AP Physics 2 Practice Test 15 – 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). Use g = 9.8 m/s2 where relevant.

Unit 9: Thermodynamics

Q1. How much heat is needed to melt ice at 0°C (specific latent heat of fusion 334,000 J/kg) if the mass involved is 2 kg?

  • (A) 668000 J
  • (B) 1.34 × 106 J
  • (C) 334000 J
  • (D) 2 × 106 J
Show Answer

Answer: A – Q = mL = 2 × 334000 = 668000 J.

Q2. A reversible heat engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. What is its maximum possible efficiency?

  • (A) 50 %
  • (B) 25 %
  • (C) 60 %
  • (D) 40 %
Show Answer

Answer: A – Carnot efficiency = 1 – Tc ÷ Th = 1 – 300 ÷ 600 = 50%.

Q3. A heat engine takes in 10000 J of heat in one cycle and does 1500 J of useful work. How much heat does it reject to the cold reservoir?

  • (A) 17000 J
  • (B) 4250 J
  • (C) 25500 J
  • (D) 8500 J
Show Answer

Answer: D – Energy conservation: Qh = W + Qc, so Qc = 10000 – 1500 = 8500 J.

Q4. A gas at 200 kPa and 300 K occupies 5 L. Its volume changes to 10 L and its temperature changes to 150 K. What is its new pressure?

  • (A) 100 kPa
  • (B) 25 kPa
  • (C) 150 kPa
  • (D) 50 kPa
Show Answer

Answer: D – p1V1 ÷ T1 = p2V2 ÷ T2, so p2 = 200 × 5 × 150 ÷ (10 × 300) = 50 kPa.

Q5. Which statement about entropy is correct for an isolated system?

  • (A) Its total entropy never decreases
  • (B) Its total entropy never increases
  • (C) Its total entropy always stays exactly constant
  • (D) Its total entropy decreases whenever work is done
Show Answer

Answer: A – The second law states that the total entropy of an isolated system never decreases.

Q6. A pressurised storage vessel holds 1 mol of an ideal gas at 250 K in a volume of 0.04 m3. What is the pressure of the gas? (Use R = 8.31 J/(mol·K).)

  • (A) 103875 Pa
  • (B) 51937.5 Pa
  • (C) 25968.75 Pa
  • (D) 155812.5 Pa
Show Answer

Answer: B – pV = nRT gives p = nRT ÷ V = (1 × 8.31 × 250) ÷ 0.04 = 51937.5 Pa.

Q7. An ideal gas sample contains 5 mol at 300 K and a pressure of 100 kPa. What is its volume? (Use R = 8.31 J/(mol·K).)

  • (A) 0.25 m3
  • (B) 0.06 m3
  • (C) 0.12 m3
  • (D) 0.37 m3
Show Answer

Answer: C – pV = nRT gives V = nRT ÷ p = (5 × 8.31 × 300) ÷ 100000 = 0.12 m3.

Q8. A ideal gas in an insulated chamber absorbs 1500 J of heat from its surroundings and does 150 J of work on its surroundings. What is the change in its internal energy?

  • (A) 2700 J
  • (B) 675 J
  • (C) 1350 J
  • (D) 4050 J
Show Answer

Answer: C – First law: ΔU = Q – W = 1500 – 150 = 1350 J.

Q9. A gas expands at a constant pressure of 500 kPa, increasing its volume by 0.1 m3. How much work does the gas do?

  • (A) 100000 J
  • (B) 25000 J
  • (C) 150000 J
  • (D) 50000 J
Show Answer

Answer: D – At constant pressure, W = pΔV = 500000 × 0.1 = 50000 J.

Unit 10: Electric Force, Field, and Potential

Q10. Two parallel plates are 0.02 m apart and have a potential difference of 200 V between them. What is the magnitude of the uniform electric field between the plates?

  • (A) 20000 N/C
  • (B) 10000 N/C
  • (C) 5000 N/C
  • (D) 30000 N/C
Show Answer

Answer: B – E = V ÷ d = 200 ÷ 0.02 = 10000 N/C.

