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Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/11

Paper 1 Multiple Choice October/November 2016

MARK SCHEME

Maximum Mark: 40

Published

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, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 2016 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2016 9702 11

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Question Number Key Question

Number Key

1 D 21 D 2 B 22 A 3 A 23 A 4 A 24 D 5 D 25 D

6 B 26 C 7 B 27 A 8 C 28 A 9 A 29 C 10 C 30 A

11 A 31 B 12 D 32 A 13 A 33 C 14 A 34 B 15 D 35 B

16 B 36 D 17 C 37 A 18 C 38 A 19 A 39 C 20 C 40 B





PHYSICS 9702/12


MARK SCHEME

Maximum Mark: 40

Published



© UCLES 2016


Number Key

1 B 21 D 2 D 22 B 3 D 23 A 4 C 24 B 5 A 25 A

6 D 26 B 7 B 27 D 8 B 28 C 9 D 29 C 10 C 30 B

11 A 31 A 12 D 32 C 13 B 33 D 14 D 34 B 15 C 35 B

16 C 36 B 17 C 37 C 18 B 38 A 19 C 39 C 20 C 40 A





PHYSICS 9702/13


MARK SCHEME

Maximum Mark: 40

Published



© UCLES 2016


Number Key

1 D 21 D 2 B 22 A 3 A 23 A 4 A 24 D 5 D 25 D

6 B 26 C 7 B 27 A 8 C 28 A 9 A 29 C 10 C 30 A

11 A 31 B 12 D 32 A 13 A 33 C 14 A 34 B 15 D 35 B

16 B 36 D 17 C 37 A 18 C 38 A 19 A 39 C 20 C 40 B





PHYSICS 9702/21

Paper 2 AS Level Structured Questions October/November 2016

MARK SCHEME

Maximum Mark: 60

Published



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1 (a) (density =) mass / volume B1 [1] (b) (i) d = [(6 × 7.5) / (π × 8100)]1/3

= 0.12(1) m A1 [1] (ii) percentage uncertainty = (4 + 5) / 3 (= 3%) or

fractional uncertainty = (0.04 + 0.05) / 3 (= 0.03) C1

absolute uncertainty (= 0.03 × 0.121) = 0.0036 C1

d = 0.121 ± 0.004 m A1 [3] 2 (a) force per unit positive charge B1 [1] (b) (i) time = 5.9 × 10–2

/ 3.7 × 107 = 1.6 × 10–9

s (1.59 × 10–9 s) A1 [1]

(ii) E = V / d C1

= 2500 / 4.0 × 10–2

= 6.3 × 104

N C–1 (6.25 × 104 or 62500 N C–1) A1 [2] (iii) a = Eq / m or F = ma and F = Eq C1

= (6.3 × 104 × 1.60 × 10–19) / 9.11 × 10–31 = 1.1 × 1016 m s–2 A1 [2]

(iv) s = ut + ½at

2

= ½ × 1.1 × 1016 × (1.6 × 10–9)2 C1 = 1.4 × 10–2 (m) C1

distance from plate = 2.0 – 1.4 = 0.6 cm (allow 1 or more s.f.) A1 [3]

(v) electric force ≫ gravitational force (on electron)/weight

or acceleration due to electric field ≫acceleration due to gravitational field B1 [1]

(vi) vX–t graph: horizontal line at a non-zero value of vX B1

vY–t graph: straight line through the origin with positive gradient B1 [2]


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3 (a) force/load is proportional to extension/compression (provided proportionality limit is not exceeded) B1 [1]

(b) (i) k = F / x or k = gradient C1

k = 600 N m–1 A1 [2]

(ii) (W =) ½kx2 or (W =) ½Fx or (W =) area under graph C1

(W =) 0.5 × 600 × (0.040)2 = 0.48 J or (W =) 0.5 × 24 × 0.040 = 0.48 J A1 [2] (iii) 1. (EK =) ½mv2 C1

= ½ × 0.025 × 6.02

= 0.45 J A1 [2]

2. (work done against resistive force =) 0.48 – 0.45 [= 0.03(0) J] C1

average resistive force = 0.030 / 0.040 C1 = 0.75 N A1 [3]

(iv) efficiency = [useful energy out / total energy in] (×100) C1

= [0.45 / 0.48] (×100) = 0.94 or 94% A1 [2]

4 (a) the number of oscillations per unit time M1

of the source/of a point on the wave/of a particle (in the medium) A1 [2] or the number of wavelengths/wavefronts per unit time (M1) passing a (fixed) point (A1)

(b) T or period = 2.5 × 250 (µs) (= 625 µs) M1

frequency = 1 / (6.25 × 10–4) or 1 / (2.5 × 250 × 10–6) = 1600 Hz A1 [2] (c) (i) for maximum frequency: fo = fsv / (v – vs)

1640 = (1600 × 330) / (330 – vs) C1 vs = 8(.0) m s–1 (8.049 m s–1) A1 [2]

(ii) loudspeaker moving towards observer causes rise in/higher frequency B1

loudspeaker moving away from observer causes fall in/lower frequency B1 [2] or repeated rise and fall/higher and then lower frequency (M1) caused by loudspeaker moving towards and away from observer (A1)


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5 (a) wave incident on/passes by or through an aperture/edge B1

wave spreads (into geometrical shadow) B1 [2] (b) nλ = d sinθ C1

substitution of θ = 90° or sinθ = 1 C1 4 × 500 × 10–9 = d × sin 90° line spacing = 2.0 × 10–6

m A1 [3] (c) wavelength of red light is longer (than 500 nm) M1

(each order/fourth order is now at a greater angle so) the fifth-order maximum cannot be formed/not formed A1 [2]

6 (a) charge

forms) other to electrical (from ed)(transformenergy or done work B1 [1]

(b) (i) 1. V = IR or E = IR C1

I = 14 / 6.0 = 2.3 (2.33) A A1 [2]

2. total resistance of parallel resistors = 8.0 Ω C1

current = 14 / (6.0 + 8.0) = 1.0 A A1 [2]

(ii) P = EI (allow P = VI) or P = V2

/ R or P = I2R C1

change in power = (14 × 2.33) – (14 × 1.0) or (142 / 6.0) – (142

/ 14) or (2.332 × 6.0) – (1.02 × 14) = 19 W (18 W if 2.3 A used) A1 [2]

(c) I = Anvq

ratio = (0.50n / n) × (1.8 A / A) or ratio = 0.50 × 1.8 C1 = 0.90 A1 [2]


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7 (a) hadron not a fundamental particle/lepton is fundamental particle or hadron made of quarks/lepton not made of quarks or strong force/interaction acts on hadrons/does not act on leptons B1 [1]

(b) (i) proton: up, up, down / uud B1

neutron: up, down, down / udd B1 [2] (ii) composition: 2(uud) + 2(udd)

= 6 up, 6 down / 6u, 6d B1 [1] (c) (i) most of the atom is empty space

or the nucleus (volume) is (very) small compared to the atom B1 [1]

(ii) nucleus is (positively) charged B1

the mass is concentrated in (very small) nucleus/small region/small volume/small core or the majority of mass in (very small) nucleus/small region/small volume/small core B1 [2]





PHYSICS 9702/22

Paper 2 AS Structured Questions October/November 2016

MARK SCHEME

Maximum Mark: 60

Published



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1 (a) (i) force / area (normal to the force) B1 [1] (ii) (p = F / A so) units: kg m s–2

/ m2 = kg m–1 s–2 A1 [1] allow use of other correct equations: e.g. (∆p = ρg∆h so) kg m–3 m s–2 m = kg m–1 s–2 e.g. (p = W / ∆V so) kg m s–2 m / m3 = kg m–1 s–2

(b) units for m: kg, t: s and ρ: kg m–3 C1 units of C: kg / s (kg m–3 kg m–1 s–2)1/2

or units of C2: kg2

/ s2 kg m–3 kg m–1 s–2 C1

units of C: m2 A1 [3] 2 (a) ∆E = mg∆h C1 = 0.030 × 9.81 × (–)0.31 = (–)0.091 J A1 [2] (b) E = ½mv 2 C1

(initial) E = ½ × 0.030 × 1.32 (= 0.0254) C1 0.5 × 0.030 × v 2 = (0.5 × 0.030 × 1.32) + (0.030 × 9.81 × 0.31) so v = 2.8 m s–1

or 0.5 × 0.030 × v 2 = (0.0254) + (0.091) so v = 2.8 m s–1 A1 [3] (c) (i) 0.096 = 0.030 (v + 2.8) C1 v = 0.40 m s–1 A1 [2] (ii) F = ∆p / (∆)t or F = ma

= 0.096 / 20 × 10–3 or 0.030 (0.40 + 2.8) / 20 × 10–3 C1 = 4.8 N A1 [2]

(d) kinetic energy (of ball and wall) decreases/changes/not conserved, so inelastic or

(relative) speed of approach (of ball and wall) not equal to/greater than (relative) speed of separation, so inelastic. B1 [1] (e) force = work done / distance moved = (0.091 – 0.076) / 0.60 C1 = 0.025 N A1 [2]


