A-Level Physics: Required Practicals and Uncertainties - Worksheets, Questions and Revision

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A-Level · Physics

ARP-Phys A-Level Physics: Required Practicals and Uncertainties

AQA 7408 · Calculator allowed · about 135 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A student is preparing to determine the resistivity of a length of constantan wire. Before assembling the circuit, they use a digital micrometer screw gauge to measure the diameter of the wire at several points along its length.

(a)With the micrometer's jaws fully closed, it reads +0.02 mm instead of zero. State the type of error this represents, and state how the student should use this reading to correct each subsequent measurement of the wire's diameter.(2)

(b)After correcting for the zero error, the student measures the diameter at five points along the wire and obtains: 0.54 mm, 0.56 mm, 0.55 mm, 0.53 mm, 0.57 mm. Calculate the mean diameter.(1)

(c)Calculate the absolute uncertainty in the mean diameter, taking it as half the range of the five readings.(2)

(d)Calculate the percentage uncertainty in the mean diameter, and explain why the percentage uncertainty in the cross-sectional area A = π x d² / 4 is twice this value.(3)

(Total for Question 1 is 8 marks)

This question tests general ideas about measurement uncertainty used throughout the required practicals.

(a)Which of the following is an example of a systematic error?
A) A student's reaction time varies each time they start a stopwatch.
B) A top-pan balance is not zeroed correctly, so every reading is 0.05 g too high.
C) The count rate from a radioactive source fluctuates randomly from one 10 s interval to the next.
D) Mains-voltage fluctuations cause small variations in a digital meter's readings.(1)

A) A student's reaction time varies each time they start a stopwatch.
B) A top-pan balance is not zeroed correctly, so every reading is 0.05 g too high.
C) The count rate from a radioactive source fluctuates randomly from one 10 s interval to the next.
D) Mains-voltage fluctuations cause small variations in a digital meter's readings.

(b)State the difference between the resolution of a measuring instrument and the accuracy of a measurement.(2)

(c)A student uses a metre ruler (resolution 1 mm) to measure the extension of a spring, obtaining a single reading of 12.0 mm. Taking the absolute uncertainty as half the resolution, calculate the percentage uncertainty in this reading.(2)

(d)The student instead measures the total extension of 10 identical springs connected in series, obtaining 118 mm, then divides by 10 to find the extension of one spring. Explain why this method reduces the percentage uncertainty in the extension of one spring.(2)

(Total for Question 2 is 7 marks)

The wire from question 1 (mean diameter 0.550 mm, length L = 0.800 m, measured with a metre ruler to ± 0.001 m) is now connected into a circuit. The student measures its resistance as R = 3.42 ohm ± 0.05 ohm.

(a)Calculate the cross-sectional area A of the wire, in m², using the mean diameter found in question 1.(2)

(b)Use resistivity = R x A / L to calculate the resistivity of the wire.(2)

(c)The percentage uncertainty in R is 1.5%, in L is 0.13%, and in d is 3.6% (from Q1(d)). Calculate the percentage uncertainty in the resistivity, given that resistivity depends on d².(3)

(d)Calculate the absolute uncertainty in the resistivity, and state the resistivity with its uncertainty to an appropriate number of significant figures.(2)

(Total for Question 3 is 9 marks)

A student determines g by a free-fall method. A steel ball bearing is held by an electromagnet a measured distance s above a trapdoor switch. Releasing the electromagnet starts an electronic timer, which stops when the ball strikes the trapdoor, giving the fall time t. The results are:
s (m): 0.200, 0.400, 0.600, 0.800, 1.000
t (s): 0.202, 0.286, 0.350, 0.404, 0.451
t² (s²): 0.0408, 0.0818, ?, 0.163, 0.203

(a)Using s = (1/2) g t² (from the Physics Equations Sheet), explain why plotting a graph of s against t² gives a more reliable value of g than calculating g from a single pair of s and t values.(2)

(b)Show that t² = 0.123 s² (3 significant figures) for s = 0.600 m, using the tabulated value of t.(2)

(c)The gradient of the line of best fit through the s against t² data is 4.93 m s^-2. Calculate g.(2)

(d)The uncertainty in the gradient, found from the steepest and shallowest lines that fit within the error bars, is ± 0.15 m s^-2. Calculate the percentage uncertainty and the absolute uncertainty in g, and state g with its uncertainty to an appropriate number of significant figures.(3)

(e)The accepted value of g is 9.81 m s^-2. State, with a reason, whether the student's result is consistent with this value, and suggest one systematic error that could affect the free-fall method, stating its likely effect on the calculated value of g.(3)

(Total for Question 4 is 12 marks)

A student determines the Young modulus of a copper wire using a simple technique: a long test wire (original length L0 = 2.500 m, mean diameter 0.28 mm, percentage uncertainty in diameter 7.1%) is clamped at one end, passed over a pulley, and loaded with slotted masses at the other end, with extension measured against a fixed reference marker.