Q11. A charge of 8 µC sits at a point where the electric potential is 200 V. What is its electric potential energy?

  • (A) 3.2 × 10-3 J
  • (B) 8 × 10-4 J
  • (C) 1.6 × 10-3 J
  • (D) 4.8 × 10-3 J
Show Answer

Answer: C – U = qV = (8 × 10-6) × 200 = 1.6 × 10-3 J.

Q12. A charge of 8 µC sits at a point where the electric potential is 500 V. What is its electric potential energy?

  • (A) 8 × 10-3 J
  • (B) 2 × 10-3 J
  • (C) 0.01 J
  • (D) 4 × 10-3 J
Show Answer

Answer: D – U = qV = (8 × 10-6) × 500 = 4 × 10-3 J.

Q13. Two small charged spheres carry charges of 2 µC and 8 µC and are 0.2 m apart in a vacuum. What is the magnitude of the electrostatic force between them? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 7.2 N
  • (B) 1.8 N
  • (C) 10.8 N
  • (D) 3.6 N
Show Answer

Answer: D – F = kq1q2 ÷ r2 = (9.0 × 109 × 2 × 10-6 × 8 × 10-6) ÷ 0.22 = 3.6 N.

Q14. What is the magnitude of the electric field 0.2 m from a point charge of 2 µC in a vacuum? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 450000 N/C
  • (B) 900000 N/C
  • (C) 225000 N/C
  • (D) 1.35 × 106 N/C
Show Answer

Answer: A – E = kQ ÷ r2 = (9.0 × 109 × 2 × 10-6) ÷ 0.22 = 450000 N/C.

Q15. What is the magnitude of the electric field 0.5 m from a point charge of 2 µC in a vacuum? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 72000 N/C
  • (B) 144000 N/C
  • (C) 36000 N/C
  • (D) 216000 N/C
Show Answer

Answer: A – E = kQ ÷ r2 = (9.0 × 109 × 2 × 10-6) ÷ 0.52 = 72000 N/C.

Q16. A small charge of 5 µC is placed in a uniform electric field of 3000 N/C. What is the magnitude of the electric force on the charge?

  • (A) 0.03 N
  • (B) 0.01 N
  • (C) 7.5 × 10-3 N
  • (D) 0.04 N
Show Answer

Answer: B – F = qE = (5 × 10-6) × 3000 = 0.01 N.

Q17. What is the electric potential 0.2 m from a point charge of 2 µC in a vacuum? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 180000 V
  • (B) 45000 V
  • (C) 270000 V
  • (D) 90000 V
Show Answer

Answer: D – V = kQ ÷ r = (9.0 × 109 × 2 × 10-6) ÷ 0.2 = 90000 V.

Q18. Two point charges of 3 µC and 6 µC are 0.1 m apart in a vacuum. What is their electric potential energy? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 3.24 J
  • (B) 1.62 J
  • (C) 0.81 J
  • (D) 4.86 J
Show Answer

Answer: B – U = kq1q2 ÷ r = (9.0 × 109 × 3 × 10-6 × 6 × 10-6) ÷ 0.1 = 1.62 J.

Unit 11: Electric Circuits

Q19. An ideal ammeter is used to measure the current through a resistor. How should it be connected?

  • (A) In parallel with the resistor
  • (B) Across the battery only
  • (C) In series with the resistor
  • (D) In place of the battery
Show Answer

Answer: C – An ammeter measures current through a branch, so it must be in series with that branch, and an ideal ammeter has negligible resistance.

Q20. A potential difference of 12 V is applied across a resistor of 6 Ω. What current flows through it?

  • (A) 4 A
  • (B) 2 A
  • (C) 1 A
  • (D) 6 A
Show Answer

Answer: B – I = V ÷ R = 12 ÷ 6 = 2 A.

Q21. Resistors of 12 Ω, 8 Ω, and 6 Ω are connected in series. What is their equivalent resistance?

