May 2002 Gp 1 1B
1.	(a)	(i)	–1 per error in circuit A [Allow incorrect polarity/range]
Correct circuit without help    (3)
Sensible values, to 0.1 mA precision    (1)
with units for both    (1)	5
(ii)	Both V values sensible and recorded to 0.01 V with unit for both    (1)
R1 correctly calculated with unit and ³ 2 s.f    (1)
[e.c.f.]
[e.c.f. including repeat of above error]
	R2 correctly calculated and > 2 s.f. and R2 ³ R1    (1)
[W seen at least once]
R constant    (1)
so Ohm’s law obeyed    (1)
	OR
	R not constant    (1)
so Ohm’s law not obeyed    (1)
	OR
	Temperature changes    (1)
so Ohm’s law not relevant/not obeyed    (1)	5
k found for each case    (2)
[Ignore units and ³2 s.f]
	% difference using either k or average as denominator
[working must be shown]    (1)
	Related to meter uncertainty    (1)	4
(ii)	Correct circuit [–1 for each error or omission of components]    (3)
[0/3 if not a potential divider]    (3)
Any 4 points    (1)
R v  [OR R2 v I] 1
Hence straight line through origin shown [or stated] 
[i.e. dependent mark]    (1)
Gradient [] [Dependent mark on correct plots]    (1)	10
[24]
	Sample results:
(a)	(i)	Imax = 57.8 mA, Imin = 52.3 mA
(ii)	Vmin, = 5.05 V
	R1 = 
	96.6 W
	Vmax = 6.04 V
	R2 = 
	= 104.5 W
	The lamp does not obey Ohm’s law because the resistance is not 
constant due to change in temperature.
(b)	(i)	k1 = = 422 WA–½
	k2 = = 422 WA–½
	% difference = × 100%
	= 3%
	So results confirm relationship within uncertainty of meters
(ii)	
	Start with lowest I [or V]; vary I [or V] with resistor or potentiometer. 
Measure I and corresponding V. Calculate values of R.
	Graph:

	k would be the gradient of the graph

May 2004 Gp2 2A

2.	(a)	(i)	W± 0.3 mm of Supervisor’s value, with unit (1)
t ± 0.3 mm of Supervisor’s value, with unit (1)
[W and t both to 0.1 mm or better]
Both from ³ 3 readings (1)
at different points along the rule (1)	4
[May score mark from diagram]
[Use unit penalty once only]
(ii)	x and y to mm precision, with unit seen at least once (1)
y by difference method (1)
W correct, 2/3 significant figures, + unit (1)
± 0.1 N of Supervisor’s value (1) (1)	5
[± 0.2 N gets 1 mark only]
(iii)	Check heights vertical by using set square on bench [may be on diagram] (1)
For x place rule on bench and note positions vertically below X and Y (1)
Statement of how it was ensured that positions were vertically below
X and Y (1)	3
[Last two marks can by scored by using Pythagoras with XY = 80 cm]
(b)	(i)	Circuit set up correctly without help (1) (1)	2
[Ignore meter polarity errors.
Polarity of cell –1]
(ii)	All units correct (1)
V » 1.5 V both to 0.01 V or better and IAB » 16 mA and
IBA » 8 mA, both to 0.1 mA or better (1)
Correct calculation of R 2/3 significant figures [ecf] (1)
RBA = 200 W ± 10 W [no ecf] (1)	4
(iii)	RBA » 200 W (1)
Hence circuit L (1)
Either
Diode arm does not conduct when B positive (1)	3
Or
For circuit J, RBA = 50 W and circuit K RBA = 100 W (1)
(iv)	Correct formula (1)
Diode arm resistance = RD + 100 W (1)
[Allow ecf from wrong circuit]
Correct calculation of diode resistance [ecf] (1)	3
[24]
	Sample results
(a)	(i)	W = 28.4, 28.1, 28.1, 28.2 mm
= 28.2 mm
t = 6.3, 6.3, 6.0, 6.2 mm
 = 6.2 mm
(ii)	x = 797 – 30 = 767 mm
y = 400 – 196 = 204 mm
W= 2 × 0.2 × 9.81 × 
= 1.04 N
(iii)	Check heights vertical by using set square on bench. For x place rule on bench and note positions vertically below X and Y.
Statement of how it was ensured that positions were vertically below X and Y.
(b)	(i)	Circuit set up correctly without help.
(ii)	
X connected
to	Y connected
to	Current/mA	Voltage/V	Resistance/W	
A	B	16.5	1.55	93.9	
B	A	7.9	1.57	199	
(iii)	RBA » 200 W
Hence circuit L
Either
Diode arm does not conduct when B positive
Or
For circuit J, RBA = 50 W and circuit K RBA = 100 W

(iv)	
[Candidates may write 1/200 for second term]
X = 178 W
Resistance of diode = X – 100
= 178 – 100
= 78 W