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3 (a) resultant force (in any direction) is zero B1 resultant moment/torque (about any point) is zero B1 [2] (b) (i) force = 33 sin 52° or 33 cos 38° = 26 N A1 [1] (ii) 26 × 0.30 or W × 0.20 or 12 × 0.40 C1 26 × 0.30 = (W × 0.20) + (12 × 0.40) C1 W = 15 N A1 [3] (c) (i) E = ∆σ / ∆ε or E = σ / ε C1 ∆σ = 2.0 × 1011 × 7.5 × 10–4 = 1.5 × 108 Pa A1 [2] (ii) ∆σ = ∆F / A or σ = F / A C1 A = 78 / 1.5 × 108 (= 5.2 × 10–7

m2) C1 5.2 × 10–7 = πd 2 / 4 d = 8.1 × 10–4

m A1 [3] 4 (a) wave incident on/passes by or through an aperture/edge B1 wave spreads (into geometrical shadow) B1 [2] (b) (i) waves (from slits) overlap (at point X) B1 path difference (from slits to X) is zero/ phase difference (between the two waves) is zero (so constructive interference gives bright fringe) B1 [2] (ii) difference in distances = λ / 2 = 580 / 2 = 290 nm A1 [1] (iii) λ = ax / D C1 D = [0.41 × 10–3 × (2 × 2.0 × 10–3)] / 580 × 10–9 C1 = 2.8 m A1 [3] (iv) same separation/fringe width/number of fringes bright fringe(s)/central bright fringe/(fringe at) X less bright dark fringe(s)/(fringe at) Y/(fringe at) Z brighter contrast between fringes decreases Any two of the above four points, 1 mark each B2 [2]


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5 (a) total/sum of electromotive forces or e.m.f.s = total/sum of potential differences or p.d.s M1 around a loop/(closed) circuit A1 [2] (b) (i) (current in battery =) current in A + current in B or IA + IB C1 (I =) 0.14 + 0.26 = 0.40 A A1 [2] (ii) E = V + Ir 6.8 = 6.0 + 0.40r or 6.8 = 0.40 (15 + r) C1 r = 2.0 Ω A1 [2] (iii) R = V / I C1 ratio (= RA / RB) = (6.0 / 0.14) / (6.0 / 0.26) = 42.9 / 23.1 or 0.26 / 0.14 = 1.9 (1.86) A1 [2] (iv) 1. P = EI or VI or P = I 2R or P = V 2 / R C1 = 6.8 × 0.40 = 0.402 × 17 = 6.82

/ 17 = 2.7 W (2.72 W) A1 [2] 2. output power = VI = 6.0 × 0.40 (= 2.40 W) C1 efficiency = (6.0 × 0.40) / (6.8 × 0.40) = 2.40 / 2.72 = 0.88 or 88% (allow 0.89 or 89%) A1 [2]


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6 (a) hadron not a fundamental particle/lepton is fundamental particle or

hadron made of quarks/lepton not made of quarks or

strong force/interaction acts on hadrons/does not act on leptons B1 [1] (b) (i) e0

1)(+ or )+(β 0

1 B1 )(e

00ν B1 [2]

(ii) weak (nuclear force / interaction) B1 [1] (iii) • mass-energy • momentum

• proton number • nucleon number • charge

Any three of the above quantities, 1 mark each B3 [3] (c) (quark structure of proton is) up, up, down or uud B1 up/u (quark charge) is (+)⅔(e), down/d (quark charge) is –⅓(e) C1 ⅔e + ⅔e – ⅓e = (+)e A1 [3]





PHYSICS 9702/23

Paper 2 AS Level Structured Questions October/November 2016

MARK SCHEME

Maximum Mark: 60

Published



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1 (a) (density =) mass / volume B1 [1] (b) (i) d = [(6 × 7.5) / (π × 8100)]1/3

= 0.12(1) m A1 [1] (ii) percentage uncertainty = (4 + 5) / 3 (= 3%) or

fractional uncertainty = (0.04 + 0.05) / 3 (= 0.03) C1

absolute uncertainty (= 0.03 × 0.121) = 0.0036 C1

d = 0.121 ± 0.004 m A1 [3] 2 (a) force per unit positive charge B1 [1] (b) (i) time = 5.9 × 10–2

/ 3.7 × 107 = 1.6 × 10–9

s (1.59 × 10–9 s) A1 [1]

(ii) E = V / d C1

= 2500 / 4.0 × 10–2

= 6.3 × 104

N C–1 (6.25 × 104 or 62500 N C–1) A1 [2] (iii) a = Eq / m or F = ma and F = Eq C1

= (6.3 × 104 × 1.60 × 10–19) / 9.11 × 10–31 = 1.1 × 1016 m s–2 A1 [2]

(iv) s = ut + ½at

2

= ½ × 1.1 × 1016 × (1.6 × 10–9)2 C1 = 1.4 × 10–2 (m) C1

distance from plate = 2.0 – 1.4 = 0.6 cm (allow 1 or more s.f.) A1 [3]

(v) electric force ≫ gravitational force (on electron)/weight

or acceleration due to electric field ≫acceleration due to gravitational field B1 [1]

(vi) vX–t graph: horizontal line at a non-zero value of vX B1

vY–t graph: straight line through the origin with positive gradient B1 [2]


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3 (a) force/load is proportional to extension/compression (provided proportionality limit is not exceeded) B1 [1]

(b) (i) k = F / x or k = gradient C1

k = 600 N m–1 A1 [2]

(ii) (W =) ½kx2 or (W =) ½Fx or (W =) area under graph C1

(W =) 0.5 × 600 × (0.040)2 = 0.48 J or (W =) 0.5 × 24 × 0.040 = 0.48 J A1 [2] (iii) 1. (EK =) ½mv2 C1

= ½ × 0.025 × 6.02

= 0.45 J A1 [2]

2. (work done against resistive force =) 0.48 – 0.45 [= 0.03(0) J] C1

average resistive force = 0.030 / 0.040 C1 = 0.75 N A1 [3]

(iv) efficiency = [useful energy out / total energy in] (×100) C1

= [0.45 / 0.48] (×100) = 0.94 or 94% A1 [2]

4 (a) the number of oscillations per unit time M1

of the source/of a point on the wave/of a particle (in the medium) A1 [2] or the number of wavelengths/wavefronts per unit time (M1) passing a (fixed) point (A1)

(b) T or period = 2.5 × 250 (µs) (= 625 µs) M1

frequency = 1 / (6.25 × 10–4) or 1 / (2.5 × 250 × 10–6) = 1600 Hz A1 [2] (c) (i) for maximum frequency: fo = fsv / (v – vs)

1640 = (1600 × 330) / (330 – vs) C1 vs = 8(.0) m s–1 (8.049 m s–1) A1 [2]

(ii) loudspeaker moving towards observer causes rise in/higher frequency B1

loudspeaker moving away from observer causes fall in/lower frequency B1 [2] or repeated rise and fall/higher and then lower frequency (M1) caused by loudspeaker moving towards and away from observer (A1)


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5 (a) wave incident on/passes by or through an aperture/edge B1

wave spreads (into geometrical shadow) B1 [2] (b) nλ = d sinθ C1

substitution of θ = 90° or sinθ = 1 C1 4 × 500 × 10–9 = d × sin 90° line spacing = 2.0 × 10–6

m A1 [3] (c) wavelength of red light is longer (than 500 nm) M1

(each order/fourth order is now at a greater angle so) the fifth-order maximum cannot be formed/not formed A1 [2]

6 (a) charge

forms) other to electrical (from ed)(transformenergy or done work B1 [1]

(b) (i) 1. V = IR or E = IR C1

I = 14 / 6.0 = 2.3 (2.33) A A1 [2]

2. total resistance of parallel resistors = 8.0 Ω C1

current = 14 / (6.0 + 8.0) = 1.0 A A1 [2]

(ii) P = EI (allow P = VI) or P = V2

/ R or P = I2R C1

change in power = (14 × 2.33) – (14 × 1.0) or (142 / 6.0) – (142

/ 14) or (2.332 × 6.0) – (1.02 × 14) = 19 W (18 W if 2.3 A used) A1 [2]

(c) I = Anvq

ratio = (0.50n / n) × (1.8 A / A) or ratio = 0.50 × 1.8 C1 = 0.90 A1 [2]


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7 (a) hadron not a fundamental particle/lepton is fundamental particle or hadron made of quarks/lepton not made of quarks or strong force/interaction acts on hadrons/does not act on leptons B1 [1]

(b) (i) proton: up, up, down / uud B1

neutron: up, down, down / udd B1 [2] (ii) composition: 2(uud) + 2(udd)

= 6 up, 6 down / 6u, 6d B1 [1] (c) (i) most of the atom is empty space

or the nucleus (volume) is (very) small compared to the atom B1 [1]

(ii) nucleus is (positively) charged B1

the mass is concentrated in (very small) nucleus/small region/small volume/small core or the majority of mass in (very small) nucleus/small region/small volume/small core B1 [2]





PHYSICS 9702/31

Paper 3 Advanced Practical Skills 1 October/November 2016

MARK SCHEME

Maximum Mark: 40

Published



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1 (b) Value for T in the range 0.60 s to 0.80 s with unit. [1]

Evidence of repeat timings (at least two recordings of nT where n ⩾ 5). [1] (c) Correct calculation of k. [1]

Value of k must be given to the same number of s.f. as (or one more than) the s.f. in the values of the raw times. [1]

(e) Six sets of readings of (h – h1) and T (with correct trend and without help from

Supervisor) scores 5 marks, five sets scores 4 marks etc. [5]

Range: [1] ∆(h – h1) ⩾ 30.0 cm.

Column headings: [1] Each column heading must contain a quantity and a unit where appropriate. The presentation of the quantity and unit must conform to accepted scientific convention, e.g. T / s or T (s).

Consistency: [1] All values of (h – h1) must be given to the nearest mm.

(f) (i) Axes: [1]

Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-linear) are not allowed. Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity that is being plotted.

Scale markings should be no more than three large squares apart.

Plotting of points: [1] All observations must be plotted on the grid. Diameter of plotted points must be ⩽ half a small square (no “blobs”). Points must be plotted to an accuracy of half a small square.

