AP Physics 2 Practice Test 3 – 60 Questions with Answers

Test yourself with 60 original AP Physics 2 practice questions covering all 7 units of the course: thermodynamics, electric force, field, and potential, electric circuits, magnetism and electromagnetism, geometric optics, waves, sound, and physical optics, and modern physics. Every answer includes the full working. All questions are algebra-based (no calculus). Take g = 9.8 m/s2 where relevant.

Unit 9: Thermodynamics

Q1. A sealed container holds 1 mol of oxygen in a volume of 80 L at a temperature of 332 K. What is the pressure of the gas? (Use R = 8.31 J/(mol.K).)

  • (A) 68973 Pa
  • (B) 34486 Pa
  • (C) 17243 Pa
  • (D) 51730 Pa
Show Answer

Answer: B – PV = nRT gives P = nRT ÷ V = (1 × 8.31 × 332) ÷ 0.08 = 34486 Pa.

Q2. A gas syringe contains gas at constant pressure with an initial volume of 40 L at 360 K. The volume changes to 24 L. What is the new temperature of the gas?

  • (A) 432 K
  • (B) 216 K
  • (C) 108 K
  • (D) 324 K
Show Answer

Answer: B – At constant pressure, V ÷ T is constant, so T2 = T1 × V2 ÷ V1 = 360 × 24 ÷ 40 = 216 K.

Q3. A gas in a piston absorbs 1150 J of heat while doing 500 J of work on its surroundings. What is the change in internal energy of the gas?

  • (A) 1300 J
  • (B) 325 J
  • (C) 975 J
  • (D) 650 J
Show Answer

Answer: D – First law: ΔU = Q – W = 1150 – 500 = 650 J.

Q4. A gas expands at a constant pressure of 250 kPa, increasing its volume by 10 L. How much work is done by the gas?

  • (A) 2500 J
  • (B) 5000 J
  • (C) 1250 J
  • (D) 3750 J
Show Answer

Answer: A – W = PΔV = 250 kPa × 10 L = 2500 J, since 1 kPa × 1 L = 1 J.

Q5. A iron plate of mass 2.5 kg is heated through a temperature rise of 15 °C. The specific heat capacity of the material is 450 J/(kg°C). How much heat is supplied?

  • (A) 16875 J
  • (B) 33750 J
  • (C) 8438 J
  • (D) 25312 J
Show Answer

Answer: A – Q = mcΔT = 2.5 × 450 × 15 = 16875 J.

Q6. A Carnot engine operates between a hot reservoir at 750 K and a cold reservoir at 250 K. What is its maximum possible efficiency?

  • (A) 100 %
  • (B) 33.33 %
  • (C) 66.67 %
  • (D) 73.337 %
Show Answer

Answer: C – Efficiency = (1 – Tc ÷ Th) × 100% = (1 – 250 ÷ 750) × 100% = 66.67%.

Q7. A heat engine takes in 1000 J of heat in each cycle and rejects 400 J to the cold reservoir. What is the efficiency of the engine?

  • (A) 40 %
  • (B) 30 %
  • (C) 60 %
  • (D) 66 %
Show Answer

Answer: C – Work per cycle W = Qh – Qc = 1000 – 400 = 600 J. Efficiency = W ÷ Qh × 100% = 600 ÷ 1000 × 100% = 60%.

Q8. A fixed volume of gas has a pressure of 240 kPa at 280 K. It is heated to 500 K at constant volume. What is its new pressure?

  • (A) 857 kPa
  • (B) 214 kPa
  • (C) 429 kPa
  • (D) 643 kPa
Show Answer

Answer: C – At constant volume, P ÷ T is constant, so P2 = P1 × T2 ÷ T1 = 240 × 500 ÷ 280 = 429 kPa.

Q9. Two samples of the same liquid, of masses 4 kg at 90 °C and 2 kg at 30 °C, are mixed in an insulated container. What is the final temperature of the mixture? (Ignore heat loss.)

  • (A) 60 °C
  • (B) 77 °C
  • (C) 70 °C
  • (D) 105 °C
Show Answer

Answer: C – Heat lost = heat gained, so for the same material Tf = (m1T1 + m2T2) ÷ (m1 + m2) = (4 × 90 + 2 × 30) ÷ 6 = 70 °C.

