 only [2] (b) (i) 1. systematic error: the average / peak is not the true value / the readings are not centred around the true value B1 [1] 2. random error: readings have positive and negative values around the peak value / values are scattered / wide range B1 [1] (ii) 1. accurate: peak / average value moves towards the true value B1 [1] 2. precise: lines are closer together / sharper peak B1 [1]

This question in 9702/23 May/June 2011

1 The volume V of liquid flowing in time t through a pipe of radius r is given by the equation Use V π Pr 4 = t 8Cl where P is the pressure difference between the ends of the pipe of length l, and C depends on the frictional effects of the liquid. An experiment is performed to determine C. The measurements made are shown in Fig. 1.1. V / 10–6 m3 s–1 P / 103 N m–2 r / mm l / m t 1.20 ± 0.01 2.50 ± 0.05 0.75 ± 0.01 0.250 ± 0.001 Fig. 1.1 (a) Calculate the value of C. C = … N s m–2 [2] (b) Calculate the uncertainty in C. uncertainty = … N s m–2 [3] (c) State the value of C and its uncertainty to the appropriate number of significant figures. C = … ± … N s m–2 [1]

6 marks

Mark scheme: V π P r 1 (a) = t 8C l C = [π × 2.5 × 103 × (0.75 × 10–3)4] / (8 × 1.2 × 10–6 × 0.25) C1 = 1.04 × 10–3 N s m–2 A1 [2] (b) 4 × %r C1 %C = %P + 4 × %r + %V/t + %l = 2% + 5.3% + 0.83% + 0.4% (= 8.6%) A1 ∆C = ± 0.089 × 10–3 N s m–2 A1 [3] (c) C = (1.04 ± 0.09) × 10–3 N s m–2 A1 [1] 2 2

This question in 9702/22 May/June 2012

2 The time T for a satellite to orbit the Earth is given by For KR3 Examiner’sUse T = c m M where R is the distance of the satellite from the centre of the Earth, M is the mass of the Earth, and K is a constant. (a) Determine the SI base units of K. SI base units of K … [2] (b) Data for a particular satellite are given in Fig. 2.1. quantity measurement uncertainty T 8.64 × 104 s ± 0.5% R 4.23 × 107 m ± 1% M 6.0 × 1024 kg ± 2% Fig. 2.1 Calculate K and its actual uncertainty in SI units. K = … ± … SI units [4]

6 marks

Mark scheme: 2 (a) SI units for T: s, R: m and M: kg (or seen clearly in formula) C1 s 2kg K = T2 M / R3 units: s2 kg m–3 (allow s2 kg / m3 or 3 ) A1 [2] m (b) % uncertainty in K: 1% (for T) + 3% (for R) + 2% (for M) OR = 6% C1 K = [(86400)2 × 6 × 1024] / (4.23 × 107)3 = 5.918 × 1011 C1 6% of K = 0.355 × 1011 C1 K = (5.9 ± 0.4) × 1011 (SI units) correct power of ten required for both A1 [4] [incorrect % value then max. 1]

This question in 9702/23 Oct/Nov 2013

2 A coin is made in the shape of a thin cylinder, as shown in Fig. 2.1. diameter thickness Fig. 2.1 Fig. 2.2 shows the measurements made in order to determine the density ρ of the material used to make the coin. quantity measurement uncertainty mass 9.6 g ± 0.5 g thickness 2.00 mm ± 0.01 mm diameter 22.1 mm ± 0.1 mm Fig. 2.2 (a) Calculate the density ρ in kg m–3. ρ = … kg m–3 [3] (b) (i) Calculate the percentage uncertainty in ρ. percentage uncertainty = … [3] (ii) State the value of ρ with its actual uncertainty. ρ = … ± … kg m–3 [1]

7 marks

Mark scheme: 2 (a) ρ = m / V C1 V = (π d 2 / 4) × t = 7.67 × 10–7 m3 ρ = (9.6 × 10–3) / [π(22.1/2 × 10–3)2 × 2.00 × 10–3] C1 ρ = 12513 kg m–3 (allow 2 or more s.f.) A1 [3] (b) (i) ∆ρ / ρ = ∆m / m + ∆t / t + 2∆d / d C1 = 5.21% + 0.50% + 0.905% [or correct fractional uncertainties] C1 = 6.6% (6.61%) A1 [3] (ii) ρ = 12 500 ± 800 kg m–3 A1 [1]

This question in 9702/22 May/June 2014

3 The resistance R of a uniform metal wire is measured for different lengths l of the wire. The variation with l of R is shown in Fig. 3.1. 4.0 3.0 R / 1 2.0 1.0 0 0 0.20 0.40 0.60 0.80 1.00 l / m Fig. 3.1 (a) The points shown in Fig. 3.1 do not lie on the best-fit line. Suggest a reason for this. … … [1] (b) Determine the gradient of the line shown in Fig. 3.1. gradient = … [2] (c) The cross-sectional area of the wire is 0.12 mm2. Use your answer in (b) to determine the resistivity of the metal of the wire. resistivity = … Ω m [3] (d) The resistance R of different wires is measured. The wires are of the same metal and same length but have different cross-sectional areas A. On Fig. 3.2, sketch a graph to show the variation with A of R. R 0 0 A Fig. 3.2 [2]