Quality: [1] All points in the table must be plotted on the grid for this mark to be awarded. All points must be within 0.01 s on the y-axis of a straight line.

(ii) Line of best fit: [1]

Judge by balance of all points on the grid about the candidate’s line (at least 5 points). There must be an even distribution of points either side of the line along the full length. Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate. There must be at least five points left after the anomalous point is disregarded. Line must not be kinked or thicker than half a small square.


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(iii) Gradient: [1] The hypotenuse of the triangle must be greater than half the length of the drawn line. The method of calculation must be correct. Do not allow ∆x / ∆y. Both read-offs must be accurate to half a small square in both the x and y directions.

y-intercept: [1] Either: Check correct read-off from a point on the line and substituted into y = mx + c. Read-off must be accurate to half a small square in both x and y directions. Or: Check read-off of the intercept directly from the graph (accurate to half a small square.

(g) Value of P = candidate’s gradient and value of Q = candidate’s intercept. [1]

Do not allow fractions.

Units for P (e.g. s m–1 or s cm–1 or s mm–1) and Q (s) correct. [1] 2 (a) (i) Value for D with unit in the range 0.14 mm to 0.16 mm. [1] (ii) Percentage uncertainty in D based on an absolute uncertainty of 0.01 mm. Correct method of calculation to obtain percentage uncertainty. [1] (c) (iii) Value of I in range 10 mA ⩽ I ⩽ 200 mA with unit (collected without help from Supervisor). [1] (iv) Value of V in range 0.4 V ⩽ V ⩽ 1.0 V with unit (collected without help from Supervisor). [1] (d) (i) Value of d > D and d < 1 mm. [1] (ii) Correct calculation of G. [1] (iii) Justification for s.f. in G linked to s.f. in D and d. [1] (f) (i) Second value of V. [1]

Quality: second value of V less than first value of V. [1] (ii) Second value of d. [1]


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(g) (i) Two values of k calculated correctly, and to at least 2 significant figures. [1]

(ii) Valid comment consistent with calculated values of k, testing against a stated numerical criterion. [1]

(h) (i) Limitations [4] (ii) Improvements [4] Do not credit

A Two readings not enough to draw a conclusion

Take many readings (for different diameters) and plot a graph/ take more readings and compare k values

Two readings not enough for accurate results Repeat readings Few readings Take more readings and calculate average k

B Difficult to measure diameter(s) with reason e.g. awkward placing micrometer round wire/ only one direction to measure diameter

Provide separate lengths of wire

C Meter readings changed in a particular direction over time/ repeat readings of I or V were often different/ contact resistance varies

Use power supply/ allow reading to reach steady value/ method of cleaning crocodile clips or wires

D Wire is very thin introducing a large percentage error in the diameter

Use thicker wire/ use digital micrometer

E Rheostat movement not precise enough – overshot I reading

Method to ensure exact current easier to produce e.g. use of screw thread adjustment





PHYSICS 9702/33


MARK SCHEME

Maximum Mark: 40

Published



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1 (a) (v) Value of I in range 20 mA ⩽ I ⩽ 200 mA with unit. [1] (c) Six sets of readings of R and y (with correct trend and without help from Supervisor) scores 5 marks, five sets scores 4 marks etc. [5]

Range: [1] Values of R must include 33 Ω.

Column headings: [1] Each column heading must contain a quantity and a unit where appropriate. The presentation of the quantity and unit must conform to accepted scientific convention, e.g. 1 / y / m–1 or 1 / y (m–1).

Consistency: [1] All values of raw y must be given to the nearest mm.

Significant figures: [1] Every value of 1 / R must be given to 2 or 3 significant figures.

Calculation: [1] 1 / y calculated correctly to the number of s.f. given by the candidate.

(d) (i) Axes: [1]

Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-linear) are not allowed. Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity that is being plotted. Scale markings should be no more than three large squares apart.

Plotting of points: [1] All observations must be plotted on the grid. Diameter of plotted points must be ⩽ half a small square (no “blobs”). Plotted points must be accurate to half a small square.

Quality: [1] All points in the table must be plotted on the grid for this mark to be awarded. All points must be within 0.001 cm–1 (0.1 m–1) (to scale) on the y-axis of a straight line.


Judge by balance of all points on the grid about the candidate’s line (at least 5 points). There must be an even distribution of points either side of the line along the full length. Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate. There must be at least five points left after the anomalous point is disregarded. Line must not be kinked or thicker than half a small square.


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(iii) Gradient: [1] The hypotenuse of the triangle must be greater than half the length of the drawn line. The method of calculation must be correct. Do not allow ∆x / ∆y. Both read-offs must be accurate to half a small square in both the x and y directions. y-intercept: [1] Either: Check correct read-off from a point on the line and substituted into y = mx + c. Read-off must be accurate to half a small square in both x and y directions. Or: Check read-off of the intercept directly from the graph (accurate to half a small square.

(e) Value of P = – candidate’s gradient and value of Q = candidate’s intercept.

Do not allow fractions. [1]

Units for P (e.g. Ω m–1) and Q (e.g. m–1) correct. [1] (f) (ii) Value of X in the range 5.0 Ω to 20.0 Ω from correct calculation. [1] 2 (b) (iii) All raw readings stated to at least 0.1 s. Value for T with unit in the range

0.50 s to 0.90 s. [1]

Evidence of repeats (at least two recordings of nT where n ⩾ 5). [1] (c) (i) Correct calculation of l. [1] (ii) Justification of s.f. in l linked to s.f. in time readings (e.g. time, raw time, nT). [1] (d) (ii) Value of d (l ± 0.050 m). [1] (iii) Absolute uncertainty in d in range 2 mm–8 mm.

If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown. Correct method of calculation to obtain percentage uncertainty. [1]

(e) (ii) Value for t in the range 2.0 s–8.0 s. [1] (f) Second value of T. [1]

Second value of t. [1]

Quality: Second value of t > first value of t. [1]


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(g) (i) Two values of k calculated correctly, and to at least 2 significant figures. [1] (ii) Valid comment consistent with calculated values of k, testing against a stated

numerical criterion. [1]



Take many readings (for different masses) and plot a graph/ take more readings and compare k values


B Difficult to set d with reason e.g. knot takes some string/knot or weight slips

Improved method to set d e.g. trial and error/glue knots to stop slipping

C Difficult to measure d with reason e.g. parallax error/cannot get ruler close/hands move holding ruler in place/difficult to judge centre of mass(es)/rule not vertical

Clamped metre rule/ measure to top and bottom of masses and find the average/ detailed method to check rule vertical/ detailed method to reduce parallax error

Eyes level Eyes perpendicular Reference to set square without detail

D Difficult to measure (n)T with reason e.g. difficult to judge complete oscillation

Use fiducial marker at centre of oscillation/

use a motion sensor placed under masses

Reaction time error Force on release

E Difficult to see when coil separation is constant or Separation of coils constant for a short time

Video with timer/ video and view frame by frame/ screen behind spring

Video timer

F Movement of spring along rod Method to fix spring to rod e.g. Blu-Tack or groove in rod

Air conditioning/draughts Unwanted modes of oscillation





PHYSICS 9702/34


MARK SCHEME

Maximum Mark: 40

Published



© UCLES 2016

1 (b) (ii) Value for x in range 24.0 cm to 26.0 cm, with unit. [1] (iv) Value for T in range 0.30 s to 1.00 s. [1] Evidence of repeat readings (at least two recordings of nT where n ⩾ 5). [1] (c) Six sets of values for x and T (with correct trend and without help from Supervisor) scores 4 marks, five sets scores 3 marks etc. [4]

Range: [1] xmin ⩽ 20.0 cm and xmax ⩾ 30.0 cm. Column headings: [1] Each column heading must contain a quantity and an appropriate unit. The presentation of the quantity and unit must conform to accepted scientific convention e.g. 1 / T2

(s–2) or 1 / T2 / 1 / s2.

Consistency: [1] All values of x must be given to the nearest mm. Significant figures: [1] Every value of 1/T2 must be given to the same number of s.f. as (or one greater than) the number of s.f. in the corresponding times. Calculation: [1] Values of 1/T

2 calculated correctly to the number of s.f. given by the candidate. (d) (i) Axes: [1]

Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-linear) are not allowed. Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions Scales must be labelled with the quantity that is being plotted. Scale markings must be no more than three large squares apart.

Plotting of points: [1] All observations in the table must be plotted on the grid. Diameter of plotted points must be ⩽ half a small square (no “blobs”). Points must be plotted to an accuracy of half a small square.

Quality: [1] All points in the table must be plotted (at least 5) for this mark to be awarded. All points must be within ± 1.0 cm (to scale) of a straight line in the x direction.


Judge by balance of all points on the grid about the candidate's line (at least 5 points). There must be an even distribution of points either side of the line along the full length. Allow one anomalous plot if clearly indicated (i.e. circled or labelled) by the candidate. There must be at least five points left after the anomalous point is disregarded. Line must not be kinked or thicker than half a small square.


© UCLES 2016


(e) Value of p = candidate’s gradient and value of q = candidate’s intercept. Do not allow fractions. [1] Units for p (e.g. cm–1 s–2) and q (e.g. s–2) correct. [1]

2 (b) L in range 19.0 cm to 21.0 cm, with unit. [1] (c) (iv) Values for x1 and x2 to nearest mm and x2 > x1. [1] Evidence of repeat readings of x1 and x2. [1] (v) Correct calculation of X. [1] (d) Absolute uncertainty in X in range 2 mm to 10 mm.

