2025 OCR A-Level Physics B (Advancing Physics) - H557 模擬試題連答案詳解

For a uniform field \(E = V/d = 2400 / (8.0 \times 10^{-3}) = 3.0 \times 10^{5}\ \text{V m}^{-1}\). B results from forgetting to convert mm to m; C inverts the ratio; D is not a sensible combination of the data.

A (1 mark). Award for correct use of \(E = V/d\) with d converted to metres (POT). Distractors: B = POT slip, C = inverted ratio, D = incorrect.

For an ideal transformer \(N_s/N_p = V_s/V_p\), so \(N_s = 2300 \times 12/230 = 120\). A comes from inverting the ratio.

B (1 mark). Correct use of the ideal transformer turns ratio. A = inverted ratio; C and D = arbitrary slips of powers of ten.

\(V = -GM/r\), so at a height \(R\) the distance is \(2R\) and \(V = -3.2 \times 10^{7}\ \text{J kg}^{-1}\). Change in potential = \(-3.2\times10^{7} - (-6.4\times10^{7}) = 3.2 \times 10^{7}\ \text{J kg}^{-1}\).

C (1 mark). Recognises r = 2R, potential halves, then takes the difference. A = surface value only; B = doubled; D = assumes inverse-square (r^2) dependence of potential, giving 3/4 of the surface value.

Percentage uncertainty in I = 2.0%, so in \(I^2\) = 4.0%. Percentage uncertainty in R = 0.4/12.0 = 3.3%. Total = 4.0 + 3.3 = 7.3%, about 7%. A forgets to double the current uncertainty.

B (1 mark). Doubling the percentage uncertainty for the squared quantity and adding to that of R. A = power ignored (2.0 + 3.3), C = doubled the total, D = current only.

\(d = 1/(600\times10^{3}) = 1.67\times10^{-6}\ \text{m}\). Maximum order: \(n = d\sin90^\circ/\lambda = 3.3\), so n = 3. Maxima exist for \(n = 0, \pm1, \pm2, \pm3\), giving 7.

C (1 mark). Uses \(n\lambda = d\sin\theta\) with \(\sin\theta \le 1\), truncates n to 3 and counts both sides plus centre. A = one side only, B = excludes the central maximum, D = rounds n up to 4.

18 hours is 3 half-lives, so the activity is \(800 \times (1/2)^3 = 100\ \text{Bq}\).

C (1 mark). A = divides by 3, B = divides by 6, D = four half-lives.

\(V_{out} = 9.0 \times 3.0/(6.0+3.0) = 3.0\ \text{V}\). B is the p.d. across the 6.0 k\(\Omega\) resistor (inverted ratio).

A (1 mark). Correct use of the potential divider equation. B = inverted ratio, C = assumes equal sharing, D = incorrect.

The reaction-time uncertainty is roughly fixed in absolute terms, so timing many oscillations makes it a much smaller fraction of the total time, and dividing by 20 gives the period with a smaller percentage uncertainty. Zero-error checking addresses a systematic error; amplitude and mass do not reduce the timing uncertainty.

D (1 mark). A = systematic error, not reaction time; B and C = do not affect the timing uncertainty (and a larger amplitude may break the small-angle assumption).

\(E = V/d = 600 / (4.0\times10^{-3}) = 1.5\times10^{5}\ \text{V m}^{-1}\). Option D multiplies instead of dividing; A and B contain powers-of-ten slips.

C (1 mark). Bald correct answer gains the mark. Distractors: D from \(V\times d\) with d in mm; B from using d in cm; A from \(V\times d\) with d in m... (power-of-ten slip).

\(V_s/V_p = N_s/N_p\), so \(V_s = 230\times 50/1150 = 10\ \text{V}\).

B (1 mark). Distractor D uses the inverted turns ratio; C is \(230/5\); A is a power-of-ten slip.

\(T = 100\times60 = 6000\ \text{s}\). \(\omega = 2\pi/T = 2\pi/6000 = 1.05\times10^{-3}\ \text{rad s}^{-1}\).

B (1 mark). A omits the factor \(2\pi\); C forgets to convert minutes to seconds; D multiplies by T.

\(d = 1/500000 = 2.0\times10^{-6}\ \text{m}\). \(n\lambda = d\sin\theta\) gives \(\sin\theta = 2\times600\times10^{-9}/2.0\times10^{-6} = 0.60\), so \(\theta = 37^{\circ}\).

B (1 mark). A is first order; C is \(\cos^{-1}0.6\); D doubles the second-order angle.

\(R = 6.0/0.50 = 12\ \Omega\). Percentage uncertainties: V 5%, I 2%; for a quotient they add: 7%. \(0.07\times12 = 0.84\approx0.8\ \Omega\).

B (1 mark). C uses the uncertainty in V only; D doubles the combined percentage; A adds the absolute uncertainties 0.3 and 0.01 and ignores the quotient.

\(I = E/(R+r) = 12/6.0 = 2.0\ \text{A}\). \(V = E - Ir = 12 - 2.0\times0.50 = 11\ \text{V}\).

B (1 mark). A ignores internal resistance; D is the lost volts \(Ir\) (1.0 V); C subtracts 2 V.

18 h is three half-lives, so the activity is \(800\times(1/2)^3 = 100\ \text{Bq}\).

D (1 mark). B is \(800/3\); C is \(800/6\); A is one half-life only.

\(pV = NkT\), so \(N = pV/kT = (1.0\times10^{5}\times2.0\times10^{-3})/(1.38\times10^{-23}\times300) = 200/4.14\times10^{-21} = 4.8\times10^{22}\).

B (1 mark). A and C are powers-of-ten slips (e.g. volume in litres); D is the inverted ratio.

\(E = V/d = 600 / 0.040 = 1.5 \times 10^{4}\ \text{V m}^{-1}\). Option A forgets to convert mm to m; D inverts the ratio; C is a slip in the arithmetic.

B (1 mark). Correct conversion of 40 mm to 0.040 m and use of E = V/d.

\(R = 6.0/0.50 = 12\ \Omega\). Percentage uncertainties: V is 1.7%, I is 4.0%. For a quotient they add: 5.7%. \(0.057 \times 12 = 0.68 \approx 0.7\ \Omega\).

C (1 mark). Method: add percentage uncertainties (1.7% + 4.0%) and convert to an absolute value. B uses only the current's uncertainty; D doubles the combined value.

Gravity provides the centripetal force: \(GMm/r^2 = mv^2/r\), so \(v^2 = GM/r\) and \(v \propto r^{-1/2}\). With \(r\) multiplied by 4, \(v\) is multiplied by \(1/2\).

D (1 mark). Method: equate gravitational force to \(mv^2/r\) to find \(v \propto 1/\sqrt{r}\). C forgets the square root; B and A invert the relationship.

At constant volume \(p \propto T\). Convert to kelvin: 300 K to 327 K. \(327/300 = 1.09 \approx 1.1\).

B (1 mark). Must use absolute temperatures. A results from doubling the Celsius value; C inverts it.

\(V_{out} = 12 \times 2.0/(4.0+2.0) = 4.0\) V.

C (1 mark). Potential divider equation with the correct resistor in the numerator. A uses the 4.0 kΩ resistor in the numerator.

\(V_s/V_p = N_s/N_p\), so \(V_s = 230 \times 60/1150 = 12\) V.

A (1 mark). D inverts the turns ratio.

