AP Physics 2 Practice Test 13 – 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 is heated at constant volume. Which quantity is zero for this process?

  • (A) The work done by the gas
  • (B) The heat transferred to the gas
  • (C) The change in internal energy of the gas
  • (D) The change in pressure of the gas
Show Answer

Answer: A – Work is pΔV, and ΔV = 0 at constant volume, so no work is done.

Q2. A rigid glass container holds 6 mol of an ideal gas at 400 K in a volume of 0.03 m3. What is the pressure of the gas? (Use R = 8.31 J/(mol·K).)

  • (A) 1.6 × 106 Pa
  • (B) 398880 Pa
  • (C) 2.39 × 106 Pa
  • (D) 797760 Pa
Show Answer

Answer: D – pV = nRT gives p = nRT ÷ V = (6 × 8.31 × 400) ÷ 0.03 = 797760 Pa.

Q3. An ideal gas sample contains 4 mol at 350 K and a pressure of 400 kPa. What is its volume? (Use R = 8.31 J/(mol·K).)

  • (A) 0.06 m3
  • (B) 0.03 m3
  • (C) 0.01 m3
  • (D) 0.09 m3
Show Answer

Answer: B – pV = nRT gives V = nRT ÷ p = (4 × 8.31 × 350) ÷ 400000 = 0.03 m3.

Q4. A ideal gas in an insulated chamber absorbs 200 J of heat from its surroundings and does 50 J of work on its surroundings. What is the change in its internal energy?

  • (A) 300 J
  • (B) 75 J
  • (C) 150 J
  • (D) 450 J
Show Answer

Answer: C – First law: ΔU = Q – W = 200 – 50 = 150 J.

Q5. A gas expands at a constant pressure of 200 kPa, increasing its volume by 0.08 m3. How much work does the gas do?

  • (A) 32000 J
  • (B) 16000 J
  • (C) 8000 J
  • (D) 48000 J
Show Answer

Answer: B – At constant pressure, W = pΔV = 200000 × 0.08 = 16000 J.

Q6. A 3 kg block of lead (specific heat capacity 240 J/(kg·K)) is heated so its temperature rises by 50 K. How much heat is supplied?

  • (A) 72000 J
  • (B) 36000 J
  • (C) 18000 J
  • (D) 108000 J
Show Answer

Answer: B – Q = mcΔT = 3 × 240 × 50 = 36000 J.

Q7. 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 0.2 kg?

  • (A) 904000 J
  • (B) 226000 J
  • (C) 1.36 × 106 J
  • (D) 452000 J
Show Answer

Answer: D – Q = mL = 0.2 × 2.26 × 106 = 452000 J.

Q8. A reversible heat engine operates between a hot reservoir at 800 K and a cold reservoir at 400 K. What is its maximum possible efficiency?

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

Answer: B – Carnot efficiency = 1 – Tc ÷ Th = 1 – 400 ÷ 800 = 50%.

Q9. A heat engine takes in 1000 J of heat in one cycle and does 250 J of useful work. How much heat does it reject to the cold reservoir?

  • (A) 1500 J
  • (B) 750 J
  • (C) 375 J
  • (D) 2250 J
Show Answer

Answer: B – Energy conservation: Qh = W + Qc, so Qc = 1000 – 250 = 750 J.

Unit 10: Electric Force, Field, and Potential

Q10. The distance between two point charges is tripled, with both charges unchanged. By what factor does the electrostatic force between them change?

  • (A) It becomes 1/3 of its original value
  • (B) It becomes 1/9 of its original value
  • (C) It becomes 3 times its original value
  • (D) It becomes 9 times its original value
Show Answer

Answer: B – Force is proportional to 1 ÷ r2, so tripling r gives a factor of 1 ÷ 32 = 1/9.

Q11. What is the magnitude of the electric field 0.5 m from a point charge of 4 µC in a vacuum? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 288000 N/C
  • (B) 72000 N/C
  • (C) 144000 N/C
  • (D) 432000 N/C
Show Answer

Answer: C – E = kQ ÷ r2 = (9.0 × 109 × 4 × 10-6) ÷ 0.52 = 144000 N/C.

Q12. A small charge of 4 µC is placed in a uniform electric field of 2000 N/C. What is the magnitude of the electric force on the charge?

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

Answer: A – F = qE = (4 × 10-6) × 2000 = 8 × 10-3 N.

Q13. What is the electric potential 0.25 m from a point charge of 6 µC in a vacuum? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 432000 V
  • (B) 108000 V
  • (C) 216000 V
  • (D) 648000 V
Show Answer

Answer: C – V = kQ ÷ r = (9.0 × 109 × 6 × 10-6) ÷ 0.25 = 216000 V.

