9702 s07 ms_all

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the May/June 2007 question paper
9702 PHYSICS
9702/01 Paper 1 (Multiple Choice), maximum raw mark 40
Mark schemes must be read in conjunction with the question papers and the report on the
examination.
• CIE will not enter into discussions or correspondence in connection with these mark schemes.
CIE is publishing the mark schemes for the May/June 2007 question papers for most IGCSE, GCE
Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Mark Scheme Syllabus Paper
GCE A/AS LEVEL – May/June 2007 9702 01
© UCLES 2007
Question
Number
Key
Question
Number
Key
1 C 21 C
2 D 22 A
3 A 23 D
4 C 24 D
5 A 25 C
6 D 26 D
7 C 27 A
8 A 28 D
9 B 29 A
10 B 30 B
11 B 31 D
12 B 32 A
13 B 33 C
14 C 34 A
15 A 35 C
16 C 36 C
17 C 37 C
18 B 38 A
19 A 39 D
20 B 40 C

9702 PHYSICS
9702/02 Paper 2 (AS Structured Questions), maximum raw mark 60
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of
the examination. It shows the basis on which Examiners were instructed to award marks. It does not
indicate the details of the discussions that took place at an Examiners’ meeting before marking began.
All Examiners are instructed that alternative correct answers and unexpected approaches in
candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills
demonstrated.
examination.

© UCLES 2007
1 (a) (i) all positions (accept 20, 40, 60, 80) marked to within ±5° B2
positions are 40°, 70°, 90° and 102°
(-1 for each error or omission)
(ii) allow 107° → 113° B1 [3]
(b) e.g. more sensitive at low volumes B1 [1]
(do not allow reference to ‘accuracy’)
2 (a) force per unit positive charge (on a small test charge) B1 [1]
(b) field strength = (210/{1.5 × 10-2
} =) 1.4 ×104
N C-1
A1 [1]
(c) (i) acceleration = Eq / m C1
= (1.4 × 104
× 1.6 × 10-19
) / (9.1 × 10-31
) C1
= 2.5 × 1015
m s-2
(2.46 × 1015
) A1
towards positive plate / upwards (and normal to plate) B1 [4]
(ii) time = 2.4 × 10-9
s A1 [1]
(d) either vertical displacement after acceleration for 2.4 × 10-9
s
= ½ × 2.46 × 1015
× (2.4 × 10-9
)2
C1
= 7.1 × 10-3
m A1
(0.71 cm < 0.75 cm and) so will pass between plates A1 [3]
i.e. valid conclusion based on a numerical value
or 0.75 × 10-2
= ½ × 2.46 × 1015
× t2
(C1)
t is time to travel ‘half-way across’ plates = 2.47 × 10-9
s (A1)
(2.4 ns < 2.47 ns) so will pass between plates (A1)
i.e. valid conclusion based on a numerical value
3 (a) mass / volume (ratio idea essential) B1 [1]
(b) (i) mass = Ahρ B1 [1]
(ii) pressure = force/area B1
weight (of liquid)/force (on base) = Ahρg B1
pressure = hρg A0 [2]
(c) (i) ratio = 1600 or 1600:1 A1 [1]
(ii) ratio = 3
√1600 C1
= 11.7 (allow 12) A1 [2]