Unit 10: Electric Force, Field, and Potential

Q10. Two charged beads carry charges of 12 µC and 2 µC and are 15 cm apart in a vacuum. What is the magnitude of the electrostatic force between them? (Use k = 9 × 109 N·m2/C2.)

  • (A) 19.2 N
  • (B) 4.8 N
  • (C) 14.4 N
  • (D) 9.6 N
Show Answer

Answer: D – F = kq1q2 ÷ r2 = (9 × 109 × 12 × 10-6 × 2 × 10-6) ÷ 0.152 = 9.6 N.

Q11. What is the magnitude of the electric field at a distance of 20 cm from a point charge of 5 µC in a vacuum? (Use k = 9 × 109 N·m2/C2.)

  • (A) 2250000 N/C
  • (B) 1125000 N/C
  • (C) 562500 N/C
  • (D) 1687500 N/C
Show Answer

Answer: B – E = kQ ÷ r2 = (9 × 109 × 5 × 10-6) ÷ 0.22 = 1125000 N/C.

Q12. Two parallel plates are 4 mm apart with a potential difference of 300 V between them. What is the magnitude of the uniform electric field between the plates?

  • (A) 150000 N/C
  • (B) 37500 N/C
  • (C) 75000 N/C
  • (D) 112500 N/C
Show Answer

Answer: C – E = V ÷ d = 300 ÷ 0.004 = 75000 N/C.

Q13. What is the electric potential at a distance of 20 cm from a point charge of 9 µC in a vacuum? (Use k = 9 × 109 N·m2/C2.)

  • (A) 810000 V
  • (B) 405000 V
  • (C) 202500 V
  • (D) 607500 V
Show Answer

Answer: B – V = kQ ÷ r = (9 × 109 × 9 × 10-6) ÷ 0.2 = 405000 V.

Q14. A charge of 8 µC is placed at a point where the electric potential is 300 V. What is its electric potential energy?

  • (A) 0.0024 J
  • (B) 0.0048 J
  • (C) 0.0012 J
  • (D) 0.0036 J
Show Answer

Answer: A – U = qV = (8 × 10-6) × 300 = 0.0024 J.

Q15. A charge of 3 µC is placed in a uniform electric field of strength 2000 N/C. What is the magnitude of the electric force on the charge?

  • (A) 0.012 N
  • (B) 0.003 N
  • (C) 0.009 N
  • (D) 0.006 N
Show Answer

Answer: D – F = qE = (3 × 10-6) × 2000 = 0.006 N.

Q16. A uniform electric field of 1200 N/C points along a straight path of length 20 cm. What is the potential difference between the ends of the path?

  • (A) 480 V
  • (B) 120 V
  • (C) 360 V
  • (D) 240 V
Show Answer

Answer: D – ΔV = Ed = 1200 × 0.2 = 240 V.

Q17. A stationary charge of 5 µC is accelerated from rest through a potential difference of 800 V. What kinetic energy does it gain? (Ignore gravity.)

  • (A) 0.008 J
  • (B) 0.002 J
  • (C) 0.006 J
  • (D) 0.004 J
Show Answer

Answer: D – Kinetic energy gained = qΔV = (5 × 10-6) × 800 = 0.004 J.

Q18. Two point charges exert a force of 36 N on each other. If the distance between them is halved while everything else stays the same, what is the new force between them?

  • (A) 288 N
  • (B) 72 N
  • (C) 216 N
  • (D) 144 N
Show Answer

Answer: D – Coulomb’s law gives F proportional to q1q2 ÷ r2. For this change the force is multiplied by 4, so F = 36 × 4 = 144 N.

Unit 11: Electric Circuits

Q19. A motor of resistance 6 Ω is connected across a 36 V battery. What current flows through it? (Ignore internal resistance.)

  • (A) 12 A
  • (B) 3 A
  • (C) 6 A
  • (D) 9 A
Show Answer

Answer: C – I = V ÷ R = 36 ÷ 6 = 6 A.