18 hours is 3 half-lives. Activity = \(8000 \times (1/2)^3 = 1000\) Bq.

D (1 mark). A divides by 3 and B divides by 6 (linear thinking); C uses four half-lives.

Field strength \(E = V/d = 400 / 0.050 = 8.0 \times 10^{3}\) V m\(^{-1}\). Charge \(q = 2e = 3.2 \times 10^{-19}\) C. Force \(F = qE = 3.2 \times 10^{-19} \times 8.0 \times 10^{3} = 2.6 \times 10^{-15}\) N. Answer B. A omits the factor of 2 in the charge; C doubles the force wrongly; D multiplies by d instead of dividing (\(qVd\)).

B (1 mark). Correct method: \(E = V/d\) with d converted to metres (0.050 m), then \(F = qE\) with \(q = 2e\). Distractors: A uses q = e; C uses q = 4e; D uses \(qVd\) (not dividing by d).

\( E = V/d = 600 / 0.020 = 3.0 \times 10^{4}\ \text{V m}^{-1} \). \( F = eE = 1.60 \times 10^{-19} \times 3.0 \times 10^{4} = 4.8 \times 10^{-15}\ \text{N} \). B results from using d = 2.0 m; C is the field strength, not the force; D results from a power-of-ten slip converting cm to m the wrong way.

A (1 mark). Method: convert 2.0 cm to 0.020 m, use E = V/d, then F = eE. Reject B, C, D.

For an ideal transformer \( I_p N_p = I_s N_s \), so \( I_p = 2.0 \times 50 / 1150 = 0.087\ \text{A} \). B uses the inverted turns ratio; C ignores the ratio; D uses 230 in place of 1150 in the ratio.

A (1 mark). Method: input power = output power gives \( I_p/I_s = N_s/N_p \). Reject B, C, D.

Gravitational force provides the centripetal force: \( GMm/r^2 = m r \omega^2 \), so \( \omega^2 = GM/r^3 \) and \( T^2 \propto r^3 \). Doubling r multiplies T by \( 2^{3/2} = 2.8 \). A is \( 2^{1/2} \), B assumes direct proportionality, D assumes \( T \propto r^2 \).

C (1 mark). Method: equate gravitational force to \( m r \omega^2 \), deduce \( T^2 \propto r^3 \), then evaluate \( 2^{1.5} \). Reject A, B, D.

Percentage uncertainties: V is \( 0.1/6.0 = 1.7\% \); I is \( 0.02/0.50 = 4.0\% \). For a quotient these add: 5.7%. \( 5.7\% \) of \( 12\ \Omega = 0.68 \approx 0.7\ \Omega \). A adds absolute uncertainties; C uses only the larger percentage; D doubles the correct value.

B (1 mark). Method: add percentage uncertainties for division, then convert to absolute. Reject A, C, D.

Slit spacing \( d = 1/(300 \times 10^{3}) = 3.33 \times 10^{-6}\ \text{m} \). From \( n\lambda = d\sin\theta \) with \( \sin\theta \le 1 \): \( n \le d/\lambda = 5.56 \), so the highest order is 5. C rounds up; D counts all maxima on both sides including the central one; A is a slip in d.

B (1 mark). Method: find d, set \( \sin\theta = 1 \), round down to an integer. Reject A, C, D.

\( I = E/(R + r) = 12/6.0 = 2.0\ \text{A} \). Terminal pd \( = IR = 2.0 \times 5.0 = 10\ \text{V} \) (equivalently \( E - Ir = 10\ \text{V} \)). B ignores internal resistance; C is the 'lost volts'; D subtracts 1.0 V incorrectly.

A (1 mark). Method: find current from total resistance, then pd across the load. Reject B, C, D.

(a) \(E = V/d = 900/0.012 = 7.5\times10^4\) V m\(^{-1}\).
(b)(i) Balance: \(qE = mg\), so \(q = mg/E = (3.7\times10^{-15}\times9.81)/(7.5\times10^4) = 4.8\times10^{-19}\) C.
(ii) \(4.84\times10^{-19}/1.60\times10^{-19} = 3.0\), so three electrons.
(c) New \(E = 1050/0.012 = 8.75\times10^4\) V m\(^{-1}\). Electric force \(= 4.8\times10^{-19}\times8.75\times10^4 = 4.2\times10^{-14}\) N. Weight \(= 3.63\times10^{-14}\) N. Net force \(\approx 5.7\times10^{-15}\) N upwards. \(a = F/m = 5.7\times10^{-15}/3.7\times10^{-15} \approx 1.5\) m s\(^{-2}\) upwards.

(a) \(E=V/d\) ✓; \(7.5\times10^4\) V m^-1 ✓ (allow N C^-1; 12 not converted to m gains 1 mark max, POT).
(b)(i) \(qE=mg\) AW ✓; \(4.8\times10^{-19}\) C ✓ (accept 4.8-4.9).
(ii) ratio to \(1.60\times10^{-19}\) gives 3.0 ✓ (ECF).
(c) new field or force \(qE\) with new p.d. ✓; net force = electric force minus weight ✓; \(a\approx1.5\)-\(1.6\) m s^-2 upwards ✓ (direction required for the mark; ECF from (b)). Bald correct answer gains all marks.

(a) \(V_s = V_p N_s/N_p = 230\times46/1150 = 9.2\) V.
(b) Output power \(= 9.2\times2.5 = 23\) W. Ideal: input power = 23 W so \(I_p = 23/230 = 0.10\) A.
(c) Input power \(= 23/0.92 = 25\) W; \(I_p = 25/230 = 0.109\) A \(\approx 0.11\) A.
(d) The changing flux induces e.m.f.s (and so eddy currents) in the core. Laminations with insulation confine the current to thin narrow paths, increasing the resistance of the path and reducing the current, so less \(I^2R\) heating.

(a) \(V_s/V_p=N_s/N_p\) ✓; 9.2 V ✓.
(b) \(P=IV\) used for secondary (23 W) or \(I_pV_p=I_sV_s\) ✓; 0.10 A ✓ (ECF from (a)).
(c) input power = output / 0.92 (25 W) ✓; 0.11 A ✓ (accept 0.109; ECF). Not 0.10 A.
(d) core is conducting so changing flux induces eddy currents (AW) ✓; laminations/insulation make current paths narrower or longer, greater resistance so smaller current / less heating ✓. Ignore references to the core being a better conductor of flux.

(a) Gravity provides the centripetal force: \(GMm/r^2 = mv^2/r\), so \(v = \sqrt{GM/r} = \sqrt{(6.67\times10^{-11}\times5.97\times10^{24})/7.20\times10^{6}} = \sqrt{5.53\times10^{7}} = 7.44\times10^{3}\) m s\(^{-1}\).
(b) \(T = 2\pi r/v = 2\pi\times7.20\times10^{6}/7.44\times10^{3} = 6.08\times10^{3}\) s.
(c) \(\Delta E = GMm(1/r_1 - 1/r_2) = 3.98\times10^{14}\times850\times(1.389\times10^{-7} - 2.370\times10^{-8}) = 3.9\times10^{10}\) J.

(a) \(GMm/r^2=mv^2/r\) ✓; rearranged to \(v=\sqrt{GM/r}\) ✓; substitution and 7.44 x 10^3 (at least 3 SF shown) ✓.
(b) \(T=2\pi r/v\) ✓; 6.1 x 10^3 s ✓ (ECF; accept 6.08 x 10^3).
(c) potential energy \(-GMm/r\) used at both radii ✓; difference of the two values (not sum) ✓; 3.9 x 10^10 J ✓ (accept 3.8-4.0 x 10^10). Allow use of \(g\) = 9.81 only if justified; otherwise 0 for that mark.