  • (A) 52 Ω
  • (B) 26 Ω
  • (C) 13 Ω
  • (D) 78 Ω
Show Answer

Answer: B – In series, Req = R1 + R2 + R3 = 12 + 8 + 6 = 26 Ω.

Q22. Resistors of 12 Ω and 6 Ω are connected in parallel. What is their equivalent resistance?

  • (A) 8 Ω
  • (B) 2 Ω
  • (C) 12 Ω
  • (D) 4 Ω
Show Answer

Answer: D – 1 ÷ Req = 1 ÷ 12 + 1 ÷ 6, so Req = (12 × 6) ÷ (12 + 6) = 4 Ω.

Q23. A current of 2 A flows through a 20 Ω resistor. What power is dissipated in the resistor?

  • (A) 80 W
  • (B) 160 W
  • (C) 40 W
  • (D) 240 W
Show Answer

Answer: A – P = I2R = 22 × 20 = 80 W.

Q24. Resistors of 12 Ω and 4 Ω are connected in parallel, and this combination is connected in series with a 5 Ω resistor. What is the equivalent resistance of the whole combination?

  • (A) 8 Ω
  • (B) 16 Ω
  • (C) 4 Ω
  • (D) 24 Ω
Show Answer

Answer: A – Parallel part: (12 × 4) ÷ (12 + 4) = 3 Ω. Adding the series resistor: 3 + 5 = 8 Ω.

Q25. A capacitor of capacitance 20 µF is connected across a 100 V battery. What charge is stored on the capacitor?

  • (A) 4000 µC
  • (B) 2000 µC
  • (C) 1000 µC
  • (D) 6000 µC
Show Answer

Answer: B – Q = CV = 20 × 100 = 2000 µC.

Q26. A 10 µF capacitor is charged to 50 V. How much energy is stored in it, in millijoules?

  • (A) 25 mJ
  • (B) 6.25 mJ
  • (C) 12.5 mJ
  • (D) 37.5 mJ
Show Answer

Answer: C – U = 0.5CV2 = 0.5 × 10 × 10-6 × 502 = 0.01 J = 12.5 mJ.

Q27. Capacitors of 10 µF and 5 µF are connected in parallel. What is their equivalent capacitance?

  • (A) 15 µF
  • (B) 30 µF
  • (C) 7.5 µF
  • (D) 45 µF
Show Answer

Answer: A – In parallel, Ceq = C1 + C2 = 10 + 5 = 15 µF.

Unit 12: Magnetism and Electromagnetism

Q28. A straight wire of length 0.5 m carries a current of 12 A perpendicular to a uniform magnetic field of 0.5 T. What is the magnitude of the magnetic force on the wire?

  • (A) 3 N
  • (B) 6 N
  • (C) 1.5 N
  • (D) 9 N
Show Answer

Answer: A – F = ILB = 12 × 0.5 × 0.5 = 3 N.

Q29. A long solenoid has 400 turns per metre and carries a current of 2 A. What is the magnetic field strength inside the solenoid, in millitesla? (Use μ0 = 4π × 10-7 T·m/A.)

  • (A) 2.01 mT
  • (B) 0.5 mT
  • (C) 3.02 mT
  • (D) 1.01 mT
Show Answer

Answer: D – B = μ0nI = (4π × 10-7) × 400 × 2 = 1.01 mT.

Q30. A long straight wire carries a current of 20 A. What is the magnetic field strength at a distance of 0.25 m from the wire, in microtesla? (Use μ0 = 4π × 10-7 T·m/A.)

  • (A) 32 µT
  • (B) 8 µT
  • (C) 48 µT
  • (D) 16 µT
Show Answer

Answer: D – B = μ0I ÷ (2πr) = (4π × 10-7 × 20) ÷ (2π × 0.25) = 16 µT.


Q31. A flat coil of area 0.02 m2 sits in a uniform magnetic field of 0.8 T. The field makes an angle of 60° with the normal to the coil. What is the magnetic flux through the coil?

  • (A) 0.02 Wb
  • (B) 4 × 10-3 Wb
  • (C) 8 × 10-3 Wb
  • (D) 0.01 Wb
Show Answer

Answer: C – Φ = BA cos θ = 0.8 × 0.02 × 0.5 = 8 × 10-3 Wb.

