AP Physics 2 Practice Test 11 – 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. A gas expands at a constant pressure of 500 kPa, increasing its volume by 0.03 m3. How much work does the gas do?

  • (A) 30000 J
  • (B) 7500 J
  • (C) 15000 J
  • (D) 45000 J
Show Answer

Answer: C – At constant pressure, W = pΔV = 500000 × 0.03 = 15000 J.

Q2. A 2.5 kg block of aluminium (specific heat capacity 900 J/(kg·K)) is heated so its temperature rises by 10 K. How much heat is supplied?

  • (A) 22500 J
  • (B) 45000 J
  • (C) 11250 J
  • (D) 67500 J
Show Answer

Answer: A – Q = mcΔT = 2.5 × 900 × 10 = 22500 J.

Q3. How much heat is needed to turn water at 100°C into steam (specific latent heat of vaporisation 2,260,000 J/kg) if the mass involved is 1.5 kg?

  • (A) 6.78 × 106 J
  • (B) 1.7 × 106 J
  • (C) 3.39 × 106 J
  • (D) 1.02 × 107 J
Show Answer

Answer: C – Q = mL = 1.5 × 2.26 × 106 = 3.39 × 106 J.

Q4. A reversible heat engine operates between a hot reservoir at 1200 K and a cold reservoir at 400 K. What is its maximum possible efficiency?

  • (A) 33.33 %
  • (B) 66.67 %
  • (C) 76.67 %
  • (D) 56.67 %
Show Answer

Answer: B – Carnot efficiency = 1 – Tc ÷ Th = 1 – 400 ÷ 1200 = 66.67%.

Q5. A heat engine takes in 10000 J of heat in one cycle and does 200 J of useful work. How much heat does it reject to the cold reservoir?

  • (A) 19600 J
  • (B) 4900 J
  • (C) 29400 J
  • (D) 9800 J
Show Answer

Answer: D – Energy conservation: Qh = W + Qc, so Qc = 10000 – 200 = 9800 J.

Q6. A gas at 100 kPa occupies 5 L at 600 K. Its pressure changes to 200 kPa and its volume changes to 2.5 L. What is its new temperature?

  • (A) 600 K
  • (B) 1200 K
  • (C) 300 K
  • (D) 1800 K
Show Answer

Answer: A – p1V1 ÷ T1 = p2V2 ÷ T2, so T2 = 200 × 2.5 × 600 ÷ (100 × 5) = 600 K.

Q7. An ideal gas expands adiabatically (no heat transfer) and does work on its surroundings. What happens to its internal energy and temperature?

  • (A) Internal energy increases and temperature increases
  • (B) Internal energy decreases and temperature decreases
  • (C) Internal energy stays constant and temperature stays constant
  • (D) Internal energy decreases and temperature increases
Show Answer

Answer: B – With Q = 0, the first law gives ΔU = -W, so doing work lowers internal energy, and for an ideal gas temperature falls with internal energy.

Q8. A pressurised storage vessel holds 1 mol of an ideal gas at 350 K in a volume of 0.03 m3. What is the pressure of the gas? (Use R = 8.31 J/(mol·K).)

  • (A) 116340 Pa
  • (B) 232680 Pa
  • (C) 58170 Pa
  • (D) 349020 Pa
Show Answer

Answer: A – pV = nRT gives p = nRT ÷ V = (1 × 8.31 × 350) ÷ 0.03 = 116340 Pa.

Q9. An ideal gas sample contains 3 mol at 400 K and a pressure of 150 kPa. What is its volume? (Use R = 8.31 J/(mol·K).)

  • (A) 0.07 m3
  • (B) 0.13 m3
  • (C) 0.03 m3
  • (D) 0.2 m3
Show Answer

Answer: A – pV = nRT gives V = nRT ÷ p = (3 × 8.31 × 400) ÷ 150000 = 0.07 m3.

Unit 10: Electric Force, Field, and Potential

Q10. Two point charges of 2 µC and 4 µC are 0.5 m apart in a vacuum. What is their electric potential energy? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 0.14 J
  • (B) 0.29 J
  • (C) 0.07 J
  • (D) 0.43 J
Show Answer

Answer: A – U = kq1q2 ÷ r = (9.0 × 109 × 2 × 10-6 × 4 × 10-6) ÷ 0.5 = 0.14 J.

Q11. A charge of 5 µC is moved through a potential difference of 200 V. How much work is done by the electric field if the charge moves to a lower potential energy state by that full amount?