If repeated readings have been taken, then absolute uncertainty can be half the range (but not zero) if working is clearly shown. Correct method of calculation to obtain percentage uncertainty. [1]

(e) Second value for L. [1] Second values for x1 and x2. [1] Quality: X smaller for larger L. [1] (f) (i) Two values of k calculated correctly. [1] (ii) Justification of s.f. in k based on the s.f. in L, x1 and x2. [1] (iii) Valid comment consistent with the calculated values of k, testing against a stated numerical criterion. [1] (g) Value for X = 50 cm. [1]


© UCLES 2016



Take more readings and plot graph/

obtain more k values and compare


B Metre rule is not parallel to bench/horizontal

Use a second rule and measure at both ends/ use a (spirit) level

C Difficult to move stands with reason e.g. friction/ bench is rough/

stands tend to stick

Guide for stands (fixed to bench)/ mount stands on rollers/ put wheels on stands/ method to reduce friction e.g. sand bench with sandpaper

Use a smooth(er) bench Use lubricant

D Difficulty with rule e.g. rule skewed/ moves sideways

Use V-shaped rods/ groove in rods/ guide for ruler with some details

Falls off

E Difficult to measure x with reason e.g parallax error/

difficult to tell point where rod touches ruler

Scale on vertical edge of rule/ draw a line on the rod/ use a thinner rod/

replace rods with sharp edges e.g. prisms

Large contact area

Do not allow ‘use a computer to improve the experiment’.





PHYSICS 9702/35


MARK SCHEME

Maximum Mark: 40

Published



© UCLES 2016

1 (a) (i) Value of VF in the range 260.0 cm3 to 400.0 cm3. [1] (c) (v) All values of raw y to the nearest mm and less than 20 cm. [1] (d) (ii) Six sets of readings of V and y (with correct trend and without help from

Supervisor) scores 5 marks, five sets scores 4 marks etc. [5]

Range: [1] Maximum value of V ⩾ 200 cm3.

Column headings: [1] Each column heading must contain a quantity and an appropriate unit. The presentation of the quantity and unit must conform to accepted scientific convention, e.g. y / m or y (cm), V / cm3 , 2VF / 3 / cm3.

Consistency: [1] All values of raw V must be given to the nearest cm3.

(e) (i) Axes: [1]

Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-linear) are not allowed. Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity that is being plotted. Scale markings should be no more than three large squares apart.

Plotting of points: [1] All observations in the table must be plotted on the grid. Diameter of plotted points must be ⩽ half a small square (no “blobs”). Points must be plotted to an accuracy of half a small square.

Quality: [1] All points in the table must be plotted on the grid (at least 5) for this mark to be awarded. All points must be within ± 0.25 cm (to scale) in the y direction from a straight line.


Judge by balance of all points on the grid about the candidate’s line (at least 5 points). There must be an even distribution of points either side of the line along the full length. Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate. There must be at least four points left after disregarding the anomalous point. Line must not be kinked or thicker than half a small square.


© UCLES 2016

(iii) Gradient: [1] The hypotenuse of the triangle must be greater than half the length of the drawn line. The method of calculation must be correct. Do not allow ∆x / ∆y. Sign of gradient on answer line must match graph drawn. Both read-offs must be accurate to half a small square in both the x and y directions.

y-intercept: [1] Either: Check correct read-off from a point on the line and substituted into y = mx + c. Read-off must be accurate to half a small square in both x and y directions. Or: Check read-off of the intercept directly from the graph (accurate to half a small square.

(f) Value of P = candidate’s gradient and value of Q = candidate’s intercept.

Do not allow fractions. [1]

Units for P (e.g. mm–2, cm–2, m–2 but not e.g. cm / cm3) and Q (e.g. m) correct. [1] (g) (i) V calculated correctly to the number of significant figures given by the candidate.

Sign of answer must be consistent with P and Q values in (f). [1] (ii) Valid comment with a comparison of volumes e.g. V is greater than VF. [1] 2 (a) (ii) Value of L with unit in range 0.790 m to 0.810 m. [1] (iii) Absolute uncertainty in L either 1 mm or 2 mm.

If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown. Correct method of calculation to obtain percentage uncertainty. [1]

(b) (ii) All values of raw d1 with unit to the nearest mm. [1] (c) (ii) d2 > d1 showing scale on rule is used correctly. [1] (iii) Correct calculation of |d2 – d1|. [1] (iv) Correct calculation of M with consistent unit. Do not allow answers to 1 s.f. [1] (d) Justification for s.f. in M linked to s.f. in m and L. [1]


© UCLES 2016

(e) Second value of d1. [1]

Second value of d2. [1]

Quality: Second value of |d2 – d1| < first value of |d2 – d1|. [1] (f) (i) Two values of k calculated correctly. [1] (ii) Valid comment consistent with the calculated values of k, testing against a

stated numerical criterion. [1]

(g) (i) Limitations [4] (ii) Improvements [4] Do not credit


Take more readings and plot a graph/ obtain more k values and compare


B Difference between d2 and d1 is small/

(d2 – d1) is small/ large % uncertainty in (d2 – d1)

Improved method to measure (d2 – d1) values e.g. travelling microscope/larger masses/clamp vernier calipers above strip

Laser Ultrasonic position sensor Longer strip Change strip Deflection of strip

C Difficult to distribute mass evenly/ gaps between masses vary/ masses do not lie in a straight line/

masses have slots

Improved method of distribution e.g. use marker or scale on wooden strip/ use smaller masses to produce the same m/ continuous strip of named material e.g. modelling clay

Masses inaccurate Stick masses to strip Masses falling Replace strip with metre rule Masses with no slots

D Difficult to measure d values with reason e.g. rule not vertical/

difficult to hold second rule still as a pointer/ parallax error/

rule not held stationary

Improved method to measure d e.g. clamp metre rule/ place a set square on the floor next to ruler (to ensure rule is vertical)/ use of pointer (to metre rule)

Strip oscillates when masses are put on it Use a set square Effects of wind Reference to L

E Strip becomes permanently bent or deformed

Method to overcome deformation e.g. check d without masses before and between readings/ lay weights on strip to flatten

Elastic properties of wood vary Use a new strip





PHYSICS 9702/36


MARK SCHEME

Maximum Mark: 40

Published



© UCLES 2016

1 (b) (i) Value for x in range 45.0 cm to 55.0 cm. [1] (iii) Value for I in range 500 µA to 1500 µA (or 0.50 mA to 1.50 mA), with unit. [1] (c) Six sets of values for x and I (with correct trend and without help from Supervisor)

scores 5 marks, five sets scores 4 marks etc. [5] Range: [1] x values must include 20 cm or less and 80 cm or more. Column headings: [1]

Each column heading must contain a quantity and an appropriate unit. The presentation of the quantity and unit must conform to accepted scientific convention e.g. I / µA.

Consistency: [1] All values of raw x must be given to the nearest mm. (d) (i) Axes: [1] Sensible scales must be used. Awkward scales (e.g. 3:10, fractions or non-linear)

are not allowed. Scales must be chosen so that the plotted points occupy at least half the graph

grid in both x and y directions. Scales must be labelled with the quantity that is being plotted. Scale markings must be no more than three large squares apart. Plotting of points: [1] All observations in the table must be plotted on the grid. Diameter of plotted points must be ⩽ half a small square (no “blobs”). Plotting of points must be accurate to half a small square. Quality: [1] All points in the table (at least 5) must be plotted on the grid. All points must be within ± 20 µA (± 0.02 mA) of a straight line in the y (I) direction. (ii) Line of best fit: [1] Judge by balance of all points on the grid about the candidate's line (at least five

points). There must be an even distribution of points either side of the line along the full length.

One anomalous plot is allowed if clearly indicated (i.e. circled or labelled). There must be at least five points left after disregarding the anomalous point.

Lines must not be kinked or thicker than half a small square.


© UCLES 2016


(e) Value of S = candidate’s gradient and value of T = candidate’s intercept. Do not allow fractions. [1] Consistent units for S (e.g. µA cm–1) and T (e.g. µA). [1] (f) Calculation: r calculated correctly to the s.f. given by the candidate. [1] Significant figures: r given to 2 or 3 s.f. [1] 2 (b) (ii) x1 in range 10.0 cm to 40.0 cm. [1] (c) Value of x2 < x1. [1] (d) (i) Second value of x1. [1] (ii) Value of x2 given to nearest mm and all other raw values of x in (b), (c) and (d) are to the nearest mm. [1] (e) (i) Two values of k calculated correctly. [1] (ii) Justification of the s.f. in k based on the s.f. in x1 and the s.f in x2. [1] (iii) Valid comment consistent with the calculated values of k, testing against a

stated numerical criterion. [1] (f) (i) Raw values of D to nearest 0.001 cm and in range 1.400 cm to 2.200 cm. [1] Evidence of repeated readings for D. [1] (ii) Absolute uncertainty in D of 0.001 cm or 0.002 cm. If repeated readings have been taken, then absolute uncertainty could be

half the range (but not zero) if working is clearly shown. Correct method of calculation to obtain percentage uncertainty. [1]


© UCLES 2016

(iii) V calculated correctly. [1] (iv) Quality: M in range 3 g to 13 g. [1]

(g) (i) Limitations [4] (ii) Improvements [4] Do not credit

A Two readings are not enough to draw a valid conclusion

Take more readings and plot graph/ take more readings and compare k values


B Empty beaker moves on bench

Fix beaker with Blu-Tack/tape/ glue

C Difficult to balance rule: rule slips on pivot/ wind disturbs balance

Make groove in rule (under 50cm mark)/ other practical method e.g. hinge/nail through rule

Blu-tack Tape Switch off fans String slips on rule

D Spheres/string/tape still wet after immersion so mass changes or string/tape adds to mass of sphere

Use waterproof string/ use wire

Dry the spheres Waterproof tape

E Difficult to measure x with reason, e.g. string too thick (so it covers graduations on rule)

