Thursday, October 16, 2014

9702 June 2014 Paper 41 43 Worked Solutions | A-Level Physics

  • 9702 June 2014 Paper 41 & 43 Worked Solutions | A-Level Physics

Paper 41 & 43


Question 1
Gravitational potential at a point is defined as the work done bringing unit mass from infinity (to the point)

Stone of mass m has gravitational potential energy EP at point X in gravitational field. Magnitude of gravitational potential at X is φ. Relation between m, EP and φ:
Ep = - mφ

Isolated spherical planet of radius R assumed to have all its mass concentrated at its centre. Gravitational potential at surface of planet is − 6.30 × 107 J kg−1. Stone of mass 1.30 kg travelling towards planet such that its distance from centre of planet changes from 6R to 5R. Change in gravitational potential energy of stone:
Gravitational potential , φ α 1/x
At a distance of 6R from the centre, potential = (6.3x107) / 6 [= 1.05x107Jkg-1]
and at 5R from the centre, potential = (6.3x107) / 5 [= 1.26x107Jkg-1]
So, change in energy in energy = (1.26 – 1.05)x107 x 1.3 = 2.7x106J
Change in gravitational potential = (1/5 – 1/6) x 6.3x107
Change in energy =   (1/5 – 1/6) x (6.3x107) x 1.3 = 2.7x106J

Question 2
The Avogadro constant is the number of atoms in 12g of carbon-12

Argon-40 (4018Ar) may be assumed to be ideal gas. Mass of 3.2 g of argon-40 has volume of 210 cm3 at temperature of 37 °C. For mass of argon-40 gas,
Amount, in mol:
Amount, in mol = 3.2 / 40 = 0.080 mol

pV = nRT
(T = 273 + 37 = 310K)
p (210x10-6) = 0.080 x 8.831 x 310
Pressure, p = 9.8x105Pa

Root-mean-square (r.m.s.) speed of argon atom:
pV = (1/3) Nm<c2>
Number of atoms, N = 0.080 (6.02x1023) [= 4.82x1022]
and  mass of 1 Ar atom, m = 40 (1.66x10-27) [= 6.64x10-26]
(9.8x105) (310x10-6) = (1/3) (4.82x1022) (6.64x10-26) <c2>
<c2> = 1.93x105
r.m.s. speed, cRMS = <c2> = 440ms-1

(Total mass of gas) Nm = 3.2x10-3
(9.8x105) (310x10-6) = (1/3) (3.2x10-3) <c2>
<c2> = 1.93x105
r.m.s. speed, cRMS = <c2> = 440ms-1

(1/2) m<c2> = (3/2) kT
0.5 (40 x 1.66x10-27) <c2> = (3/2) (1.38x10-23) (310)
<c2> = 1.93x105
r.m.s. speed, cRMS = <c2> = 440ms-1

Question 3
Volume of 1.00 kg of water in liquid state at 100 °C is 1.00 × 10−3 m3. Volume of 1.00 kg of water vapour at 100 °C and atmospheric pressure 1.01 × 105 Pa is 1.69 m3.
Show that the work done against the atmosphere when 1.00 kg of liquid water becomes water vapour is 1.71 × 105 J:
EITHER change in the volume = 1.69 – (1.00x10-3)
OR the liquid volume << volume of the vapour
So, work done (= pΔV) = (1.01x105) x 1.69 = 1.71x105J

First law of thermodynamics may be given by expression ΔU = + q + w where ΔU is increase in internal energy of system. What is meant by
+q: heating of the system / thermal energy supplied to the system

+w: work done on the system

Specific latent heat of vaporisation of water at 100 °C is 2.26 × 106 J kg−1. Mass of 1.00 kg of liquid water becomes water vapour at 100 °C. Using answer in (a), increase in internal energy of mass of water during vaporization:
ΔU = (2.26x106) – (1.71x105) = 2.09x106J