Grating spacing \(d = 1/(600\times10^{3}) = 1.67\times10^{-6}\) m.
(a) \(n\lambda = d\sin\theta\): \(\sin\theta = 2\times633\times10^{-9}/1.67\times10^{-6} = 0.760\), so \(\theta = 49.4^\circ\).
(b) Maximum order \(n < d/\lambda = 2.63\), so \(n = 0, \pm1, \pm2\): 5 maxima.
(c) First order: \(\sin\theta = 0.380\), \(\theta = 22.3^\circ\). \(x = 1.50\tan22.3^\circ = 0.62\) m.

(a) \(d=1/600000\) m ✓ (POT); \(\sin\theta=n\lambda/d=0.760\) ✓; 49.4° ✓ (accept 49-50).
(b) \(\sin\theta\le1\) so \(n\le2.6\) so maximum order 2 ✓; 5 maxima in total (both sides plus central) ✓. Bald 5 gains both marks; 2 or 3 only gains 1 mark.
(c) \(\theta_1=22.3^\circ\) ✓; \(x=L\tan\theta\) ✓; 0.62 m ✓ (accept 0.61-0.62). Not \(L\sin\theta\) (0.57 m) for the final mark.

(a) \(A = \pi d^2/4 = \pi(0.32\times10^{-3})^2/4 = 8.04\times10^{-8}\) m\(^2\). \(\rho = RA/L = 4.6\times8.04\times10^{-8}/1.200 = 3.1\times10^{-7}\) \(\Omega\) m.
(b) \(\rho = R\pi d^2/(4L)\). Percentage uncertainties: \(d\): 3.1% (doubled: 6.3%); \(R\): 2.2%; \(L\): 0.2%. Total \(= 8.6\%\). Absolute uncertainty \(= 0.086\times3.08\times10^{-7} = 0.3\times10^{-7}\) \(\Omega\) m. So \(\rho = (3.1\pm0.3)\times10^{-7}\) \(\Omega\) m.
(c) The diameter, because its percentage uncertainty is squared and largest. Reduce by measuring with a micrometer at several places along the wire and in perpendicular directions, then averaging.

(a) \(A=\pi d^2/4\) with d in metres ✓; \(\rho=RA/L\) ✓; \(3.1\times10^{-7}\) \(\Omega\) m ✓ (accept 3.08).
(b) % uncertainties of d, R and L found ✓; d's % uncertainty doubled (power) and all added ✓; absolute uncertainty \(\approx0.3\times10^{-7}\) \(\Omega\) m ✓ (accept 0.26-0.27 x 10^-7; ECF). If d's uncertainty is not doubled, max 2 marks.
(c) diameter identified (ECF from (b)) ✓; valid method: repeat readings at different positions/orientations and average, or use a more precise instrument (micrometer with 0.001 mm resolution) ✓.

(a)(i) \(I = V/R = 5.2/8.0 = 0.65\) A.
(ii) Lost volts \(= 6.0-5.2 = 0.8\) V; \(r = 0.8/0.65 = 1.2\) \(\Omega\).
(b) \(V_{out} = 6.0\times4.7/(4.7+R_T)\). At 20 \(^\circ\)C: \(6.0\times4.7/14.7 = 1.9\) V. At 60 \(^\circ\)C: \(6.0\times4.7/5.9 = 4.8\) V.
(c) Raising the temperature gives more thermal energy, freeing more charge carriers (electrons) from the lattice atoms. The number density of mobile carriers rises, so the conductance increases and the resistance falls.

(a)(i) 0.65 A ✓.
(ii) lost volts \(=0.8\) V (or \(E=I(R+r)\)) ✓; \(r\approx1.2\) \(\Omega\) ✓ (ECF; accept 1.23).
(b) potential divider equation with correct resistor in numerator ✓; 1.9 V ✓; 4.8 V ✓ (accept 4.78). Use of thermistor in numerator (4.1 V and 1.2 V) gains 1 mark only.
(c) more energy/thermal excitation releases more charge carriers ✓; number density of carriers increases (and this outweighs any increase in lattice scattering) so resistance falls ✓.

(a) \(V = V_0 e^{-t/RC}\). Taking natural logs: \(\ln V = \ln V_0 - t/RC\), which is of the form \(y = c + mx\) with gradient \(-1/RC\).
(b) Gradient \(= (0.470-2.079)/(80-0) = -0.0201\) s\(^{-1}\). \(RC = 1/0.0201 = 49.7\) s. \(C = 49.7/(1.00\times10^{5}) = 4.97\times10^{-4}\) F \(\approx 500\ \mu\)F.
(c) The voltmeter is in parallel with the resistor-capacitor circuit path, so the effective resistance discharging the capacitor is about \(100\text{ k}\Omega \parallel 10\text{ M}\Omega \approx 99\) k\(\Omega\), less than 100 k\(\Omega\). The measured time constant is therefore slightly smaller, so \(C = \tau/100\) k\(\Omega\) is underestimated by about 1%.

(a) exponential decay \(V=V_0e^{-t/RC}\) stated or used ✓; natural log taken giving \(\ln V=\ln V_0-t/RC\), compared with \(y=mx+c\) ✓.
(b) gradient calculated using two widely separated points ✓; magnitude \(0.020\) s^-1 (accept 0.0195-0.0205) with correct sign (negative) ✓; \(RC=1/\)gradient ✓; \(C\approx5.0\times10^{-4}\) F ✓ (accept 4.9-5.1 x 10^-4 F; ECF).
(c) extra parallel path lowers the total resistance through which the capacitor discharges / time constant is smaller ✓; so calculated C is slightly too small (about 1%) ✓.

(a) \(pV = nRT\): \(p = 0.080\times8.31\times300/2.0\times10^{-3} = 9.97\times10^{4}\) Pa \(\approx1.0\times10^{5}\) Pa.
(b) \(E = \tfrac32kT = 1.5\times1.38\times10^{-23}\times300 = 6.2\times10^{-21}\) J.
(c) \(\tfrac12m\langle c^2\rangle = 6.21\times10^{-21}\), so \(\langle c^2\rangle = 1.87\times10^{6}\) m\(^2\) s\(^{-2}\) and \(c_{rms} = 1.37\times10^{3}\) m s\(^{-1}\).
(d) Volume constant, so \(p\propto T\): \(p = 9.97\times10^{4}\times450/300 = 1.5\times10^{5}\) Pa. Higher temperature means greater average kinetic energy and speed; atoms hit the walls more often and with greater momentum change per collision, so the force per unit area is greater.

(a) \(pV=nRT\) ✓; \(1.0\times10^{5}\) Pa ✓ (accept 9.97 x 10^4).
(b) \(\tfrac32kT\) with T in kelvin ✓; \(6.2\times10^{-21}\) J ✓.
(c) equating \(\tfrac12m\langle c^2\rangle\) to the energy from (b) ✓; \(1.4\times10^{3}\) m s^-1 ✓ (accept 1.37 x 10^3; ECF).
(d) \(p\propto T\) used, \(1.5\times10^{5}\) Pa ✓ (ECF); atoms have higher average speed/kinetic energy so more frequent and harder collisions with walls (greater rate of change of momentum) ✓. Not 'atoms collide more with each other'.