  • (A) 2 × 10-3 J
  • (B) 1 × 10-3 J
  • (C) 5 × 10-4 J
  • (D) 3 × 10-3 J
Show Answer

Answer: B – W = qΔV = (5 × 10-6) × 200 = 1 × 10-3 J.

Q12. Two parallel plates are 0.1 m apart and have a potential difference of 600 V between them. What is the magnitude of the uniform electric field between the plates?

  • (A) 12000 N/C
  • (B) 3000 N/C
  • (C) 18000 N/C
  • (D) 6000 N/C
Show Answer

Answer: D – E = V ÷ d = 600 ÷ 0.1 = 6000 N/C.

Q13. Which quantity is the same at every point on an equipotential surface?

  • (A) Electric field strength
  • (B) Electric force on a test charge
  • (C) Electric potential
  • (D) Electric potential energy of any charge regardless of its sign
Show Answer

Answer: C – An equipotential surface is defined by constant electric potential.

Q14. A charge of 5 µC sits at a point where the electric potential is 1000 V. What is its electric potential energy?

  • (A) 1 × 10-2 J
  • (B) 5 × 10-3 J
  • (C) 2.5 × 10-3 J
  • (D) 0.01 J
Show Answer

Answer: B – U = qV = (5 × 10-6) × 1000 = 5 × 10-3 J.

Q15. Two small charged spheres carry charges of 10 µC and 3 µC and are 0.25 m apart in a vacuum. What is the magnitude of the electrostatic force between them? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 4.32 N
  • (B) 8.64 N
  • (C) 2.16 N
  • (D) 12.96 N
Show Answer

Answer: A – F = kq1q2 ÷ r2 = (9.0 × 109 × 10 × 10-6 × 3 × 10-6) ÷ 0.252 = 4.32 N.

Q16. One of two point charges is doubled and the distance between them is halved. By what factor does the electrostatic force change?

  • (A) It becomes 2 times its original value
  • (B) It becomes 4 times its original value
  • (C) It becomes 1/4 of its original value
  • (D) It becomes 8 times its original value
Show Answer

Answer: D – Force is proportional to q ÷ r2, so the factor is 2 ÷ (0.5)2 = 2 ÷ 0.25 = 8.

Q17. What is the magnitude of the electric field 0.25 m from a point charge of 2 µC in a vacuum? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 576000 N/C
  • (B) 288000 N/C
  • (C) 144000 N/C
  • (D) 864000 N/C
Show Answer

Answer: B – E = kQ ÷ r2 = (9.0 × 109 × 2 × 10-6) ÷ 0.252 = 288000 N/C.

Q18. A small charge of 3 µC is placed in a uniform electric field of 1000 N/C. What is the magnitude of the electric force on the charge?

  • (A) 3 × 10-3 N
  • (B) 6 × 10-3 N
  • (C) 1.5 × 10-3 N
  • (D) 9 × 10-3 N
Show Answer

Answer: A – F = qE = (3 × 10-6) × 1000 = 3 × 10-3 N.

Unit 11: Electric Circuits

Q19. Capacitors of 6 µF and 3 µF are connected in series. What is their equivalent capacitance?

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

Answer: A – 1 ÷ Ceq = 1 ÷ 6 + 1 ÷ 3, so Ceq = (6 × 3) ÷ (6 + 3) = 2 µF.

Q20. A 36 V battery is connected across resistors of 12 Ω and 4 Ω in series. What is the potential difference across the 12 Ω resistor?

  • (A) 54 V
  • (B) 27 V
  • (C) 13.5 V
  • (D) 81 V
Show Answer

Answer: B – Total resistance = 12 + 4 = 16 Ω. Current = 36 ÷ 16 = 2.25 A. Voltage across the first resistor = 2.25 × 12 = 27 V.

Q21. Two identical lamps are connected in parallel across a battery, then a third identical lamp is added in parallel. What happens to the total current drawn from the battery?

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

Answer: B – Adding a parallel branch lowers the equivalent resistance, so the total current from the battery increases.

Q22. A potential difference of 12 V drives a current of 6 A through a resistor. What is its resistance?

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

Answer: B – R = V ÷ I = 12 ÷ 6 = 2 Ω.

Q23. Resistors of 3 Ω, 5 Ω, and 10 Ω are connected in series. What is their equivalent resistance?

  • (A) 36 Ω
  • (B) 18 Ω
  • (C) 9 Ω
  • (D) 54 Ω
Show Answer

Answer: B – In series, Req = R1 + R2 + R3 = 3 + 5 + 10 = 18 Ω.

