y13 assessment mark scheme

1. (a) (i) Cp = 2 + 4 = 6 μF A1


(ii) 1/C = 1/2 + ¼ C1

Cs = 4/3 =1.33 μF A1

(b) (i) 6.0 V A1

(ii) Q = CpV C1

= 6 × 6 = 36 μC A1

(c) E = ½ CsV2 C1

= 24 × 10–6 A1

(d) (i) The capacitors discharge through the voltmeter. B1

(ii) V = V0e–t/CR

1/4 =e–t/(6×12) C1

ln 4 = t / 72 C1

t = 72 ln 4 ≈ 100 s A1

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2. (a) (i) Q = VC; W = ½ VC.V ( = ½ CV2) (2)

(ii) parabolic shape passing through origin (1)

plotted accurately as W = 1.1 V2 (1) 4

(b) (i) T = RC; = 6.8 × 103 × 2.2 = 1.5 × 104 s = 4.16 h 2

(ii) ΔW = ½ C(V12 –V22) = 1.1(25 – 16) ; = 9.9 (J) 2

(iii) 4 = 5 exp(–t/1.5 × 104) ; giving t = 1.5 × 104 × ln 1.25 = 3.3 × 103 (s) 2

(iv) P = ΔW/Δt = 9.9/3.3 × 103 = 3.0 mW ecf b(ii) and (iii) 1

allow P = Vav2 /R = 4.52/6.8 × 103 = 2.98 mW

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3 (a) (i)

capacitor capacitance / µF charge / µC p.d. / V energy / µJ

X 5 30 = Q/C

= 6 (V) (1) = ½ CV 2(1)

= ½ × 5 × 62

= 90 (1)

Y 25 = CV

= 25 × 6

=150 (µC) (1) = 6 (V) (1) = 450 (1)

Z 10 30 + 150 =

180 (µC) (1) = Q/C

= 180/10

= 18 (V) (1) = 1620 (1)

Each box correctly calculated scores (1) + (1) for ½ CV2 9

(ii) 1 18 V + 6 V = 24 (V) (1)

2 180 (µC) (1)

3 180 / 24 = 7.5 (1)

4 90 + 450 + 1620 = 2160 (µJ) (1) 4

(b) (i) Kirchhoff’s second law OR conservation of energy (1) 1

(ii) Kirchhoff’s first law OR conservation of charge (1) 1

(c) (i) time constant = CR (1)

= 7.5 × 10–6 × 200 000 = 1.5 (s) (1) 2

(ii) (1)

Q/Qo = e–4 = 0.0183 (1) 2

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4. (a) Positive as E-field is downwards/top plate is positive/like charges repel/AW (1) 1

(b) (i) k.e. = QV; = 300 × 1.6 × 10–19 = (4.8 × 10–17 J) (2) 2

(ii) 1/2mv2 = 4.8 × 10–17; = 0.5 × 2.3 × 10–26 × v2 so v2 = 4.17 × 109;

(giving v = 6.46 × 104 m s–1) (2) 2

(c) E = V/d; so d = V/E = 600/4 × 104 = 0.015 m (2) 2

Y12 assessment mark scheme

Note Some symbols have not been copied across accurately

1. (a) either (If in parallel) when one bulb fails, other bulbs stay on


or (If in parallel) can identify which bulb has failed; (1) 1

(b) (i) P = VI (1)

0.5 = 240 I

I = 2.1 × 10–3 A 1 s.f. in answer (–1) once only (1) 2

(ii) R = V/I (1)

= 240/(2.1 × 10–3)

= 1.14 × 105 Ω or 1.15 × 105 Ω ans

accept (1.1 to 1.2) × 105 Ω. (1) 2

(iii) A = ρ l / R (1)

= 1.1 × 10–6 × 6.0 × 10–3 / (1.14 × 105) (= 5.79 × 10–14 m2)

A = πr2 (1)