© UCLES 2007
(d) (i) density of solids and liquids are (about) equal B1 [1]
(ii) strong forces: fixed volume B1
rigid forces: retains shape / does not flow / little deformation B1 [2]
(allow 1 mark for fixed volume, fixed shape)
4 (a) (i) (change in) potential energy = mgh C1
= 0.056 × 9.8 × 16
= 8.78 J (allow 8.8) A1 [2]
(ii) (initial) kinetic energy = ½mv2
C1
= ½ × 0.056 × 182
= 9.07 J (allow 9.1) C1
total kinetic energy = 8.78 + 9.07 = 17.9 J A1 [3]
(b) kinetic energy = ½mv2
17.9 = ½ × 0.056 × v2
and v = 25(.3) m s-1
B1 [1]
(c) horizontal velocity = 18 m s-1
B1 [1]
(d) (i) correct shape of diagram
(two sides of right-angled triangle with correct orientation) B1
(ii) angle = 41° → 48° (allow trig. solution based on diagram) A2 [3]
(for angle 38°→ 41° or 48°→ 51°, allow 1 mark)
5 (a) (i) vibrations (in plane) normal to direction of energy propagation B1 [1]
(ii) vibrations in one direction (normal to direction of propagation) B1 [1]
(b) (i) at (displacement) antinodes / where there are no heaps, wave has
maximum amplitude (of vibration) B1
at (displacement) nodes/where there are heaps, amplitude of vibration is
zero/minimum B1
dust is pushed to / settles at (displacement) nodes B1 [3]
(ii) 2.5λ = 39 cm C1
v = fλ C1
v = 2.14 × 103
× 15.6 × 10-2
= 334 m s-1
(allow 330, not 340) A1 [3]
(c) Stationary wave formed by interference / superposition / overlap of B1
either wave travelling down tube and its reflection
or two waves of same (type and) frequency travelling in opposite directions B1
speed is the speed of the incident / reflected waves B1 [3]

© UCLES 2007
6 (a) (i) 1 total resistance = 0.16 Ω A1
2 e.m.f. = either (14 – E) or (E – 14) A1 [2]
(ii) either 14 – E = 42 × 0.16 or (E – 14) = -42 × 0.16 C1
E = 7.3 V A1 [2]
(b) (i) charge = It C1
= 12.5 × 4 × 60 × 60
= 1.8 × 105
C A1 [2]
(ii) either energy = EQ or energy = Eit C1
either energy = 14 × 1.8 × 105
or energy = 14 × 12.5 × 4 × 3600
= 2.52 × 106
J A1 [2]
(iii) energy = I2
Rt or Vit and V = IR C1
= 12.52
× 0.16 × 4 × 3600
= 3.6 × 105
J A1 [2]
(c) efficiency = (2.52 × 106
– 3.6 × 105
)/(2.52 × 106
) C1
= 86% A1 [2]
7 (a) β(-decay) B1 [1]
(b) γ(-decay) B1
either any two of Z, N and A do not change
or it is loss of energy only
or it is an electromagnetic wave B1 [2]
Allow ‘α(-decay) as change of 4 in the nucleon number cannot be shown on the
diagram’ (B2)
Do not give credit for a ‘bald’ α(-decay)

9702 PHYSICS
9702/31 Paper 31 (Advanced Practical Skills), maximum raw mark 40
demonstrated.
examination.

© UCLES 2007
1 Manipulation, measurement and observation
Successful collection of data
(b) Measurements [6]
One mark for each set of readings for V and n.
(b) Apparatus set up without help from supervisor. [1]
Range and distribution of values
(c) n = 1 or 2 and n = 10 or 11 must be included and no more than a gap of three. [1]
Quality of data
(Graph) Judge by scatter of points about the best fit line. [1]
At least 5 plots are needed on the trend line for this mark to be scored.
Presentation of data and observations
Table: layout
(b) Column headings (V/V, 1/V / V -1
only). Ignore n column. [1]
Each column heading must contain a quantity and a unit where appropriate.
Ignore units in the body of the table.
There must be some distinguishing mark between the quantity and the unit.
Table: raw data
(b) Consistency of presentation of raw readings. [1]
All values of V must be given to the same number of decimal places.
Table: calculated quantities
(b) Significant figures [1]
Apply to 1/V.
If V is given to 2 s.f., then accept 1/V to 2 or 3 s.f.
(b) Values of 1/V correct. [1]
Check a value. If incorrect, write in the correct value. Allow small rounding errors.
Graph: layout
(Graph) Axes [1]
Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed.
Scales must be chosen so that the plotted points must occupy at least half the graph grid
in both x and y directions. Indicate false origin with FO.
Scales must be labelled with the quantity which is being plotted. Ignore units.

