M1.(a)     extension of wire Q = 2.7 (mm)

ignore any precision given eg ± 0.1 mm

if > 2 sf condone if this rounds to 2.7

1

(b)     mass = 5.8 (kg)

allow ce for incorrect 0.1.1 (only look at 01.1 if answer here is incorrect)

allow ± 0.1 kg

1

(c)     0.51 (mm)

ignore any precision given eg ± 0.005 mm

1

(d)     method 1: for 1 expect to see some substitution of numerical data correct use of diameter for 2; ignore power of ten error; expect CSA = 2.0(4) × 10−7; allow ce from 01.3 (eg for d = 0.37 mm CSA = 1.0(8) × 10−7 m2) penalise use of g = 10 N kg−1 value of ∆l must correspond to Figure 2 value of m; answers to 01.1 and 01.2 are acceptable

expect l = 1.82 m but condone 182 etc; accept mixed units for l and ∆l

MAX 3

method 2:

evidence of from Figure 2 to calculate gradient   1

expect gradient between 0.45 to 0.48 mm kg−1 2 3

missing g loses 3

substitution of l = 1.82 m   4

condone 182 etc 4

cross-sectional area from 5

correct use of diameter for 2; ignore power of ten error; expect CSA = 2.0(4) × 10−7; allow ce from 01.3 (eg for d = 0.37 mm CSA = 1.0(8) × 10−7 m2)

MAX 3

result in range 1.84 × 1011 to 1.91 × 1011 5

condone 1.9 × 1011

5 mark requires correct working and no power of ten errors: allow ce for error(s) in 01.1, 01.2 and for false/incorrect CSA (eg for d = 0.37 mm allow result in range 3.49 × 1011 to 3.63 × 1011, 3.5 × 1011 or 3.6 × 1011)

1

(e)     (smaller diameter) produces larger extensions 1

reduces (percentage) uncertainty (in extension and in result for Young Modulus) 2

(smaller diameter) increases (percentage) uncertainty in diameter or cross sectional area is smaller or increases (percentage) uncertainty in cross sectional area 3

increases (percentage) uncertainty (in result for Young Modulus) 4

(smaller diameter) increases likelihood of wire reaching limit of proportionality or of wire snapping or reduces range of readings 5

increases (percentage) uncertainty (in result for Young Modulus) 6

outcome and correct consequence for 2 marks, ie 1 followed by 2, 3 followed by 4 etc

dna ‘error’ for ‘uncertainty’

no mark for consequence if outcome not sensible, eg ‘it gets longer and reduces uncertainty’ earns no mark for ‘diameter smaller so uncertainty greater’ award 1 (need to see further mention of uncertainty to earn 2)

MAX 4



M2.(a)     2.9%

Allow 3%

1

(b) seen

1

0.29 mm or 2.9 x 10-4 m must see 2 sf only

1

(c)     ± 0.01 mm

1

(d)     Clear indication that at least 10 spaces have been measured to give a spacing = 5.24 mm

spacing from at least 10 spaces
Allow answer within range ±0.05

1

(e)     Substitution in d sinθ = nλ

The 25 spaces could appear here as n with sin θ as 0.135 / 2.5

1

d = 0.300 x 10-3 m so
number of lines = 3.34 x10
3

Condone error in powers of 10 in substitution

Allow ecf from 1-4 value of spacing

1

(f)     Calculates % difference (4.6%)

1

and makes judgement concerning agreement

Allow ecf from 1-5 value

1

(g)     care not to look directly into the laser beam
OR
care to avoid possibility of reflected laser beam

OR
warning signs that laser is in use outside the laboratory

ANY ONE

1



M3.(a)     (i)      5.1 and 7.1

1

(ii)     Both plotted points to nearest mm
Best line of fit to points

The line should be a straight line with approximately an equal number of points on either side of the line

2

(iii)    Large triangle drawn at least 8 cm × 8 cm

Correct values read from graph
Gradient value in range 0.190 to 0.210 to 2 or 3 sf

3

(iv)    (R = ) = 5.0          Must have unit

Allow ecf from gradient value

No sf penalty

1

(b)      (i)    5.04 () or 5.0 ()

(Allow also 5.06 or 5.1 , obtained by intermediate rounding up of 3.502)