Q14. Two point charges of 2 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) 2.16 J
  • (B) 0.54 J
  • (C) 3.24 J
  • (D) 1.08 J
Show Answer

Answer: D – U = kq1q2 ÷ r = (9.0 × 109 × 2 × 10-6 × 6 × 10-6) ÷ 0.1 = 1.08 J.

Q15. A charge of 4 µC is moved through a potential difference of 50 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) 4 × 10-4 J
  • (B) 1 × 10-4 J
  • (C) 6 × 10-4 J
  • (D) 2 × 10-4 J
Show Answer

Answer: D – W = qΔV = (4 × 10-6) × 50 = 2 × 10-4 J.

Q16. Two parallel plates are 0.1 m apart and have a potential difference of 240 V between them. What is the magnitude of the uniform electric field between the plates?

  • (A) 4800 N/C
  • (B) 2400 N/C
  • (C) 1200 N/C
  • (D) 7200 N/C
Show Answer

Answer: B – E = V ÷ d = 240 ÷ 0.1 = 2400 N/C.

Q17. A charged particle moves along the direction of a uniform electric field. What happens to its electric potential?

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

Answer: C – Potential decreases in the direction of the electric field, since a positive charge loses potential energy moving that way.

Q18. A charge of 2 µC sits at a point where the electric potential is 500 V. What is its electric potential energy?

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

Answer: D – U = qV = (2 × 10-6) × 500 = 1 × 10-3 J.

Unit 11: Electric Circuits

Q19. A current of 3 A flows through a 25 Ω resistor. What power is dissipated in the resistor?

  • (A) 450 W
  • (B) 112.5 W
  • (C) 225 W
  • (D) 675 W
Show Answer

Answer: C – P = I2R = 32 × 25 = 225 W.

Q20. Resistors of 12 Ω and 6 Ω are connected in parallel, and this combination is connected in series with a 3 Ω resistor. What is the equivalent resistance of the whole combination?

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

Answer: D – Parallel part: (12 × 6) ÷ (12 + 6) = 4 Ω. Adding the series resistor: 4 + 3 = 7 Ω.

Q21. A capacitor of capacitance 4 µF is connected across a 50 V battery. What charge is stored on the capacitor?

  • (A) 400 µC
  • (B) 100 µC
  • (C) 200 µC
  • (D) 600 µC
Show Answer

Answer: C – Q = CV = 4 × 50 = 200 µC.

Q22. A 2 µF capacitor is charged to 50 V. How much energy is stored in it, in millijoules?

  • (A) 5 mJ
  • (B) 1.25 mJ
  • (C) 2.5 mJ
  • (D) 7.5 mJ
Show Answer

Answer: C – U = 0.5CV2 = 0.5 × 2 × 10-6 × 502 = 2.5 × 10-3 J = 2.5 mJ.

Q23. Capacitors of 12 µF and 4 µF are connected in series. What is their equivalent capacitance?

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

Answer: B – 1 ÷ Ceq = 1 ÷ 12 + 1 ÷ 4, so Ceq = (12 × 4) ÷ (12 + 4) = 3 µF.

Q24. A 60 V battery is connected across resistors of 8 Ω and 8 Ω in series. What is the potential difference across the 8 Ω resistor?

  • (A) 60 V
  • (B) 15 V
  • (C) 30 V
  • (D) 90 V
Show Answer

Answer: C – Total resistance = 8 + 8 = 16 Ω. Current = 60 ÷ 16 = 3.75 A. Voltage across the first resistor = 3.75 × 8 = 30 V.

Q25. A resistor is added in series with a lamp in a circuit with a fixed battery voltage. What happens to the brightness of the lamp?

  • (A) It increases
  • (B) It decreases
  • (C) It stays the same
  • (D) It goes out completely
Show Answer

Answer: B – The series resistor raises total resistance and lowers the current, so the lamp dissipates less power and dims.

Q26. A potential difference of 24 V is applied across a resistor of 24 Ω. What current flows through it?

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

Answer: B – I = V ÷ R = 24 ÷ 24 = 1 A.

Q27. Resistors of 10 Ω, 15 Ω, and 20 Ω are connected in series. What is their equivalent resistance?

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

Answer: B – In series, Req = R1 + R2 + R3 = 10 + 15 + 20 = 45 Ω.

Unit 12: Magnetism and Electromagnetism

Q28. A coil with 20 turns experiences a change in magnetic flux of 0.1 Wb in 0.01 s. What is the magnitude of the induced emf in the coil?

  • (A) 400 V
  • (B) 100 V
  • (C) 200 V
  • (D) 600 V
Show Answer

Answer: C – Magnitude of emf = N ΔΦ ÷ Δt = 20 × 0.1 ÷ 0.01 = 200 V.