(a)Calculate the cross-sectional area of the wire, and state the percentage uncertainty in this area.(3)

(b)The gradient of the student's stress-strain graph (the Young modulus) is 1.25 x 10¹¹ Pa, with an uncertainty (from the steepest and shallowest lines of best fit) of ± 0.06 x 10¹¹ Pa. Calculate the percentage uncertainty in the Young modulus.(2)

(c)State three precautions used in this experiment to improve the reliability or precision of the result, other than repeating readings.(3)

(d)A second, thicker wire made from the same copper is tested using the same method. Predict, with a reason, whether the gradient of its stress-strain graph would differ from that of the first wire.(2)

(Total for Question 5 is 10 marks)

A student investigates stationary waves on a stretched string. A vibration generator at one end and a fixed pulley at the other keep a constant tension in the string; for each length L between the generator and the pulley, the student adjusts the frequency until a clear standing-wave pattern (one loop) is seen, and records the resonant frequency f.
L (m): 0.20, 0.30, 0.40, 0.50, 0.60
f (Hz): 150, 100, 75, 60, 50

(a)Explain, in terms of superposition of waves, how a stationary wave pattern is produced on the string.(2)

(b)Using f = v / (2L) (from the Physics Equations Sheet, for the fundamental mode), calculate f x L for each pair of values in the table, and comment on what this shows about the relationship between f and L.(3)

(c)The student plots f (y-axis) against 1/L (x-axis). State the shape of this graph and what physical quantity is given by its gradient.(2)

(d)The wave speed on a stretched string is given by v = √T/&μ; (Physics Equations Sheet), where T is the tension and μ is the mass per unit length. Using v = 60 m s^-1 found in part (b) and a tension of T = 4.5 N, calculate μ.(3)

(e)State and explain the effect on the resonant frequency, for a fixed length L, of increasing the tension in the string.(2)

(Total for Question 6 is 12 marks)

A student investigates the I-V characteristic of a 6 V filament lamp, using a variable power supply with an ammeter in series and a voltmeter across the lamp.
V (V): 0.50, 1.00, 1.50, 2.00, 2.50
I (A): 0.20, 0.32, 0.39, 0.44, 0.48

(a)Describe the shape of the I-V graph for the filament lamp, and explain this shape in terms of the resistance of the filament.(3)

(b)Calculate the resistance of the lamp when V = 2.00 V.(2)

(c)The voltmeter has a percentage uncertainty of 1.0% and the ammeter a percentage uncertainty of 2.5% at this reading. Calculate the percentage and absolute uncertainty in the resistance found in part (b), and state the result to an appropriate number of significant figures.(3)

(d)Both the ammeter and voltmeter have a small but non-zero resistance. Explain why, for a low-resistance component such as this lamp, connecting the voltmeter directly across the lamp only (so the ammeter also carries the voltmeter's own small current) introduces a smaller systematic error than the alternative arrangement.(3)

(Total for Question 7 is 11 marks)

A student investigates radioactive decay using a sealed protactinium generator and a GM tube connected to a counter. The background count rate (measured with no source present) is 20 counts per minute (cpm). Measured (raw) and corrected (background-subtracted) count rates are:
t (s): 0, 30, 60, 90, 120, 150, 180
Measured count rate (cpm): 320, 232, 170, 126, 95, 73, 58
Corrected count rate (cpm): 300, 212, 150, 106, 75, 53, 38

(a)Explain why the background count rate must be measured separately and subtracted from each reading before the decay is analysed.(2)

(b)Show that the corrected count rate at t = 90 s is 106 cpm.(2)

(c)The student plots ln(corrected count rate) against t and obtains a straight line of best fit with gradient -0.01155 s^-1. Use this to determine the decay constant and the half-life of the sample.(3)

(d)Suggest one advantage of finding the half-life from the gradient of a ln(count rate)-time graph, rather than reading a single half-life directly off a raw count rate-time graph.(2)

(Total for Question 8 is 9 marks)

A second student determines g using a simple pendulum, timing 20 oscillations for a single length of L = 1.000 m, obtaining g = 9.75 m s^-2 with a percentage uncertainty of 1.8%. Compare this with the free-fall method of question 4, which gave g = (9.9 ± 0.3) m s^-2, a percentage uncertainty of about 3.0%.