Q32. A coil with 10 turns experiences a change in magnetic flux of 0.04 Wb in 0.05 s. What is the magnitude of the induced emf in the coil?

  • (A) 8 V
  • (B) 16 V
  • (C) 4 V
  • (D) 24 V
Show Answer

Answer: A – Magnitude of emf = N ΔΦ ÷ Δt = 10 × 0.04 ÷ 0.05 = 8 V.

Q33. A conducting rod of length 0.2 m moves at 5 m/s perpendicular to a uniform magnetic field of 0.8 T. What is the motional emf induced between its ends?

  • (A) 1.6 V
  • (B) 0.8 V
  • (C) 0.4 V
  • (D) 2.4 V
Show Answer

Answer: B – Motional emf = BLv = 0.8 × 0.2 × 5 = 0.8 V.

Q34. A rectangular coil with 20 turns, area 0.05 m2, carries 2 A in a magnetic field of 0.5 T. The angle in the torque formula is 30°. What is the magnitude of the magnetic torque on the coil?

  • (A) 1 N·m
  • (B) 0.25 N·m
  • (C) 1.5 N·m
  • (D) 0.5 N·m
Show Answer

Answer: D – τ = NIAB sin θ = 20 × 2 × 0.05 × 0.5 × 0.5 = 0.5 N·m.

Q35. A proton moves parallel to a uniform 0.9 T magnetic field. What is the magnitude of the magnetic force acting on the proton?

  • (A) Maximum and in the direction of motion
  • (B) Maximum and opposite to the direction of motion
  • (C) Zero
  • (D) Equal to qvB in the field direction
Show Answer

Answer: C – F = qvB sin θ, and θ = 0 for motion parallel to the field, so sin θ = 0 and the force is zero.

Unit 13: Geometric Optics

Q36. A ray travels from air (refractive index 1) into water (refractive index 1.33) with sin of the angle of incidence equal to 0.8. What is sin of the angle of refraction?

  • (A) 0.8
  • (B) 0.3
  • (C) 0.4
  • (D) 0.6
Show Answer

Answer: D – Snell: n1 sin θ1 = n2 sin θ2, so sin θ2 = 1 × 0.8 ÷ 1.33 = 0.6.

Q37. Light travels at 2.25 × 108 m/s in a sample of diamond-like material. What is the refractive index of the sample? (Use c = 3.00 × 108 m/s.)

  • (A) 2.67
  • (B) 0.67
  • (C) 4
  • (D) 1.33
Show Answer

Answer: D – n = c ÷ v = (3.00 × 108) ÷ (2.25 × 108) = 1.33.

Q38. A transparent material has refractive index 2. At what speed does light travel in it, in units of 108 m/s? (Use c = 3.00 × 108 m/s.)

  • (A) 3 × 108 m/s
  • (B) 1.5 × 108 m/s
  • (C) 0.75 × 108 m/s
  • (D) 4.5 × 108 m/s
Show Answer

Answer: B – v = c ÷ n = 3.00 ÷ 2 = 1.5, so the speed is 1.5 × 108 m/s.

Q39. Light in a medium of refractive index 2 strikes the boundary with a medium of refractive index 1. What is sin of the critical angle for total internal reflection?

  • (A) 0.25
  • (B) 0.5
  • (C) 0.7
  • (D) 0.6
Show Answer

Answer: B – sin θc = n2 ÷ n1 = 1 ÷ 2 = 0.5.

Q40. A spherical mirror has a radius of curvature of 46 cm. What is its focal length?

  • (A) 46 cm
  • (B) 23 cm
  • (C) 11.5 cm
  • (D) 69 cm
Show Answer

Answer: B – f = R ÷ 2 = 46 ÷ 2 = 23 cm.

Q41. A converging lens of focal length 15 cm forms an image 45 cm from the lens. How far from the lens is the object?