5.79 × 10–14 = πr2 so r = 1.4 × 10–7 m (1) 3

(iv) filament too thin / fragile to be manufactured / used without damage;

allow ecf from (iii). (1) 1



4

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2. (a) current  p.d / voltage (for a metallic conductor) M1

as long as temperature is constant / physical conditions remain constant A1

(b) (i) (R =) (= 0.0349) B1

(ii) R = (Allow any subject) C1

ρ = = C1

resistivity = 5.6  10–5 A1

unit: ohm metre /  m (Allow V m A–1) A1

(5.6  10–n without unit or incorrect unit and n  5 or 3 – can score 2/4)

(5.6  10–3  m – can score 3/4)

(5.6  10–3  cm – can score 4/4)



3. (a) Line crosses ‘y-axis’ at 1.4 (V) / V = E or 1.4(V) when I = 0

V = E – Ir; since I = 0 (Hence V = E or 1.4(V)) B1

(b) (i) (Graph extrapolated to give) current = 2.0 (A)

(Allow tolerance  0.1A) B1

(ii) E = I(max) r gradient = r (Ignore sign) C1

(r = ) (Attempt made to find gradient)

r = 0.7(0) () r = 0.7(0) () (Possible ecf) A1

(iii) (excessive) heating of cell / energy wasted internally /

cell might ‘explode’ / cell goes ‘flat’ (quickly) B1

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4. (a) Energy (transformed by a device working) at 1 kW for 1 hour B1

(b) E = Pt / 5.8 = 0.12  time / (time =) 48.3 (hr) C1

(time =) 1.74  105  1.7  105 (s) A1

[3]



5. (i) v = I / nAe = 0.0025 / (8.5 × 1028 × 1.1 × 10–7 × 1.6 × 10–19) (1)

= 1.67 × 10–6 m s–1 (1) 2

(ii) Free electron concentration (or wtte) is much smaller in the thermistor than

in the wire. 1

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6. (a) R = R1 + R2 / R = 200 + 120 / R = 320 C1

current = C1

current = 2.5  10–2 (A) A0

(b) V = 25  10–3  120 / V =

V = 3.0 (V) (Possible ecf) B1

(c) p.d. across the 360 () resistor = p.d. across the 120 () resistor /

There is no current between A and B / in the voltmeter B1

(Allow ‘A & B have same voltage’ - BOD)

The p.d. calculated across 360  resistor is shown to be 3.0 V /

The ratio of the resistances of the resistors is shown to be the same. B1

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7. (i) evidence of knowledge of:

full/max transmission when the (transmission axis of) polarising sheet is

parallel to the light’s plane of polarisation/vibrations B1

no transmission when the (transmission axis of) polarising sheet is at right

angles to light’s plane of polarisation/vibrations B1

(ii) reflected light from surface is partially plane polarised B1

polarising sheet is placed at right angles to reflected light’s polarisation B1

plane/AW

[4]



8. (a) (i) amplitude = 1.2 (mm) B1

(ii) period = 2.4 (ms) B1

{allow 2.4 × 10–3 ms if 2.4 × 10–3 is correctly used in

substitution in (b)(i)}

(b) (i) frequency = 1/period C1

1/0.0024 = 417Hz (OR 420) A1

{1/2.4 = 0.417 OR 0.42 OR 0.4 scores 1 mark}

{allow ecf from cand’s period value}

(ii) recall of v = f  OR c = f  OR  = vT OR 1500 = 417  C1

 = 3.6 m A1

{ecf for cand’s f: e.g. λ= 1500/0.417 = 3600 m scores 2 marks

OR λ = 1500/0.4 = 3750 m scores 2 marks}

λ = 1500/0.42 = 3571 m scores 2 marks

(iii) valid scale for cand’s λ shown on position axis AND at least two full

‘sine’ waves drawn (waves can be very rough but not square waves } B1

amp. shown as 1.2 mm +/– ½ sq.: check first peak + trough only B1

first wavelength correct as 3.6 m +/– 1 sq.{allow ecf from (b)(ii)} B1

{NB If there is no scale on the position axis the 1st and 3rd marks cannot be scored}

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