Physics 9702/42 October/November 2025: question paper and mark scheme
Cambridge AS & A Level Physics (9702), Paper 4 A Level Structured Questions, October/November 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The Earth may be considered as a uniform sphere of radius 6.37 × 106 m.
Cambridge is at a point on the Earth’s surface that has a latitude of 52.2° north of the Equator, as shown in Fig. 1.1.
North Pole
Earth
Cambridge
52.2°
Equator
axis
South Pole
Fig. 1.1
As the Earth spins on its axis, Cambridge moves in a circle that is parallel to the Equator but with a smaller radius.
(a) (i) Show that the radius of the circle around which Cambridge moves is 3.90 × 106 m.
[1]
(ii) Calculate the speed at which Cambridge moves around the circle.
speed = …… m s–1 [3]
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(b) A student of mass 58.6 kg stands on horizontal ground in Cambridge.
(i) Determine the magnitude of the resultant force that acts to cause the circular motion of the student.
resultant force = …… N [2]
(ii) On Fig. 1.2, draw an arrow to show the direction of the resultant force that acts on the student.
student
Cambridge
Earth’s surface
Fig. 1.2 (not to scale)
[1]
(iii) On Fig. 1.3, draw labelled arrows from the student to show the directions of the forces that act on the student to cause the resultant force in (b)(ii).
Fig. 1.3 (not to scale)
[2]
[Total: 9]
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Mark scheme
1(a)(i) radius = 6.37 × 106 × cos 52.2° = 3.90 × 106 m A1
1(a)(ii) period = 24 hours C1
v = 2πr / T C1
or
v = rω and ω = 2π / T
v = (2π × 3.90 × 106) / (24 × 60 × 60) A1
= 280 m s–1
1(b)(i) F = mv2 / r C1
= (58.6 × 2802) / (3.90 × 106) A1
= 1.2 N
1(b)(ii) arrow pointing horizontally to the left B1
1(b)(iii) arrow from student pointing along the dotted line, labelled ‘weight’ B1
upwards arrow from student pointing in a direction to the left of normal and above the tangent to the Earth, labelled ‘contac t B1
force’
(a) State Newton’s law of gravitation.
…… [2]
(b) One of the basic assumptions of the kinetic theory of gases is that there are no forces exerted between the molecules of the gas except during collisions.
State two other basic assumptions of the kinetic theory of gases.
1 ……
2 ……
[2]
(c) Hydrogen gas consists of molecules that each have a mass of 3.34 × 10–27 kg. Hydrogen
may be considered to be an ideal gas.
A spherical balloon contains 0.0160 mol of hydrogen gas at a temperature of 282 K. At this temperature, the volume of gas in the balloon is 1.87 × 10– 4 m3.
(i) Determine the pressure of the gas.
pressure = …… Pa [2]
(ii) Estimate the average separation of the hydrogen molecules in the gas.
average separation = …… m [2]
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(d) (i) Use your answer in (c)(ii) to calculate the average gravitational force between adjacent molecules in hydrogen gas.
average force = …… N [2]
(ii) By considering the weight of a molecule, suggest with a reason whether your answer in
(d)(i) is consistent with the assumption of the kinetic theory of gases that there are no forces exerted between molecules.
…… [1]
[Total: 11]
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Mark scheme
2(a) (gravitational) force is (directly) proportional to product of masses B1
force (between point masses) is inversely proportional to the square of their separation B1
2(b) Any two points from: B2
• molecules are in continuous random motion
• molecules have negligible volume compared with volume of gas
• collisions (involving molecules) are (perfectly) elastic
• collisions (of molecules) are instantaneous
2(c)(i) pV = nRT C1
p = (0.0160 × 8.31 × 282) / (1.87 × 10–4) A1
= 2.01 × 105 Pa
2(c)(ii) number of molecules = 0.0160 × 6.02 × 1023 C1
separation = 3√[(1.87 × 10–4) / (0.0160 × 6.02 × 1023)] A1
= 2.7 × 10–9 m (allow any answer that is 3 × 10–9 m to one significant figure)
2(d)(i) F = 6.67 × 10–11 × (3.34 × 10–27)2 / (2.7 × 10–9)2 C1
= 1.0 × 10–46 N A1
2(d)(ii) numerical comparison between 10 –46 N (F) and 10–26 N (the weight of molecule) leading to a conclusion that the assumption B1
is supported
(a) State what is meant by two objects being in thermal equilibrium.