(a) \(A = \lambda N = 2.0\times10^{-3}\times5.0\times10^{10} = 1.0\times10^{8}\) Bq.
(b) \(T_{1/2} = \ln2/\lambda = 0.693/2.0\times10^{-3} = 347\) s \(\approx3.5\times10^{2}\) s.
(c) \(A = A_0e^{-\lambda t} = 1.0\times10^{8}\times e^{-1.8} = 1.0\times10^{8}\times0.165 = 1.7\times10^{7}\) Bq.
(d) The student is wrong. Each nucleus has a fixed probability per unit time of decaying, determined by the nucleus itself, and this is not affected by temperature or other external conditions, so the activity (which depends only on \(\lambda N\)) does not change.

(a) \(A=\lambda N\) ✓; \(1.0\times10^{8}\) Bq ✓.
(b) \(T_{1/2}=\ln2/\lambda\) ✓; 347 s (accept 3.5 x 10^2 s) ✓.
(c) \(A=A_0e^{-\lambda t}\) with \(\lambda t=1.8\) ✓; \(1.7\times10^{7}\) Bq ✓ (accept 1.65 x 10^7; ECF from (a)). Allow use of 2.6 half-lives.
(d) decay is random / has a fixed probability per nucleus ✓; this is not affected by temperature (or other physical conditions) so activity is unchanged ✓.

Lower orbit radius \(r_1 = 6.37\times10^6 + 0.80\times10^6 = 7.17\times10^6\) m.

Gravity provides the centripetal force: \(\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}\), so \(v = \sqrt{GM/r}\). \(GM = 3.98\times10^{14}\) m\(^3\) s\(^{-2}\).

\(v_1 = \sqrt{3.98\times10^{14}/7.17\times10^6} = 7.45\times10^3\) m s\(^{-1}\).
\(T_1 = 2\pi r_1/v_1 = 6.0\times10^3\) s.

Higher orbit \(r_2 = 4.22\times10^7\) m: \(v_2 = \sqrt{3.98\times10^{14}/4.22\times10^7} = 3.07\times10^3\) m s\(^{-1}\); \(T_2 = 2\pi r_2/v_2 = 8.6\times10^4\) s (about 24 hours).

Total energy in a circular orbit \(E = \tfrac12 mv^2 - GMm/r = -GMm/(2r)\).
\(E_1 = -1.25\times10^{10}\) J; \(E_2 = -2.12\times10^{9}\) J.
Minimum energy supplied \(= E_2 - E_1 = 1.04\times10^{10}\) J \(\approx 1.0\times10^{10}\) J.

Kinetic energy falls from \(\tfrac12(450)(7.45\times10^3)^2 = 1.25\times10^{10}\) J to \(\tfrac12(450)(3.07\times10^3)^2 = 2.1\times10^{9}\) J, a decrease of about \(1.04\times10^{10}\) J. The gravitational potential energy rises (becomes less negative) by \(2\times 1.04\times10^{10} = 2.07\times10^{10}\) J. The energy supplied is the net gain in total energy, which equals the gain in potential energy minus the loss in kinetic energy.

Levels of response (6 marks). Bald correct values in the indicative content are credited as the relevant scientific points.

Level 3 (5-6 marks): Calculations of speed, period and energy change are all attempted and are largely correct (allow one slip or an ECF). Explanation of the kinetic energy decrease is complete and correct, linking the rise in gravitational potential energy to the net energy supplied. Reasoning is clear, logical and coherent.

Level 2 (3-4 marks): Two of the three strands (speed/period comparison, energy calculation, explanation of the kinetic energy change) are covered with some correct calculation, or all three are attempted with significant errors or omissions. Some reasoning is shown but may lack coherence.

Level 1 (1-2 marks): One strand is attempted, or isolated correct statements/formulae are given with little connection. Reasoning is limited.

Level 0 (0 marks): No creditable response.

Indicative scientific points:
- Orbit radius \(r_1 = 7.17\times10^6\) m (must add Earth radius to the height) (M1)
- Equating \(GMm/r^2 = mv^2/r\) to get \(v = \sqrt{GM/r}\) (M1)
- \(v_1 \approx 7.45\times10^3\) m s\(^{-1}\) and \(v_2 \approx 3.07\times10^3\) m s\(^{-1}\) (A1 each; allow 3.1 and 7.5)
- \(T = 2\pi r/v\) giving \(T_1 \approx 6.0\times10^3\) s and \(T_2 \approx 8.6\times10^4\) s (about 24 h); the higher orbit has lower speed and longer period
- Total energy \(E = -GMm/2r\) (or GPE + KE calculated separately)
- \(E_1 = -1.25\times10^{10}\) J, \(E_2 = -2.1\times10^{9}\) J; energy needed \(\approx 1.0\times10^{10}\) J (allow 1.04 x 10^10; allow 1.0-1.1)
- Kinetic energy decreases because speed is lower in the higher orbit
- Gravitational potential energy increases by more than the KE lost (by twice the KE decrease, about \(2.1\times10^{10}\) J); energy supplied = increase in total energy = gain in GPE - loss in KE

Guidance: Using height 800 km as the orbit radius loses the first mark only and ECF applies once. Not: energy supplied equals the change in GPE alone (this ignores the fall in KE) unless the KE change is then included. Ignore rounding to 2 s.f. or more. Significant-figure penalty applies at most once.

Option 1: \(I = P/V = 1.5\times10^6/5.0\times10^3 = 300\) A. Power lost \(= I^2R = 300^2\times0.80 = 7.2\times10^4\) W, which is \(7.2\times10^4/1.5\times10^6 = 4.8\%\) of the generated power.

Option 2: \(I = 1.5\times10^6/5.0\times10^4 = 30\) A. Power lost \(= 30^2\times0.80 = 7.2\times10^2\) W, which is \(0.048\%\). Raising the p.d. by a factor 10 reduces the current by 10 and the loss by 100.

Turns: for an ideal transformer \(N_s/N_p = V_s/V_p\), so \(N_s = 150\times50/5.0 = 1500\).

Operation: the alternating current in the primary coil produces an alternating magnetic flux in the iron core, which links the secondary coil. The changing flux linkage in the secondary induces an e.m.f. \(= -\mathrm{d}(N\Phi)/\mathrm{d}t\), proportional to the number of turns.

Core: a changing flux induces e.m.f.s and hence circulating (eddy) currents in the core itself, which heat it and waste energy. Thin insulated laminations break up the conducting paths so the eddy currents are much smaller, while the iron still guides the flux.

Conclusion: Option 2 wastes about 100 times less power, so is far better.

Levels of response (6 marks).

Level 3 (5-6 marks): Both power losses are calculated correctly (allow one slip/ECF) and the secondary turns found. Explanation of transformer action AND the purpose of lamination is clear and correct, with a reasoned conclusion comparing the options. Reasoning is coherent and logical.

Level 2 (3-4 marks): Calculations for both options attempted with at least one correct, or a correct calculation plus a partial explanation of transformer action or eddy currents. Some reasoning but incomplete or not fully linked.

Level 1 (1-2 marks): Isolated correct statements or a single correct step (e.g. \(I = P/V\) or turns ratio) with little explanation.

Level 0 (0 marks): No creditable response.