Question 4
Student investigates energy changes of mass oscillating on vertical spring, as shown. Student draws graph of variation with displacement x of energy E of oscillation, as shown.
State whether energy E represents total energy, potential energy or kinetic energy of the oscillations:
Kinetic energy / KE / EK

Student repeats investigation but with smaller amplitude. Maximum value of E now found to be 1.8 mJ. Use Fig to determine change in amplitude:
EITHER Change in energy = 0.60mJ
OR The maximum value of E is proportional (amplitude)2 / equivalent numerical working
The new amplitude is 1.3cm
Change in amplitude = 0.2cm

Question 5
Isolated solid metal sphere of radius r is given positive charge. Distance from centre of sphere is x.
Electric potential at surface of sphere is V0. On axes of Fig, sketch graph to show variation with distance x of electric potential due to charged sphere, for values of x from x = 0 to x = 4r:
The graph is a straight line at constant potential = V0 from x = 0 to x = r. Then, it is a curve with a decreasing gradient passing through (2r, 0.50V0) and (4r, 0.25V0)

Electric field strength at surface of sphere is E0. On axes of Fig, sketch graph to show variation with distance x of electric field strength due to the charged sphere, for values of x from x = 0 to x = 4r:
The graph is a straight line at E = 0 from x = 0 to x = r. Then, it is a curve with a decreasing gradient from (r, E0) passing through (2r, ¼E0)

Question 6
Uncharged capacitor is connected in series with battery, switch and resistor, as shown. Battery has e.m.f. 9.0 V and negligible internal resistance. Capacitance of capacitor is 4700 μF. Switch is closed at time t = 0. During time interval t = 0 to t = 4.0 s, charge passing through resistor is 22 mC.
Energy transfer in battery during time interval t = 0 to t = 4.0 s:
Energy = EQ = 9.0 x (22x10-3) = 0.20J

For capacitor at time t = 4.0s,
Potential difference V across capacitor:
Capacitance, C = Q / V
Potential difference, V = (22x10-3) / (4700x10-6) = 4.7V

Energy stored in capacitor:
EITHER Energy, E = ½CV2 = ½(4700x10-6)(4.72) = 5.1x10-2J  
OR E = ½QV = ½(22x10-3)(4.7) = 5.1x10-2J  
OR E = ½Q2/C = ½(22x10-3)2 / (4700x10-6) = 5.1x10-2J  

Why answers in (a)(i) and (a)(ii) part 2 are different:
There is energy lost (as thermal energy) in the resistance / wires / battery / resistor

Question 7
{Detailed explanations for this question is available as Solution 1008 at Physics 9702 Doubts | Help Page 210 -}

Question 8
Quantisation of charge is the discrete and equal amounts (of charge)

Student carries out experiment to determine elementary charge. Charged oil drop is positioned between two horizontal metal plates, as shown. Plates are separated by distance of 7.0 mm. Lower plate is earthed. Potential of upper plate is gradually increased until drop is held stationary. Potential for drop to be stationary is 680 V. Weight of oil drop, allowing for upthrust of air, is 4.8 × 10−14 N. Value for charge on oil drop:
(Electric force = qV/d)
Weight of oil drop = qV / d
4.8x10-14 = q (680) / (7x10-3)
Charge, q = 4.9x10-19C

Student repeats experiment and determines the following values for charge on oil drops. 3.3 × 10−19C                  4.9 × 10−19C                      9.7 × 10−19C                3.4 × 10−19C
Use values to suggest value for elementary charge:
The elementary charge = 1.6x10-19C
EITHER The values are (approximately) multiples of this value OR It is a common factor. It is the highest common factor

Question 9
For particular metal surface, it is observed that there is a minimum frequency of light below which photoelectric emission does not occur. This observation provides evidence for particulate nature of electromagnetic radiation.
3 further observations from photoelectric emission that provide evidence for particulate nature of electromagnetic radiation:
Choose any 3:
There is no time delay between illumination and emission
The maximum (kinetic) energy of the electron is dependent on frequency
The maximum (kinetic) energy of the electron is independent of intensity
The rate of emission of electrons is dependent on / proportional to the intensity