Q20. Three resistors of 20 Ω, 5 Ω and 30 Ω are connected in series. What is their combined resistance?

  • (A) 110 Ω
  • (B) 27.5 Ω
  • (C) 55 Ω
  • (D) 82.5 Ω
Show Answer

Answer: C – For series resistors, R = R1 + R2 + R3 = 20 + 5 + 30 = 55 Ω.

Q21. Resistors of 20 Ω and 5 Ω are connected in parallel. What is their combined resistance?

  • (A) 4 Ω
  • (B) 8 Ω
  • (C) 2 Ω
  • (D) 6 Ω
Show Answer

Answer: A – 1 ÷ R = 1 ÷ 20 + 1 ÷ 5, so R = (20 × 5) ÷ (20 + 5) = 4 Ω.

Q22. A fan motor operates at 24 V and draws a current of 2 A. What power does it use?

  • (A) 48 W
  • (B) 96 W
  • (C) 24 W
  • (D) 72 W
Show Answer

Answer: A – P = VI = 24 × 2 = 48 W.

Q23. A washing machine rated at 750 W runs for 5 hours. How much electrical energy does it use, in kilowatt-hours?

  • (A) 7.5 kWh
  • (B) 3.75 kWh
  • (C) 1.875 kWh
  • (D) 5.625 kWh
Show Answer

Answer: B – Energy = Pt = 750 W × 5 h = 3750 Wh = 3.75 kWh.

Q24. A cell of emf 24 V and internal resistance 3 Ω is connected to an external resistor of 14 Ω. What current flows in the circuit?

  • (A) 2.824 A
  • (B) 1.412 A
  • (C) 0.705882 A
  • (D) 2.118 A
Show Answer

Answer: B – I = emf ÷ (R + r) = 24 ÷ (14 + 3) = 24 ÷ 17 = 1.412 A.

Q25. In a parallel circuit, one branch of resistance 4 Ω is connected across a 48 V supply. What current flows through that branch?

  • (A) 24 A
  • (B) 12 A
  • (C) 6 A
  • (D) 18 A
Show Answer

Answer: B – The branch has the full supply voltage, so I = V ÷ R = 48 ÷ 4 = 12 A.

Q26. A wire has a resistance of 20 Ω. If its length is doubled while the material and temperature stay the same, what is its new resistance?

  • (A) 80 Ω
  • (B) 20 Ω
  • (C) 60 Ω
  • (D) 40 Ω
Show Answer

Answer: D – R = ρL ÷ A, so resistance scales as described: R = 20 × 2 = 40 Ω.

Q27. Resistors of 20 Ω and 5 Ω are connected in parallel, and this combination is connected in series with a 6 Ω resistor. What is the total resistance of the circuit?

  • (A) 20 Ω
  • (B) 10 Ω
  • (C) 5 Ω
  • (D) 15 Ω
Show Answer

Answer: B – Parallel part = (20 × 5) ÷ (20 + 5) = 4 Ω. Total = 4 + 6 = 10 Ω.

Unit 12: Magnetism and Electromagnetism

Q28. A charge of 4 µC moves at 4000 m/s perpendicular to a uniform magnetic field of 0.8 T. What is the magnitude of the magnetic force on the charge?

  • (A) 0.0128 N
  • (B) 0.0256 N
  • (C) 0.0064 N
  • (D) 0.0192 N
Show Answer

Answer: A – F = qvB sin θ with θ = 90° gives F = (4 × 10-6) × 4000 × 0.8 = 0.0128 N.

Q29. A straight wire of length 0.5 m carries a current of 3 A perpendicular to a uniform magnetic field of 0.4 T. What is the magnitude of the magnetic force on the wire?

  • (A) 1.2 N
  • (B) 0.3 N
  • (C) 0.9 N
  • (D) 0.6 N
Show Answer

Answer: D – F = BIL sin θ with θ = 90° gives F = 0.4 × 3 × 0.5 = 0.6 N.

Q30. A particle of mass 0.04 kg and charge 2 C moves at 20 m/s perpendicular to a uniform magnetic field of 0.5 T. What is the radius of its circular path?