…… [2]
(b) Fig. 3.1 shows a type of thermometer called a constant volume gas thermometer.
vacuum
fixed glass tube scale movable glass tube
Y
Δh gas
X
liquid
rubber tube glass bulb
Fig. 3.1 (not to scale)
The thermometer is used to determine the thermodynamic temperature T of the gas in the glass bulb.
The glass bulb is immersed in the environment for which the temperature is to be measured.
The height of the movable glass tube is then adjusted so that the level of the liquid on the left-hand side aligns with the reference line X marked on the fixed glass tube. The reference line Y is marked on the side of the movable glass tube. The level of the liquid at Y is higher than at X as a result of the pressure of the gas in the glass bulb.
The difference in height Δh between the liquid levels at X and Y is then measured using the scale. The thermodynamic temperature T of the gas is directly proportional to the pressure of the gas. This pressure is directly proportional to Δh.
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(i) The value of Δh can be used to calculate the pressure of the gas. In order to do this, the gravitational field strength is used, along with a property of the liquid.
State the property of the liquid that is used to calculate the pressure.
…… [1]
(ii) Before the measurement of Δh can be made, the glass bulb needs to reach thermal equilibrium with the environment for which the temperature is to be measured.
State two disadvantages of using a constant volume gas thermometer to measure
temperature.
1 ……
2 ……
[2]
(iii) Suggest one situation in which a constant volume gas thermometer would be an
appropriate type of thermometer to choose for measuring temperature.
…… [1]
(iv) Level X aligns with 2.31 cm on the scale. At 0 °C, level Y aligns with 8.69 cm.
At temperature θ, level Y aligns with 7.83 cm on the scale.
Determine a value for θ in °C.
θ = …… °C [3]
[Total: 9]
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Mark scheme
3(a) same temperature B1
no net transfer of thermal energy (between them) B1
3(b)(i) density B1
3(b)(ii) Any two points from: B2
• large response time / large time to reach equilibrium
or
cannot measure rapidly changing temperatures
• reaching equilibrium requires (significant) transfer of energy
or
changes temperature of environment being measured
or
cannot measure temperature of small objects
• bulky / difficult to set up
or
difficult to take readings / scale not calibrated to read temperature
or
cannot measure temperature of solid objects
3(b)(iii) substance with large mass B1
or
temperature that is constant (over time)
or
to calibrate other thermometers (in a laboratory)
3(b)(iv) 0 °C = 273 K C1
T = 273 × (7.83 – 2.31) / (8.69 – 2.31) C1
( = 236 K)
θ = 236 – 273 A1
= – 37 °C
A cylinder contains a fixed mass of an ideal gas at pressure 2Y and volume 6X.
The gas undergoes a sequence of changes from its initial state A, through states B, C and D, then finally back to its initial state A, as shown in Fig. 4.1.
6Y
pressure C D
4Y
2Y B A
0
0 2X 4X 6X 8X volume
Fig. 4.1
Fig. 4.2 shows the variation with time of the internal energy of the gas.
60XY
internal energy
D
40XY
20XY
A A
C
B
0 time
Fig. 4.2
(a) State the first law of thermodynamics.
…… [2]
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(b) (i) Use Fig. 4.1 and Fig. 4.2 to determine the general expression for the internal energy U of the gas when it has pressure p and volume V.
U = …… [1]
(ii) An ideal gas at thermodynamic temperature T contains N molecules.