8 marks

Mark scheme: 3 (a) random error (in the measurements) of the length OR resistance B1 [1] (b) gradient = (3.6 – 1.9 ) / (0.8 – 0.4) C1 = 4.25 A1 [2] (c) R = ρl / A C1 ρ = gradient × area = 4.25 × 0.12 × 10–6 C1 = 5.1(0) × 10–7 Ω m A1 [3] (d) resistance decreasing with increasing area B1 correct shape with curve being asymptote to both axes B1 [2]

This question in 9702/21 Oct/Nov 2014

2 (a) Describe the effects, one in each case, of systematic errors and random errors when using a micrometer screw gauge to take readings for the diameter of a wire. systematic errors: … … random errors: … … [2] (b) Distinguish between precision and accuracy when measuring the diameter of a wire. precision: … … accuracy: … … [2] [Total: 4]

4 marks

Mark scheme: 2 (a) systematic: the reading is larger or smaller than (or varying from) the true reading by a constant amount B1 random: scatter in readings about the true reading B1 [2] (b) precision: the size of the smallest division (on the measuring instrument) or 0.01 mm for the micrometer B1 accuracy: how close (diameter) value is to the true (diameter) value B1 [2]

This question in 9702/23 May/June 2016

1 (a) Determine the SI base units of stress. Show your working. base units … [2] (b) A beam PQ is clamped so that the beam is horizontal. A mass M of 500 g is hung from end Q and the beam bends slightly, as illustrated in Fig. 1.1. clamp R l horizontal P Q M Fig. 1.1 The length l of the beam from the edge of the clamp R to end Q is 60.0 cm. The width b of the beam is 30.0 mm and the thickness d of the beam is 5.00 mm. The material of the beam has Young modulus E. The mass M is made to oscillate vertically. The time period T of the oscillations is 0.58 s. The period T is given by the expression 4 Ml 3 T = 2π 3 . Ebd (i) Determine E in GPa. E = … GPa [3] (ii) The quantities used to determine E should be measured with accuracy and with precision. 1. Explain the difference between accuracy and precision. accuracy: … … precision: … … [2] 2. In a particular experiment, the quantities l and T are measured with the same percentage uncertainty. State and explain which of these two quantities contributes more to the uncertainty in the value of E. … … … [1] [Total: 8]

8 marks

Mark scheme: 1(a) B1 = kg m–1 s–2 A1 1(b)(i) 0.58 = 2π × [(4 × 0.500 × 0.6003 ) / (E × 0.0300 × 0.005003)]0.5 C1 E = [4π2 × 4 × 0.500 × (0.600)3] / [(0.58)2 × 0.0300 × (0.00500)3] = 1.35 × 1010 (Pa) C1 = 14 (13.5) GPa A1 1(b)(ii)1. (accuracy determined by) the closeness of the value(s)/measurement(s) to the true value B1 (precision determined by) the range of the values/measurements B1 1(b)(ii)2. l is (cubed so) 3 × (percentage/fractional) uncertainty and T is (squared so) 2 × (percentage / fractional) uncertainty and (so) l contributes more B1

This question in 9702/21 May/June 2017

1 One end of a wire is connected to a fixed point. A load is attached to the other end so that the wire hangs vertically. The diameter d of the wire and the load F are measured as d = 0.40 ± 0.02 mm, F = 25.0 ± 0.5 N. (a) For the measurement of the diameter of the wire, state (i) the name of a suitable measuring instrument, … [1] (ii) how random errors may be reduced when using the instrument in (i). … … … [2] (b) The stress σ in the wire is calculated by using the expression 4F σ = . πd 2 (i) Show that the value of σ is 1.99 × 108 N m–2. [1] (ii) Determine the percentage uncertainty in σ. percentage uncertainty = … % [2] (iii) Use the information in (b)(i) and your answer in (b)(ii) to determine the value of σ, with its absolute uncertainty, to an appropriate number of significant figures. σ = … ± … N m–2 [2] [Total: 8]

8 marks

Mark scheme: 1(a)(i) micrometer (screw gauge)/digital calipers B1 1(a)(ii) take several readings (and average) M1 along the wire or around the circumference A1 1(b)(i) σ = 4 × 25 / [π × (0.40 × 10–3)2] = 1.99 × 108 N m–2 or σ = 25 / [π × (0.20 × 10–3)2] = 1.99 × 108 N m–2 A1 1(b)(ii) %F = 2% and %d = 5% or ∆F / F = 0.5 25 and ∆d / d = 0.02 0.4 C1 %σ = 2% + (2 × 5%) or %σ = [0.02 + (2 × 0.05)] × 100 %σ = 12% A1 1(b)(iii) absolute uncertainty = (12 / 100) × 1.99 × 108 = 2.4 × 107 C1 σ = 2.0 × 108 ± 0.2 × 108 N m–2 or 2.0 ± 0.2 × 108 N m–2 A1