(a)Compare the precision of the two methods, referring to their percentage uncertainties.(2)

(b)Given that the accepted value is g = 9.81 m s^-2, state which method's result is closer to the accepted value, then explain, using these two results, why a method can be precise but not accurate (or vice versa).(3)

(c)Suggest one modification to the pendulum method that would reduce a systematic error in its value of g.(1)

(Total for Question 9 is 6 marks)

A student investigates how the period T of a simple pendulum depends on its length L, to test the relationship T = k Lⁿ and, from it, to find a value for g.
L (m): 0.20, 0.40, 0.60, 0.80, 1.00
T (s): 0.897, 1.269, 1.554, 1.794, 2.006
log10(L): -0.699, -0.398, -0.222, -0.097, 0
log10(T): -0.047, 0.103, 0.191, 0.254, 0.302

(a)Show that taking logarithms of both sides of T = k Lⁿ gives a straight-line relationship between log10(T) and log10(L), and state what the gradient and the y-intercept of this line represent.(3)

(b)Show that, taking g = 9.81 m s^-2 and using the Physics Equations Sheet equation for the period of a simple pendulum, T = 1.55 s (3 significant figures) when L = 0.600 m.(2)

(c)The line of best fit through the log10(T) against log10(L) data has gradient 0.500 and y-intercept 0.302. Determine the value of n, and hence use the y-intercept to find k and a value for g. (Hint: since T = k Lⁿ with n = 0.5, k = 2 π / √g.)(5)

(d)Explain why the period is timed over many oscillations (e.g. 20) rather than a single swing, and suggest, without changing the apparatus, one way the percentage uncertainty in the final value of g could be further reduced.(3)

(Total for Question 10 is 13 marks)

In question 5, a student calculated a percentage uncertainty of 4.8% in the Young modulus of copper, based only on the uncertainty in the gradient of their stress-strain graph. Evaluate whether the overall percentage uncertainty in the final value of the Young modulus is likely to be greater than or smaller than this 4.8%, and suggest specific improvements to the method that would most effectively reduce the overall uncertainty in the result.

(Total for Question 11 is 6 marks)

Mark scheme · ARP-Phys A-Level Physics: Required Practicals and Uncertainties

Question 1

(a) B1 this is a systematic (zero) error
(a) B1 subtract 0.02 mm from every raw diameter reading taken with this micrometer oe
(a) Answer: Systematic (zero) error; subtract 0.02 mm from every raw reading.
(b) B1 0.550 mm cao
(b) Answer: 0.550 mm
(c) M1 range = 0.57 - 0.53 = 0.04 mm, so half range = 0.02 mm (correct method)
(c) A1 ± 0.02 mm cao
(c) Answer: ± 0.02 mm
(d) M1 (0.02/0.550) x 100 (correct substitution)
(d) A1 3.6% (awrt 3.6) cao
(d) B1 A depends on d² (d multiplied by itself), so when two quantities are multiplied their percentage uncertainties add, giving 2 x the percentage uncertainty in d oe
(d) Answer: 3.6%; percentage uncertainty in A = 2 x 3.6% = 7.3%, since A depends on d².

Question 2

(a) B1 B cao
(a) Answer: B
(b) B1 resolution is the smallest change in a quantity that gives a perceptible/detectable change in an instrument's reading (the smallest scale division it can distinguish) oe
(b) B1 accuracy is how close a measurement is to the true (accepted) value of the quantity being measured oe
(b) Answer: Resolution = smallest detectable change in a reading; accuracy = how close a reading is to the true value.
(c) M1 absolute uncertainty = 0.5 mm; (0.5/12.0) x 100 (correct substitution)
(c) A1 4.2% (awrt 4.2) cao
(c) Answer: 4.2%
(d) B1 the absolute uncertainty in reading the ruler is still about 0.5 mm (fixed by the ruler's resolution), but this is now spread over a ten times larger measured length oe
(d) B1 so the percentage uncertainty in the total (0.5/118 x 100 = 0.42%) is about ten times smaller than for a single spring (4.2%), and dividing by 10 to get one spring's extension does not change this percentage oe
(d) Answer: The fixed 0.5 mm reading uncertainty is spread over a ten times longer length, so the percentage uncertainty falls from about 4.2% to about 0.4%.