  • (A) 1.6 m
  • (B) 0.4 m
  • (C) 0.8 m
  • (D) 1.2 m
Show Answer

Answer: C – r = mv ÷ qB = (0.04 × 20) ÷ (2 × 0.5) = 0.8 m.


Q31. A long straight wire carries a current of 12 A. What is the magnitude of the magnetic field at a distance of 20 cm from the wire?

  • (A) 0.000024 T
  • (B) 0.000006 T
  • (C) 0.000012 T
  • (D) 0.000018 T
Show Answer

Answer: C – B = µ0I ÷ 2πr = 2 × 10-7 × 12 ÷ 0.2 = 0.000012 T.

Q32. A solenoid has 500 turns per metre and carries a current of 4 A. What is the magnetic field strength inside the solenoid, far from its ends? (Use µ0 = 4π × 10-7 T·m/A.)

  • (A) 0.005027 T
  • (B) 0.002513 T
  • (C) 0.001257 T
  • (D) 0.00377 T
Show Answer

Answer: B – B = µ0nI = (4π × 10-7) × 500 × 4 = 0.002513 T.

Q33. A flat loop of area 0.02 m2 sits in a uniform magnetic field of 0.4 T, with the field making an angle of 0° with the normal to the loop. What is the magnetic flux through the loop?

  • (A) 0.016 Wb
  • (B) 0.004 Wb
  • (C) 0.012 Wb
  • (D) 0.008 Wb
Show Answer

Answer: D – Flux = BA cos θ = 0.4 × 0.02 × 1 = 0.008 Wb.

Q34. A coil of 200 turns and area 0.05 m2 experiences a magnetic field change of 0.5 T in 0.5 s, with the field perpendicular to the coil. What is the magnitude of the induced emf in the coil?

  • (A) 10 V
  • (B) 20 V
  • (C) 5 V
  • (D) 15 V
Show Answer

Answer: A – emf = N Δflux ÷ Δt = 200 × 0.05 × 0.5 ÷ 0.5 = 10 V.

Q35. A conducting rod of length 0.5 m moves at 8 m/s perpendicular to a uniform magnetic field of 0.5 T, sliding on rails as in a motional emf setup. What is the induced emf across the rod?

  • (A) 4 V
  • (B) 1 V
  • (C) 3 V
  • (D) 2 V
Show Answer

Answer: D – emf = BLv = 0.5 × 0.5 × 8 = 2 V.

Unit 13: Geometric Optics

Q36. An object is placed 36 cm in front of a concave mirror of focal length 12 cm. What is the image distance? (A positive value means a real image in front of the mirror.)

  • (A) 36 cm
  • (B) 18 cm
  • (C) 9 cm
  • (D) 27 cm
Show Answer

Answer: B – 1 ÷ f = 1 ÷ do + 1 ÷ di gives 1 ÷ di = 1 ÷ 12 – 1 ÷ 36, so di = 18 cm.

Q37. An object placed 30 cm from a mirror forms an image 30 cm from the mirror. What is the lateral magnification?

  • (A) -1
  • (B) -2
  • (C) -0.5
  • (D) -1.5
Show Answer

Answer: A – m = -di ÷ do = -30 ÷ 30 = -1.

Q38. An object is placed 100 cm in front of a converging lens of focal length 25 cm. What is the image distance? (A positive value means a real image on the opposite side of the lens.)

  • (A) 33.333 cm
  • (B) 66.667 cm
  • (C) 16.667 cm
  • (D) 50 cm
Show Answer

Answer: A – 1 ÷ f = 1 ÷ do + 1 ÷ di gives 1 ÷ di = 1 ÷ 25 – 1 ÷ 100, so di = 33.333 cm.

Q39. A light ray in air (n = 1) strikes a glass block at an angle of incidence of 60° and refracts at an angle of 45° to the normal. What is the refractive index of the glass? (Use sin 60° = 0.866 and sin 45° = 0.707.)

  • (A) 1.84
  • (B) 0.82
  • (C) 1.22
  • (D) 0.61
Show Answer

Answer: C – Snell’s law: n1 sin θ1 = n2 sin θ2, so n2 = 0.866 ÷ 0.707 = 1.22.