Q24. Resistors of 20 Ω and 5 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 ÷ 20 + 1 ÷ 5, so Req = (20 × 5) ÷ (20 + 5) = 4 Ω.

Q25. A current of 4 A flows through a 20 Ω resistor. What power is dissipated in the resistor?

  • (A) 640 W
  • (B) 320 W
  • (C) 160 W
  • (D) 960 W
Show Answer

Answer: B – P = I2R = 42 × 20 = 320 W.

Q26. Resistors of 10 Ω and 10 Ω are connected in parallel, and this combination is connected in series with a 2 Ω resistor. What is the equivalent resistance of the whole combination?

  • (A) 7 Ω
  • (B) 14 Ω
  • (C) 3.5 Ω
  • (D) 21 Ω
Show Answer

Answer: A – Parallel part: (10 × 10) ÷ (10 + 10) = 5 Ω. Adding the series resistor: 5 + 2 = 7 Ω.

Q27. A capacitor of capacitance 10 µF is connected across a 30 V battery. What charge is stored on the capacitor?

  • (A) 300 µC
  • (B) 600 µC
  • (C) 150 µC
  • (D) 900 µC
Show Answer

Answer: A – Q = CV = 10 × 30 = 300 µC.

Unit 12: Magnetism and Electromagnetism

Q28. The current in a straight wire is tripled and its length in a magnetic field is halved. By what factor does the magnetic force on the wire change?

  • (A) It becomes 3 times its original value
  • (B) It becomes 6 times its original value
  • (C) It stays the same
  • (D) It becomes 1.5 times its original value
Show Answer

Answer: D – F is proportional to IL, so the factor is 3 × 0.5 = 1.5.

Q29. A particle with charge 4 µC moves at 8000 m/s perpendicular to a uniform magnetic field of 0.5 T. What is the magnitude of the magnetic force on the particle, in millinewtons?

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

Answer: B – F = qvB = (4 × 10-6) × 8000 × 0.5 = 0.02 N = 16 mN.

Q30. A straight wire of length 0.4 m carries a current of 4 A perpendicular to a uniform magnetic field of 0.1 T. What is the magnitude of the magnetic force on the wire?

  • (A) 0.32 N
  • (B) 0.08 N
  • (C) 0.48 N
  • (D) 0.16 N
Show Answer

Answer: D – F = ILB = 4 × 0.4 × 0.1 = 0.16 N.


Q31. A long solenoid has 400 turns per metre and carries a current of 3 A. What is the magnetic field strength inside the solenoid, in millitesla? (Use μ0 = 4π × 10-7 T·m/A.)

  • (A) 3.02 mT
  • (B) 0.75 mT
  • (C) 4.52 mT
  • (D) 1.51 mT
Show Answer

Answer: D – B = μ0nI = (4π × 10-7) × 400 × 3 = 1.51 mT.

Q32. A long straight wire carries a current of 5 A. What is the magnetic field strength at a distance of 0.05 m from the wire, in microtesla? (Use μ0 = 4π × 10-7 T·m/A.)

  • (A) 40 µT
  • (B) 10 µT
  • (C) 20 µT
  • (D) 60 µT
Show Answer

Answer: C – B = μ0I ÷ (2πr) = (4π × 10-7 × 5) ÷ (2π × 0.05) = 20 µT.

Q33. A flat coil of area 0.05 m2 sits in a uniform magnetic field of 2 T. The field makes an angle of 0° with the normal to the coil. What is the magnetic flux through the coil?

  • (A) 0.2 Wb
  • (B) 0.05 Wb
  • (C) 0.1 Wb
  • (D) 0.3 Wb
Show Answer

Answer: C – Φ = BA cos θ = 2 × 0.05 × 1 = 0.1 Wb.

Q34. A coil with 50 turns experiences a change in magnetic flux of 0.02 Wb in 0.02 s. What is the magnitude of the induced emf in the coil?

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

Answer: B – Magnitude of emf = N ΔΦ ÷ Δt = 50 × 0.02 ÷ 0.02 = 50 V.

Q35. A conducting rod of length 0.5 m moves at 5 m/s perpendicular to a uniform magnetic field of 0.2 T. What is the motional emf induced between its ends?

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

Answer: B – Motional emf = BLv = 0.2 × 0.5 × 5 = 0.5 V.

Unit 13: Geometric Optics

Q36. An object is 30 cm from a mirror and its image forms 10 cm from the mirror. What is the lateral magnification?

  • (A) -0.67
  • (B) -0.17
  • (C) -1
  • (D) -0.33
Show Answer

Answer: D – m = -di ÷ do = -10 ÷ 30 = -0.33.