Question 3

(a) M1 A = π x (0.550 x 10^-3)² / 4 (correct substitution, ft from Q1(b))
(a) A1 2.38 x 10^-7 m² (awrt 2.38) cao
(a) Answer: 2.38 x 10^-7 m²
(b) M1 ρ = 3.42 x 2.38 x 10^-7 / 0.800 (correct substitution, ft their A)
(b) A1 1.02 x 10^-6 ohm m (awrt 1.02) cao
(b) Answer: 1.02 x 10^-6 ohm m
(c) M1 identifies that percentage uncertainties combine by addition for a product/quotient, with the d term doubled: %R + %L + 2 x %d
(c) M1 1.5 + 0.13 + 2(3.6) (correct substitution)
(c) A1 8.8% (awrt 8.8) cao
(c) Answer: 8.8%
(d) M1 0.088 x 1.02 x 10^-6 (correct method, ft their percentage uncertainty and ρ)
(d) A1 ρ = (1.02 ± 0.09) x 10^-6 ohm m oe
(d) Answer: ρ = (1.02 ± 0.09) x 10^-6 ohm m

Question 4

(a) B1 using several data points and a line of best fit averages out random error/timing variation across all the readings, rather than relying on one possibly anomalous pair oe
(a) B1 the gradient method also allows any anomalous points to be identified and excluded, and does not require the line to pass exactly through the origin, so it is not affected by a fixed (systematic) timing offset in the same way a single calculation would be oe
(a) Answer: Using the gradient of a graph of many points averages out random timing errors and allows anomalies to be spotted, unlike a single calculation.
(b) M1 0.350² (correct substitution of the tabulated t)
(b) A1 0.1225, which rounds to 0.123 s² cso (answer given; accept 0.122 if the more precise unrounded t is used)
(b) Answer: 0.123 s²
(c) M1 g = 2 x gradient (correct rearrangement of s = (1/2) g t²)
(c) A1 9.86 m s^-2 (awrt 9.86) cao
(c) Answer: 9.86 m s^-2
(d) M1 percentage uncertainty in gradient = (0.15/4.93) x 100 = 3.0%; this is unchanged by doubling to find g, since multiplying by a constant does not change a percentage uncertainty
(d) A1 absolute uncertainty = 0.030 x 9.86 = 0.3 m s^-2 (awrt 0.3)
(d) B1 g = (9.9 ± 0.3) m s^-2 oe
(d) Answer: g = (9.9 ± 0.3) m s^-2
(e) B1 the range (9.6 to 10.2) m s^-2 includes 9.81 m s^-2 oe
(e) B1 so the result is consistent with the accepted value
(e) B1 a valid systematic error with correct direction, e.g. air resistance on the ball reduces its true acceleration below g, making the measured/calculated g too low; or a delay between the electromagnet switching off and the ball actually starting to fall means the recorded t is too short, making the calculated g too high (any one, oe)
(e) Answer: Consistent, since 9.81 lies within (9.9 ± 0.3) m s^-2; e.g. air resistance would make the measured g too low.

Question 5

(a) M1 A = π x (0.28 x 10^-3)² / 4 (correct substitution)
(a) A1 6.16 x 10^-8 m² (awrt 6.16) cao
(a) B1 14% (= 2 x 7.1%, since A depends on d²) cao
(a) Answer: A = 6.16 x 10^-8 m²; percentage uncertainty = 14%
(b) M1 (0.06/1.25) x 100 (correct substitution)
(b) A1 4.8% (awrt 4.8) cao
(b) Answer: 4.8%
(c) B1 the wire is loaded and unloaded to check it returns to its original length (no permanent extension), confirming the elastic limit has not been exceeded oe
(c) B1 a long, thin wire is used, which gives a larger, more easily measured extension for a given load, reducing the percentage uncertainty in the extension oe
(c) B1 a second, unloaded reference wire of the same material is mounted alongside the test wire, so any sagging of the support or change in length due to a temperature change affects both wires equally and can be compensated for oe
(c) Answer: Check for elastic behaviour by unloading; use a long thin wire for a larger extension; use a reference wire to compensate for sagging/temperature effects.
(d) B1 the gradient (Young modulus) would be the same/unchanged
(d) B1 because dividing by cross-sectional area and original length in stress and strain already accounts for the wire's dimensions, so the Young modulus is a property of the material only, not of the sample's size oe
(d) Answer: The gradient (Young modulus) would be unchanged, since it is a property of the material, independent of the wire's dimensions.