Indicative scientific points:
- Option 1: \(I = P/V = 300\) A (C1)
- Power loss \(= I^2R = 7.2\times10^4\) W (about 72 kW), 4.8% of output (A1)
- Option 2: \(I = 30\) A and loss \(= 7.2\times10^2\) W, 0.048% (A1)
- Current falls by factor 10 so loss falls by factor 100 (as \(P \propto I^2\)); Option 2 preferred (conclusion)
- \(N_s = N_pV_s/V_p = 1500\) turns (A1)
- Alternating current in primary produces changing flux in the core; flux links the secondary; induced e.m.f. is the rate of change of flux linkage
- Changing flux induces eddy currents in a solid core; these cause heating/energy loss
- Insulated thin laminations increase the resistance of eddy-current paths / reduce eddy currents; core still guides flux

Guidance: Allow loss calculated as \(V^2/R\) only if the p.d. across the cable is used correctly; Not: using 5.0 kV or 50 kV as the p.d. across the cable (gains no credit for that mark; ECF for later conclusion). Allow AW for 'flux linkage' wording. Ignore references to the transformer being not 100% efficient. Significant-figure penalty applies at most once.

(a) \(E = V/d = 1500 / 0.012 = 1.25 \times 10^{5}\) V m\(^{-1}\), which rounds to \(1.3 \times 10^{5}\).

(b) The droplet is in equilibrium so \(QE = mg\). \(Q = mg/E = (4.0 \times 10^{-15} \times 9.81)/(1.25 \times 10^{5}) = 3.1 \times 10^{-19}\) C.

(c) \(n = 3.14 \times 10^{-19} / 1.60 \times 10^{-19} = 1.96\), so 2 electrons.

(d) Electric force doubles to \(2mg\) upwards. Resultant = \(2mg - mg = mg\) upwards, so \(a = g = 9.81\) m s\(^{-2}\) upwards (about 9.8).

(e) The charges are multiples of a smallest value (about \(1.6 \times 10^{-19}\) C), so charge comes in discrete packets, the charge of one electron, and cannot take any continuous value.

(a) \(E = V/d\) with correct substitution giving 1.25 x 10^5 (must show unrounded value or 1.25) [1]

(b) Electric force equals weight / \(QE = mg\) [1]; substitution \(Q = mg/E\) [1]; \(3.1 \times 10^{-19}\) C (allow 3.14, 3.2 x 10^-19 if 1.3 x 10^5 used) [1]. Bald correct answer gains all marks. Allow ECF from (a).

(c) Divide by \(1.60 \times 10^{-19}\) C [1]; 2 electrons (must be a whole number; ECF from (b)) [1]

(d) New electric force is \(2mg\) so resultant is \(mg\) / \(a = g\) [1]; 9.8 m s^-2 (allow 9.81) upwards [1]. Direction needed for second mark.

(e) Charge is discrete / quantised [1]; there is a smallest unit of charge / charges are multiples of the electron charge (AW) [1]. Not: 'charge is conserved'.

(a) Gravitational force provides the centripetal force: \(GMm/r^{2} = mv^{2}/r\), so \(v = \sqrt{GM/r} = \sqrt{(6.67 \times 10^{-11} \times 5.97 \times 10^{24})/(6.80 \times 10^{6})} = 7.65 \times 10^{3}\) m s\(^{-1}\), about \(7.7 \times 10^{3}\).

(b) \(T = 2\pi r / v = 2\pi \times 6.80 \times 10^{6} / 7.65 \times 10^{3} = 5.6 \times 10^{3}\) s (about 93 minutes).

(c) Kinetic energy \(= \tfrac{1}{2}mv^{2} = 0.5 \times 450 \times 5.86 \times 10^{7} = 1.32 \times 10^{10}\) J. Gravitational potential energy \(= -GMm/r = -450 \times 5.86 \times 10^{7} = -2.64 \times 10^{10}\) J. Total energy \(= -1.32 \times 10^{10}\) J. Energy to reach infinity with zero kinetic energy \(= 1.3 \times 10^{10}\) J.

(d) As the satellite moves to a smaller radius the gravitational potential energy decreases (more negative) by twice as much as the gain in kinetic energy, so kinetic energy, and hence speed, increases; the rest is dissipated as thermal energy by drag. Equivalently, \(v = \sqrt{GM/r}\), so smaller \(r\) gives larger \(v\).

(a) Gravitational force = centripetal force / \(GMm/r^2 = mv^2/r\) [1]; rearranged to \(v = \sqrt{GM/r}\) [1]; correct substitution giving 7.65 x 10^3 (must show more than 2 s.f.) [1]

(b) \(T = 2\pi r/v\) (or \(2\pi/\omega\) with \(\omega = v/r\)) [1]; \(5.6 \times 10^{3}\) s (allow 5.58 x 10^3; ECF from (a)) [1]

(c) Total energy = KE + GPE with either term correct [1]; total = \(-GMm/2r = -1.3 \times 10^{10}\) J [1]; energy needed = \(1.3 \times 10^{10}\) J (allow 1.32 x 10^10) positive [1]. Bald correct answer gains all marks. Allow \(\tfrac{1}{2}mv^2\) with v = 7.7 x 10^3 giving 1.33 x 10^10.

(d) Gravitational potential energy decreases / work done by gravity as it falls [1]; part of this becomes kinetic energy (rest is dissipated by drag) / \(v \propto 1/\sqrt{r}\) (AW) [1]. Not: 'drag speeds it up'.

(a) \(0.01/0.28 \times 100 = 3.6\%\).

(b) \(\rho = \pi (0.28 \times 10^{-3})^{2} \times 2.4 / (4 \times 0.800) = 1.8 \times 10^{-7}\) \(\Omega\) m.

(c) Percentage uncertainties: \(d\): 3.6%, so \(d^{2}\): 7.1%; \(R\): \(0.1/2.4 = 4.2\%\); \(L\): \(0.002/0.800 = 0.25\%\). Total \(= 7.1 + 4.2 + 0.25 = 11.6\% \approx 12\%\). Absolute uncertainty \(= 0.116 \times 1.85 \times 10^{-7} = 0.2 \times 10^{-7}\), so \(\rho = (1.8 \pm 0.2) \times 10^{-7}\) \(\Omega\) m.

(d) The diameter contributes the largest share (7.1% because it is squared). Measure the diameter at several positions along the wire and at right angles to each other, then take the mean, which reduces the random uncertainty; also check the micrometer for zero error.

(e) The range \(1.6\) to \(2.0 \times 10^{-7}\) \(\Omega\) m does not include \(4.9 \times 10^{-7}\), so the result is not consistent with the data book value. A possible cause is an uncorrected systematic error, for example a zero error in the micrometer or meter, or the contact resistance at the clips adding to R; the wire may also be a different material.

(a) 3.6% (allow 4%) [1]

(b) Correct substitution with d converted to metres (POT) [1]; \(1.8 \times 10^{-7}\) \(\Omega\) m (allow 1.85 x 10^-7) [1]. Bald correct answer gains both marks.

(c) Diameter percentage uncertainty doubled (squared quantity) [1]; percentage uncertainties of d^2, R and L added [1]; 12% (allow 11.6%) and \(\rho = (1.8 \pm 0.2) \times 10^{-7}\) \(\Omega\) m [1]. ECF from (a). Allow 1 mark for 4.2% + 3.6% + 0.25% = 8% if the square is omitted.

(d) Diameter (d) identified as the largest source [1]; repeat measurements at different positions / orientations and average (or use a micrometer zeroed before use) [1]. Not: 'repeat the experiment' alone, not 'use a better meter'.