Q37. An object is placed 90 cm from a converging lens of focal length 30 cm. What is the image distance?

  • (A) 90 cm
  • (B) 22.5 cm
  • (C) 45 cm
  • (D) 135 cm
Show Answer

Answer: C – 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ di = 1 ÷ 30 – 1 ÷ 90, giving di = 45 cm.

Q38. A ray travels from air (refractive index 1) into glass (refractive index 1.5) with sin of the angle of incidence equal to 0.9. What is sin of the angle of refraction?

  • (A) 0.9
  • (B) 0.3
  • (C) 0.4
  • (D) 0.6
Show Answer

Answer: D – Snell: n1 sin θ1 = n2 sin θ2, so sin θ2 = 1 × 0.9 ÷ 1.5 = 0.6.

Q39. Light travels at 1 × 108 m/s in a sample of glass. What is the refractive index of the sample? (Use c = 3.00 × 108 m/s.)

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

Answer: C – n = c ÷ v = (3.00 × 108) ÷ (1 × 108) = 3.

Q40. A transparent material has refractive index 1.33. At what speed does light travel in it, in units of 108 m/s? (Use c = 3.00 × 108 m/s.)

  • (A) 4.51 × 108 m/s
  • (B) 1.13 × 108 m/s
  • (C) 6.77 × 108 m/s
  • (D) 2.26 × 108 m/s
Show Answer

Answer: D – v = c ÷ n = 3.00 ÷ 1.33 = 2.26, so the speed is 2.26 × 108 m/s.

Q41. Light in a medium of refractive index 1.5 strikes the boundary with a medium of refractive index 1. What is sin of the critical angle for total internal reflection?

  • (A) 0.33
  • (B) 0.87
  • (C) 0.77
  • (D) 0.67
Show Answer

Answer: D – sin θc = n2 ÷ n1 = 1 ÷ 1.5 = 0.67.

Q42. A spherical mirror has a radius of curvature of 44 cm. What is its focal length?

  • (A) 44 cm
  • (B) 11 cm
  • (C) 66 cm
  • (D) 22 cm
Show Answer

Answer: D – f = R ÷ 2 = 44 ÷ 2 = 22 cm.

Q43. A converging lens of focal length 10 cm forms an image 30 cm from the lens. How far from the lens is the object?

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

Answer: C – 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ do = 1 ÷ 10 – 1 ÷ 30, giving do = 15 cm.

Unit 14: Waves, Sound, and Physical Optics

Q44. Two tuning forks of frequencies 600 Hz and 605 Hz sound together. What is the beat frequency heard?

  • (A) 5 Hz
  • (B) 10 Hz
  • (C) 2.5 Hz
  • (D) 15 Hz
Show Answer

Answer: A – Beat frequency = |f1 – f2| = |600 – 605| = 5 Hz.

Q45. In a double-slit experiment, light of wavelength 500 nm passes through slits 0.1 mm apart and falls on a screen 1 m away. How far from the central maximum is the second bright fringe, in millimetres?

  • (A) 20 mm
  • (B) 10 mm
  • (C) 5 mm
  • (D) 30 mm
Show Answer

Answer: B – y = mλL ÷ d = 2 × 500 × 10-9 × 1 ÷ (0.1 × 10-3) = 10 mm.

Q46. Light of wavelength 650 nm passes through double slits 0.26 mm apart onto a screen 2 m away. What is the spacing between adjacent bright fringes, in millimetres?

  • (A) 5 mm
  • (B) 10 mm
  • (C) 2.5 mm
  • (D) 15 mm
Show Answer

Answer: A – Fringe spacing Δy = λL ÷ d = 650 × 10-9 × 2 ÷ (0.26 × 10-3) = 5 mm.

Q47. A source emitting 400 Hz moves away from a stationary observer at 68 m/s. The speed of sound is 340 m/s. What frequency does the observer hear?

  • (A) 666.67 Hz
  • (B) 333.33 Hz
  • (C) 166.67 Hz
  • (D) 1000 Hz
Show Answer

Answer: B – For a source moving away, f′ = f × v ÷ (v + vs) = 400 × 340 ÷ (340 + 68) = 333.33 Hz.

Q48. A wave completes one full oscillation in 0.5 s. What is its frequency?

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

Answer: A – f = 1 ÷ T = 1 ÷ 0.5 = 2 Hz.

Q49. In a standing wave on a string fixed at both ends, what is true at a node?

  • (A) The displacement is always maximum
  • (B) The string moves fastest there
  • (C) The displacement is always zero
  • (D) The wavelength is zero there
Show Answer

Answer: C – A node is a point of zero displacement at all times in a standing wave.