Use your answer in (b)(i) and the equation of state for an ideal gas to deduce an expression for U in terms of N and T. Identify any other symbols you use.
U = …… [2]
(c) Determine expressions, in terms of X and Y, for the work W done on the gas during:
(i) change AB
W = …… [1]
(ii) change CD.
W = …… [1]
(d) Use your answers in (c) and the first law of thermodynamics to determine an expression, in terms of X and Y, for the net thermal energy Q supplied to the gas during one full cycle
ABCDA. Explain your reasoning.
Q = …… [3]
[Total: 10]
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Mark scheme
4(a) change in internal energy = work done + energy transfer by heating C1
increase in internal energy = work done on system + energy transferred to the system by heating A1
4(b)(i) U = (3 / 2) pV A1
4(b)(ii) pV = NkT and k identified as Boltzmann constant B1
U = (3 / 2) NkT A1
4(c)(i) W = (+)8XY A1
4(c)(ii) W = –20XY A1
4(d) work done during stages BC and DA = 0 B1
change in internal energy (over complete cycle) = 0 C1
thermal energy supplied = 20XY – 8XY A1
= (+)12XY
A steel ball on the end of a thin string oscillates with small oscillations, as shown in Fig. 5.1.
thin string
equilibrium position steel ball
x
oscillations
Fig. 5.1 (not to scale)
The displacement of the centre of the ball from its equilibrium position is x.
(a) Fig. 5.2 shows the variation with x of the acceleration a of the ball.
15 a / cm s–2
10
5
0
–2 –1 0 1 2 x / cm
–5
– 10
– 15
Fig. 5.2
(i) Explain how Fig. 5.2 shows that the oscillations of the ball are simple harmonic.
…… [2]
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(ii) Determine the period T of the oscillations.
T = …… s [3]
(b) At time t = 0, when the displacement of the ball has its maximum value, the ball is immersed in a trough containing thick oil so that the ball is just below the surface of the oil. This results in the subsequent motion of the ball being heavily damped.
(i) State what is meant by damping.
…… [2]
(ii) On Fig. 5.3, sketch a possible variation of the displacement x of the ball with t between t = 0 and t = 2T.
1.5
1.0 x / cm
0.5
0
0 T 2T t
– 0.5
– 1.0
– 1.5
Fig. 5.3
[3]
[Total: 10]
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Mark scheme
5(a)(i) straight line through the origin shows that a is proportional to x B1
negative gradient shows that a is always in the opposite direction to x B1
5(a)(ii) a0 = ω2x0 C1
ω = 2π / T C1
T = 2π √(x0 / a0) A1
= 2π √ (1.2 / 13)
= 1.9 s
5(b)(i) loss of energy of oscillations B1
due to resistive force(s) B1
5(b)(ii) line starting from x = ±1.2 cm at t = 0 B1
line starting from non-zero value of x from t = 0 to t = 2T that is entirely either above or below the t-axis B1
curve from t = 0 starting from non-zero x value, with both magnitude of x value and magnitude of gradient continuously B1
decreasing
(a) Define electric field at a point.
…… [1]
(b) An isolated conducting sphere in a vacuum has a capacitance of 69 pF.
The charge on the sphere is +83 pC.
(i) On Fig. 6.1, draw field lines to represent the electric field outside the sphere due to the charge on the sphere.
Fig. 6.1
[2]
(ii) Calculate the electric potential at the surface of the sphere.
electric potential = …… V [2]
(iii) Determine the radius of the sphere.
radius = …… m [2]
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(iv) Calculate the electric field strength E at the surface of the sphere. Give a unit with your answer.