Use thin(ner) string

Parallax problems

F Marble not round Improved method of finding V (e.g. liquid displacement)

Repeat readings and average





PHYSICS 9702/41

Paper 4 A Level Structured Questions October/November 2016

MARK SCHEME

Maximum Mark: 100

Published



© UCLES 2016

1 (a) gravitational force provides/is the centripetal force B1 GMm / r

2 = mv 2

/ r or GMm / r 2 = mrω2

and v = 2πr / T or ω = 2π / T M1 with algebra to T

2 = 4π2r 3

/ GM A1 [3] or acceleration due to gravity is the centripetal acceleration (B1) GM / r

2 = v 2

/ r or GM / r 2 = rω2

and v = 2πr / T or ω = 2π / T (M1) with algebra to T

2 = 4π2r3 / GM (A1)

(b) (i) equatorial orbit/orbits (directly) above the equator B1 from west to east B1 [2] (ii) (24 × 3600)2 = 4π2r

3 / (6.67 × 10–11 × 6.0 × 1024) C1

r

3 = 7.57 × 1022

r = 4.2 × 107

m A1 [2] (c) (T / 24)2 = (2.64 × 107) / (4.23 × 107)3 B1 = 0.243 T = 12 hours A1 [2] or k (= T

2 / r

3) = 242 / (4.23 × 107)3 (B1)

k = 7.61 × 10–21 T

2 (= kr 3) = 7.61 × 10–21 × (2.64 × 107)3

= 140 T = 12 hours (A1) 2 (a) (i) p ∝ T or pV / T = constant or pV = nRT C1 T (= 5 × 300 =) 1500 K A1 [2] (ii) pV = nRT 1.0 × 105 × 4.0 × 10–4 = n × 8.31 × 300 or 5.0 × 105 × 4.0 × 10–4 = n × 8.31 × 1500 C1 n = 0.016 mol A1 [2]


© UCLES 2016

(b) (i) 1. heating/thermal energy supplied B1 2. work done on/to system B1 [2] (ii) 1. 240 J A1 2. same value as given in 1. (= 240 J) and zero given for 3. A1 3. zero A1 [3] 3 (a) 2k/m = ω2 M1 ω = 2πf M1 (2 × 64 / 0.810) = (2π × f)2 leading to f = 2.0 Hz A1 [3] (b) v0 = ωx0 or v0 = 2πfx0 or v = ω(x0

2 – x2)1/2 and x = 0 C1 v0 = 2π × 2.0 × 1.6 × 10–2

= 0.20 m s–1 A1 [2] (c) frequency: reduced/decreased B1 maximum speed: reduced/decreased B1 [2] 4 (a) (i) noise/distortion is removed (from the signal) B1 the (original) signal is reformed/reproduced/recovered/restored B1 [2] or signal detected above/below a threshold creates new signal (B1)

of 1s and 0s (B1) (ii) noise is superposed on the (displacement of the) signal/cannot be distinguished or analogue/signal is continuous (so cannot be regenerated)

or analogue/signal is not discrete (so cannot be regenerated) B1 noise is amplified with the signal B1 [2]


© UCLES 2016

(b) (i) gain/dB = 10 lg (P2 / P1) 32 = 10 lg [PMIN / (0.38 × 10–6)] or –32 = 10 lg (0.38 × 10–6

/ PMIN) C1 PMIN = 6.0 × 10–4 W A1 [2] (ii) attenuation = 10 lg [(9.5 × 10–3) / (6.02 × 10–4)] C1 = 12 dB attenuation per unit length (= 12/58) = 0.21 dB km–1 A1 [2] 5 (a) in an electric field, charges (in a conductor) would move B1 no movement of charge so zero field strength or charge moves until F = 0 / E = 0 B1 [2] or charges in metal do not move (B1) no (resultant) force on charges so no (electric) field (B1) (b) at P, EA = (3.0 × 10–12) / [4πε0(5.0 × 10–2)2] (= 10.79 N C–1) M1 at P, EB = (12 × 10–12) / [4πε0(10 × 10–2)2] (= 10.79 N C–1) M1 or (3.0 × 10–12) / [4πε0(5.0 × 10–2)2] – (12 × 10–12) / [4πε0(10 × 10–2)2] = 0 or (3.0 × 10–12) / [4πε0(5.0 × 10–2)2] = (12 × 10–12) / [4πε0(10 × 10–2)2] (M2) fields due to charged spheres are (equal and) opposite in direction, so E = 0 A1 [3] (c) potential = 8.99 × 109 (3.0 × 10–12) / (5.0 × 10–2) + (12 × 10–12) / (10 × 10–2) C1 = 1.62 V A1 [2] (d) ½mv2 = qV EK = ½ × 107 × 1.66 × 10–27 × v

2 C1 qV = 47 × 1.60 × 10–19 × 1.62 C1 v

2 = 1.37 × 108 v = 1.2 × 104

m s–1 A1 [3]


© UCLES 2016

6 (a) reference to input (voltage) and output (voltage) B1 there is no time delay between change in input and change in output B1 [2] or reference to rate at which output voltage changes (B1) infinite rate of change (of output voltage) (B1) (b) (i) 2.00 / 3.00 = 1.50 / R C1 or V+ = (3.00 × 4.5) / (2.00 + 3.00) = 2.7 2.7 = 4.5 × R / (R + 1.50) (C1) resistance = 2.25 kΩ A1 [2] (ii) 1. correct symbol for LED M1 two LEDs connected with opposite polarities between VOUT and earth A1 [2] 2. below 24 °C, RT > 1.5 kΩ or resistance of thermistor increases/high B1 V– < V+ or V– decreases/low (must not contradict initial statement) M1 VOUT is positive/+5 (V) and LED labelled as ‘pointing’ from VOUT to earth A1 [3] 7 (a) region (of space) where a force is experienced by a particle B1 [1] (b) (i) gravitational B1 (ii) gravitational and electric B1 (iii) gravitational, electric and magnetic B1 [3] (c) (i) force (always) normal to direction of motion M1 (magnitude of) force constant or speed is constant/kinetic energy is constant M1 magnetic force provides/is the centripetal force A1 [3] (ii) mv2

/ r = Bqv B1 momentum or p or mv = Bqr B1 [2]


© UCLES 2016

8 strong uniform magnetic field B1 nuclei precess/rotate about field (direction) (1) radio-frequency pulse (applied) B1 R.F. or pulse is at Larmor frequency/frequency of precession (1) causes resonance/excitation (of nuclei)/nuclei absorb energy B1 on relaxation/de-excitation, nuclei emit r.f./pulse B1 (emitted) r.f./pulse detected and processed (1) non-uniform magnetic field B1 allows position of nuclei to be located B1 allows for location of detection to be changed/different slices to be studied (1) any two of the points marked (1) B2 [8] 9 (a) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 [2] (b) flux linkage = BAN = π × 10–3 × 2.8 × π × (1.6 × 10–2)2 × 85 = 6.0 × 10–4 Wb B1 [1] (c) e.m.f. = ∆NΦ / ∆t e.m.f. = (6.0 × 10–4 × 2) / 0.30 C1 e.m.f. = 4.0 mV A1 [2] (d) sketch: E = 0 for t = 0 → 0.3 s, 0.6 s → 1.0 s, 1.6 s → 2.0 s B1 E = 4 mV for t = 0.3 s → 0.6 s (either polarity) B1 E = 2 mV for t = 1.0 s → 1.6 s B1 with opposite polarity B1 [4]


© UCLES 2016

10 (a) electromagnetic radiation/photons incident on a surface B1 causes emission of electrons (from the surface) B1 [2] (b) E = hc / λ = (6.63 × 10–34 × 3.00 × 108) / (436 × 10–9) C1 = 4.56 × 10–19

J (4.6 × 10–19 J) A1 [2]

(c) (i) Φ = hc / λ0 λ0 = (6.63 × 10–34 × 3.00 × 108) / (1.4 × 1.60 × 10–19) C1 = 890 nm A1 [2] (ii) λ0 = (6.63 × 10–34 × 3.00 × 108) / (4.5 × 1.60 × 10–19) = 280 nm A1 [1] (d) caesium: wavelength of photon less than threshold wavelength (or v.v.) or λ0 = 890 nm > 436 nm so yes A1 tungsten: wavelength of photon greater than threshold wavelength (or v.v.) or λ0 = 280 nm < 436 nm so no A1 [2] 11 in metal, conduction band overlaps valence band/no forbidden band/no band gap B1 as temperature rises, no increase in number of free electrons/charge carriers B1 as temperature rises, lattice vibrations increase M1 (lattice) vibrations restrict movement of electrons/charge carriers M1 (current decreases) so resistance increases A1 [5]


© UCLES 2016

12 (a) (i) time for number of atoms/nuclei or activity to be reduced to one half M1 reference to (number of…) original nuclide/single isotope or reference to half of original value/initial activity A1 [2] (ii) A = A0 exp(–λt) and either t = t½, A = ½A0 or ½A0 = A0 exp(–λt½) M1 so ln 2 = λt½ (and ln 2 = 0.693), hence 0.693 = λt½ A1 [2] (b) A = λN N = 200 / (2.1 × 10–6) C1 = 9.52 × 107 C1 mass = (9.52 × 107 × 222 × 10–3) / (6.02 × 1023) or mass = 9.52 × 107 × 222 × 1.66 × 10–27 C1 = 3.5 × 10–17

kg A1 [4]