This question in 9702/22 Oct/Nov 2017

1 (a) An analogue voltmeter is used to take measurements of a constant potential difference across a resistor. For these measurements, describe one example of (i) a systematic error, … … [1] (ii) a random error. … … [1] (b) The potential difference across a resistor is measured as 5.0 V ± 0.1 V. The resistor is labelled as having a resistance of 125 Ω ± 3%. (i) Calculate the power dissipated by the resistor. power = … W [2] (ii) Calculate the percentage uncertainty in the calculated power. percentage uncertainty = … % [2] (iii) Determine the value of the power, with its absolute uncertainty, to an appropriate number of significant figures. power = … ± … W [2] [Total: 8]

8 marks

Mark scheme: 1(a)(i) zero error or wrongly calibrated scale B1 1(a)(ii) reading scale from different angles or wrongly interpolating between scale readings/divisions B1 1(b)(i) P = V 2 / R or P = VI and V = IR C1 P = 5.02 / 125 or 5.0 × 0.04 or (0.04)2 × 125 = 0.20 W A1 1(b)(ii) %V = 2% or ∆V / V = 0.02 C1 %P = (2 × 2%) + 3% or %P = (2 × 0.02 + 0.03) × 100 = 7% A1 1(b)(iii) absolute uncertainty in P = (7 / 100) × 0.20 = 0.014 C1 power = 0.20 ± 0.01 W or (2.0 ± 0.1) × 10–1 W A1

This question in 9702/23 May/June 2018

1 (a) Define velocity. … … [1] (b) The speed v of a sound wave through a gas of pressure P and density ρ is given by the equation kP v = ρ where k is a constant that has no units. An experiment is performed to determine the value of k. The data from the experiment are shown in Fig. 1.1. quantity value uncertainty v 3.3 × 102 m s−1 ± 3% P 9.9 × 104 Pa ± 2% ρ 1.29 kg m−3 ± 4% Fig. 1.1 (i) Use data from Fig. 1.1 to calculate k. k = … [2] (ii) Use your answer in (b)(i) and data from Fig. 1.1 to determine the value of k, with its absolute uncertainty, to an appropriate number of significant figures. k = … ± … [3] [Total: 6]

6 marks

Mark scheme: 1(a) (velocity =) change in displacement / time (taken) B1 1(b)(i) k = [1.29 × (3.3 × 102)2] / 9.9 × 104 C1 = 1.4 A1 1(b)(ii) percentage uncertainty = (3 × 2) + 4 + 2 (= 12%) or fractional uncertainty = (0.03 × 2) + 0.04 + 0.02 (= 0.12) C1 ∆k = 0.12 × 1.42 = 0.17 (allow to 1 significant figure) C1 k = 1.4 ± 0.2 A1

This question in 9702/21 May/June 2019

1 (a) The diameter d of a cylinder is measured as 0.0125 m ± 1.6%. Calculate the absolute uncertainty in this measurement. absolute uncertainty = … m [1] (b) The cylinder in (a) stands on a horizontal surface. The pressure p exerted on the surface by the cylinder is given by 4 W p = 2 . π d The measured weight W of the cylinder is 0.38 N ± 2.8%. (i) Calculate the pressure p. p = … N m−2 [1] (ii) Determine the absolute uncertainty in the value of p. absolute uncertainty = … N m−2 [2] [Total: 4]

4 marks

Mark scheme: 1(a) = 2 × 10–4 m A1 1(b)(i) p = (4 × 0.38) / (π × 0.01252) = 3100 N m–2 A1 1(b)(ii) percentage uncertainty = 2.8 + (2 × 1.6) (= 6%) or fractional uncertainty = 0.028 + (2 × 0.016) (= 0.06) C1 absolute uncertainty = 0.06 × 3100 = 190 N m–2 (allow to 1 significant figure) A1

This question in 9702/22 May/June 2019

1 (a) Make estimates of: (i) the mass, in g, of a new pencil mass = … g [1] (ii) the wavelength of ultraviolet radiation. wavelength = … m [1] (b) The period T of the oscillations of a mass m suspended from a spring is given by m T = 2π k where k is the spring constant of the spring. The manufacturer of a spring states that it has a spring constant of 25 N m–1 ± 8%. A mass of 200 × 10–3 kg ± 4 × 10–3 kg is suspended from the end of the spring and then made to oscillate. (i) Calculate the period T of the oscillations. T = … s [1] (ii) Determine the value of T, with its absolute uncertainty, to an appropriate number of significant figures. T = … ± … s [3] [Total: 6]