(e) Value is not consistent / does not agree, as the data book value lies outside the range 1.6 to 2.0 x 10^-7 (ECF) [1]; valid cause such as systematic zero error, contact resistance, or a different alloy (AW) [1]. Ignore human error.

(a) \(E = V/d = 450 / 8.0\times10^{-3} = 5.6\times10^4\) V m\(^{-1}\).
(b) \(V = \tfrac{4}{3}\pi r^3 = 4.19\times10^{-18}\) m\(^3\); \(m = \rho V = 880\times4.19\times10^{-18} = 3.7\times10^{-15}\) kg.
(c) Stationary so \(qE = mg\): \(q = mg/E = (3.69\times10^{-15}\times9.81)/(5.625\times10^4) = 6.4\times10^{-19}\) C. \(n = 6.4\times10^{-19}/1.60\times10^{-19} = 4\).
(d) Electric force on droplet must be upwards to balance weight. Negative charge is forced opposite to the field, so field points downwards, so the top plate is positive.
(e) \(qE\) is proportional to V, so upward force becomes \(mg\times400/450\). Net downward force \(= mg(1-400/450) = mg/9\). \(a = g/9 = 1.09\) m s\(^{-2}\), downwards.
(f) Charges on all droplets are whole-number multiples of a smallest value of about \(1.6\times10^{-19}\) C; no values lie between; if charge were continuous, any value would be found.

(a) \(E = V/d\) with 450 and 8.0e-3 correctly used giving 5.6e4 (accept 56 250 seen) ✓ [1]
(b) Volume \(= \tfrac43\pi r^3\) = 4.2e-18 m^3 ✓; mass = 3.7e-15 kg ✓ (ECF on volume) [2]
(c) \(qE = mg\) ✓; \(q = 6.4\times10^{-19}\) C ✓ (allow 6.3-6.5e-19); \(n = 4\) ✓ (ECF; allow 'about 4'). Bald correct answer gains all marks. [3]
(d) Electric force upward / opposes weight ✓; negative charge so field downward, hence top plate positive ✓ [2]
(e) Recognises \(F\propto V\) / calculates new \(qE\) = 3.2e-14 N ✓; net force = mg − qE′ (≈4e-15 N) ✓; \(a = 1.09\) m s^-2 (allow 1.0-1.2 from rounding) downward ✓ [3]
(f) Measure charge on many droplets ✓; all values are integer multiples of a smallest value ✓; smallest value ≈1.6e-19 C (the electron charge) ✓; no intermediate values / continuous distribution would give any value ✓ [4]
Not: 'charge is quantised' without any supporting evidence for the final mark.

(a) Taking natural logs: \(\ln V = \ln V_0 - t/(RC)\). Compared with \(y = mx + c\) this is linear in t, with gradient \(-1/(RC)\) and intercept \(\ln V_0\).
(b) \(\ln 1.68 = 0.519\); \(\ln 0.72 = -0.329\).
(c) Gradient \(= (-0.329 - 1.792)/(100-0) = -0.0212\ \text{s}^{-1}\). \(RC = 1/0.0212 = 47.2\) s. \(C = 47.2/47000 = 1.0\times10^{-3}\) F.
(d) \(0.01/0.72\times100 = 1.4\%\). \(\Delta(\ln V) = \Delta V/V = 0.014\).
(e) Half range of gradients \(= 0.0008\); mean \(0.0212\); percentage uncertainty \(= 3.8\%\). Add 2% for R: 5.8%. \(C = 1.00\times10^{-3}\) F \(\pm 0.06\times10^{-3}\) F. The marked value 1000 \(\mu\)F lies in this range (and within its own 20% tolerance), so consistent.

(a) \(\ln V = \ln V_0 - t/RC\) (by taking logs) ✓; compares with \(y=mx+c\) / linear in t ✓; gradient \(=-1/RC\) ✓ [3]
(b) 0.519 ✓ and −0.329 ✓ (allow ±0.001) [2]
(c) Change in ln V = −2.121 ✓; gradient = −0.0212 (s^-1) ✓ (allow −0.0210 to −0.0214); \(\tau = 1/|\text{gradient}|\) = 47 s ✓; \(C = \tau/R = 1.0\times10^{-3}\) F ✓ (ECF) [4]
(d) \(0.01/0.72 = 1.4\%\) ✓; \(\Delta\ln V = 0.014\) ✓ (ECF) [2]
(e) % uncertainty in gradient = 3.8% (allow 3.7–3.8) ✓; add 2% for R (percentages added for a quotient) ✓; \(C = (1.00\pm0.06)\times10^{-3}\) F ✓ (ECF); consistent because 1000 µF lies within range / within ±20% tolerance ✓ [4]
Ignore sf slips if within 2-3 sf. ECF allowed once for the gradient.

(a) \(N_s = N_p V_s/V_p = 1150\times12/230 = 60\).
(b) (i) \(I_s = P/V_s = 36/12 = 3.0\) A. (ii) \(I_p = 36/230 = 0.157\) A.
(c) Input power \(= 36/0.92 = 39.1\) W; \(I_p = 39.1/230 = 0.170\) A.
(d) See mark scheme.
(e) \(V_{peak} = 12\sqrt2 = 17.0\) V; \(\omega = 2\pi\times50 = 314\) rad s\(^{-1}\). Peak e.m.f. \(= N\omega\Phi_0\), so \(\Phi_0 = 17.0/(60\times314) = 9.0\times10^{-4}\) Wb.

(a) \(N_s/N_p = V_s/V_p\) ✓; 60 ✓ [2]
(b) (i) 3.0 A ✓ (ii) 0.157 A ✓ (allow 0.16) [2]
(c) Input power 39.1 W ✓; 0.170 A ✓ (allow 0.17) [2]
(d)* Levels of response [6]
Level 3 (5–6 marks): Clear, coherent explanation covering both the operation and loss reduction, with correct physics linking flux linkage, induced e.m.f. and eddy currents.
Level 2 (3–4 marks): Explains one strand well and the other partly, or both with limited reasoning.
Level 1 (1–2 marks): Isolated correct statements with little linkage.
0 marks: no creditworthy response.
Indicative points:
• alternating current in the primary produces changing flux in the core
• iron core has high permeance and guides flux so it links the secondary
• changing flux linkage in the secondary induces an e.m.f. (induced e.m.f. = rate of change of flux linkage)
• e.m.f. alternates at same frequency as the input
• changing flux induces eddy currents in the core, which heat it
• lamination (insulated thin sheets) raises resistance to eddy current paths, reducing the losses
• thick copper wire / low-resistance windings reduce \(I^2R\) heating
• closed core / coils wound close reduces flux leakage
• soft iron reduces hysteresis losses
(e) \(V_{peak} = 12\sqrt2 = 17.0\) V and \(\omega = 314\) rad s^-1 ✓; peak e.m.f. \(= N\omega\Phi_0\) (from \(-N\,d\Phi/dt\)) ✓; \(\Phi_0 = 9.0\times10^{-4}\) Wb ✓ [3]
Not: using 12 V as the peak value (max 2 of 3 marks, ECF).