Question 6

(a) B1 a progressive wave travels along the string and is reflected at the fixed end (pulley), so two waves of the same frequency and amplitude travel in opposite directions along the string oe
(a) B1 these waves superpose; at the resonant frequency this produces a stable pattern of fixed nodes (no displacement) and antinodes (maximum displacement) oe
(a) Answer: Waves travelling in opposite directions (incident and reflected) superpose to form a fixed pattern of nodes and antinodes.
(b) M1 calculates f x L for at least three pairs: 0.20x150=30, 0.30x100=30, 0.40x75=30, 0.50x60=30, 0.60x50=30
(b) A1 all five products are equal (constant at 30 m s^-1, within experimental variation)
(b) A1 this confirms f is inversely proportional to L (f = v/(2L)), with wave speed v = 2 x 30 = 60 m s^-1
(b) Answer: fL = 30 m/s for every row, confirming f is inversely proportional to L; wave speed v = 60 m/s
(c) B1 a straight line through the origin
(c) B1 gradient = v/2, so the wave speed v on the string is twice the gradient oe
(c) Answer: A straight line through the origin; gradient = v/2, so v is twice the gradient.
(d) M1 rearranges to μ = T / v²
(d) M1 4.5 / 60² (correct substitution, ft their v)
(d) A1 1.25 x 10^-3 kg m^-1 (awrt 1.25) cao
(d) Answer: 1.25 x 10^-3 kg m^-1
(e) B1 increasing T increases v, since v = √T/&μ;
(e) B1 with L fixed, f = v/(2L), so increasing v increases the resonant frequency f oe
(e) Answer: The resonant frequency increases, because increasing tension increases the wave speed v, and f = v/(2L) with L fixed.

Question 7

(a) B1 the graph curves, with the gradient (proportional to 1/R) decreasing as V (and I) increase oe
(a) B1 as current increases, the filament's temperature increases oe
(a) B1 the increased temperature increases the amplitude of vibration of the lattice ions, increasing the frequency of collisions with the conduction (charge-carrying) electrons, which increases resistance oe
(a) Answer: The graph curves and flattens as V increases, because heating the filament raises its resistance (more ion vibration, more electron collisions).
(b) M1 R = 2.00/0.44 (correct substitution)
(b) A1 4.5 ohm (awrt 4.5) cao
(b) Answer: 4.5 ohm
(c) M1 %R = %V + %I = 1.0 + 2.5 = 3.5% (correct combining rule for a quotient)
(c) A1 absolute uncertainty = 0.035 x 4.5 = 0.16 ohm (awrt 0.16, ft their R)
(c) B1 R = (4.5 ± 0.2) ohm oe
(c) Answer: R = (4.5 ± 0.2) ohm
(d) B1 a real ammeter has a small non-zero resistance and a real voltmeter has a large but finite (not infinite) resistance, so either arrangement introduces some systematic error oe
(d) B1 with the voltmeter directly across the lamp, the ammeter reading is slightly too high (it includes the small current taken by the voltmeter), but since the voltmeter's resistance is very much greater than the lamp's resistance, this extra current is negligible oe
(d) B1 in the alternative arrangement (ammeter directly next to the lamp, voltmeter across both), the voltmeter reading would include the (larger, comparable) voltage across the ammeter's own resistance, giving a proportionally bigger systematic error for a low-resistance component oe
(d) Answer: Placing the voltmeter directly across the (low-resistance) lamp gives only a negligible ammeter error, since the voltmeter's resistance is much greater than the lamp's; the alternative arrangement would add the ammeter's comparable resistance into the voltage reading.

Question 8

(a) B1 background radiation (e.g. from cosmic rays, rocks/building materials, other radioactive sources nearby) is detected by the GM tube even with no sample present oe
(a) B1 this background is unrelated to the decay of the sample, so it must be subtracted to avoid a systematic error/bias in the corrected count rate used to find the decay constant oe
(a) Answer: Background radiation is always present and unrelated to the source's decay, so it must be subtracted to avoid a systematic error.
(b) M1 126 - 20 (correct method: measured minus background)
(b) A1 106 cpm cso (answer given)
(b) Answer: 106 cpm
(c) M1 decay constant, λ = magnitude of gradient = 0.01155 s^-1 (since ln C = ln C0 - λ t, gradient = -λ)
(c) M1 half-life T(1/2) = ln(2) / λ (correct rearrangement)
(c) A1 T(1/2) = 60.0 s (= 1.00 minute) (awrt 60.0) cao
(c) Answer: λ = 0.01155 s^-1; half-life = 60.0 s (1.00 minute)
(d) B1 the gradient method uses every data point (via a line of best fit), averaging out the random fluctuations inherent in radioactive decay, rather than relying on one possibly anomalous reading oe
(d) B1 taking logarithms converts the exponential decay curve into a straight line, making it much easier to identify anomalous points and to determine the constant (decay rate) precisely from the gradient oe
(d) Answer: Using the gradient of a straight-line log graph averages out random count-rate fluctuations across all readings and makes anomalies easier to spot, rather than relying on a single reading from a curve.