PHYSICS 9702/42


MARK SCHEME

Maximum Mark: 100

Published



© UCLES 2016

1 (a) force per unit mass B1 [1] (b) (i) radius/diameter/size (of Proxima Centauri) ≪/is much less than 4.0 × 1013

km/separation (of Sun and star) or (because) it is a uniform sphere B1 [1] (ii) 1. field strength = GM / x2 = (6.67 × 10–11 × 2.5 × 1029) / (4.0 × 1013 × 103)2 C1

= 1.0 × 10–14

N kg–1 A1 [2] 2. force = field strength × mass = 1.0 × 10–14 × 2.0 × 1030 C1 or force = GMm / x2

= (6.67 × 10–11 × 2.5 × 1029 × 2.0 × 1030) / (4.0 × 1013 × 103)2 (C1) = 2.0 × 1016

N A1 [2] (c) force (of 2 × 1016 N) would have little effect on (large) mass of Sun B1 would cause an acceleration of Sun of 1.0 × 10–14 m s–2/very small/negligible acceleration B1 [2] or many stars all around the Sun (B1) net effect of forces/fields is zero (B1) 2 (a) (i) number of moles/amount of substance B1 [1] (ii) kelvin temperature/absolute temperature/thermodynamic temperature B1 [1] (b) pV = nRT 4.9 × 105 × 2.4 × 103 × 10–6 = n × 8.31 × 373 B1 n = 0.38 (mol) C1 number of molecules or N = 0.38 × 6.02 × 1023 = 2.3 × 1023 A1 [3]


© UCLES 2016

or pV = NkT (C1) 4.9 × 105 × 2.4 × 103 × 10–6 = N × 1.38 × 10–23 × 373 (M1) number of molecules or N = 2.3 × 1023 (A1) (c) volume occupied by one molecule = (2.4 × 103) / (2.3 × 1023) C1 = 1.04 × 10–20 cm3 mean spacing = (1.04 × 10–20)1/3 C1 = 2.2 × 10–7 cm (allow 1 s.f.) A1 [3] (allow other dimensionally correct methods e.g. V = (4/3)πr3) 3 (a) (sum of/total) potential energy and kinetic energy of (all) molecules/particles M1 reference to random (distribution) A1 [2] (b) (i) no heat enters (gas)/leaves (gas)/no heating (of gas) B1 work done by gas (against atmosphere as it expands) M1 internal energy decreases A1 [3] (ii) volume decreases so work done on ice/water B1 (allow work done negligible because ∆V small) heating of ice (to break rigid forces/bonds) M1 internal energy increases A1 [3] 4 (a) (i) 0.225 s and 0.525 s A1 [1] (ii) period or T = 0.30 s and ω = 2π / T C1 ω = 2π / 0.30 ω = 21 rad s–1 A1 [2] (iii) speed = ωx0 or ω(x0

2 – x2)1/2 and x = 0 C1 = 20.9 × 2.0 × 10–2 = 0.42 m s–1 A1 [2]


© UCLES 2016

or use of tangent method: correct tangent shown on Fig. 4.2 (C1) working e.g. ∆y / ∆x leading to maximum speed in range (0.38–0.46) m s–1 (A1) (b) sketch: reasonably shaped continuous oval/circle surrounding (0,0) B1 curve passes through (0, 0.42) and (0, –0.42) B1 curve passes through (2.0, 0) and (–2.0, 0) B1 [3] 5 (a) transducer/transmitter can be also be used as the receiver or transducer both transmits and receives receives reflected pulses between the emitted pulses (needs to be pulsed) in order to measure/determine depth(s) (needs to be pulsed) to determine nature of boundaries Any three of the above marking points, 1 mark each B2 [2] (b) (i) product of speed of (ultra)sound and density (of medium) M1 reference to speed of sound in medium A1 [2] (ii) if Z1 and Z2 are (nearly) equal, IT / I0 (nearly) equal to 1/unity/(very) little reflection/mostly transmission B1 if Z1 ≫ Z2 or Z1 ≪ Z2 or the difference between Z1 and Z2 is (very) large, then IT / I0 is small/zero/mostly reflection/little transmission B1 [2] 6 (a) E = 0 or EA = (–)EB (at x = 11 cm) B1 QA / x2 = QB / (20 – x)2 = 112

/ 92 C1 QA / QB or ratio = 1.5 A1 [3] or E ∝ Q because r same or E = Q / 4πε0r2 and r same (B1) QA / QB = 48 / 32 (C1) QA / QB or ratio = 1.5 (A1)


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(b) (i) for max. speed, ∆V = (0.76 – 0.18) V or ∆V = 0.58 V C1 q∆V = ½mv2 2 × (1.60 × 10–19) × 0.58 = ½ × 4 × 1.66 × 10–27 × v2 C1 v2 = 5.59 × 107 v = 7.5 × 103

m s–1 A1 [3] (ii) ∆V = 0.22 V C1 2 × (1.60 × 10–19) × 0.22 = ½ × 4 × 1.66 × 10–27 × v2 v2 = 2.12 × 107

v = 4.6 × 103

m s–1 A1 [2] 7 (a) (i) charge / potential (difference) or charge per (unit) potential (difference) B1 [1] (ii) (V = Q / 4πε0r and C = Q / V) for sphere, C (= Q / V) = 4πε0r C1 C = 4π × 8.85 × 10–12 × 12.5 × 10–2 = 1.4 × 10–11

F A1 [2] (b) (i) 1 / CT = 1 / 3.0 + 1 / 6.0 CT = 2.0 µF A1 [1] (ii) total charge = charge on 3.0 µF capacitor = 2.0 (µ) × 9.0 = 18 (µC) C1 potential difference = Q / C = 18 (µ)C / 3.0 (µ)F = 6.0 V A1 [2] or argument based on equal charges: 3.0 × V = 6.0 × (9.0 – V) (C1) V = 6.0 V (A1) (iii) potential difference (= 9.0 – 6.0) = 3.0 V C1 charge (= 3.0 × 2.0 (µ)) = 6.0 µC A1 [2]


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8 (a) P shown between earth symbol and voltmeter B1 [1] (b) (i) gain = (50 × 103) / 100 = 500 C1 VIN (= 5.0 / 500) = 0.010 V A1 [2] (ii) VIN (= 5.0 / 5.0) = 1.0 V A1 [1] (c) e.g. multi-range (volt)meter c.r.o. sensitivity control amplifier channel selector B1 [1] 9 (a) (by Newton’s third law) force on wire is up(wards) M1 by (Fleming’s) left-hand rule/right-hand slap rule to give current A1 in direction left to right shown on diagram A1 [3] (b) force ∝ current or F = BIL or B (= 0.080 / 6.0L) = 1 / 75L C1 maximum current = 2.5 × √2 C1

= 3.54 A maximum force in one direction = (3.54 / 6.0) × 0.080 C1 = 0.047 N difference (= 2 × 0.047) = 0.094 N or force varies from 0.047 N upwards to 0.047 N downwards A1 [4] 10 nuclei emitting r.f. (pulse) B1 Larmor frequency/r.f. frequency emitted/detected depends on magnitude of magnetic field B1 nuclei can be located (within a slice) B1 changing field enables position of detection (slice) to be changed B1 [4] 11 (a) (induced) e.m.f. proportional/equal to rate M1 of change of (magnetic) flux (linkage) A1 [2] (b) (for same current) iron core gives large(r) (rates of change of) flux (linkage) B1 e.m.f induced in solenoid is greater (for same current) M1 induced e.m.f. opposes applied e.m.f. so current smaller/acts to reduce current A1 [3]


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or same supply so same induced e.m.f. balancing it (B1) (rate of change of) flux linkage is same (M1) smaller current for same flux when core present (A1) (c) e.g. (heating due to) eddy currents in core (heating due to current in) resistance of coils hysteresis losses/losses due to changing magnetic field in core Any two of the above marking points, 1 mark each B2 [2] 12 (a) (i) electron diffraction/electron microscope (allow other sensible suggestions) B1 [1] (ii) photoelectric effect/Compton scattering (allow other sensible suggestions) B1 [1] (b) (i) arrow clear from –0.54 eV to –3.40 eV B1 [1] (ii) E = hc / λ or E = hf and c = fλ C1 λ = (6.63 × 10–34 × 3.00 × 108) / [(3.40 – 0.54) × 1.60 × 10–19] = 4.35 × 10–7 m A1 [2] (c) (i) wavelength associated with a particle M1 that is moving/has momentum/has speed/has velocity A1 [2] (ii) λ = h / mv v = (6.63 × 10–34) / (9.11 × 10–31 × 4.35 × 10–7) C1 = 1.67 × 103

m s–1 A1 [2] 13 X-ray image of a (single) slice/cross-section (through the patient) M1 taken from different angles/rotating X-ray (beam) A1 computer is used to form/process/build up/store image B1 2D image (of the slice) B1 repeated for many/different (neighbouring) slices M1 to build up 3D image A1 [6]


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14 (a) (i) He42 or α4

2 B1 [1] (ii) n1

0 B1 [1] (b) (i) ∆m = (29.97830 +1.00867) – (26.98153 + 4.00260) C1 = 30.98697 – 30.98413 = 2.84 × 10–3 u C1 [2] (ii) E = c2∆m or mc2 C1 = (3.0 × 108)2 × 2.84 × 10–3 × 1.66 × 10–27

= 4.2 × 10–13 J A1 [2] (c) mass of products is greater than mass of Al plus α or reaction causes (net) increase in (rest) mass (of the system) B1 α-particle must have at least this amount of kinetic energy B1 [2]





PHYSICS 9702/43


MARK SCHEME

Maximum Mark: 100

Published



© UCLES 2016

1 (a) gravitational force provides/is the centripetal force B1 GMm / r

2 = mv 2

/ r or GMm / r 2 = mrω2

and v = 2πr / T or ω = 2π / T M1 with algebra to T

2 = 4π2r 3

/ GM A1 [3] or acceleration due to gravity is the centripetal acceleration (B1) GM / r