(a) Work done on ion = kinetic energy gained: \( eV = \tfrac{1}{2}mv^2 \), so \( v = \sqrt{2eV/m} \).
(b) \( m = 131 \times 1.66 \times 10^{-27} = 2.17 \times 10^{-25} \) kg. \( v = \sqrt{2 \times 1.60 \times 10^{-19} \times 1200 / 2.17 \times 10^{-25}} = 4.2 \times 10^{4} \) m s\(^{-1}\).
(c) \( F = \Delta(mv)/\Delta t = (2.0 \times 10^{-6})(4.2 \times 10^{4}) = 0.084 \) N.
(d) \( a = F/m = 0.084/450 = 1.9 \times 10^{-4} \) m s\(^{-2}\).

(a) Equates electrical work/energy eV to kinetic energy \( \tfrac{1}{2}mv^2 \) and rearranges to given result [1].
(b) Converts 131 u to kg (\( 2.17 \times 10^{-25} \) kg) [1]; correct substitution and answer \( 4.2 \times 10^{4} \) m s\(^{-1}\) [1]. Allow 4.1-4.2 x 10^4. Bald correct answer gains both marks. Not POT error.
(c) Thrust = rate of change of momentum = \( 2.0 \times 10^{-6} \times v \) = 0.084 N [1]. ECF from (b).
(d) a = F/m = \( 1.9 \times 10^{-4} \) m s\(^{-2}\) [1]. ECF from (c).

(a) \( GMm/r^2 = m r \omega^2 \) with \( \omega = 2\pi/T \) gives \( GM/r^2 = 4\pi^2 r/T^2 \), so \( T^2 = 4\pi^2 r^3/(GM) \).
(b) \( T^2 = 4\pi^2 (9.38\times10^6)^3 / (6.67\times10^{-11} \times 6.42\times10^{23}) = 7.6 \times 10^{8} \) s\(^2\); T = \( 2.76 \times 10^4 \) s = 7.7 h.

(a) Equates \( GMm/r^2 \) to \( mr\omega^2 \) (or \( mv^2/r \)) [1]; substitutes \( \omega = 2\pi/T \) (or \( v = 2\pi r/T \)) and rearranges correctly to the given expression [1].
(b) Correct substitution giving \( T = 2.76 \times 10^4 \) s [1]; division by 3600 giving 7.7 h (allow 7.6-7.7) [1]. Not forgiven: failing to convert to hours (max 1). Bald correct answer gains both marks.

(a) \( A = \pi d^2/4 = \pi(0.32\times10^{-3})^2/4 = 8.04\times10^{-8} \) m\(^2\). \( \rho = 12.4 \times 8.04\times10^{-8}/1.500 = 6.6\times10^{-7} \) \( \Omega \) m.
(b) %R = 1.6%, %L = 0.33%, %d = 3.1% so %A = 6.3%. Total = 1.6 + 0.33 + 6.3 = 8.2%.
(c) The diameter (6.3% contribution). Take several readings of d at different places and orientations and average them.

(a) Area from \( \pi d^2/4 \) with d in metres [1]; \( \rho = 6.6 \times 10^{-7} \) \( \Omega \) m [1]. Allow 6.6-6.7.
(b) Percentage uncertainties of R, L, d calculated (1.6%, 0.33%, 3.1%) [1]; d doubled for area and all added giving 8% (8.2%) [1]. Allow 8-8.3%. Not using d once (gives 5%).
(c) Diameter identified AND one valid improvement (repeat readings at several positions and average / use instrument with better resolution) [1].

(a) \( N B A = 250 \times 0.30 \times 0.040 = 3.0 \) Wb turns.
(b) \( \omega = 2\pi \times 2.0 = 12.6 \) rad s\(^{-1}\). Differentiating, peak \( \varepsilon = NBA\omega = 3.0 \times 12.6 = 38 \) V.
(c) Peak e.m.f. is proportional to \( \omega \), so it doubles to about 75 V.

(a) 3.0 Wb turns (allow 3.0 Wb with N included; unit Wb turns or Wb) [1].
(b) \( \omega = 12.6 \) rad s\(^{-1}\) [1]; peak e.m.f. = \( NBA\omega \) = 38 V (37.7) [1]. ECF from (a). Bald correct answer gains both marks.
(c) 75 V (allow 76) [1]. ECF from (b), doubling.

(a) Mean \( = (0.31+0.32+0.30+0.31+0.32)/5 = 0.312 \) mm. Half range \( = (0.32-0.30)/2 = 0.01 \) mm.

(b) \( A = \pi d^2/4 = \pi (0.312\times10^{-3})^2/4 = 7.65\times10^{-8} \) m\(^2\). Percentage uncertainty in \( d = 0.01/0.312 = 3.2\% \); \( A \propto d^2 \) so percentage uncertainty in \( A = 6.4\% \).

(c) \( \rho = RA/L = 6.2 \times 7.65\times10^{-8}/1.000 = 4.74\times10^{-7} \) \( \Omega \) m. Percentage uncertainties: \( R \): 0.1/6.2 = 1.6%; \( L \): 0.2%; \( A \): 6.4%. Total = 8.2%. Absolute = 0.39e-7, so \( \rho = (4.7 \pm 0.4)\times10^{-7} \) \( \Omega \) m.

(d) Range is 4.3e-7 to 5.1e-7, which includes 4.9e-7, so the result is consistent with the alloy. However, the largest uncertainty is in diameter (6.4%), so a different alloy with similar resistivity cannot be excluded. Other factors: heating of the wire by current changes the resistance; contact resistance and lead resistance; micrometer zero error; wire not uniform or not circular; kinks changing length; one value of R only.

(a) Mean = 0.312 mm (allow 0.31 mm) [1]; uncertainty 0.01 mm [1].

(b) Substitution into \( \pi d^2/4 \) giving 7.6e-8 (must show at least 2 sf more, e.g. 7.65e-8) [1]; % uncertainty in d = 3.2% (allow 3%) [1]; doubled to 6.4% (ECF from (a)) [1].

(c) Use of \( \rho = RA/L \) [1]; \( \rho = 4.7\times10^{-7} \) \( \Omega \) m (4.74) [1]; sum of percentage uncertainties 1.6+6.4+0.2 = 8.2% (ECF) [1]; absolute uncertainty ≈ 0.4e-7 \( \Omega \) m, quoted to 1 sf [1]. Unit penalty once only.

(d) Levels of response.
Level 3 (5-6 marks): a reasoned conclusion that includes a numerical range comparison with 4.9e-7, identifies the diameter as the dominant uncertainty and discusses at least two further procedural factors, coherently linked.
Level 2 (3-4 marks): comparison with the book value made and a conclusion given, plus at least one other relevant factor; reasoning partly incomplete.
Level 1 (1-2 marks): isolated statements, e.g. a factor named without link to validity, or value compared without using uncertainty.
0 marks: no creditable response.
Indicative content: range 4.3-5.1e-7 includes 4.9e-7 so consistent; not proof, other alloys similar; diameter dominates uncertainty, use thicker wire/more readings/microscope; heating effect, use low current and switch off between readings; contact and lead resistance, use four-terminal method or subtract lead resistance; zero error check; repeat R at several lengths and plot R against L to find gradient.

(a) \( \ln 1.46 = 0.378 \); \( \ln 0.95 = -0.051 \).

(b) Taking logs: \( \ln V = \ln V_0 - t/(RC) \). This is \( y = c + mx \) with gradient \( m = -1/(RC) \) and intercept \( \ln V_0 \).

(c) Gradient \( = (-0.051 - 2.079)/100 = -0.0213 \) s\(^{-1}\). \( RC = 1/0.0213 = 46.9 \) s. \( C = 46.9/1.00\times10^5 = 4.69\times10^{-4} \) F.