6 marks

Mark scheme: 1(a)(i) A1 1(a)(ii) wavelength in range 1 × 10–8 m to 4 × 10–7 m A1 1(b)(i) T = 2π × (200 × 10–3 / 25)0.5 = 0.56 s A1 1(b)(ii) percentage uncertainty = (2% + 8%) / 2 (= 5%) or fractional uncertainty = (0.02+0.08) / 2 (= 0.05) C1 ∆T = 0.56 × 0.05 = 0.028 (s) C1 T = (0.56 ± 0.03) s A1

This question in 9702/21 Oct/Nov 2019

1 (a) Length, mass and temperature are all SI base quantities. State two other SI base quantities. 1. … 2. … [2] (b) The acceleration of free fall g may be determined from an oscillating pendulum using the equation 4π2l g = T 2 where l is the length of the pendulum and T is the period of oscillation. In an experiment, the measured values for an oscillating pendulum are l = 1.50 m ± 2% and T = 2.48 s ± 3%. (i) Calculate the acceleration of free fall g. g = … m s–2 [1] (ii) Determine the percentage uncertainty in g. percentage uncertainty = … % [2] (iii) Use your answers in (b)(i) and (b)(ii) to determine the absolute uncertainty of the calculated value of g. absolute uncertainty = … m s–2 [1] [Total: 6]

6 marks

Mark scheme: 1(a) time (electric) current allow amount of substance allow luminous intensity any two of the above quantities, 1 mark each B2 1(b)(i) g = (4π2 × 1.50) / (2.482) = 9.63 m s–2 A1 1(b)(ii) percentage uncertainty = 2 + (3 × 2) or fraction uncertainty = 0.02 + (0.03 × 2) C1 percentage uncertainty = 8% A1 1(b)(iii) absolute uncertainty = 0.08 × 9.6 = 0.8 m s–2 A1

This question in 9702/22 Feb/March 2020

1 (a) State one similarity and one difference between distance and displacement. similarity: … … difference: … … [2] (b) A student takes several measurements of the same quantity. This set of measurements has high precision, but low accuracy. Describe what is meant by: (i) high precision … … [1] (ii) low accuracy. … … [1] [Total: 4]

4 marks

Mark scheme: 1(a) similarity: both have magnitude B1 difference: distance is a scalar/does not have direction or displacement is a vector/has direction B1 1(b)(i) the measurements have a small range B1 1(b)(ii) the (average of the) measurements is not close to the true value B1

This question in 9702/23 May/June 2020

1 (a) An electromagnetic wave has a wavelength of 85 μm. (i) State the wavelength, in m, of the wave. wavelength = … m [1] (ii) Calculate the frequency, in THz, of the wave. frequency = … THz [2] (iii) State the name of the region of the electromagnetic spectrum that contains this wave. … [1] (b) The current I in a coil of wire produces a magnetic field. The energy E stored in the magnetic field is given by I 2 L E = 2 where L is a constant. The manufacturer of the coil states that the value of L, in SI base units, is 7.5 × 10–6 ± 5%. The current I in the coil is measured as (0.50 ± 0.02) A. The values of L and I are used to calculate E. Determine the percentage uncertainty in the value of E. percentage uncertainty = … % [2] [Total: 6]

6 marks

Mark scheme: 1(a)(i) wavelength = 8.5 × 10–5 m A1 1(a)(ii) f = v / λ or c / λ C1 = 3.0 × 108 / 8.5 × 10–5 (= 3.5 × 1012) = 3.5 THz A1 1(a)(iii) infrared B1 1(b) (implied) percentage uncertainty in I = 4% or (implied) fractional uncertainty in I = 0.04 C1 percentage uncertainty in E = 5% + (4% × 2) = 13% A1

This question in 9702/23 Oct/Nov 2020

1 (a) Define density. … … [1] (b) A smooth pebble, made from uniform rock, has the shape of an elongated sphere as shown in Fig. 1.1. r L Fig. 1.1 The length of the pebble is L. The cross-section of the pebble, in the plane perpendicular to L, is circular with a maximum radius r. A student investigating the density of the rock makes measurements to determine the values of L, r and the mass M of the pebble as follows: L = (0.1242 ± 0.0001) m r = (0.0420 ± 0.0004) m M = (1.072 ± 0.001) kg. (i) State the name of a measuring instrument suitable for making this measurement of L. … [1] (ii) Determine the percentage uncertainty in the measurement of r. percentage uncertainty = … % [1] (c) The density ρ of the rock from which the pebble in (b) is composed is given by Mr n ρ = kL where n is an integer and k is a constant, with no units, that is equal to 2.094. (i) Use SI base units to show that n is equal to –2. [2] (ii) Calculate the percentage uncertainty in ρ. percentage uncertainty = … % [3] (iii) Determine ρ with its absolute uncertainty. Give your values to the appropriate number of significant figures. ρ = ( … ± … ) kg m–3 [3] [Total: 11]