2 = v 2

/ r or GM / r 2 = rω2

and v = 2πr / T or ω = 2π / T (M1) with algebra to T

2 = 4π2r3 / GM (A1)

(b) (i) equatorial orbit/orbits (directly) above the equator B1 from west to east B1 [2] (ii) (24 × 3600)2 = 4π2r

3 / (6.67 × 10–11 × 6.0 × 1024) C1

r

3 = 7.57 × 1022

r = 4.2 × 107

m A1 [2] (c) (T / 24)2 = (2.64 × 107) / (4.23 × 107)3 B1 = 0.243 T = 12 hours A1 [2] or k (= T

2 / r

3) = 242 / (4.23 × 107)3 (B1)

k = 7.61 × 10–21 T

2 (= kr 3) = 7.61 × 10–21 × (2.64 × 107)3

= 140 T = 12 hours (A1) 2 (a) (i) p ∝ T or pV / T = constant or pV = nRT C1 T (= 5 × 300 =) 1500 K A1 [2] (ii) pV = nRT 1.0 × 105 × 4.0 × 10–4 = n × 8.31 × 300 or 5.0 × 105 × 4.0 × 10–4 = n × 8.31 × 1500 C1 n = 0.016 mol A1 [2]


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(b) (i) 1. heating/thermal energy supplied B1 2. work done on/to system B1 [2] (ii) 1. 240 J A1 2. same value as given in 1. (= 240 J) and zero given for 3. A1 3. zero A1 [3] 3 (a) 2k/m = ω2 M1 ω = 2πf M1 (2 × 64 / 0.810) = (2π × f)2 leading to f = 2.0 Hz A1 [3] (b) v0 = ωx0 or v0 = 2πfx0 or v = ω(x0

2 – x2)1/2 and x = 0 C1 v0 = 2π × 2.0 × 1.6 × 10–2

= 0.20 m s–1 A1 [2] (c) frequency: reduced/decreased B1 maximum speed: reduced/decreased B1 [2] 4 (a) (i) noise/distortion is removed (from the signal) B1 the (original) signal is reformed/reproduced/recovered/restored B1 [2] or signal detected above/below a threshold creates new signal (B1)

of 1s and 0s (B1) (ii) noise is superposed on the (displacement of the) signal/cannot be distinguished or analogue/signal is continuous (so cannot be regenerated)

or analogue/signal is not discrete (so cannot be regenerated) B1 noise is amplified with the signal B1 [2]


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(b) (i) gain/dB = 10 lg (P2 / P1) 32 = 10 lg [PMIN / (0.38 × 10–6)] or –32 = 10 lg (0.38 × 10–6

/ PMIN) C1 PMIN = 6.0 × 10–4 W A1 [2] (ii) attenuation = 10 lg [(9.5 × 10–3) / (6.02 × 10–4)] C1 = 12 dB attenuation per unit length (= 12/58) = 0.21 dB km–1 A1 [2] 5 (a) in an electric field, charges (in a conductor) would move B1 no movement of charge so zero field strength or charge moves until F = 0 / E = 0 B1 [2] or charges in metal do not move (B1) no (resultant) force on charges so no (electric) field (B1) (b) at P, EA = (3.0 × 10–12) / [4πε0(5.0 × 10–2)2] (= 10.79 N C–1) M1 at P, EB = (12 × 10–12) / [4πε0(10 × 10–2)2] (= 10.79 N C–1) M1 or (3.0 × 10–12) / [4πε0(5.0 × 10–2)2] – (12 × 10–12) / [4πε0(10 × 10–2)2] = 0 or (3.0 × 10–12) / [4πε0(5.0 × 10–2)2] = (12 × 10–12) / [4πε0(10 × 10–2)2] (M2) fields due to charged spheres are (equal and) opposite in direction, so E = 0 A1 [3] (c) potential = 8.99 × 109 (3.0 × 10–12) / (5.0 × 10–2) + (12 × 10–12) / (10 × 10–2) C1 = 1.62 V A1 [2] (d) ½mv2 = qV EK = ½ × 107 × 1.66 × 10–27 × v

2 C1 qV = 47 × 1.60 × 10–19 × 1.62 C1 v

2 = 1.37 × 108 v = 1.2 × 104

m s–1 A1 [3]


© UCLES 2016

6 (a) reference to input (voltage) and output (voltage) B1 there is no time delay between change in input and change in output B1 [2] or reference to rate at which output voltage changes (B1) infinite rate of change (of output voltage) (B1) (b) (i) 2.00 / 3.00 = 1.50 / R C1 or V+ = (3.00 × 4.5) / (2.00 + 3.00) = 2.7 2.7 = 4.5 × R / (R + 1.50) (C1) resistance = 2.25 kΩ A1 [2] (ii) 1. correct symbol for LED M1 two LEDs connected with opposite polarities between VOUT and earth A1 [2] 2. below 24 °C, RT > 1.5 kΩ or resistance of thermistor increases/high B1 V– < V+ or V– decreases/low (must not contradict initial statement) M1 VOUT is positive/+5 (V) and LED labelled as ‘pointing’ from VOUT to earth A1 [3] 7 (a) region (of space) where a force is experienced by a particle B1 [1] (b) (i) gravitational B1 (ii) gravitational and electric B1 (iii) gravitational, electric and magnetic B1 [3] (c) (i) force (always) normal to direction of motion M1 (magnitude of) force constant or speed is constant/kinetic energy is constant M1 magnetic force provides/is the centripetal force A1 [3] (ii) mv2

/ r = Bqv B1 momentum or p or mv = Bqr B1 [2]


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8 strong uniform magnetic field B1 nuclei precess/rotate about field (direction) (1) radio-frequency pulse (applied) B1 R.F. or pulse is at Larmor frequency/frequency of precession (1) causes resonance/excitation (of nuclei)/nuclei absorb energy B1 on relaxation/de-excitation, nuclei emit r.f./pulse B1 (emitted) r.f./pulse detected and processed (1) non-uniform magnetic field B1 allows position of nuclei to be located B1 allows for location of detection to be changed/different slices to be studied (1) any two of the points marked (1) B2 [8] 9 (a) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 [2] (b) flux linkage = BAN = π × 10–3 × 2.8 × π × (1.6 × 10–2)2 × 85 = 6.0 × 10–4 Wb B1 [1] (c) e.m.f. = ∆NΦ / ∆t e.m.f. = (6.0 × 10–4 × 2) / 0.30 C1 e.m.f. = 4.0 mV A1 [2] (d) sketch: E = 0 for t = 0 → 0.3 s, 0.6 s → 1.0 s, 1.6 s → 2.0 s B1 E = 4 mV for t = 0.3 s → 0.6 s (either polarity) B1 E = 2 mV for t = 1.0 s → 1.6 s B1 with opposite polarity B1 [4]


© UCLES 2016

10 (a) electromagnetic radiation/photons incident on a surface B1 causes emission of electrons (from the surface) B1 [2] (b) E = hc / λ = (6.63 × 10–34 × 3.00 × 108) / (436 × 10–9) C1 = 4.56 × 10–19

J (4.6 × 10–19 J) A1 [2]

(c) (i) Φ = hc / λ0 λ0 = (6.63 × 10–34 × 3.00 × 108) / (1.4 × 1.60 × 10–19) C1 = 890 nm A1 [2] (ii) λ0 = (6.63 × 10–34 × 3.00 × 108) / (4.5 × 1.60 × 10–19) = 280 nm A1 [1] (d) caesium: wavelength of photon less than threshold wavelength (or v.v.) or λ0 = 890 nm > 436 nm so yes A1 tungsten: wavelength of photon greater than threshold wavelength (or v.v.) or λ0 = 280 nm < 436 nm so no A1 [2] 11 in metal, conduction band overlaps valence band/no forbidden band/no band gap B1 as temperature rises, no increase in number of free electrons/charge carriers B1 as temperature rises, lattice vibrations increase M1 (lattice) vibrations restrict movement of electrons/charge carriers M1 (current decreases) so resistance increases A1 [5]


© UCLES 2016

12 (a) (i) time for number of atoms/nuclei or activity to be reduced to one half M1 reference to (number of…) original nuclide/single isotope or reference to half of original value/initial activity A1 [2] (ii) A = A0 exp(–λt) and either t = t½, A = ½A0 or ½A0 = A0 exp(–λt½) M1 so ln 2 = λt½ (and ln 2 = 0.693), hence 0.693 = λt½ A1 [2] (b) A = λN N = 200 / (2.1 × 10–6) C1 = 9.52 × 107 C1 mass = (9.52 × 107 × 222 × 10–3) / (6.02 × 1023) or mass = 9.52 × 107 × 222 × 1.66 × 10–27 C1 = 3.5 × 10–17

kg A1 [4]





PHYSICS 9702/51

Paper 5 Planning, Analysis and Evaluation October/November 2016

MARK SCHEME

Maximum Mark: 30

Published



© UCLES 2016

Question Answer Marks

1 Defining the problem

B is the independent variable and v is the dependent variable, or vary B measure v.

1

Keep starting position of magnet constant/magnet always released from rest.

1

Methods of data collection

Labelled diagram showing a magnet and the vertical copper tube supported. 1

Method to ensure that copper tube is vertical, e.g. set square, spirit level, plumb line.

1

Method to determine time at bottom of tube e.g. use of light gate(s)/motion sensor attached to timer/datalogger/computer or distance between two fixed marks at bottom of tube and stopwatch. Do not allow time over length of tube.