Q40. Light inside a material of refractive index 1.36 strikes the boundary with air. What is the critical angle for total internal reflection?

  • (A) 42.7 °
  • (B) 47.3 °
  • (C) 23.7 °
  • (D) 71 °
Show Answer

Answer: B – sin θc = 1 ÷ n = 1 ÷ 1.36 = 0.735294, so θc = 47.33°.

Q41. Light travels at 225000000 m/s in a transparent solid. Taking the speed of light in a vacuum as 3 × 108 m/s, what is the refractive index of the solid?

  • (A) 2.667
  • (B) 0.666667
  • (C) 1.333
  • (D) 2
Show Answer

Answer: C – n = c ÷ v = (3 × 108) ÷ 225000000 = 1.333.

Q42. A spherical mirror has a radius of curvature of 60 cm. What is its focal length?

  • (A) 30 cm
  • (B) 60 cm
  • (C) 15 cm
  • (D) 45 cm
Show Answer

Answer: A – f = R ÷ 2 = 60 ÷ 2 = 30 cm.

Q43. An object is placed 5 cm in front of a converging lens of focal length 10 cm, inside the focal point. What is the image distance? (A negative value means a virtual image on the same side as the object.)

  • (A) 10 cm
  • (B) -20 cm
  • (C) -10 cm
  • (D) -5 cm
Show Answer

Answer: C – 1 ÷ di = 1 ÷ f – 1 ÷ do = 1 ÷ 10 – 1 ÷ 5 = -0.1, so di = -10 cm.

Unit 14: Waves, Sound, and Physical Optics

Q44. A wave has a frequency of 100 Hz and a wavelength of 0.5 m. What is its speed?

  • (A) 100 m/s
  • (B) 25 m/s
  • (C) 75 m/s
  • (D) 50 m/s
Show Answer

Answer: D – v = fλ = 100 × 0.5 = 50 m/s.

Q45. An oscillator completes one full cycle in 0.02 s. What is its frequency?

  • (A) 50 Hz
  • (B) 100 Hz
  • (C) 25 Hz
  • (D) 75 Hz
Show Answer

Answer: A – f = 1 ÷ T = 1 ÷ 0.02 = 50 Hz.

Q46. A sound source radiates 10 W uniformly in all directions. What is the sound intensity at a distance of 5 m from the source? (Use π = 3.14.)

  • (A) 0.063662 W/m2
  • (B) 0.015915 W/m2
  • (C) 0.047746 W/m2
  • (D) 0.031831 W/m2
Show Answer

Answer: D – I = P ÷ 4πr2 = 10 ÷ (4 × 3.14 × 52) = 0.031831 W/m2.

Q47. A string of length 0.5 m, fixed at both ends, supports waves travelling at 400 m/s. What is the frequency of its 1th harmonic?

  • (A) 800 Hz
  • (B) 400 Hz
  • (C) 200 Hz
  • (D) 600 Hz
Show Answer

Answer: B – For a string fixed at both ends, fn = nv ÷ 2L = 1 × 400 ÷ (2 × 0.5) = 400 Hz.

Q48. An air column of length 1.7 m, closed at one end, has a fundamental frequency given by v ÷ 4L with v = 343 m/s. What is its fundamental frequency?

  • (A) 101 Hz
  • (B) 50.441 Hz
  • (C) 25.221 Hz
  • (D) 75.662 Hz
Show Answer

Answer: B – f1 = v ÷ 4L = 343 ÷ (4 × 1.7) = 50.441 Hz.

Q49. Two tuning forks of frequencies 256 Hz and 260 Hz sound together. What is the beat frequency heard?

  • (A) 8 Hz
  • (B) 2 Hz
  • (C) 4 Hz
  • (D) 6 Hz
Show Answer

Answer: C – Beat frequency = |f1 – f2| = |256 – 260| = 4 Hz.

Q50. In a double-slit experiment, light of wavelength 650 nm passes through slits 0.1 mm apart, with a screen 1.5 m away. What is the distance from the central bright fringe to the first bright fringe on the screen?