© UCLES 2007
Graph: plotting of points
(Graph) All observations must be plotted. [1]
Ring and check a suspect plot. Tick if correct. Re-plot if incorrect.
Work to an accuracy of half a small square.
Graph: trend line
(Graph) Line of best fit (of 5 or 6) [1]
Judge by scatter of points about the candidate’s line.
There must be a fair scatter of points either side of the line.
Indicate best line if candidate’s line is not the best line.
Analysis, conclusions and evaluation
Interpretation of graph
(c) Gradient [1]
The hypotenuse of the ∆ must be greater than half the length of the drawn line.
Read-offs must be accurate to half a small square.
Check for ∆y/∆x (i.e. do not allow ∆x/∆y).
(c) y-intercept from graph or substitute correct read-offs into y = mx + c. [1]
If a false origin has been used then label FO.
Drawing conclusions
(d) Value for E. [1]
Expect between 4–5V. Should be 1/y-intercept.
Check the value. Unit required. 2/3 s.f.
(d) Value for R1/R2. [1]
Expect 0.19–0.23 unless supervisor has used different resistors.
Method of working must be correct.
If a unit is given then this mark cannot be scored. 2/3 s.f.
[Total: 20]

© UCLES 2007
(a) (iii) Position of centre of mass of ball at equilibrium [1]
(Value < 1m and appropriate unit. No more than 1 d.p. in cm.)
(b) (i) Position of centre of mass of ball when displaced [1]
(ii) Position of centre of mass of ball at maximum height [1]
(d) Second position of centre of mass of ball when displaced [1]
(d) Second position of centre of mass of ball at maximum height [1]
(b)/(d) Repeated measurements for maximum height [1]
Quality of data
(d) Bigger x gives bigger h [1]
Display of calculation and reasoning
(b), (d) Values of x calculated correctly. (Displaced – equilibrium position) [1]
Both values required. Unit need not be stated but must be consistent.
Calculations must be checked.
(b), (d) Values of h calculated correctly. (Max height – equilibrium position) [1]
Both values required. Unit need not be stated but must be consistent.
Calculations must be checked.
(e) Correct calculation to check proportionality [1]
Possibilities include: Two calculations of x2
/h or ratio of x2
values and ratio of h values both
calculated.
Drawing conclusions
(e) Conclusion [1]
Sensible comments supported by calculations and suggested relation.
Incorrect ideas score zero.
Estimating uncertainties
(c) (ii) Percentage uncertainty in h. [1]
Uncertainty in h is 2–10 mm. Whole numbers only.
If repeated readings have been done then the uncertainty could be half the range.
Correct ratio idea required, ×100 stated/implied.

© UCLES 2007
Identifying limitations
(f) (i) Relevant points must be underlined and ticked. [4]
Some of these might be:
A Ruler not vertical.
B Locating the centre of the ball (when reading ruler).
C Parallax error.
D Establishing when the ball is at its maximum displacement.
E Only two displacements (are not enough to validate the conclusion).
F Difficulty in the release of the mass (reference to force/vertical plane).
Suggesting improvements
(f) (ii) Relevant points must be underlined and ticked. [4]
A Sensible method to ensure ruler vertical.
B Place the rule as close as possible to the mass/mark the centre of the ball with mark
or pointer/use the bottom/top of the ball.
C Measure at eye level/repeat to get eye in the right place/place the rule as close as
possible to the mass.
D Use video camera (play back) frame by frame/slow motion/position sensor above or
below.
E Need a wider range of displacements and plot a graph/find mean k.
F Use a clamp/electromagnet to release the mass.
Do not allow ‘repeated readings’, ‘human error’.
Do not allow ‘use a computer to improve the experiment’.
[Total: 20]