E = …… unit …… [2]
(c) The sphere in (b) is discharged by connecting it to earth (0 V) through a resistor of resistance
120 MΩ.
Calculate the time taken for the charge to fall to 26 pC.
time = …… s [2]
[Total: 11]
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Mark scheme
6(a) force per unit positive charge B1
6(b)(i) radial lines B1
arrows pointing away from the sphere B1
6(b)(ii) C =Q/V C1
V = 83 / 69 A1
= (+)1.2 V
6(b)(iii) V = Q / 4πε0r C1
r = (83 × 10–12) / (4π × 8.85 × 10–12 × 1.2)
= 0.62 m A1
6(b)(iv) E = Q / 4πε0r2 C1
= (83 × 10–12) / (4π × 8.85 × 10–12 × 0.622) A1
= 1.9 N C–1
6(c) 26 = 83 exp [– t / (120 × 106 × 69 × 10–12)] C1
t = 9.6 × 10–3 s A1
An alternating voltage V varies with time t according to
V = 18 cos 40 πt
where V is in V and t is in s.
(a) For the alternating voltage:
(i) show that the period is 0.050 s
[1]
(ii) determine the root-mean-square (r.m.s.) voltage.
r.m.s. voltage = …… V [1]
(b) On Fig. 7.1, sketch the variation of V with t for values of t from t = 0 to t = 100 ms.
20
V /V
0
0 25 50 75 100 t / ms
– 20
Fig. 7.1
[3]
(c) The alternating voltage is rectified to produce an output voltage across a load resistor R, as shown in Fig. 7.2.
rectification R
V output voltage circuit
Fig. 7.2
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Fig. 7.3 shows the variation with t of the power P in the load resistor.
30
P /W
20
10
0
0 25 50 75 100 t / ms
Fig. 7.3
State three conclusions that can be drawn from Fig. 7.3. The conclusions may be qualitative or quantitative. Use the space for any working.
1 ……
2 ……
3 ……
[3]
[Total: 8]
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Mark scheme
7(a)(i) T = 2π / 40π = 0.050 s A1
7(a)(ii) Vr.m.s. = 18 / √2 A1
= 13 V
7(b) sinusoidal curve of period 50 ms from t = 0 to t = 100 ms B1
correct phase (VMAX at t = 0, 50, 100 ms and –VMAX at 25, 75 ms etc.) B1
maximum and minimum voltages shown as ±18 V B1
7(c) Any three points from: B3
• rectification is full-wave
• mean power = 14 W
• resistance of R = 12 Ω
• peak current in R = 1.6 A
or
r.m.s. current in R = 1.1 A
• period of output voltage / power = 25 ms
or
frequency of output voltage / power = 40 Hz
or
angular frequency of output voltage / power = 250 rad s–1
Fig. 8.1 shows the three lowest-frequency lines in the part of the emission spectrum for hydrogen that relates to electron transitions to the ground state (level n = 1).
3.09 2.92 2.47
Fig. 8.1 (not to scale)
The numbers represent the frequencies, in 1015 Hz, associated with the spectral lines.
(a) Use the photon model of electromagnetic radiation to explain how the existence of spectral lines in the emission spectrum provides evidence for discrete electron energy levels in the hydrogen atom.
…… [3]
(b) The energy of the ground state (level n = 1) in a hydrogen atom is –13.6 eV.
(i) Calculate the energy, in J, of the ground state.
energy = …… J [1]
(ii) Show that the energy difference between levels n = 1 and n = 2 is 10.2 eV.
[2]
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(iii) Complete Table 8.1 to show the energy differences from the ground state, and the energies of the levels up to n = 4, in the hydrogen atom. Use the space for any working.
Table 8.1
(energy difference level energy / eV from n = 1) / eV
n=4
n=3
n=2 10.2
n=1 0.0 –13.6
[4]
[Total: 10]
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Mark scheme
8(a) Any three points from: B3
• electrons moving between levels emit a single photon
• energy of photon = difference between energy levels
• energy of photon depends on frequency
• discrete frequencies (in spectrum) so differences between electron energies must be discrete
• discrete differences between electron energies means energy levels must be discrete
8(b)(i) energy = – (13.6 × 1.60 × 10–19) A1
= – 2.18 × 10–18 J
8(b)(ii) ΔE = hf C1
= (6.63 × 10–34 × 2.47 × 1015) / (1.60 × 10–19) = 10.2 eV A1
8(b)(iii) n = 2 energy level = – 3.4 eV A1
n = 3 energy difference = 12.1 eV A1
n = 4 energy difference = 12.8 eV A1
n = 3 energy level = – 1.5 eV and n = 4 energy level = – 0.8 eV A1
(a) State what is meant by the mass defect of a nucleus.