1

Method to measure B, e.g. Hall probe. 1

Method of analysis

Plot a graph of ln v against B. 1

λ = – gradient 1

v0 = ey-intercept 1


© UCLES 2016


Additional detail including safety considerations 6

1. Keep mass of magnet constant. 2. Measurement of an appropriate length to determine v at bottom of tube,

e.g. use ruler to measure distance between light gates/length of magnet/between two fixed marks.

3. v = d / t for appropriate lengths (not length of tube)4. Adjust Hall probe until maximum reading obtained/perpendicular to

field/pole or Use Hall probe to take readings for both poles and average.

5. Method to calibrate Hall probe using a known field.6. Safety precaution linked to falling magnets/use sand tray/cushion to

soften fall. 7. Repeat experiment with magnets reversed and average

or Repeat v (or t) for same B and average.

8. ln v = –λB + ln v0 9. Relationship is valid if the graph is a straight line.10. Method to vary B, e.g. re-magnetise in a coil.


© UCLES 2016


2 (a) gradient = 1

YE

y-intercept = 1E

1

(b)

2.7 or 2.70 0.833 or 0.8333

1.4 or 1.35 0.513 or 0.5128

0.90 or 0.900 0.426 or 0.4255

0.68 or 0.675 0.364 or 0.3636

0.54 or 0.540 0.345 or 0.3448

0.45 or 0.450 0.328 or 0.3279 All first column correct. Allow a mixture of significant figures. All second column correct. Allow a mixture of significant figures. Uncertainties in X from ± 0.4 to ± 0.07 (± 0.1). Allow more than one significant figure.

1

1

1

(c) (i) Six points plotted correctly. Must be within half a small square. No “blobs”.

1

All error bars in X plotted correctly. All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical.

1

(ii) Line of best fit drawn. Line must not be drawn from top point to bottom point. The lower end of line should pass between (0.95, 0.45) and (1.1, 0.45) and upper end of line should pass between (2.10, 0.70) and (2.25, 0.70).

1

Worst acceptable line drawn correctly. Steepest or shallowest possible line that passes through all the error bars. Mark scored only if all error bars are plotted.

1

(iii) Gradient determined with a triangle that is at least half the length of the drawn line. Read-offs must be accurate to half a small square.

1

Method of determining absolute uncertainty. uncertainty = gradient of line of best fit – gradient of worst acceptable line or uncertainty = ½(steepest worst line gradient – shallowest worst line gradient)

1

(iv) y-intercept determined correctly by substitution into y = mx + c. Read-offs must be accurate to half a small square.

1


© UCLES 2016


Method of determining absolute uncertainty. uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line or uncertainty = ½(steepest worst line y-intercept – shallowest worst line y-intercept) No ECF from false origin method.

1

(d) (i) E = 1/y-intercept and given to 2 or 3 s.f. 1

gradient1

×=

EY or

gradientintercept−y

Y in the range (0.90 to 1.20) × 10–3 F.

Appropriate unit required. Correct substitution of numbers must be seen.

1

(ii) Percentage uncertainty in Y ∆ ∆ = + ×

100m c

m c or

∆ ∆ = + × 100m E

m E or

∆= × 100Y

Y

Maximum/minimum methods:

gradient mininterceptmax

gradient min min1max −

=×

=y

EY

gradientmax intercept min

gradientmax max 1min −

=×

=y

EY

1





PHYSICS 9702/52


MARK SCHEME

Maximum Mark: 30

Published



© UCLES 2016



p is the independent variable and B is the dependent variable, or vary p and measure B.

1

Keep the current/I (in the electromagnet) constant. 1


Labelled diagram showing Hall probe correctly positioned (along p) and ruler correctly positioned and either Hall probe or rule supported.

1

Correct circuit diagram to include d.c. power supply in series with coil and ammeter. Must be a workable circuit diagram to measure current through the coil.

1

Measure p with ruler. 1

Method to determine an accurate value of p. Examples include: Height of P above bench – height of electromagnet Height of P measured from ruler across the top of the electromagnet

1

Method of analysis

Plot a graph of ln B against p. 1

α = – gradient 1

IN

ky intercepte −

= 1


© UCLES 2016



1. Keep the number of turns/N constant. 2. Use large number of turns/current (to increase B). 3. Avoid overheating the coil/do not touch hot coil. 4. Use of variable resistor to keep ammeter reading constant. 5. Method to ensure that Hall probe is equidistant from the poles, e.g.

determine centre of electromagnet and use of plumb line/ruler and spirit level/set square.

6. Adjust Hall probe until maximum reading obtained/perpendicular to field. 7. Repeat each experiment for the same value of p and reverse the

current/Hall probe and average

8. ln B = –αp + ln kNI 9. Relationship is valid if the graph is a straight line. 10. Calibrate Hall probe using a known field.


© UCLES 2016


2 (a) gradient = q y-intercept = lg p

1

(b) 2.80 or 2.799 or 2.7993 0.28 or 0.279

2.79 or 2.792 or 2.7924 0.30 or 0.301

2.77 or 2.771 or 2.7709 0.36 or 0.362

2.72 or 2.716 or 2.7160 0.49 or 0.491

2.69 or 2.690 or 2.6902 0.57 or 0.568

2.67 or 2.672 or 2.6721 0.61 or 0.613

All first column correct – either 2 and 3 decimal places or 3 and 4 decimal places. All second column correct. Allow a mixture of decimal places. Uncertainties in lg (V / V) from ± 0.02 to ± 0.01. Allow more than one significant figure.

1

1

1


1

All error bars in lg (V / V) plotted correctly. All error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.

1

(ii) Line of best fit drawn. Line must not be drawn from top point to bottom point unless points are balanced. Upper end of line should pass between (2.694, 0.55) and (2.700, 0.55) and lower end of line should pass between (2.770, 0.35) and (2.776, 0.35).

1


1

(iii) Gradient determined with a triangle that is at least half the length of the drawn line. Read-offs must be accurate to half a small square. Gradient must be negative.

1


1


© UCLES 2016


(iv) y-intercept determined by substitution into y = mx + c. Read-offs must be accurate to half a small square.

1


1

(d) Use of p = 10answer to 2(c)(iv) or lg p = answer to 2(c)(iv)

1

q = gradient and in the range –2.50 to –2.70 and given to 2 or 3 s.f. 1

(e) Use of V = p × 950q

or lg V = q lg 950 + lg p or lg V = q lg 950 + y-intercept Correct substitution of numbers must be seen to give V.

1





PHYSICS 9702/53


MARK SCHEME

Maximum Mark: 30

Published



© UCLES 2016



B is the independent variable and v is the dependent variable, or vary B measure v.

1

Keep starting position of magnet constant/magnet always released from rest.

1


Labelled diagram showing a magnet and the vertical copper tube supported. 1

Method to ensure that copper tube is vertical, e.g. set square, spirit level, plumb line.

1

Method to determine time at bottom of tube e.g. use of light gate(s)/motion sensor attached to timer/datalogger/computer or distance between two fixed marks at bottom of tube and stopwatch. Do not allow time over length of tube.

1

Method to measure B, e.g. Hall probe. 1

Method of analysis

Plot a graph of ln v against B. 1

λ = – gradient 1

v0 = ey-intercept 1


© UCLES 2016



1. Keep mass of magnet constant. 2. Measurement of an appropriate length to determine v at bottom of tube,

e.g. use ruler to measure distance between light gates/length of magnet/between two fixed marks.

3. v = d / t for appropriate lengths (not length of tube)4. Adjust Hall probe until maximum reading obtained/perpendicular to

field/pole or Use Hall probe to take readings for both poles and average.

5. Method to calibrate Hall probe using a known field.6. Safety precaution linked to falling magnets/use sand tray/cushion to

soften fall. 7. Repeat experiment with magnets reversed and average

or Repeat v (or t) for same B and average.

8. ln v = –λB + ln v0 9. Relationship is valid if the graph is a straight line.10. Method to vary B, e.g. re-magnetise in a coil.


© UCLES 2016


2 (a) gradient = 1

YE

y-intercept = 1E

1

(b)

2.7 or 2.70 0.833 or 0.8333

1.4 or 1.35 0.513 or 0.5128

0.90 or 0.900 0.426 or 0.4255

0.68 or 0.675 0.364 or 0.3636

0.54 or 0.540 0.345 or 0.3448

0.45 or 0.450 0.328 or 0.3279 All first column correct. Allow a mixture of significant figures. All second column correct. Allow a mixture of significant figures. Uncertainties in X from ± 0.4 to ± 0.07 (± 0.1). Allow more than one significant figure.

1

1

1


1

All error bars in X plotted correctly. All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical.

1

(ii) Line of best fit drawn. Line must not be drawn from top point to bottom point. The lower end of line should pass between (0.95, 0.45) and (1.1, 0.45) and upper end of line should pass between (2.10, 0.70) and (2.25, 0.70).

1


1

(iii) Gradient determined with a triangle that is at least half the length of the drawn line. Read-offs must be accurate to half a small square.

1


1

(iv) y-intercept determined correctly by substitution into y = mx + c. Read-offs must be accurate to half a small square.

1


© UCLES 2016



1

(d) (i) E = 1/y-intercept and given to 2 or 3 s.f. 1

gradient1

×=

EY or

gradientintercept−y

Y in the range (0.90 to 1.20) × 10–3 F.

Appropriate unit required. Correct substitution of numbers must be seen.

1

(ii) Percentage uncertainty in Y ∆ ∆ = + ×

100m c

m c or

∆ ∆ = + × 100m E

m E or

∆= × 100Y

Y

Maximum/minimum methods:

gradient mininterceptmax

gradient min min1max −

=×

=y

EY

gradientmax intercept min

gradientmax max 1min −

=×

=y

EY

1

Education

9702 w16 ms_all