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IGCSE Physics 0625 Paper 4 Extended Structured Mock Paper (10 questions, 80 marks)
IGCSE Physics 0625 Paper 4
IGCSE Physics 0625 Paper 4 Extended structured mock paper: 10 questions, 80 marks, 1 hour 15 minutes, with mechanics, thermal physics, circuits, waves, transformers, radioactivity and gas laws, plus full answers and marking notes.
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Paper details Level: IGCSE | Subject: Physics | Time: 1 hour 15 minutes | Marks: 80 | Questions: 10
Section A: mechanics, matter and thermal physics (42 marks)
Structured questions: show the formula used, the substitution, the numerical answer and the unit. Working must be shown, since an answer alone earns at most half the marks. 10 questions, 80 marks, 1 hour 15 minutes, calculator allowed.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A car starts from rest and accelerates uniformly, reaching a speed of 24.0 m/s in 8.0 s. It then travels at a constant speed of 24.0 m/s for 12.0 s, and finally decelerates uniformly to rest in 4.0 s. (a) Calculate the acceleration during the first 8.0 s. (2 marks) (b) Calculate the distance travelled during the first 8.0 s. (2 marks) (c) Calculate the total distance travelled. (2 marks) (d) Calculate the average speed for the whole journey. (2 marks) |
Answer: (a) a = (24.0 – 0) / 8.0 = 3.0 m/s^2. (b) s = 0.5 x 3.0 x 8.0^2 = 96 m. (c) Deceleration stage: s = 0.5 x 4.0 x 24.0 = 48 m; constant-speed stage: s = 24.0 x 12.0 = 288 m; total = 96 + 288 + 48 = 432 m. (d) Average speed = 432 / 24.0 = 18.0 m/s. Marking: 1 mark for the formula used and 1 mark for the correct value with unit in each part; part (c) may also be solved from the area under the speed-time graph; part (d) must use total distance over total time, not the mean of the two speeds. |
| 2 | A uniform beam is pivoted at a point. A 2.5 kg mass hangs 0.40 m to the left of the pivot and the beam is balanced by a downward force F applied to the right of the pivot. Take g = 9.8 N/kg. (a) Calculate the weight of the mass. (2 marks) (b) If F acts 0.25 m to the right of the pivot, calculate the size of F. (4 marks) (c) If the same mass is instead balanced by a force acting 0.50 m to the right of the pivot, calculate the force needed and explain why it is smaller. (2 marks) |
Answer: (a) W = mg = 2.5 x 9.8 = 24.5 N. (b) Taking moments about the pivot: F x 0.25 = 24.5 x 0.40, so F = 9.8 / 0.25 = 39.2 N. (c) F = 9.8 / 0.50 = 19.6 N. Doubling the distance from the pivot doubles the moment produced by a given force, so half the force is needed to produce the same moment. Marking: (b) 2 marks for stating the principle of moments, 2 marks for correct substitution and answer; (c) 1 mark for the value, 1 mark for the explanation. |
| 3 | A rectangular metal block measures 0.20 m x 0.10 m x 0.05 m and has a mass of 1.2 kg. (a) Calculate the volume of the block. (2 marks) (b) Calculate the density of the metal. (2 marks) (c) Calculate the weight of the block (take g = 9.8 N/kg). (2 marks) (d) The block rests on the table on its smallest face. Calculate the pressure it exerts on the table. (2 marks) |
Answer: (a) V = 0.20 x 0.10 x 0.05 = 1.00e-03 m^3. (b) density = m / V = 1.2 / 1.00e-03 = 1200 kg/m^3. (c) W = mg = 1.2 x 9.8 = 11.76 N. (d) Contact area A = 0.10 x 0.05 = 0.005 m^2, so p = 11.76 / 0.005 = 2352 Pa. Marking: a unit conversion error (for example treating centimetres as metres without converting) loses the marks in that part; part (d) must use the smallest face, so using the largest face loses both marks; 2.4 kPa is also accepted. |
| 4 | A 350 W heater warms 0.50 kg of water from 20 degrees C to 60 degrees C. The specific heat capacity of water is 4200 J/(kg degrees C). (a) Calculate the energy absorbed by the water. (3 marks) (b) Assuming no energy is lost, calculate the time needed. (3 marks) (c) In the experiment 300 s was actually needed. Calculate the efficiency of the arrangement. (2 marks) |
Answer: (a) E = mc dT = 0.50 x 4200 x 40 = 84,000 J. (b) t = E / P = 84,000 / 350 = 240 s. (c) efficiency = useful output / total input = 240 / 300 x 100% = 80%. Marking: (a) 1 mark for the formula, 1 mark for the substitution, 1 mark for the value and unit; using 60 instead of 40 for the temperature rise scores zero; (c) the ideal time divided by the actual time is the correct quantity. |
| 5 | A 6.0 V battery is connected in series with resistors R1 = 12 ohm and R2 = 4.0 ohm. (a) Calculate the total resistance of the circuit. (2 marks) (b) Calculate the current in the circuit. (2 marks) (c) Calculate the potential difference across R2. (2 marks) (d) Calculate the power dissipated in R2. (2 marks) (e) Calculate the energy dissipated in R2 in 5.0 minutes. (2 marks) |
Answer: (a) R = 12 + 4.0 = 16 ohm. (b) I = V / R = 6.0 / 16 = 0.375 A. (c) V2 = I R2 = 0.375 x 4.0 = 1.5 V. (d) P = V2 I = 1.5 x 0.375 = 0.5625 W, about 0.56 W. (e) E = Pt = 0.5625 x 300 = 168.8 J. Marking: adding the series resistances earns 2 marks; the current must be the supply voltage divided by the total resistance (6.0 / 4.0 = 1.5 A scores zero); 0.56 W or 0.5625 W is accepted in part (d). |
Section B: electricity, waves, magnetism and gases (38 marks)
Structured questions: show the formula used, the substitution, the numerical answer and the unit. Working must be shown, since an answer alone earns at most half the marks. 10 questions, 80 marks, 1 hour 15 minutes, calculator allowed.