Question 9

(a) B1 the pendulum method (1.8%) has a smaller percentage uncertainty than the free-fall method (3.0%), so it is more precise
(a) B1 valid supporting reason, e.g. timing 20 oscillations and dividing reduces the effect of the (fixed) reaction-time uncertainty far more than timing a single, much shorter, free-fall event oe
(a) Answer: The pendulum method is more precise (1.8% versus 3.0%), largely because timing many oscillations reduces the relative effect of reaction-time uncertainty.
(b) B1 the pendulum result (9.75 m s^-2, 0.06 from 9.81) is closer to the accepted value than the free-fall result (9.9 m s^-2, 0.09 from 9.81), since 0.06 < 0.09; the difference is small, so a comment that they are comparable is also credited provided the correct differences (0.06 vs 0.09) are used
(b) B1 precision describes the spread/repeatability of a set of measurements (a small percentage uncertainty means high precision), while accuracy describes how close a result is to the true value oe
(b) B1 a method can give a small, consistent percentage uncertainty (be precise) yet still be inaccurate if there is an unaccounted systematic error affecting every reading in the same way, so precision alone does not guarantee accuracy oe
(b) Answer: The pendulum result (9.75 m s^-2) is closer to 9.81 m s^-2 (0.06 away) than the free-fall result (9.9 m s^-2, 0.09 away). Precision (small % uncertainty) and accuracy (closeness to the true value) are different: a method can be precise yet inaccurate if it has an uncorrected systematic error, so a smaller % uncertainty does not guarantee a result closer to 9.81 m s^-2.
(c) B1 any valid modification, e.g. keep the angle of swing small (less than about 10 degrees) so the motion approximates simple harmonic motion (the equation T = 2 π √L/g assumes small-angle oscillations); or measure L to the centre of mass of the bob rather than to its top or bottom oe
(c) Answer: Keep the swing angle small (under about 10 degrees) so the pendulum's motion satisfies the small-angle approximation used in T = 2 π √L/g.

Question 10

(a) M1 log10(T) = log10(k) + n log10(L) (correct application of log rules to T = k Lⁿ)
(a) A1 gradient of the log10(T) against log10(L) graph = n
(a) A1 y-intercept (value of log10(T) at log10(L) = 0) = log10(k)
(a) Answer: log10(T) = log10(k) + n log10(L); gradient = n, y-intercept = log10(k).
(b) M1 T = 2 π √0.600/9.81 (correct substitution)
(b) A1 1.55 s cso (answer given)
(b) Answer: 1.55 s
(c) B1 n = 0.500 (the gradient), confirming T is proportional to the square root of L
(c) M1 k = 10^y-intercept (correct rearrangement of log10(k) = y-intercept)
(c) A1 k = 2.00 s m^-0.5 (awrt 2.00) cao
(c) M1 rearranges k = 2 π / √g to give g = (2 π / k)², and substitutes their k
(c) A1 g = 9.9 m s^-2 (accept any value in the range 9.7 to 10.0 m s^-2, due to rounding in k)
(c) Answer: n = 0.500; k = 2.00 s m^-0.5; g is approximately 9.9 m s^-2 (accept 9.7 to 10.0 m s^-2).
(d) B1 the fixed absolute uncertainty from human reaction time is divided by the number of oscillations timed, greatly reducing the percentage uncertainty in the period T oe
(d) B1 use the largest practical range of pendulum lengths L (including the longest achievable length) when taking readings for the graph oe
(d) B1 a longer L gives a longer T for the same fixed reaction-time uncertainty, reducing its percentage uncertainty, and a wider range of L values on the log-log graph reduces the effect of any single anomalous point on the gradient (and hence on g) oe
(d) Answer: Timing many oscillations divides the fixed reaction-time uncertainty across all of them, reducing its percentage effect; using the widest practical range of L (including longer lengths) further reduces the percentage uncertainty in g.