  • (A) 1.95 cm
  • (B) 0.4875 cm
  • (C) 0.975 cm
  • (D) 1.462 cm
Show Answer

Answer: C – y = λL ÷ d = (650 × 10-9 × 1.5) ÷ (0.1 × 10-3) = 0.00975 m = 0.975 cm.

Q51. A siren emitting 800 Hz moves toward a stationary observer at 60 m/s. Taking the speed of sound as 343 m/s, what frequency does the observer hear?

  • (A) 681 Hz
  • (B) 970 Hz
  • (C) 800 Hz
  • (D) 1164 Hz
Show Answer

Answer: B – f’ = f × v ÷ (v – vs) = 800 × 343 ÷ (343 – 60) = 970 Hz.

Unit 15: Modern Physics

Q52. What is the energy, in electron-volts, of an orange photon of wavelength 600 nm? (Use hc = 1240 eV·nm.)

  • (A) 4.13 eV
  • (B) 2.07 eV
  • (C) 1.03 eV
  • (D) 3.1 eV
Show Answer

Answer: B – E = hc ÷ λ = 1240 ÷ 600 = 2.07 eV.

Q53. Light of photon energy 2.5 eV strikes a zinc surface whose work function is 1.5 eV. What is the maximum kinetic energy of an emitted photoelectron?

  • (A) 2 eV
  • (B) 0.5 eV
  • (C) 1.5 eV
  • (D) 1 eV
Show Answer

Answer: D – KEmax = hf – work function = 2.5 – 1.5 = 1 eV.

Q54. An electron of mass 9.11 × 10-31 kg moves at 500000 m/s. What is its de Broglie wavelength? (Use h = 6.63 × 10-34 J·s.)

  • (A) 1.456 nm
  • (B) 2.912 nm
  • (C) 0.728 nm
  • (D) 2.184 nm
Show Answer

Answer: A – λ = h ÷ mv = (6.63 × 10-34) ÷ (9.11 × 10-31 × 500000) = 1.456 nm.

Q55. What is the energy of a hydrogen atom in the n = 1 energy level? (Ground-state energy is -13.6 eV.)

  • (A) -13.6 eV
  • (B) -14.96 eV
  • (C) 13.6 eV
  • (D) -16.728 eV
Show Answer

Answer: A – En = -13.6 ÷ n2 = -13.6 ÷ 1 = -13.6 eV.

Q56. A hydrogen atom drops from the n = 5 level to the n = 2 level. What is the energy of the photon emitted?

  • (A) 1.43 eV
  • (B) 5.71 eV
  • (C) 10.74 eV
  • (D) 2.86 eV
Show Answer

Answer: D – ΔE = 13.6 × (1 ÷ 4 – 1 ÷ 25) = 2.86 eV.

Q57. A nucleus has a mass defect of 0.1 u. What is its total binding energy in MeV? (1 u corresponds to 931.5 MeV.)

  • (A) 93.15 MeV
  • (B) 186 MeV
  • (C) 46.575 MeV
  • (D) 140 MeV
Show Answer

Answer: A – Binding energy = mass defect × 931.5 MeV/u = 0.1 × 931.5 = 93.15 MeV.

Q58. An carbon-14 sample of mass 160 g decays through 3 half-lives. What mass remains?

  • (A) 40 g
  • (B) 10 g
  • (C) 20 g
  • (D) 30 g
Show Answer

Answer: C – After 3 half-lives the fraction remaining is (1 ÷ 2)3 = 0.125, so mass = 160 × 0.125 = 20 g.

Q59. A clock moving at 0.8c relative to an observer ticks with a proper time interval of 6 s between two events. What time interval does the stationary observer measure?

  • (A) 3.6 s
  • (B) 6 s
  • (C) 15 s
  • (D) 10 s
Show Answer

Answer: D – With v = 0.8c, the Lorentz factor is 1.667, so Δt = 1.667 × 6 = 10 s.

Q60. A rod of proper length 50 m moves at 0.6c relative to an observer. What length does the observer measure?

  • (A) 40 m
  • (B) 62.5 m
  • (C) 50 m
  • (D) 20 m
Show Answer

Answer: A – L = L0 ÷ gamma = 50 ÷ 1.25 = 40 m.

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