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | A tuning fork of frequency 512 Hz sounds in air where the speed of sound is 340 m/s. (a) Calculate the wavelength of the sound. (3 marks) (b) The echo from a cliff is heard 1.20 s after the sound leaves the fork. Calculate the distance between the fork and the cliff. (3 marks) (c) If the distance to the cliff is doubled, what is the new time for the echo to return? (2 marks) |
Answer: (a) v = f x wavelength, so wavelength = v / f = 340 / 512 = 0.66 m. (b) The sound travels twice the distance: d = v t / 2 = 340 x 1.20 / 2 = 204 m. (c) With twice the distance, 408 m, t = 2 d / v = 2.40 s, because the time is proportional to the distance. Marking: forgetting to halve in (b), giving 408 m, earns 1 mark only; using f / v in (a) scores zero. |
| 7 | A transformer has its primary coil connected to a 240 V supply and has 1200 turns. The secondary output is 12 V. Assume an ideal transformer. (a) Calculate the number of turns on the secondary coil. (3 marks) (b) If the output power is 24 W, calculate the secondary current. (3 marks) (c) Calculate the primary current. (2 marks) (d) Give one reason why a real transformer is less than 100% efficient. (2 marks) |
Answer: (a) Ns / Np = Vs / Vp, so Ns = 1200 x 12 / 240 = 60 turns. (b) Is = P / Vs = 24 / 12 = 2.0 A. (c) For an ideal transformer input power equals output power, so Ip = 24 / 240 = 0.1 A. (d) The coils have resistance and become warm (copper losses), and the core loses energy through eddy currents and hysteresis as it is repeatedly magnetised. Marking: (a) 3 marks for the turns ratio being proportional to the voltage ratio; (c) must use equal power, so computing the current from the turns ratio loses the marks. |
| 8 | A radioactive sample has a half-life of 6.0 hours and an initial activity of 800 Bq. (a) What is the activity after 18 hours? (3 marks) (b) How many half-lives have passed in 18 hours? (2 marks) (c) How long does it take for the activity to fall to 50 Bq? (3 marks) |
Answer: (a) The activity halves each time: 800, 400, 200, 100 Bq, so after 18 hours it is 100 Bq. (b) 18 / 6.0 = 3 half-lives. (c) 50 Bq is 1/16 of the initial activity, that is (1/2)^4, so four half-lives are needed: 4 x 6.0 = 24 hours. Marking: (a) the halving steps (or 800 x (1/2)^3) should be shown, an answer alone earns 1 mark; (c) repeated halving or logarithms are both accepted. |
| 9 | A fixed mass of gas occupies 300 cm3 at 300 K and a pressure of 100 kPa. It is heated to 350 K while being compressed to a pressure of 150 kPa. (a) Calculate the new volume of the gas. (4 marks) (b) Name the gas law you used and explain the effect of raising the pressure from 100 kPa to 150 kPa on the volume. (2 marks) |
Answer: (a) Using the combined gas law, p1 V1 / T1 = p2 V2 / T2, so V2 = p1 V1 T2 / (T1 p2) = 100 x 300 x 350 / (300 x 150) = 233 cm3. (b) The combined gas law (Boyle law combined with Charles law). A higher pressure tends to reduce the volume while a higher temperature tends to increase it; here the temperature rises by about 16.7% but the pressure rises by 50%, so the volume falls. Marking: (a) 1 mark for the equation, 1 mark for using kelvin, 1 mark for substitution, 1 mark for the answer in cm3. |
| 10 | A hydraulic jack has a small piston of area 4.0 cm2 and a large piston of area 100 cm2. A force of 50 N is applied to the small piston. (a) Calculate the pressure transmitted through the liquid. (3 marks) (b) Calculate the force that can be lifted by the large piston. (3 marks) |
Answer: (a) A1 = 4.0 cm2 = 4.0e-04 m2, so p = F / A1 = 50 / 4.0e-04 = 125,000 Pa. (b) By Pascal principle the pressure is the same throughout, and A2 = 100 cm2 = 1.0e-02 m2, so F2 = p A2 = 125,000 x 1.0e-02 = 1,250 N, a force multiplication of 25. Marking: the areas must be converted to m2 (4.0 cm2 = 4.0 x 10^-4 m2); substituting in cm2 directly loses all marks. |
A TutorZone practice paper; original questions and passages, not publisher material.