9702 PHYSICS
9702/32 Paper 32 (Advanced Practical Skills 2),
maximum raw mark 40
demonstrated.
examination.

© UCLES 2007
(a) (i) Measurement of e.m.f. of power supply [1]
(b) Measurements [5]
Five marks for six sets of readings for I and R3, four for five sets, etc.
(b) Circuit set up without help from Supervisor [1]
Range and distribution of values
(b) R3 = 33 or 47Ω and R3 = 560 or 680Ω must be included [1]
Quality of data
(Graph) Judge by scatter of points about the best fit line. Trend must be correct. [1]
At least 5 plots are needed for this mark to be scored.
Table: layout
(b) Column headings [1]
Each column heading must contain a quantity and a unit where appropriate.
Ignore units in the body of the table.
There must be some distinguishing mark between the quantity and the unit
(i.e. solidus is expected, but accept, for example, I (A)).
Table: raw data
(b) Consistency of presentation of raw readings [1]
All values of I must be given to the same number of decimal places.
Table: calculated quantities
(b) Significant figures [1]
Apply to 1/I only.
If I is given to 2 sf, then accept 1/ I to 2 or 3 sf.
(b)Values of 1/ I correct. [1]
Check a value. If incorrect, write in the correct value.

© UCLES 2007
Graph: layout
(Graph) Axes [1]
Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed.
There should not be more than three large squares between axis labels.
Scales must be chosen so that the plotted points must occupy at least half
the graph grid in both x and y directions.
Scales must be labelled with the quantity which is being plotted. Ignore units.
Do not penalise reversed axes or if the wrong graph has been plotted.
Graph: plotting of points
(Graph) All observations must be plotted. [1]
Ring and check a suspect plot. Tick if correct. Re-plot if incorrect (and re-check quality mark).
Work to an accuracy of half a small square.
Graph: trend line
(Graph) Line of best fit (must be 5 or more plots) [1]
Judge by scatter of points about the candidate's line.
There must be a fair scatter of points either side of the line.
Indicate best line if candidate's line is not the best line.
Interpretation of graph
(c) (iii) Gradient [1]
The hypotenuse must be greater than half the length of the drawn line.
Read-offs must be accurate to half a small square.
Check for ∆y/∆x (i.e. do not allow ∆x/∆y).
(c) (iii) y-intercept [1]
The value must be read to the nearest half square.
The value can be calculated using ratios or y = mx + c.
If a false origin has been used then label FO.
Drawing conclusions
(d) Must be in range 40.0 to 55.0 Ω. [1]
Value for R1 obtained from y-intercept x E.
2 or 3 sf. Unit required
(d) Value for R2 [1]
Should be 220 Ω ± 50 Ω unless Supervisor has used different resistors to those specified.
Method of working must be correct.
2 or 3 sf. Unit required.
[Total for Question 1: 20]

© UCLES 2007
(a) (ii) First value of d (less than 40 cm) no more precise than 1 mm. [1]
(a) (ii) First value of h (less than d) [1]
(a) (iii) Method of measuring h accurately [1]
e.g. Use of set squares to indicate height / repeat to refine position.
Do not accept repeated readings for this mark
Do not accept just ‘use a set square’
(b) Second value of d (less than 40 cm) [1]
(b) Second value of h (less than d) [1]
(b) Evidence of repeated measurements for h (first or second reading) [1]
Quality of data
(b) Values of e within 10% of each other [1]
Display of calculation and reasoning
(b) Values of e calculated correctly [2]
One mark each
Calculations must be checked
(c) Consideration of the percentage uncertainty in h from (a)(iv) is expected. [1]
Knowledge of error propagation methods is not required.

© UCLES 2007
Drawing conclusions
(c) Conclusion [1]
Sensible comments relating to values of e.
Incorrect ideas score zero.
Estimating uncertainties
(a) (iv) Percentage uncertainty in h [1]
If repeated readings have been done then the uncertainty must be half the range.
Absolute uncertainty must be 2 to 10 mm.
Correct ratio idea required.
Identifying limitations
(d) (i) Relevant points must be underlined and ticked. [4]
A Two sets of readings not enough (to draw valid conclusion)
B Hard to judge rebound height, with reason
C Parallax (error in measuring h)
D Difficult to release without applying a force
E Rule may not be vertical / perpendicular
F Only cm divisions on rule (if borne out by readings)
G Inconsistent bounce
Suggesting improvements
(d) (ii) Relevant points must be underlined and ticked. [4]
A Take several d values and plot graph/compare e values
B Use video and play back slowly/position sensor
C Method of reducing parallax problem (adjustable marker/drop many times to refine
value of h/assistant to drop ball/ensure measurement taken at eye level)
D Mechanical method of release/hold ball against stop
E Method of making rule vertical
G Use flat surface/turn off fans
Do not allow ‘repeated readings’ (unless qualified by ‘plot a graph’)
Do not allow ‘use a computer to improve the experiment’
Do not allow ‘increase d’
[Total for Question 2: 20]