From R = 1

(ii)     (Uncertainty in V = 0.29% )
Uncertainty in V
2 = 0.57%, 0.58% or 0.6%

From uncertainty in V = 0.01 / 3.50 × 100%

Uncertainty in P = 2.1%

From uncertainty in P = 0.05 / 2.43 × 100% = 2.1%

Uncertainty in R =2.6%, 2.7% or 3%
Answer to 1 or 2 sf only

2.1 % + uncty in V2 (0.6%) = 2.7%
Allow ecf from incorrect uncertainty for V
2 or P

3

(iii)     (Absolute) uncertainty in R is ( ± ) 0.14 or just 0.1 (using 2.6%)
(or 0.15 or 0.2
using 3%)

Must have unit ()
Must be to 1 or 2 sf and must be consistent with sf used from (ii)
No penalty for omitting ± sign

1

(iv)     Works out possible range of values of R based on uncertainty in
(iii), e.g. R is in range 5.0 to 5.2
using uncertainty of ± 0.1

No credit for statement to effect that the values are or are not consistent, without any reference to uncertainty

Allow ecf from (iii)

Value from (a)(iv) is within the calculated range (or not depending on figures, allowing ecf)

Allow ecf from (a)(iv)

2



M4.(a)     Straight line of best fit passing through all error bars Look for reasonable distribution of points on either side

1

(b)     h0 = 165 ± 2 mm

1

(c)     Clear attempt to determine gradient

1

Correct readoffs (within ½ square) for points on line more than 6 cm apart and correct substitution into gradient equation

1

h0k gradient =( -) 0.862 mm K-1 and
negative sign quoted

Condone negative sign
Accept range -0.95 to -0.85

1

(d)     k = = 5.2 x 10-3

Allow ecf from candidate values

1

K-1

Accept range 0.0055 to 0.0049

1

(e)     for h = 8000 mm, d-1 = 1

d = 1.8 x 10-3 mm

1

(f)     Little confidence in this answer because
One of
It is too far to take extrapolation

OR
This is a very small diameter

1



M5.(a)     Clear identification of distance from centre of sphere to right hand end of mark, or to near r.h.end of mark

1

(b)     0.393 (s)

Accept 0.39 (s)

1

(c)     For 10 oscillations percentage uncertainty = = 0.00637 ≡ 0.64%

same for the ¼ period

2

(d)     Identifies anomaly [0.701] and calculates mean distance = 0.759 (m)

Allow 1 max if anomaly included in calculation giving 0.750 (m)

2

(e)     Largest to smallest variation = 0.026 (m)

Absolute uncertainty = 0.013 (m)

1

(f)     Use of g = leading to

9.83 (m s-2)

Allow 9.98 (m s-2) if 0.39 used

Ecf if anomaly included in mean in (d)

percentage uncertainty in distance = 1.7%

Total percentage uncertainty =

1.7 + 2 x 0.64 = 3.0%

Absolute uncertainty = 0.30 (m s-2)

[g = 10.0 ± 0.3 m s-2]

Expressed sf must be consistent with uncertainty calculations

3

(g)     suggests one change

Sensible comment about change or its impact on uncertainty

eg

Use pointed mass not sphere

Because this will give better defined mark OR because the distance determination has most impact on uncertainty

OR

Time more swings / oscillations

As this reduces the percentage uncertainty in timing

OR

longer / heavier bar would take a greater time to swing to the vertical increasing t and s and reducing the percentage uncertainty in each

If data logger proposed, it must be clear what sensors are involved and how the data are used.

2

(h) plot graph of s against t 2 or √s against t

the gradient is g / 2 or √(g / 2)

Accept: plot s against t 2 / 2 or plot 2s against t 2:

in both cases gradient gives g

Allow 1 max for answer that evaluates g for each data point and averages

3



Background count = 20 count/minute unit required

Ignore any –ve sign for background count
Must be written to 2 sf

2

2

(b)     Correct line of best fit

The line must be a straight line (as instructed), with approximately
an equal number of points on either side of the line.

1

(c)     Triangle drawn with smallest side at least 8 cm
(or 8 grid squares)
correct values read from graph

gradient = – 0.00698 (± 0.00030) min
–1
must have –ve sign and must be to 2 or 3 sf

Gradient must lie within limits stated. No ecf from incorrectly read
values unless it falls within stated limits. No unit penalty.