11 marks

Mark scheme: 1(a) mass / volume B1 1(b)(i) (vernier/digital) calipers B1 1(b)(ii) percentage uncertainty = (0.0004 / 0.0420) × 100 = 1% A1 1(c)(i) kg m–3 = kg × mn / m or kg m–3 = kg × mn × m–1 M1 –3 = n – 1 and (so) n = –2 A1 1(c)(ii) (Δρ / ρ) = (ΔM / M) + 2(Δr / r) + (ΔL / L) C1 percentage uncertainty = [(0.001 / 1.072) + 2 × (0.0004 / 0.0420) + (0.0001 / 0.1242)] (× 100) C1 = 0.09% + 2 × 0.95% + 0.08% = 2% A1 1(c)(iii) ρ = (1.072 × 0.0420–2) / (2.094 × 0.1242) = 2337 (kg m–3) C1 ∆ρ = 0.021 × 2337 = 49 (kg m–3) C1 ρ = (2340 ± 50) kg m–3 A1

This question in 9702/21 Oct/Nov 2021

1 (a) A unit may be stated with a prefix that represents a power-of-ten multiple or submultiple. Complete Table 1.1 to show the name and symbol of each prefix and the corresponding power-of-ten multiple or submultiple. Table 1.1 power-of-ten multiple prefix or submultiple kilo (k) 103 tera (T) ( ) 10–12 [2] (b) In the following list, underline all the units that are SI base units. ampere coulomb metre newton [1] (c) The potential difference V between the two ends of a uniform metal wire is given by 4ρLI V = 2 πd where d is the diameter of the wire, I is the current in the wire, L is the length of the wire, and ρ is the resistivity of the metal. For a particular wire, the percentage uncertainties in the values of some of the above quantities are listed in Table 1.2. Table 1.2 quantity percentage uncertainty d ± 3.0% I ± 2.0% L ± 2.5% V ± 3.5% The quantities listed in Table 1.2 have values that are used to calculate ρ as 4.1 × 10–7 Ω m. For this value of ρ, calculate: (i) the percentage uncertainty percentage uncertainty = … % [2] (ii) the absolute uncertainty. absolute uncertainty = … Ω m [1] [Total: 6]

6 marks

Mark scheme: 1(a) 1012 B1 pico (p) B1 1(b) ampere and metre both underlined (and no other units underlined) B1 1(c)(i) percentage uncertainty = 3.5 + (3.0 × 2) + 2.5 + 2.0 C1 = 14% A1 1(c)(ii) absolute uncertainty = 4.1 × 10–7 × 14 / 100 = 6 × 10–8 Ω m A1

This question in 9702/22 Oct/Nov 2021

1 (a) In the following list, underline all units that are SI base units. ampere degree Celsius kilogram newton [1] (b) Fig. 1.1 shows a horizontal beam clamped at one end with a block attached to the other end. block direction of oscillations clamp beam Fig. 1.1 The block is made to oscillate vertically. The Young modulus E of the material of the beam is given by kM E = T2 where M is the mass of the block, T is the period of the oscillations and k is a constant. A student determines the values and percentage uncertainties of k, M and T. Table 1.1 lists the percentage uncertainties. Table 1.1 percentage quantity uncertainty k ± 2.1% M ± 0.6% T ± 1.5% The student uses the values of k, M and T to calculate the value of E as 8.245 × 109 Pa. (i) Calculate the percentage uncertainty in the value of E. percentage uncertainty = … % [2] (ii) Use your answer in (b)(i) to determine the value of E, with its absolute uncertainty, to an appropriate number of significant figures. E = ( … ± … ) × 109 Pa [2] [Total: 5]

5 marks

Mark scheme: 1(a) only ampere and kilogram underlined B1 1(b)(i) percentage uncertainty = 2.1 + 0.6 + (1.5  2) C1 = 5.7% A1 1(b)(ii) absolute uncertainty = (5.7 / 100)  8.245  109 ( = 4.7  108 Pa or 0.47  109 Pa) C1 E = (8.2 ± 0.5)  109 Pa A1

This question in 9702/22 May/June 2022

1 A solid metal sphere has a diameter of (3.42 ± 0.02) cm and a mass of (67 ± 2) g. (a) Calculate the density, in g cm–3, of the metal. density = … g cm–3 [3] (b) Determine the percentage uncertainty in the density. percentage uncertainty = … % [2] [Total: 5]

5 marks

Mark scheme: 1(a) C1 V = (4 / 3)    r3 = (4 / 3)    (3.42 / 2)3 ( = 20.9 cm3) C1  = 67 / 20.9 = 3.2 g cm−3 A1 1(b) % = %m + 3  %d = [(2 / 67)  100] + [3  (0.02 / 3.42)  100] C1 = 3.0% + 3  0.58% = 4.7% or 5% A1

