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}
m^{2})
✔
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}✔
crosssectional 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}
m^{2})
MAX 3
result in range 1.84 × 10^{11}
to 1.91 × 10^{11}
_{
5}✔
condone 1.9 × 10^{11}
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 × 10^{11}
to 3.63 × 10^{11},
3.5 × 10^{11}
or 3.6 × 10^{11})
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
[11]
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 14 value of spacing
1
(f) Calculates % difference (4.6%)
✓
1
and
makes judgement concerning agreement
✓
Allow ecf from 15 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
[10]
M3.(a)
(i) 5.1 and 7.1
✓
Exact answers only
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
Ω Must
have unit
✓
Allow ecf from gradient value
No sf penalty
1
(b) (i) 5.04 (Ω)
or 5.0 (Ω)
ઙ
(Allow also 5.06
Ω
or 5.1
Ω,
obtained by intermediate rounding up of 3.50^{2})
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 V^{2}
(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
Ω
(using 2.6%)
(or 0.15 or 0.2
Ω
using 3%)
✓
Must have unit (Ω)
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
Ω
using uncertainty of ± 0.1
Ω
✓
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
[14]
M4.(a)
Straight line of best fit passing through all error bars
✓
Look for reasonable distribution of points on either side
1
(b) h_{0}
= 165 ± 2
mm✓
1
(c) Clear attempt to determine gradient
✓
1
Correct readoffs (within ˝ square) for points on line more than 6 cm
apart and correct substitution into gradient equation
✓
1
h_{0}k
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
[10]
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 Ľ 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
✓
calculate the gradient
✓
the gradient is
g
/ 2 or √(g
/ 2)
✓
Accept: plot
s
against t^{
2}
/ 2 or plot 2s
against t^{
2}:
calculate the gradient
in both cases gradient gives
g
Allow 1 max for answer that evaluates g for each data point and averages
3
[15]
M6.(a)
Background radiation
✔
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}
✔
T_{˝}
= 99 minutes to 2 or 3 sf
✔
For 1^{st}
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
[13]
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
✓
Hence 6 readings
✓
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 halftime to discharge
✓
1 /
RC
= ln 2 /
t_{˝}
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 2^{nd}
mark, credit use of datalogger for recording
V
or
I.
4
[15]
M8.(a)
n
changes by 4 units, 2 units, 1 unit for each change in 100
nm
✓
OR
this is not halflife 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 readoff for points from triangle at least half length of
line and readings taken
✓
Substitution correct✓
(–) (1.5 ± 0.2 ) x 10^{4}
and
m^{1}
✓
Condone power of ten error in first two marks
3
(c) Column heading correct
✓
All calculations correct
✓
appropriate (3) sfs✓

1 / λ^{2} / 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 /
λ^{2}
/ m^{2}.
Alternative acceptable labelling includes 1 /
λ^{2}
(m^{2}),
1 /
λ^{2}
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
✓
Bestfit 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.
2^{nd}
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
[15]