3

(d)     Recognises gradient = (–) λ      or
Uses gradient for value of
λ = 7.0 (± 0.30) × 10–3 minute–1
T½ = 99 minutes      to 2 or 3 sf

For 1st mark accept evidence that value of gradient has been substituted into correct formula for half life. No penalty for missing –ve sign. Allow ect from incorrect gradient value.
Unit penalty if half life has been calculated in different unit (to minutes stated in question)

2

(e)     Random

1

(f)     (i)      Uncertainty = ( ± √429) = ± 21
No sf penalty

Details of calculation not required. Marks can be awarded for correct numerical answers. Also no penalty for quoting uncertainty or % uncertainty without ‘±’.

1

(ii)     % uncertainty = ± 4.9%
No sf penalty. (Note that % uncertainty in total count is same as % uncertainty in corresponding count rate.)

Accept also 4.8% (number achieved keeping all sig figs in calculator)
No penalty for omitting % or ‘±’.
No sf penalty

1

(iii)    % uncertainty for 84 counts is ± 10.9%
Taking data over larger time period / larger total count will have smaller percentage uncertainty.

Accept ±11%
No penalty for omission of ± sign. No sf penalty for estimated % uncertainties.

2



M7.(a)     Capacitor must not lose charge through the meter

1

(b)     Position on scale can be marked / easier to read quickly etc

1

(c)     Initial current = = 60.0 μA

100 μA or 200 μA (250 probably gives too low a reading)

Give max 1 mark if 65 μA (from 2.6) used and 100 μA meter chosen

2

(d)     0.05 V

1

(e)     Total charge = 6.0 x 680 x 10-6 (C) (= 4.08 mC)

Time = 4.08 x 10-3 / 60.0 x 10-6 = 68 s

3

(f)     Recognition that total charge = 65 t μC and final pd = 0.098 t

so C = 65μ / 0.098

660 μF

Allow 663 μF

2

(g)     (yes) because it could lie within 646 – 714 to be in tolerance

OR

it is 97.5 % of quoted value which is within 5%

1

(h)     Suitable circuit drawn

Charge C then discharge through R and record V or I at 5 or 10 s intervals

Plot ln V or ln I versus time

gradient is 1 / RC

OR

Suitable circuit drawn

Charge C then discharge through R and record V or I at 5 or 10 s intervals

Use V or I versus time data to deduce half-time to discharge

1 / RC = ln 2 / t½ quoted

OR

Suitable circuit drawn

Charge C then discharge through R and record V or I at 5 or 10 s intervals

Plot V or I against t and find time T for V or I to fall to 0.37 of initial value

T = CR Either A or V required

For 2nd mark, credit use of datalogger for recording V or I.

4



M8.(a)     n changes by 4 units, 2 units, 1 unit for each change in 100 nm

OR

this is not half-life behaviour because graph has false origin for n

OR

the magnitude of n does not halve every interval

1

(b)     Sensible long (> 8 cm) tangent drawn, correct read-off for points from triangle at least half length of line and readings taken

Substitution correct

(–) (1.5 ± 0.2 ) x 104 and m-1

Condone power of ten error in first two marks

3

(c)     Column heading correct

All calculations correct

appropriate (3) sfs

 1 / λ2 / 10-12 m-2 11.1 8.16 6.25 4.94 4.00 3.31 2.78

Accept if calculated in nm-2 instead of m-2

11.1 × 10-6 nm-2 etc

Units as follows: 1 / λ2 / m-2. Alternative acceptable labelling includes 1 / λ2 (m-2), 1 / λ2 in m-2. The 10-12 can be in the body of the table or at top of column.

3

(d)     Graph axes labelled correctly and sensible axes

Plots correct to within half a square

Best-fit line by eye

Suitably large graph scale (do not award if scale on axis could have been doubled) Scale must be sensible divisions which can be easily read. eg scales in multiples of 3, 6, 7, 9 etc are unsatisfactory.

2nd mark is independent mark i.e. if candidates have used an unsuitable scale they can still achieve marks for accurately plotting the points.

The line of best fit should have an approximately equal distribution of points on either side of the line.

Check bottom 3 plots.

3

(e)     Intercept correct to within half a square

[1.6014]

1

(f)     The value of refractive index at infinite / very long wavelength

1

(g)     states that log n = log c + d log λ

plot log n versus log λ

d is the gradient of the graph

3