This question in 9702/23 May/June 2022

1 (a) The boxes in Fig. 1.1 contain terms on the left-hand side and examples of these terms on the right-hand side. Draw a line between each term on the left and the correct example on the right. base quantity coulomb base unit electric current derived quantity force derived unit kilogram Fig. 1.1 [2] (b) A set of experimental measurements is described as precise and not accurate. State what is meant by: (i) precise … … [1] (ii) not accurate. … … [1] (c) An object of mass m travels with speed v in a circle of radius r. The force F acting on the object is given by mv2 F = . r The percentage uncertainties of three of the quantities are given in Table 1.1. Table 1.1 quantity percentage uncertainty F ± 3% m ± 4% r ± 5% The value of v is determined from F, m and r. (i) Calculate the percentage uncertainty in v. percentage uncertainty = … % [2] (ii) The value of v is 15.0 m s–1. Calculate the absolute uncertainty in v. absolute uncertainty = … m s–1 [1] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) C1 any two joined correctly all four joined correctly A1 1(b)(i) the measurements have a small range B1 1(b)(ii) (average of the) measurements not close to the true value B1 1(c)(i) percentage uncertainty = (3 + 5 + 4) / 2 C1 = 6% A1 1(c)(ii) absolute uncertainty = (6 / 100)  15.0 A1 = 0.9 ms–1

This question in 9702/21 Oct/Nov 2022

1 (a) State what is meant by work done. … … [1] (b) Use the answer to (a) to determine the SI base units of power. SI base units … [2] (c) The maximum useful output power P of a car travelling on a horizontal road is given by P = v 3 b where v is the maximum speed of the car and b is a constant. For the car, P = 84 kW ± 5% and b = 0.56 ± 7% in SI units. (i) Calculate the value of v. v = … m s–1 [2] (ii) Determine the absolute uncertainty in the value of v. absolute uncertainty = … m s–1 [2] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) force  displacement in the direction of the force B1 1(b) P = Fs / t C1 = (kg m s–2  m) / s = kg m2 s–3 A1 1(c)(i) 84  103 = v 3  0.56 C1 v = 53 m s–1 A1 1(c)(ii) percentage uncertainty = (5% + 7%) / 3 (= 4%) C1 or fractional uncertainty = (0.05 + 0.07) / 3 (= 0.04) absolute uncertainty = 0.04  53 A1 = () 2 m s–1

This question in 9702/22 Oct/Nov 2022

1 The rate of flow Q of a liquid along a narrow pipe of length L and radius r is given by αr 4 Q = L where α is a constant. An experiment is carried out to determine the value of α. The data from the experiment are shown in Table 1.1. Table 1.1 quantity value percentage uncertainty Q 2.72 × 10–8 m3 s–1 ± 3% r 7.1 × 10–5 m ± 2% L 2.5 × 10–2 m ± 4% (a) Use information in Table 1.1 to show that the SI base unit of α is s–1. [1] (b) Show that the percentage uncertainty in α is 15%. [1] (c) Calculate α with its absolute uncertainty. Give your answer to an appropriate number of significant figures. α = ( … ± … ) × 107 s–1 [3] [Total: 5]

5 marks

Mark scheme: Question Answer Marks 1(a) (SI base unit of =) m3 s–1  m / m4 = s–1 A1 1(b) (percentage uncertainty =) 3 + 4 + 2  4 = 15 (%) A1 1(c)  = QL / r 4 C1 = 2.72  10–8  2.5  10–2 / (7.1  10–5)4 = 2.7  107 absolute uncertainty = 0.15  [2.7  107] C1 = 0.4  107 = (2.7 ± 0.4)  107 s–1 A1

This question in 9702/23 Oct/Nov 2022

1 (a) Underline all the SI base units in the following list. ampere coulomb current kelvin newton [1] (b) A toy car moves in a horizontal straight line. The displacement s of the car is given by the equation v 2 s = 2a where a is the acceleration of the car and v is its final velocity. State two conditions that apply to the motion of the car in order for the above equation to be valid. 1 … 2 … [2] (c) An experiment is performed to determine the acceleration of the car in (b). The following measurements are obtained: s = 3.89 m ± 0.5% v = 2.75 m s–1 ± 0.8%. (i) Calculate the acceleration a of the car. a = … m s–2 [1] (ii) Determine the percentage uncertainty, to two significant figures, in a. percentage uncertainty = … % [2] (iii) Use your answers in (c)(i) and (c)(ii) to determine the absolute uncertainty in the calculated value of a. absolute uncertainty = … m s–2 [1] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) only ampere and kelvin underlined B1 1(b) initial speed / velocity is zero B1 (non-zero magnitude of) acceleration is constant / uniform (and in a straight line) B1 1(c)(i) a = 2.752 / (2  3.89) A1 = 0.97 m s–2 1(c)(ii) percentage uncertainty = (2  0.8) + 0.5 C1 = 2.1% A1 1(c)(iii) absolute uncertainty = (2.1 / 100)  0.97 A1 = 0.02 m s–2