9702 PHYSICS
9702/04 Paper 4 (A2 Structures Questions), maximum raw mark 100
demonstrated.
examination.

© UCLES 2007
1 (a) (region of space) where a mass experiences a force B1 [1]
(b) (i) potential energy = (–)GMm / x C1
∆EP = GMm/2R – GMm/3R M1
= GMm/6R A0 [2]
(ii) EK = ½m (76002
– 73202
) M1
= (2.09 × 106
)m A0 [1]
(c) (i) 2.09 × 106
= (6.67 × 10–11
M)/(6 × 3.4 × 106
) C1
M = 6.39 × 1023
kg A1 [2]
(ii) e.g. no energy dissipated due to friction with atmosphere/air
rocket is outside atmosphere
not influenced by another planet etc. B1 [1]
2 (a) (on melting,) bonds between molecules are broken/weakened
or molecules further apart/are able to slide over one another B1
kinetic energy unchanged so no temperature change B1
potential energy increased/changed so energy required B1 [3]
(b) thermal energy/heat required to convert unit mass of solid to liquid M1
with no change in temperature/ at its normal boiling point A1 [2]
(c) (i) thermal energy lost by water = 0.16 × 4.2 x 100
= 67.2 kJ C1
67.2 = 0.205 × L C1
L = 328 kJ kg–1
A1 [3]
(ii) more energy (than calculated) melts ice M1
so, (calculated) L is lower than the accepted value A1 [2]
3 (a) field strength = potential gradient M1
correct sign OR directions discussed A1 [2]
(b) area is 21.2 cm2
± 0.4 cm2
C2
(if outside ± 0.4 cm2
but within ± 0.8 cm2
, allow 1 mark)
1.0 cm2
represents (1.0 × 10–2
× 2.5 × 103
=) 25 V C1
potential difference = 530 V A1 [4]
(c) ½mv2
= qV
½ × 9.1 × 10–31
× v2
= 1.6 × 10–19
× 530 C1
v = 1.37 × 107
ms–1
A1 [2]
(d) (i) d = 0 B1 [1]
(ii) acceleration decreases then increases B1
some quantitative analysis (e.g. minimum at 4.0 cm) B1 [2]
(any suggestion that acceleration becomes zero or that there is a
deceleration scores 0/2)

© UCLES 2007
4 (a) r.m.s. output = 9/√2 or peak input = 230√2 C1
NS/NP = VS/VP C1
NS = 138 → 140 turns A1 [3]
(b) (i) four diodes correctly positioned regardless of output polarity M1
giving correct output polarity (all ‘point to left’) A1 [2]
(ii) capacitor shown in parallel with R B1 [1]
(c) (i) time t1 to time t2 B1 [1]
(ii) sketch: same peak values M1
ripple reduced and reasonable shape A1 [2]
5 (a) (i) packet/discrete quantity/quantum (of energy) of e.m. radiation B1 [1]
(ii) either E = (6.63 × 10–34
× 3 × 108
)/(350 × 10–9
)
or E = (6.63 × 10–34
× 8.57 × 1014
) M1
E = 5.68 × 10–19
J A0 [1]
(iii) 0.5 B1 [1]
(b) (i) energy of photon M1
to cause emission of electron from surface
either with zero k.e or photon energy is minimum A1 [2]
(ii) correct conversion eV → J or J → eV seen once B1
photon energy must be greater than work function C1
350 nm wavelength and potassium metal A1 [3]
6 (a) probability of decay M1
of a nucleus per unit time A1 [2]
(allow 1 mark for A = λN, with symbols explained)
(b) (i) λ = ln2/(28 × 365 × 24 × 3600) C1
= 7.85 × 10–10
s–1
A1 [2]
(ii) A = (–)λN
N = (6.4 × 109
)/(7.85 × 10–10
) C1
= 8.15 × 1018
C1
mass = (8.15 × 1018
× 90)/(6.02 × 1023
) (e.c.f. for value of N) C1
= 1.22 × 10–3
g A1 [4]
(iii) volume = (1.22 × 10–3
/2.54 =) 4.8 × 10–4
cm3
A1 [1]
(c) either very small volume of Strontium-90 has high activity
or dust can be highly radioactive B1
breathing in dust presents health hazard B1 [2]