Q29. A conducting rod of length 1.5 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) 3 V
  • (B) 0.75 V
  • (C) 4.5 V
  • (D) 1.5 V
Show Answer

Answer: D – Motional emf = BLv = 0.2 × 1.5 × 5 = 1.5 V.

Q30. A rectangular coil with 50 turns, area 0.02 m2, carries 2 A in a magnetic field of 0.2 T. The angle in the torque formula is 30°. What is the magnitude of the magnetic torque on the coil?

  • (A) 0.4 N·m
  • (B) 0.1 N·m
  • (C) 0.6 N·m
  • (D) 0.2 N·m
Show Answer

Answer: D – τ = NIAB sin θ = 50 × 2 × 0.02 × 0.2 × 0.5 = 0.2 N·m.


Q31. A bar magnet is pushed toward a coil. What happens to the induced emf if the magnet is pushed in twice as fast?

  • (A) It approximately doubles
  • (B) It halves
  • (C) It stays the same
  • (D) It becomes four times larger
Show Answer

Answer: A – Faster motion gives a faster rate of change of flux, and emf is proportional to that rate, so it approximately doubles.

Q32. A particle with charge 4 µC moves at 1000 m/s perpendicular to a uniform magnetic field of 1 T. What is the magnitude of the magnetic force on the particle, in millinewtons?

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

Answer: A – F = qvB = (4 × 10-6) × 1000 × 1 = 4 × 10-3 N = 4 mN.

Q33. A particle with charge 10 µC moves at 1000 m/s perpendicular to a uniform magnetic field of 0.4 T. What is the magnitude of the magnetic force on the particle, in millinewtons?

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

Answer: A – F = qvB = (10 × 10-6) × 1000 × 0.4 = 4 × 10-3 N = 4 mN.

Q34. A straight wire of length 0.8 m carries a current of 8 A perpendicular to a uniform magnetic field of 0.2 T. What is the magnitude of the magnetic force on the wire?

  • (A) 2.56 N
  • (B) 1.28 N
  • (C) 0.64 N
  • (D) 3.84 N
Show Answer

Answer: B – F = ILB = 8 × 0.8 × 0.2 = 1.28 N.

Q35. A long solenoid has 1000 turns per metre and carries a current of 4 A. What is the magnetic field strength inside the solenoid, in millitesla? (Use μ0 = 4π × 10-7 T·m/A.)

  • (A) 10.05 mT
  • (B) 2.51 mT
  • (C) 5.03 mT
  • (D) 15.08 mT
Show Answer

Answer: C – B = μ0nI = (4π × 10-7) × 1000 × 4 = 5.03 mT.

Unit 13: Geometric Optics

Q36. A spherical mirror has a radius of curvature of 20 cm. What is its focal length?

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

Answer: A – f = R ÷ 2 = 20 ÷ 2 = 10 cm.

Q37. A converging lens of focal length 20 cm forms an image 60 cm from the lens. How far from the lens is the object?

  • (A) 60 cm
  • (B) 30 cm
  • (C) 15 cm
  • (D) 90 cm
Show Answer

Answer: B – 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ do = 1 ÷ 20 – 1 ÷ 60, giving do = 30 cm.

Q38. An object is placed inside the focal length of a magnifying converging lens. What kind of image is seen?

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

Answer: D – An object inside F of a converging lens gives a virtual, upright, magnified image on the same side as the object.

Q39. An object is placed 45 cm from a concave mirror of focal length 15 cm. What is the image distance?

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

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

Q40. An object is 60 cm from a mirror and its image forms 60 cm from the mirror. What is the lateral magnification?

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

Answer: C – m = -di ÷ do = -60 ÷ 60 = -1.

Q41. An object is placed 30 cm from a converging lens of focal length 15 cm. What is the image distance?

  • (A) 60 cm
  • (B) 15 cm
  • (C) 90 cm
  • (D) 30 cm
Show Answer

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

Q42. A ray travels from air (refractive index 1) into glass (refractive index 1.5) with sin of the angle of incidence equal to 0.6. What is sin of the angle of refraction?

  • (A) 0.4
  • (B) 0.6
  • (C) 0.8
  • (D) 0.5
Show Answer

Answer: A – Snell: n1 sin θ1 = n2 sin θ2, so sin θ2 = 1 × 0.6 ÷ 1.5 = 0.4.

Q43. Light travels at 2.5 × 108 m/s in a sample of glass. What is the refractive index of the sample? (Use c = 3.00 × 108 m/s.)

  • (A) 2.4
  • (B) 0.6
  • (C) 3.6
  • (D) 1.2
Show Answer

Answer: D – n = c ÷ v = (3.00 × 108) ÷ (2.5 × 108) = 1.2.

Unit 14: Waves, Sound, and Physical Optics

Q44. 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 third harmonic? (Only odd harmonics exist for this pipe.)