  • (A) 22.5 cm
  • (B) 45 cm
  • (C) 11.25 cm
  • (D) 67.5 cm
Show Answer

Answer: A – 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ do = 1 ÷ 15 – 1 ÷ 45, giving do = 22.5 cm.

Q42. An object is placed beyond the centre of curvature of a concave mirror. What kind of image is formed?

  • (A) Virtual, upright, and magnified
  • (B) Real, inverted, and diminished
  • (C) Real, upright, and magnified
  • (D) Virtual, inverted, and diminished
Show Answer

Answer: B – For an object beyond C, a concave mirror forms a real, inverted image between F and C that is smaller than the object.

Q43. An object is placed 40 cm from a concave mirror of focal length 20 cm. What is the image distance?

  • (A) 80 cm
  • (B) 20 cm
  • (C) 40 cm
  • (D) 120 cm
Show Answer

Answer: C – 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ di = 1 ÷ 20 – 1 ÷ 40, giving di = 40 cm.

Unit 14: Waves, Sound, and Physical Optics

Q44. Light of wavelength 500 nm passes through double slits 0.25 mm apart onto a screen 2 m away. What is the spacing between adjacent bright fringes, in millimetres?

  • (A) 8 mm
  • (B) 4 mm
  • (C) 2 mm
  • (D) 12 mm
Show Answer

Answer: B – Fringe spacing Δy = λL ÷ d = 500 × 10-9 × 2 ÷ (0.25 × 10-3) = 4 mm.

Q45. A source emitting 500 Hz moves away from a stationary observer at 85 m/s. The speed of sound is 340 m/s. What frequency does the observer hear?

  • (A) 800 Hz
  • (B) 200 Hz
  • (C) 400 Hz
  • (D) 1200 Hz
Show Answer

Answer: C – For a source moving away, f′ = f × v ÷ (v + vs) = 500 × 340 ÷ (340 + 85) = 400 Hz.

Q46. A wave completes one full oscillation in 0.25 s. What is its frequency?

  • (A) 4 Hz
  • (B) 8 Hz
  • (C) 2 Hz
  • (D) 12 Hz
Show Answer

Answer: A – f = 1 ÷ T = 1 ÷ 0.25 = 4 Hz.

Q47. Sound travels from air into water. What happens to its frequency?

  • (A) It increases
  • (B) It decreases
  • (C) It stays the same
  • (D) It becomes zero
Show Answer

Answer: C – Frequency is set by the source and does not change when a wave enters a new medium; speed and wavelength change instead.

Q48. A pipe closed at one end has length 0.5 m and sound travels at 340 m/s inside it. What is the frequency of its fifth harmonic? (Only odd harmonics exist for this pipe.)

  • (A) 1700 Hz
  • (B) 425 Hz
  • (C) 850 Hz
  • (D) 2550 Hz
Show Answer

Answer: C – For a closed pipe, fn = nv ÷ (4L) = 5 × 340 ÷ (4 × 0.5) = 850 Hz.

Q49. A sound wave travels at 400 m/s with frequency 400 Hz. What is its wavelength?

  • (A) 2 m
  • (B) 0.5 m
  • (C) 3 m
  • (D) 1 m
Show Answer

Answer: D – λ = v ÷ f = 400 ÷ 400 = 1 m.

Q50. A string of length 0.6 m is fixed at both ends. Waves travel along it at 400 m/s. What is the frequency of its third harmonic?

  • (A) 2000 Hz
  • (B) 500 Hz
  • (C) 1000 Hz
  • (D) 3000 Hz
Show Answer

Answer: C – For a string fixed at both ends, fn = nv ÷ (2L) = 3 × 400 ÷ (2 × 0.6) = 1000 Hz.

Q51. An open pipe of length 0.5 m is filled with air in which sound travels at 340 m/s. What is its fundamental frequency?

  • (A) 680 Hz
  • (B) 170 Hz
  • (C) 1020 Hz
  • (D) 340 Hz
Show Answer

Answer: D – For an open pipe, f1 = v ÷ (2L) = 340 ÷ (2 × 0.5) = 340 Hz.