Q50. A pipe closed at one end has length 0.85 m and sound travels at 340 m/s inside it. What is the frequency of its third harmonic? (Only odd harmonics exist for this pipe.)

  • (A) 600 Hz
  • (B) 150 Hz
  • (C) 300 Hz
  • (D) 900 Hz
Show Answer

Answer: C – For a closed pipe, fn = nv ÷ (4L) = 3 × 340 ÷ (4 × 0.85) = 300 Hz.

Q51. A sound wave travels at 400 m/s with frequency 1000 Hz. What is its wavelength?

  • (A) 0.8 m
  • (B) 0.2 m
  • (C) 1.2 m
  • (D) 0.4 m
Show Answer

Answer: D – λ = v ÷ f = 400 ÷ 1000 = 0.4 m.

Unit 15: Modern Physics

Q52. An electron in a hydrogen atom drops from level n = 5 to level n = 2. What is the energy of the photon emitted?

  • (A) 2.86 eV
  • (B) 5.72 eV
  • (C) 1.43 eV
  • (D) 8.58 eV
Show Answer

Answer: A – Photon energy = Ehigh – Elow = (-13.6 ÷ 52) – (-13.6 ÷ 22) = 2.86 eV.

Q53. A radioactive sample initially has mass 200 g. After 4 half-lives, what mass remains undecayed?

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

Answer: C – After 4 half-lives the fraction remaining is (1/2)4 = 0.06, so the mass remaining is 200 × 0.06 = 12.5 g.

Q54. How much energy is equivalent to a mass of 0.25 kg? (Use c = 3.00 × 108 m/s.)

  • (A) 4.5 × 1016 J
  • (B) 1.12 × 1016 J
  • (C) 6.75 × 1016 J
  • (D) 2.25 × 1016 J
Show Answer

Answer: D – E = mc2 = 0.25 × (3.00 × 108)2 = 2.25 × 1016 J.

Q55. A photon has energy 4.64 × 10-19 J. What is its frequency, in units of 1014 Hz? (Use h = 6.63 × 10-34 J·s.)

  • (A) 7 × 1014 Hz
  • (B) 14 × 1014 Hz
  • (C) 3.5 × 1014 Hz
  • (D) 21 × 1014 Hz
Show Answer

Answer: A – f = E ÷ h = (4.64 × 10-19) ÷ (6.63 × 10-34) = 7 × 1014 Hz, so the value is 7 in units of 1014 Hz.

Q56. A radioactive sample has a half-life of 10 days. After 20 days, what fraction of the original nuclei remains undecayed?

  • (A) 1/2
  • (B) 1/8
  • (C) 1/4
  • (D) 3/4
Show Answer

Answer: C – 20 days is 2 half-lives, so the fraction remaining is (1/2)2 = 1/4.

Q57. A photon has frequency 5 × 1014 Hz. What is its energy? (Use h = 6.63 × 10-34 J·s.)

  • (A) 6.63 × 10-19 J
  • (B) 1.66 × 10-19 J
  • (C) 9.95 × 10-19 J
  • (D) 3.31 × 10-19 J
Show Answer

Answer: D – E = hf = (6.63 × 10-34) × (5 × 1014) = 3.31 × 10-19 J.

Q58. What is the energy of a photon of wavelength 400 nm? (Use h = 6.63 × 10-34 J·s and c = 3.00 × 108 m/s.)

  • (A) 9.94 × 10-19 J
  • (B) 2.49 × 10-19 J
  • (C) 4.97 × 10-19 J
  • (D) 1.49 × 10-18 J
Show Answer

Answer: C – E = hc ÷ λ = (6.63 × 10-34 × 3.00 × 108) ÷ (400 × 10-9) = 4.97 × 10-19 J.

Q59. Light of photon energy 4.2 eV falls on a metal with work function 1.4 eV. What is the maximum kinetic energy of the emitted photoelectrons?

  • (A) 5.6 eV
  • (B) 1.4 eV
  • (C) 2.8 eV
  • (D) 8.4 eV
Show Answer

Answer: C – Kmax = hf – Φ = 4.2 – 1.4 = 2.8 eV.

Q60. A particle has momentum 19.89 × 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) 0.67
  • (B) 0.17
  • (C) 0.33
  • (D) 1
Show Answer

Answer: C – λ = h ÷ p = (6.63 × 10-34) ÷ (19.89 × 10-24) = 0.33 × 10-10 m, so the value is 0.33 in units of 10-10 m.

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