…… [2]
(b) The nuclear fusion reaction for the formation of helium-4 from deuterium is represented by
2 2 4
1H + 1H 2He.
Table 9.1 shows the masses of the nuclides involved in this reaction.
Table 9.1
nuclide nuclide mass / u
2 2.013 553
1H
4 4.001 505
2He
Calculate the energy released in the formation of 1.00 mol of helium-4.
energy = …… J [4]
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(c) The star Sirius has a radius of 1.19 × 109 m and loses mass due to nuclear fusion at a rate of
1.09 × 1011 kg s–1. Assume that the power of the radiation emitted by the star is equal to the power released by this process.
(i) Determine a value for the luminosity of Sirius. Give a unit with your answer.
luminosity = …… unit …… [2]
(ii) Use your answer in (c)(i) to determine the surface temperature of Sirius.
surface temperature = …… K [2]
(d) Explain how cosmologists use standard candles to estimate the distance of a galaxy from the
Earth.
…… [3]
[Total: 13]
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Mark scheme
9(a) difference between mass of nucleus and mass of (constituent) nucleons M1
when nucleons are separated to infinity A1
9(b) Δm = (2 × 2.013553) – (4.001505) (u) C1
( = 0.025601 u)
E = c2Δm C1
energy from one He-4 nucleus= 0.025601 × 1.66 × 10–27 × (3.00 × 108)2 C1
(= 3.82 × 10–12 J)
energy to form 1.00 mol= 3.82 × 10–12 × 6.02 × 1023 A1
= 2.30 × 1012 J
9(c)(i) L = 1.09 × 1011 × (3.00 × 108)2 C1
= 9.81 × 1027 W A1
9(c)(ii) L = 4πσ r2T 4 C1
9.81 × 1027 = 4π × 5.67 × 10–8 × (1.19 × 109)2 × T 4
T = 9930 K A1
9(d) standard candles have known luminosity B1
radiant flux intensity (from star) measured (on the Earth) B1
distance found from F = L / (4πd2) B1
(a) State what is meant by contrast in an X-ray image.
…… [1]
(b) X-rays of intensity I0 are incident normally on a structure, as shown in Fig. 10.1.
material P 2.1 cm material Q
A
incident X-rays, detected intensity I0 X-rays
B
5.8 cm
Fig. 10.1
Material P has a linear attenuation coefficient of 0.35 cm–1.
The X-rays emerging from the structure in region A have an intensity of 0.053I0.
(i) Show that the intensity of the X-rays emerging in region B is 0.13I0.
[1]
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(ii) Determine the linear attenuation coefficient μ of material Q.
μ = …… cm–1 [3]
(iii) Use the information in (b)(i) to suggest why the X-rays emerging from the structure form an image that has poor contrast.
…… [1]
(c) Explain how X-rays are used in computed tomography (CT) scanning to produce a three-dimensional image of an internal structure.
…… [3]
[Total: 9]
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Mark scheme
10(a) difference in degrees of blackening B1
10(b)(i) I = I0 exp (–μx) A1
= I0 exp (– 5.8 × 0.35) = 0.13 I0
10(b)(ii) use of exp {–(0.35 × 3.7)} factor C1
0.053I0 = I0 exp {–[(0.35 × 3.7) + 2.1μ]} C1
μ = 0.78 cm–1 A1
10(b)(iii) factor of only 2.5 between the (detected) intensities (so not good contrast) B1
10(c) (structure) scanned in (thin) sections B1
(many) scans (of each section) taken from different angles B1
scanning repeated for all sections and (data) compiled (to form 3D image) B1
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