Unit 15: Modern Physics

Q52. How much energy is equivalent to a mass of 2 kg? (Use c = 3.00 × 108 m/s.)

  • (A) 3.6 × 1017 J
  • (B) 1.8 × 1017 J
  • (C) 9 × 1016 J
  • (D) 5.4 × 1017 J
Show Answer

Answer: B – E = mc2 = 2 × (3.00 × 108)2 = 1.8 × 1017 J.

Q53. A photon has energy 3.98 × 10-19 J. What is its frequency, in units of 1014 Hz? (Use h = 6.63 × 10-34 J·s.)

  • (A) 12 × 1014 Hz
  • (B) 3 × 1014 Hz
  • (C) 6 × 1014 Hz
  • (D) 18 × 1014 Hz
Show Answer

Answer: C – f = E ÷ h = (3.98 × 10-19) ÷ (6.63 × 10-34) = 6 × 1014 Hz, so the value is 6 in units of 1014 Hz.

Q54. A photon has frequency 6 × 1014 Hz. What is its energy? (Use h = 6.63 × 10-34 J·s.)

  • (A) 7.96 × 10-19 J
  • (B) 1.99 × 10-19 J
  • (C) 1.19 × 10-18 J
  • (D) 3.98 × 10-19 J
Show Answer

Answer: D – E = hf = (6.63 × 10-34) × (6 × 1014) = 3.98 × 10-19 J.

Q55. A photon has frequency 8 × 1014 Hz. What is its energy? (Use h = 6.63 × 10-34 J·s.)

  • (A) 5.3 × 10-19 J
  • (B) 1.06 × 10-18 J
  • (C) 2.65 × 10-19 J
  • (D) 1.59 × 10-18 J
Show Answer

Answer: A – E = hf = (6.63 × 10-34) × (8 × 1014) = 5.3 × 10-19 J.

Q56. What is the energy of a photon of wavelength 650 nm? (Use h = 6.63 × 10-34 J·s and c = 3.00 × 108 m/s.)

  • (A) 6.12 × 10-19 J
  • (B) 3.06 × 10-19 J
  • (C) 1.53 × 10-19 J
  • (D) 9.18 × 10-19 J
Show Answer

Answer: B – E = hc ÷ λ = (6.63 × 10-34 × 3.00 × 108) ÷ (650 × 10-9) = 3.06 × 10-19 J.

Q57. Light of photon energy 3.5 eV falls on a metal with work function 1.4 eV. What is the maximum kinetic energy of the emitted photoelectrons?

  • (A) 2.1 eV
  • (B) 4.2 eV
  • (C) 1.05 eV
  • (D) 6.3 eV
Show Answer

Answer: A – Kmax = hf – Φ = 3.5 – 1.4 = 2.1 eV.

Q58. A particle has momentum 13.26 × 10-24 kg·m/s. What is its de Broglie wavelength, in units of 10-10 m? (Use h = 6.63 × 10-34 J·s.)

  • (A) 1
  • (B) 0.25
  • (C) 0.5
  • (D) 1.5
Show Answer

Answer: C – λ = h ÷ p = (6.63 × 10-34) ÷ (13.26 × 10-24) = 0.5 × 10-10 m, so the value is 0.5 in units of 10-10 m.

Q59. In the Bohr model, the energy of level n is En = -13.6 ÷ n2 eV. What is the energy of level n = 2?

  • (A) -3.4 eV
  • (B) -6.8 eV
  • (C) -1.7 eV
  • (D) -10.2 eV
Show Answer

Answer: A – E2 = -13.6 ÷ 22 = -13.6 ÷ 4 = -3.4 eV.

Q60. An electron in a hydrogen atom drops from level n = 4 to level n = 2. What is the energy of the photon emitted?

  • (A) 5.1 eV
  • (B) 1.27 eV
  • (C) 2.55 eV
  • (D) 7.65 eV
Show Answer

Answer: C – Photon energy = Ehigh – Elow = (-13.6 ÷ 42) – (-13.6 ÷ 22) = 2.55 eV.

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