This question in 9702/22 Feb/March 2023

1 (a) (i) Define pressure. … … [1] (ii) Use the answer to (a)(i) to show that the SI base units of pressure are kg m–1 s–2. [1] (b) A horizontal pipe has length L and a circular cross‑section of radius R. A liquid of density ρ flows through the pipe. The mass m of liquid flowing through the pipe in time t is given by π(p2 – p1)R 4ρt m = 8kL where p1 and p2 are the pressures at the ends of the pipe and k is a constant. Determine the SI base units of k. SI base units … [3] (c) An experiment is performed to determine the value of k by measuring the values of the other quantities in the equation in (b). The values of L and R each have a percentage uncertainty of 2%. State and explain, quantitatively, which of these two quantities contributes more to the percentage uncertainty in the calculated value of k. … … … [1] [Total: 6]

6 marks

Mark scheme: 1(a)(i) force / area (normal to the force) B1 1(a)(ii) (p = F / A so units are) kg m s–2 / m2 = kg m–1 s–2 A1 1(b) unit of R: m and unit of t: s and unit of L: m C1 unit of : kg m–3 or  = m / V C1 base units of k: (kg m–1 s–2  m4  kg m–3  s) / (kg  m) = kg m–1 s–1 A1 1(c) R contributes 4  2% or 8% (and L contributes 2%) so R contributes more (to the percentage uncertainty in k) B1

This question in 9702/22 May/June 2023

1 A well has a depth of 36 m from ground level to the surface of the water in the well, as shown in Fig. 1.1. ground 36 m well surface of water Fig. 1.1 (not to scale) A student wishes to find the depth of the well. The student plans to drop a stone down the well and record the time taken from releasing the stone to hearing the splash made by the stone as it enters the water. (a) Assume that air resistance is negligible and that the stone is released from rest. Calculate the time taken for the stone to fall from ground level to the surface of the water. time = … s [2] (b) The time recorded by the student using a stop-watch is not equal to the time in (a). Suggest three possible reasons, other than the effect of air resistance, for this difference. 1 … … 2 … … 3 … … [3] (c) The student repeats the experiment three times and uses the results to calculate the depth of the well. The values are shown in Table 1.1. Table 1.1 1st experiment 2nd experiment 3rd experiment depth / m 54.4 53.9 54.1 The true depth of the well is 36.0 m. Explain why these results may be described as precise but not accurate. … … … … [2] [Total: 7]

7 marks

Mark scheme: 1(a) t = √(2s / g) = √[(2  36) / 9.81] C1 = 2.7 s A1 1(b)  reaction time between hearing the splash and stopping the stop-watch  the sound (of the splash) takes time to reach the student or the stone hits the water at a different time to the sound being heard or the sound (of the splash) has to travel to the student  the student might not let go of the stone from ground level  the student might not let go of the stone and start the stop-watch at the same time  stop-watch may not be properly calibrated / has a zero error  (local value of) g is not (exactly) 9.81 (m s2)  stone given initial velocity / initial velocity not zero  stone does not fall (exactly) vertically / in a straight line Any three points, 1 mark each B3 1(c) precise: results are close together / have little scatter B1 not accurate: the values are not close to / 50% different / (very) different from the true value B1

This question in 9702/23 May/June 2023

1 (a) Compare scalar and vector quantities. … … … [2] (b) The radius of a small sphere is determined from a measurement of the volume of the sphere. The sphere is submerged in water, displacing some of the water into a measuring cylinder as shown in Fig. 1.1. measuring cylinder sphere displaced water Fig. 1.1 (not to scale) The measured volume of displaced water is (28.0 ± 0.5) cm3. Calculate: (i) the radius, in cm, of the sphere radius = … cm [1] (ii) the percentage uncertainty in the radius of the sphere. percentage uncertainty = … % [2] [Total: 5]

5 marks

Mark scheme: Question Answer Mark 1(a) scalar and vector have magnitude B1 vector has direction (and scalar does not have direction) B1 1(b)(i) r = [(3  28) / 4]1/3 A1 = 1.9 cm 1(b)(ii) percentage uncertainty in V = (0.5 / 28)  100 C1 ( = 1.79%) percentage uncertainty in r = 1.79 / 3 A1 = 0.6%