2B	(a)	Sensible volume with unit, from difference method or a series of
small additions (1)
Temperature recorded to better dm 1°C with unit (1)	2
(b)	Sensible values to at least 1 d.p. plus units (1)
Table with units (1)
Readings at least every 30 s/0.5 min
Some attempt at better than 1°C (1)
Straight line of S shaped curve (1)	5
(c)	Sensible scale with data occupying more than half page in both directions (1)
Labelled and units (1)
Plots and straight line (1)	3
(d)	Dx > 6 cm (1)
Correct calculation 2/3 s.f. + unit (1)
Correct calculation ³ 2 s.f. + unit (1)
Correct re-arrangement [ignore units] (1)
Correct conversion of  to °C s–1 here or above (1)
Correct calculation 2/3 s.f + unit	6
(e)	Not all of the power supplied by the heater goes to the water (ie p to water is
smaller than above value) OR heat is lost to the surroundings (1)
Hence, value is too high (1)
Improved accuracy, any two from:
·	Lag the beaker
·	Cover the beaker
·	Increase the power supply voltage to give larger temperature
rise
·	Take account of s.h.c. of beaker
·	Increase the time of the experiment (1) (1)	4
(f)	All of circuit diagram correct (1) (1)
[–1 each error or omission]
[No penalty if thermistor and resistor are reversed]
Block diagram:
Temperature sensor [Do not accept thermometer] (1)
Datalogger and computer (1)	4
[24]

	Sample results
(a)	V= 100 – 57 = 43 ml
Filled cylinder to 100 ml and poured water into beaker until wire just covered.
Recorded amount of water left in the cylinder.
q0 = 19.0 °C
(b)	I = 1.07 A V = 6.72 V
t/s	q/°C	
0
20
40
60
80
100
120
140
160
180
200
220
240
260
280
300	19.0
19.5
20.0
20.5
21.0
21.0
21.5
22.0
22.5
23.0
23.5
24.0
24.3
24.6
25.0
25.5	

(c)	Graph

(d)	
= 2.23 × 10–2 °C s–1
	P = IV = 1.07 × 6.72
= 7.19 W
cW = 
= 7500 J kg–1°C–1

(e)	Not all of the power supplied by the heater goes to the water (ie P to water is
smaller than above value) or heat is lost to the surroundings.
Hence value is too high
·	Lag the beaker
·	Cover the beaker
·	Increase the power supply voltage to give larger temperature rise
·	Take account of s.h.c. of beaker
·	Increase the time of the experiment
(f)	Block diagram:

May 2005 G1 

1A	(a)	(i)	Approximately 1.5 V measured to 0.01 V or better + unit (1)	1
(ii)	Circuit set up correctly without help	2
(iii)	≈ 0.8 V to 0.01 V or better + unit (1)
7.0 mA → 12.0 mA to 0.1 mA or better + unit (1)
Correct calculation to 2 to 4 significant figures + unit (1)	3
[allow ecf]
	[Apply unit penalty once only for each quantity (I, V and R)
in part (a)]
[R values to 2 to 4 significant figures else –1 once only]
(iv)	Value + unit (1)
Correct percentage difference with expected value as denominator (1)	2
(v)	Sensible values (V2 < V1, I2 > I1) with V to 0.01 V or better (1)
and I to 0.1 mA or better + units
Correct calculation to 2 to 4 significant figures + unit (1)
[allow ecf from wrong I and V or R1]
Value + unit (1)	3
(vi)	Any 3 of:
Resistor values will have a tolerance;
Potential difference across connecting wires / wires in the
circuit will have resistance;
Ammeter will have resistance / potential difference across
ammeter;
Cell has / potential difference across internal resistance of 
cell;
Error in meters including not accurate;
Voltmeter has finite resistance / current in voltmeter (1) (1) (1)	Max 3
[Do not accept resistors heating up]

(b)	(i)	di = ± 0.03 cm from Supervisor’s value from repeat
measurements (1) (1)
[± 0.05 cm from repeats (1) or ± 0.03 cm from single
reading (1)]
	de = ± 0.03 cm from Supervisor’s value from repeat
measurements (1) (1)
	[± 0.05 cm from repeats (1) or ± 0.03 cm from single reading (1)]
	Checked in perpendicular directions (1)	)	Max 
[Allow different directions]	)	1	5
Zero error check here or in (ii) (1)	)
	[Allow centre average value if no Supervisor’s data]
(ii)	± 0.03 mm from Supervisor’s value from repeat
measurements (1) (1)
[± 0.05 mm from repeats (1) or ± 0.03 mm from single reading (1)]	2
(iii)	Correct calculation ≥ 2 significant figures + unit (1)
Correct substitution into mass / volume, (1)
Value ≥ 2 significant figures + unit and in range
6.0 → 10.0 g cm–3 or ± 2.0 g cm–3 of Supervisor’s density (1)	3
[24]
	Sample results 
(a)	(i)	E = 1.52 V
(iii)	V1 = 0.76 V
I1 = 10.62 mA
R1 = 0.76 / 10.62 × 10–3
= 71.6 Ω
(iv)	Expected value = 68 Ω
(71.6 − 68 / 68) × 100%
= 5.3%
(v)	V2 = 0.38 V
I2 = 15.93 mA
R2 = 0.38 / 15.93 × 10–3
= 23.9 Ω
Expected value
= 1/3 × 68 = 22.7 Ω
(b)	(i)	di = 1.66 cm, 1.66 cm
average di = 1.66 cm
de = 3.47 cm, 3.47 cm
average de = 3.47 cm
(ii)	Thickness 1.81 mm, 1.79 mm, 1.85 mm
average t = 1.81 mm
(iii)	V = ¼ π (3.472 − 1.662) × 0.181
= 1.32 cm3
m = 10.32 g
Density = 10.32 / 1.32
             = 7.8 g cm–3