(d) Gradient uncertainty \( = (0.0220-0.0206)/2 = 0.0007 \), i.e. 3.3%. Add 5% for \( R \): 8.3%. \( 8.3\% \) of \( 4.69\times10^{-4} = 0.39\times10^{-4} \). \( C = (4.7 \pm 0.4)\times10^{-4} \) F.

(e) The voltmeter is in parallel with \( R \) during discharge, so the effective resistance is \( (100\times1000)/1100 = 91 \) k\( \Omega \) and the true time constant is shorter. The measured \( \tau \) corresponds to \( R_{eff}C \), so using 100 k\( \Omega \) overestimates \( C \) by about 10%. Use a voltmeter (e.g. digital or oscilloscope/data logger) with resistance much greater than \( R \), or use the effective resistance in the calculation.

(a) 0.378 [1]; -0.051 [1] (allow -0.05).

(b) Takes natural logs of both sides / \( \ln V = \ln V_0 - t/RC \) [1]; compares with \( y = mx + c \), gradient = \( -1/RC \) [1]. Not: gradient = \( 1/RC \) without sign.

(c) Gradient calculation using change in ln V over change in t [1]; \( -0.0213 \) s\(^{-1}\) (allow -0.021 to -0.0215) [1]; \( RC = 1/\) gradient (= 46.9 s) [1]; \( C = 4.7\times10^{-4} \) F (ECF) [1].

(d) % uncertainty in gradient 3.3% (half difference / mean) [1]; add 5% for R giving 8.3% [1]; \( C = (4.7 \pm 0.4)\times10^{-4} \) F (ECF) [1].

(e) Voltmeter in parallel with the resistor / provides second discharge path [1]; effective resistance lower than 100 k\( \Omega \) (about 91 k\( \Omega \)) [1]; so time constant is shorter and C calculated with \( R = 100 \) k\( \Omega \) is too large (about 10%) [1]; use a meter of resistance >> R (e.g. oscilloscope/ data logger/ electrometer) [1]. Allow use of a larger R-independent correction.

(a) \( A = \pi d^2/4 = \pi(0.38\times10^{-3})^2/4 = 1.13\times10^{-7} \) m\(^2\).

(b) \( F = 2.00\times9.81 = 19.6 \) N. Stress \( = 19.6/1.13\times10^{-7} = 1.73\times10^{8} \) Pa. Strain \( = 2.2\times10^{-3}/2.50 = 8.8\times10^{-4} \). \( E = 1.73\times10^8/8.8\times10^{-4} = 1.97\times10^{11} \) Pa.

(c) % in \( d \) = 0.01/0.38 = 2.6%, so 5.3% in \( A \). % in extension = 0.1/2.2 = 4.5%. % in \( L \) = 0.002/2.50 = 0.08%. Total = 9.9% (≈10%). Absolute = 0.2e11 Pa. \( E = (2.0 \pm 0.2)\times10^{11} \) Pa.

(d) Beyond the elastic limit the deformation is plastic: the wire does not return to its original length and stress is no longer proportional to strain, so the equation/measurement is invalid.

(e) See marking scheme.

(a) Substitution into \( \pi d^2/4 \) with d in metres [1]; 1.13e-7 (must show more than 2 sf) [1].

(b) Stress = F/A using \( F = mg \) = 19.6 N [1]; strain = 8.8e-4 [1]; \( E = \) stress/strain = 1.97e11 Pa (allow 2.0e11) [1]. Unit Pa or N m\(^{-2}\).

(c) % uncertainty in area = 2 x 2.6% = 5.3% [1]; % in extension 4.5%, and L 0.08% summed with area giving 9.9% (allow 10%) [1]; absolute 0.2e11 Pa, with \( E = (2.0 \pm 0.2)\times10^{11} \) Pa (ECF) [1].

(d) Wire would deform plastically / not return to original length / not obey Hooke's law (stress not proportional to strain) [1].

(e) Levels of response.
Level 3 (5-6 marks): a coherent method covering all three strands: several loads and extensions measured, graph of stress/force against strain/extension with E from the gradient (with how), and at least two specific ways of reducing uncertainty, linked to the cause.
Level 2 (3-4 marks): two strands covered with reasonable detail, or all three in a limited way.
Level 1 (1-2 marks): isolated correct points, e.g. 'repeat readings' with no detail.
0 marks: no creditable response.
Indicative content: add loads in steps (e.g. 0.50 kg to 3.00 kg) and remove again to check elasticity; plot force against extension, gradient = EA/L, so \( E = \) gradient x L/A; or stress against strain with E = gradient; use of a vernier scale/travelling microscope/reference wire to measure extension more precisely, reducing the 4.5% uncertainty; measure d at several points along the wire and in two perpendicular directions with a micrometer, then average; measure L from clamp to marker with a metre rule; use a longer/thinner wire to increase extension; check for kinks and straighten the wire with a small initial load; safety: goggles, sand box beneath the mass.

(a) \( eV = hf \) and \( f = c/\lambda \), so \( eV = hc/\lambda \) and \( V = (hc/e)(1/\lambda) \).

(b) \( 1/(525\times10^{-9}) = 1.905\times10^{6} \) m\(^{-1}\); \( 1/(470\times10^{-9}) = 2.128\times10^{6} \) m\(^{-1}\).

(c) Gradient \( = (2.62-1.95)/((2.128-1.587)\times10^{6}) = 0.67/0.541\times10^{6} = 1.24\times10^{-6} \) V m. \( h = \text{gradient}\times e/c = 1.24\times10^{-6}\times1.60\times10^{-19}/3.00\times10^{8} = 6.6\times10^{-34} \) J s.

(d) Difference \( = (6.63-6.60)/6.63 \approx 0.5\% \).

(e) See marking scheme.

(a) \( E = hf \) and \( c = f\lambda \) used [1]; \( eV = hc/\lambda \) rearranged to the required form [1].

(b) 1.905 [1]; 2.128 [1] (allow 1.91, 2.13).

(c) Change in V (0.67 V) and change in 1/λ (0.541e6 m^-1) correctly identified [1]; gradient = 1.24e-6 V m (allow 1.2e-6 to 1.3e-6) [1]; \( h = \) gradient x e/c [1]; \( h = 6.6\times10^{-34} \) J s (allow 6.5-6.7) [1].

(d) 0.5% (ECF from (c)) [1].

(e) Levels of response.
Level 3 (5-6 marks): evaluates both V and λ sources of uncertainty with reasons, links to the effect on the gradient and h, and gives at least two sensible improvements; coherent.
Level 2 (3-4 marks): identifies at least two relevant sources with some explanation and at least one improvement.
Level 1 (1-2 marks): lists points with little explanation.
0 marks: no creditable response.
Indicative content: threshold at which light is first seen is subjective, because LED emission rises gradually; use a darkened room, a light sensor, or plot I-V and extrapolate the linear region to zero current; only four points, so more LEDs and a best-fit line with uncertainty bars would reduce random error; nominal wavelengths are from the manufacturer and LEDs emit a spread of wavelengths (peak, not mean) and a tolerance of several nm, so measure λ with a diffraction grating; voltmeter resolution/ systematic zero error; temperature heating changing V; infrared LED can be included with a detector; agreement within 0.5% is likely partly coincidental given the uncertainties, so repeat to check precision.