© UCLES 2007
7 (a) (i) oscillations are damped/amplitude decreases B1
as magnet moves, flux is cut by coil B1
e.m.f./current is induced in the coil B1
causing energy loss in load OR force on magnet B1
energy is derived from oscillations of magnet
OR force opposes motion of magnet B1 [5]
(ii) T = 0.60 s C1
ω0 (= 2π/T) = 10.5 rad s–1
A1 [2]
(b) sketch: sinusoidal wave with period unchanged or slightly smaller M1
same initial displacement, less damping A1 [2]
(c) (i) sketch: general shape – peaked curve M1
peak at ω0 and amplitude never zero A1 [2]
(ii) resonance B1 [1]
(iii) useful: e.g. child on swing, microwave oven heating B1
avoid: e.g. vibrating panels, vibrating bridges B1 [2]
(for credit, stated example must be put in context)
Section B
8 (a) e.g. infinite (voltage) gain
infinite input impedance
zero output impedance
infinite bandwidth
infinite slew rate
(any three, 1 each) B3 [3]
(b) (i) negative (feedback) B1 [1]
(ii) 1 gain (= 5.8/0.069) = 84 B1 [1]
(ii) 2 gain = 1 + 120/X C1
84 = 1 + 120/X
X = 1.45 kΩ A1 [2]
(iii) gain increases OR bandwidth reduced OR output increases B1 [1]

© UCLES 2007
9 (a) X-ray beam directed through body onto detector (plate) B1
different tissues absorb/attenuate beam by different amounts B1
giving ‘shadow’ image of structures B1
any other detail e.g. comment re sharpness or contrast B1 [4]
(b) X-ray image is flat OR 2-dimensional (1)
CT scan takes many images of a slice at different angles (1)
these build up an image of a slice through the body (1)
series of images of slices is made (1)
so that 3D image can be built up (1)
image can then be rotated (1)
1 mark for each point, max 5 B5 [5]
10 (a) correct values of 2, 5, 10, 15 and 4 (–1 each error) B2
graph drawn as a series of steps M1
steps occurring at correct times A1 [4]
(b) sample more frequently B1
greater number of bits B1 [2]
11 (a) modulator and oscillator identified B1
both amplifiers identified correctly B1
ADC and parallel-to serial converter identified B1 [3]
(b) computer at cellular exchange B1
monitors signal strength B1
switches call from one base station to another B1
to maintain maximum signal strength B1 [4]