This question in 9702/21 Oct/Nov 2023

1 (a) The list below shows some SI quantities. Underline the quantity that is not an SI base quantity. charge current length time [1] (b) A square solar panel with sides of length 1300 mm is shown in Fig. 1.1. incident light solar panel 1300 mm 1300 mm Fig. 1.1 (not to scale) Light is incident normally on the solar panel. (i) The power of the light incident on the solar panel is 750 W. Calculate the intensity of the light. intensity = … W m–2 [3] (ii) The percentage uncertainty in the incident power is ± 3%. The uncertainty in the length of each side is ± 5 mm. Calculate the percentage uncertainty in the intensity of the light. percentage uncertainty = … % [2] (iii) The useful power output of the solar panel is 160 W. Calculate the percentage efficiency of the solar panel. efficiency = … % [1] (iv) Another square solar panel is placed so that light of the same intensity is incident normally on it. The new panel has shorter sides than the original panel. The new panel has the same power output as the original panel. State and explain whether the efficiency of the new panel is greater than, less than or the same as the efficiency of the original panel. … … … … … [3] [Total: 10]

10 marks

Mark scheme: 1(a) charge underlined (and no others) B1 1(b)(i) I = P / A C1 = 750 / (1300  10–3)2 C1 = 440 W m–2 A1 1(b)(ii) percentage uncertainty = 3 + 2  (5 / 1300)  100 C1 = 3 + 2  0.38 = (±) 4% A1 1(b)(iii) efficiency = useful output power / total input power = (160 / 750)  100 = 21% A1 1(b)(iv) area (of the new panel) is less B1 input power (of the new panel) is less (than the input power of the original panel) (as intensity is constant) B1 (useful power output is unchanged so) efficiency is greater (than the original panel) B1

This question in 9702/22 May/June 2024

1 (a) Explain what is meant by the accuracy of a measured value. … … [1] (b) Two solid cubes, A and B, are measured to determine the density of their materials. Table 1.1 shows the measurements for cube A. Table 1.1 quantity measurement length of side (1.53 ± 0.01) cm mass (31.3 ± 0.5) g (i) Show that the calculated density of the material of cube A is 8.7 × 103 kg m–3. [2] (ii) Calculate the percentage uncertainty in the density of the material of cube A. percentage uncertainty = … % [2] (iii) The density of the material of cube B is determined to be 9.2 × 103 kg m–3 ± 6%. State and explain whether cube A and cube B could be made from the same material. … … … … [2] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) how close the (measured) value is to the true value (of the quantity) A1 1(b)(i) (=) m / V C1 = 31.3  10–3 / (1.53  10–2)3 = 8.7  103 (kg m–3) A1 1(b)(ii) (0.5 / 31.3) or (0.01 / 1.53) C1 % uncertainty = (0.016  100) + 3  (0.0065  100) = 4% A1 1(b)(iii) The ranges (of the densities of A and B) overlap or M1 the (calculated) density of A is within the range of the density of B or the difference in densities is within the uncertainty (of B) (so) they could be the same A1

This question in 9702/22 Feb/March 2025

1 Scientists are investigating the variation in air pressure at different locations on a mountain. (a) The scientists take measurements of several physical quantities at each location. Complete Table 1.1 by stating the SI base unit for each quantity and identifying with a tick (3) whether each quantity is a scalar or a vector. Use the space for any working. Table 1.1 quantity measured SI base unit scalar vector air temperature air pressure [2] (b) (i) At one location, the density of the air is 1.1 kg m–3. A spherical weather balloon is filled with a gas and released from rest. The balloon has radius 0.90 m. Calculate the upthrust acting on the balloon when it is released. upthrust = … N [2] (ii) Explain why an upthrust acts on the balloon. … … … … [2] (iii) The balloon has weight 19 N. Calculate the magnitude of the initial acceleration of the balloon. acceleration = … m s–2 [3] (c) A quantity c relating to the motion of the balloon is calculated from three measured quantities k, F and v using the formula 2kF c = . v 2 The percentage uncertainties in the measured quantities are given in Table 1.2. Table 1.2 measured quantity percentage uncertainty k 5% F 3% v 4% The calculated value of c is 1.8. Determine the absolute uncertainty in c. absolute uncertainty = … [2] [Total: 11]

11 marks

Mark scheme: Question Answer Marks 1(a) air temperature: K and air pressure: kg m–1 s–2 B1 scalar only ticked for both air temperature and air pressure B1 1(b)(i) upthrust = 1.1  9.81  (4  0.903 / 3) C1 = 33 N A1 1(b)(ii) (due to difference in height / depth there is a) difference in pressure between top and bottom (of balloon) B1 (due to pressure difference, upwards) B1 force on bottom of balloon is greater (than downwards force on top of balloon, so resultant force is upwards) 1(b)(iii) ()F = 33 – 19 C1 (= 14 N) m = 19 / 9.81 C1 ( = 1.94 kg) a = (33 – 19) / (19 / 9.81) A1 = 7.2 m s–2 1(c) 5 + 3 + (2  4) C1 (= 16%) absolute uncertainty in c = 1.8  0.16 A1 = (±) 0.3

This question in 9702/22 Oct/Nov 2025