Some data for variation with frequency f of maximum kinetic energy EMAX of electrons emitted from metal surface are shown.
Why emitted electrons may have kinetic energy less than maximum at any particular frequency:
The (photon) interaction with the electron may be below the surface. So, energy is required to bring the electron to the surface

Use Fig to determine
Threshold frequency:
(x-intercept) Threshold frequency = 5.8x1014Hz                

Work function, in eV, of metal surface:
Work function, ϕ = hf0 = (6.63x10-34) (5.8x1014) = 3.84x10-19J
Work function, ϕ = (3.84x10-19) / (1.6x10-19) = 2.4eV
Use equation: hf = ϕ + EMAX
Choose a point on the line and substitute values of EMAX, f and h into the equation with the units of hf converted from J to eV
Work function, ϕ = 2.4eV

Question 10
The binding energy of a nucleus is the energy required to separate the nucleons (in a nucleus) to infinity

Data for masses of some particles are given.
Proton                                     1.00728
Neutron                                   1.00867
tritium (31H) nucleus               3.01551
polonium (21084Po) nucleus      209.93722

Energy equivalent of 1.0 u is 930 MeV
Binding energy, in MeV, of tritium (31H) nucleus:
Change in mass, Δm = 2(1.00867) + 1.00728 – 3.01551 = 9.11x10-3u
Binding energy = (9.11x10-3) x 930 = 8.47MeV

Total mass of separate nucleons that make up polonium-210 (21084Po) nucleus is 211.70394 u. Binding energy per nucleon of polonium-210:
Change in mass, Δm = 211.70394 – 209.93722 = 1.76672u
Binding energy per nucleon = (1.76672 x 930) / 210 = 7.82MeV

1 possible fission reaction is 23592U + 10n - - -> 14156Ba  +  9236Kr  +  310n
Reference to binding energy, why this fission reaction is energetically possible:
The total binding energy of barium and krypton is greater than the binding energy of uranium


Question 11
Circuit incorporating ideal operational amplifier (op-amp) is shown.
Name of this circuit:
Inverting amplifier

Why point P is referred to as virtual earth:
The gain is very large / infinite. Since V+ is earthed / zero, for the amplifier not to saturate, point P must be (almost) earth / zero.

Circuit of Fig. is modified, as shown. Voltmeter has infinite resistance and its full-scale deflection is 1.0 V. Input potential to circuit is VIN. Switch position may be changed in order to have different values of resistance in circuit.
Input potential VIN and switch position are varied. For each switch position, reading of voltmeter is 1.0 V. Complete Fig. for switch positions shown.
Switch position           VIN/mV            Resistance
A                                 10                    RA = 100kΩ
B                                 100                  RB = 10kΩ
C                                 1000                RC = 1.0kΩ
(VOUT/VIN = - RF/RIN. So, RF = (VOUT/VIN) RIN. VOUT = (-) 1.0V and RIN = 1.0kΩ {VIN needs to be taken in V not mV})

Reference to answers in (i), suggest use for circuit of Fig:
It can be used as a variable range meter

Question 12
Outline briefly principles of CT scanning:
A series of X-ray images (for one section / slice) is taken from different angles to give an image of the section / slice. This is repeated for many slices to build up a three-dimensional image (of the whole object)

In model for CT scanning, section is divided into 4 voxels. Pixel numbers P, Q, R and S of voxels are shown in Fig.
P          Q
S          R
Section is viewed from 4 directions D1, D2, D3 and D4. Detector readings for each direction are noted. Detector readings are summed as shown.
49        61
73        55
Background reading is 34. Determine pixel numbers P, Q, R and S as shown in Fig:
The background must be deducted from the readings
(15       27)
(39       21)
Divide the values by 3
5          9
13        7

Question 13
Signal from microphone is to be transmitted in digital form. Block diagram of part of transmission system is shown.
2 advantages of transmission of signal in digital form rather than in analogue form:
Choose any 2:
Noise can be eliminated / the waveform can be regenerated
Some extra bits of data can be added to check for errors
It is cheaper / more reliable
It has a greater rate of transfer of data

Function of parallel-to-serial converter:
A parallel-to-serial converter receives the bits all at one time and transmits the bits one after another.