1B	(a)	Box suspended [allow box on rule] and 10 g used [on 
diagram or in calculation] (1)
Lengths to centres of mass (1)
b found (490 mm to 510 mm) (245 mm to 255 mm for ½ (1)
metre rule)
x ≥ 400 mm (≥ 200 mm for ½ metre rule) (1)
Repeat (1)
Correct moments used to give (1)
m0 + unit (1)	7
(b)	

	h shown accurately from bench to bottom of ramp (1)
	l or x shown accurately (1)
	Set square shown (1)
	h and l (or x) recorded to mm or better (1)
	Repeat / eye level with reading (1)
	Correct calculation of µ and µ in range 0.20 → 0.50 and
≥ 2 significant figures + no unit (1)
	Correct calculation (1)	7
(c)	Values of ms found with unit (1)
Repeat or attempt at interpolation(1)
Correct calculation of µ and µ in range 0.20 → 0.50 and ≥ 2 
significant figures + no unit [allow ecf] (1)	3
(d)	Sensible ∆ms (2 g → 10 g) (1)
Correct calculation of percentage (1)	2
(e)	(i)	Change M (by adding a further 10 g mass to the box) (1)
Find (corresponding) ms (for box to slide) (1)
(ii)	Plot ms against M (1)
Or
F against R
Calculate F from F = msg and calculate R from R = Mg and (1)
plot F against R
Straight line through origin stated or shown (1)
(iii)	Gradient = µ (1)	5
[24]

	Sample results
(a)	

a = 8 mm
b = 498 mm
c = 610 mm
m0 = 10 × 
     = 2.3 g
(b)	

	h = 169 mm, 155 mm    average h = 162 mm
	l = 458 mm, 458 mm
	sin α = 162 / 458
	        = 0.354
	\ tan α = 0.38
	Percentage uncertainty =  × 100 = 5.3 %
(c)	ms = 20 g – no movement
ms = 30 g – slides easily
\ms = 25 g
µ = 
= 0.40
(d)	Percentage uncertainty =  × 100 = 20 %
(e)	(i)	Change M (by adding a further 10 g mass to the box)
Find (corresponding) ms (for box to slide)

(ii)	

(iii)	Gradient = µ

May 2004 Gp1 1B

1B.	(a)	All temperatures recorded with units (1)
Dq » (5 ® 15) K [Need not be calculated] (1)
Some attempt at temperatures better than 1°C (1)
Precautions (Max 2): (1) (1)	5
·	stirred water
·	took qi just before water added
·	recorded qf immediately after rapid fall
·	equilibrium between water and thermometer
(b)	Correct re-arrangement and substitution (1)
Correct calculation, 2/3 significant figures and no unit (1)
Value 0.15 to 0.30 or centre value if > 0.30 (1)	3
(c)	Discard 20 cm3 into the waste beaker (1)
Add further 20 cm3 into the hot/warm water (1)
Temperatures recorded, with units and qi < qf from previous value (1)
Correct calculation ³ 2 significant figures and no unit (1)
Value ± 0.04 of previous k (1)	5
[Dependent mark]
(d)	Correct calculation with average as denominator (1)
Sensible comparison with 10% (1)
Sensible conclusion [Dependent mark] (1)	3
(e)	(i)	Description:
·	Keep volume of hot water constant at 100 ml (1)
·	Volume of added water constant at 20 ml (1)
·	Discard (20 ml) each time so that initial volume remains constant (1)
·	(Measure Dq) by measuring qi and qf (1)
·	for several values of qi (1)
	Any two sensible precautions (eg lagging/lid/larger
volumes/check q0 (1) (1)	(Max 5)

(ii)	Axes labelled correctly (1)
Straight line through origin (1)
OR
if Dq against qf straight line negative intercept (1)
(iii)	k = gradient [Dependent mark] (1)	Max 8
[24]
	Sample results
(a)	qo = 18.5 °C
qi = 85.0 °C
qf = 75.0 °C
Dq = 10.0°C
(b)	k =  =
= 0.177
(c)	Discard 20 cm3 into the wastebeaker
Add further 20 cm3 into the hot/warm water
qi = 57.5 °C
qf = 51.5 °C
Dq = 6.0 °C
k = = 0.182
(d)	% difference = 
=3.6%
	Difference is less than 10% \ relationship is supported
(e)	(i)	·	Keep volume of hot water constant at 100 ml
·	Volume of added water constant at 20 ml
·	Discard (20 ml) each time so that initial volume remains constant
·	(Measure Dq) by measuring qI and q
·	for several values of qI
	Any two sensible precautions: eg lagging/lid/larger volumes/check q0
(ii)	

(iii)	k = gradient