9702 PHYSICS
9702/05 Paper 5 (Planning, Analysis and Evaluation),
maximum raw mark 30
demonstrated.
examination.

© UCLES 2007
1 Planning (15 marks)
Defining the problem (3 marks)
P1 r is the independent variable or vary r (accept diameter but not mass or size). [1]
P2 v is the dependent variable or determine v (accept speed) [1]
P3 A controlled variable – accept temperature, [1]
distance when time is measured, or time when distance measured.
Do not accept volume/height of oil.
Methods of data collection (5 marks)
M1 Diagram of a workable arrangement including a deep container of oil, ball and some
measurement indicated for either time or distance. [1]
M2 Measure diameter by using a micrometer (screw gauge)/vernier callipers (and halving to obtain
radius). Accept from diagram. Accept travelling microscope. [1]
M3 Measure the time for the ball to fall a set distance in oil (or distance for a set time). [1]
M4 Measure the (constant) distance fallen (constant time) and show how v is calculated. [1]
M5 Evidence that ball has reached terminal velocity (e.g. starting mark well below surface of oil) [1]
Reject equations of uniform acceleration ideas.
Method of analysis (2 marks)
A1 Plot a graph of v against r2
or logarithmic equivalent. [1]
A2 Relationship is correct if graph is a straight line through the origin. [1]
An explicit statement is required.
If lg v against lg r is plotted gradient should equal 2.
Safety considerations (1 mark)
S1 Relevant safety precaution related to the oil, [1]
e.g. mop up spillages of oil/wear gloves with reason/keep away from flames.
Do not accept vague answers e.g. goggles/spills/washing hands but allow credit for detailed
reasoning e.g. drop ball near surface to avoid splashing.
Additional detail (4 marks)
D1/2/3/4 Relevant points might include: [4]
Allow oil to stand so that air bubbles escape/ball may trap air bubbles.
Wash and dry steel balls/handle steel balls with tweezers/gloves.
Distance marks should be as far apart as possible or use long tube.
Large distance to reduce percentage uncertainty.
Wide tube to reduce edge effects/method to keep long tube vertical.
Discussion of parallax for stop watch methods.
Method of ensuring that terminal velocity has been reached.
Retrieve steel balls using a magnet.
Use clear oil.
Repeat diameter measurements and average.
An additional variable kept constant.
[Total: 15]

© UCLES 2007
2 Analysis, conclusions and evaluation (15 marks)
Approach to data analysis (1 mark)
(a) 0
2
R
V
R +=
lρ
and a correct comment.
This mark is not scored for R being proportional to l2
. [1]
Table of results (2 marks)
(b) Column heading for l2
. Allow l2
/ cm2
and l2
(cm2
) (or equivalent units). [1]
(b) Values of l2
. [1]
36, 100, 196, 324, 484, 676
3 significant figures needed (except 1st
row). Allow 4sf. All correct for one mark.
Graph (3 marks)
(c) (i) Points plotted correctly. [1]
All six required for this mark and must be Ğ half a small square. Indicate an error.
Ecf from (b)
(c) (ii) Line of best fit. [1]
Must be within tolerances. Do not allow a line forced through the origin.
(c) (iii) Worst acceptable straight line. [1]
Must be within tolerances. Line should be clearly labelled. Allow broken line.
Conclusion (4 marks)
(c) (iii) gradient of best-fit line [1]
Gradient should be in the range 0.550 to 0.560.
If (b) and/or (c)(i) and/or (ii) are incorrect then the triangle used should be greater than
half the length of the drawn line. Check the read offs and ratio to be correct. Work to
half a small square.
(d) Value of ρ
Candidate’s gradient value = ρ/V. May be implicit from working. [1]
ρ in range 10.3 -10.6 [1]
(d) Unit of ρ. Must be consistent with previous answer e.g. Ω cm [1]

© UCLES 2007
Treatment of errors (5 marks)
(b) Errors in l2
[1]
± 4.6 – 5.0
± 7.8 – 8.2
± 11.0 – 11.4
± 14.2 - 14 or 15
± 17 or 18
± 20 or 21
(c) (i) error bars in l2
plotted correctly [1]
Must be within tolerances. For ecf check first and last point
(c) (iii) error in gradient [1]
Check method e.g. gradient of best-fit line – gradient of worst acceptable line
(d) correct method for determining error in ρ (e.g. (worst gradient × volume) - ρ) [1]
Value for error in ρ in the range ± 0.4 to ± 0.6. [1]
Last mark is zero if vertical error bars plotted or wrong worst acceptable line plotted.
[Total: 15]

9702 s07 ms_all

Recomendados

Recomendados

Más contenido relacionado

La actualidad más candente

La actualidad más candente (20)

Similar a 9702 s07 ms_all

Similar a 9702 s07 ms_all (20)

Más de Sajit Chandra Shakya

Más de Sajit Chandra Shakya (20)

Último

Último (20)

9702 s07 ms_all