In particular telephone system, sampling frequency is 8 kHz. In manufacture of compact disc, sampling frequency is approximately 44 kHz. Explain why sampling frequency is much higher for compact disc:
The sampling frequency must be higher than / (at least) twice the frequency to be sampled.
EITHER Higher (range of) frequencies are reproduced on the disc OR Lower (range of) frequencies on the phone.
Additionally, EITHER a higher quality (of sound) may be needed on the disc OR high quality (of sound) is not required for the phone

Question 14
{Detailed explanations for this question is available as Solution 611 at Physics 9702 Doubts | Help Page 121 -}


  1. can you please graph 2c from w14 qp 43?
    2b (same paper) is it a straight line passing through origin because f=bqv where v is proportional to f?

    1. Check question 624 at

  2. Replies
    1. For question 4, check solution 632 at


  3. in 10 b i) can you explain"op-amp has very large / infinite gain and so saturates" this statement?whats saturation and infinite gain ?

    1. Firstly, I would like to point out that if you read the application booklet carefully, these are already explained well.

      But for a quick explanation:
      What saturation means?
      The supply line usually provide a potential of about 9.0V, 6.0V, ... (like these values) to the op-amp. If there is a significant difference between the input potentials (inverting and non-inverting or V- and V+ here), the output voltage would be very much larger than the voltages supplied by the supply line. (see application booklet for examples)

      From the conservation of energy, this is not possible - a supply voltage of 9.0V cannot give an output voltage of, say, 10 000V. SO, then the output would be equal to the supply voltage, 9.0V here. THat is what we call saturation. Theoretically, the output should be 10 000V, but in reality this is not the case. We say that the op-amp has been saturated.

      Infinite gain is that very large value of output obtained theoretically, when compared to the input potential.

      However, saturation does not occur if the inputs are approximately the same. V+ is connected to the ground, so its potential is 0.0V. For the op-amp not to saturate, V- should be approximately equal to V+. Since V+ is zero, V- should be approximately equal to zero. That's why it's called virtual earth

    2. Can you explain to me the answers in 14)c) in detail? Thanks :)

    3. Can you explain to me the answer for 14c) in detail?
      Do you mind explaining to me answer for O/N 2014/41 question 5b)ii) as well?
      Your help is much appreciated :)

    4. Details have been updated

    5. For Nov 2014 P41 Q5, see question 619 at

    6. for Nov 2014 P41 Q5 b i, how is electric potential constant? as you go to the right,the potential decreases. Moreover, even if the potential is constant. why would that prevent the radius from being any bigger?

    7. Even though this not directly shown in the graph, it is known that the electric potential starts to decrease from the surface of the sphere from further away and that the electric potential is constant inside the sphere. Since the graph starts at x = 1cm (and the part before x = 1cm is not shown [it is not be zero]), we can infer that before x = 1cm, we are inside the sphere.

      If the radius was greater than 1.0cm, the electric potential should be constant for some values of x after x = 1cm.

  4. 11 ciii) why is it multiplied instead of dividing?and 12 c it says greater number of voltage and then smaller voltage which doesnt make sense for w 14 qp 43. Thank you very much.

    1. For the first one, see solution 615 at

  5. Hye can you explain question 7 June 2014

  6. Where are the graphs for question 5
    May June 2014 41

    1. follow the instructions given above. The graph will take that form.

      It's important that you understand what is happening (from the explanation) instead of having the final answer directly.



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