  • (A) 510 Hz
  • (B) 1020 Hz
  • (C) 255 Hz
  • (D) 1530 Hz
Show Answer

Answer: A – For a closed pipe, fn = nv ÷ (4L) = 3 × 340 ÷ (4 × 0.5) = 510 Hz.

Q45. A sound wave travels at 400 m/s with frequency 170 Hz. What is its wavelength?

  • (A) 4.71 m
  • (B) 2.35 m
  • (C) 1.18 m
  • (D) 7.06 m
Show Answer

Answer: B – λ = v ÷ f = 400 ÷ 170 = 2.35 m.

Q46. 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 first harmonic?

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

Answer: C – For a string fixed at both ends, fn = nv ÷ (2L) = 1 × 400 ÷ (2 × 0.6) = 333.33 Hz.

Q47. A pipe closed at one end has length 1.7 m. Sound travels at 340 m/s in the air inside it. What is its fundamental frequency?

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

Answer: A – For a pipe closed at one end, f1 = v ÷ (4L) = 340 ÷ (4 × 1.7) = 50 Hz.

Q48. Two tuning forks of frequencies 600 Hz and 606 Hz sound together. What is the beat frequency heard?

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

Answer: A – Beat frequency = |f1 – f2| = |600 – 606| = 6 Hz.

Q49. In a double-slit experiment, light of wavelength 650 nm passes through slits 0.25 mm apart and falls on a screen 1.5 m away. How far from the central maximum is the second bright fringe, in millimetres?

  • (A) 7.8 mm
  • (B) 15.6 mm
  • (C) 3.9 mm
  • (D) 23.4 mm
Show Answer

Answer: A – y = mλL ÷ d = 2 × 650 × 10-9 × 1.5 ÷ (0.25 × 10-3) = 7.8 mm.

Q50. Light of wavelength 450 nm passes through double slits 0.15 mm apart onto a screen 1 m away. What is the spacing between adjacent bright fringes, in millimetres?

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

Answer: C – Fringe spacing Δy = λL ÷ d = 450 × 10-9 × 1 ÷ (0.15 × 10-3) = 3 mm.

Q51. A source emitting 300 Hz moves toward 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) 1200 Hz
  • (D) 400 Hz
Show Answer

Answer: D – For a source moving toward the observer, f′ = f × v ÷ (v – vs) = 300 × 340 ÷ (340 – 85) = 400 Hz.

Unit 15: Modern Physics

Q52. What is the energy of a photon of wavelength 500 nm? (Use h = 6.63 × 10-34 J·s and c = 3.00 × 108 m/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 = hc ÷ λ = (6.63 × 10-34 × 3.00 × 108) ÷ (500 × 10-9) = 3.98 × 10-19 J.

Q53. Light of photon energy 5.6 eV falls on a metal with work function 2.8 eV. What is the maximum kinetic energy of the emitted photoelectrons?

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

Answer: A – Kmax = hf – Φ = 5.6 – 2.8 = 2.8 eV.

Q54. A particle has momentum 6.63 × 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) 2
  • (C) 0.5
  • (D) 3
Show Answer

Answer: A – λ = h ÷ p = (6.63 × 10-34) ÷ (6.63 × 10-24) = 1 × 10-10 m, so the value is 1 in units of 10-10 m.

Q55. In the Bohr model, the energy of level n is En = -13.6 ÷ n2 eV. What is the energy of level n = 3?

  • (A) -1.51 eV
  • (B) -3.02 eV
  • (C) -0.76 eV
  • (D) -4.53 eV
Show Answer

Answer: A – E3 = -13.6 ÷ 32 = -13.6 ÷ 9 = -1.51 eV.

Q56. An electron in a hydrogen atom drops from level n = 3 to level n = 1. What is the energy of the photon emitted?

  • (A) 12.09 eV
  • (B) 24.18 eV
  • (C) 6.04 eV
  • (D) 36.27 eV
Show Answer

Answer: A – Photon energy = Ehigh – Elow = (-13.6 ÷ 32) – (-13.6 ÷ 12) = 12.09 eV.

Q57. A radioactive sample initially has mass 120 g. After 4 half-lives, what mass remains undecayed?

  • (A) 15 g
  • (B) 3.75 g
  • (C) 7.5 g
  • (D) 22.5 g
Show Answer

Answer: C – After 4 half-lives the fraction remaining is (1/2)4 = 0.06, so the mass remaining is 120 × 0.06 = 7.5 g.

Q58. How much energy is equivalent to a mass of 0.5 kg? (Use c = 3.00 × 108 m/s.)

  • (A) 9 × 1016 J
  • (B) 2.25 × 1016 J
  • (C) 4.5 × 1016 J
  • (D) 1.35 × 1017 J
Show Answer

Answer: C – E = mc2 = 0.5 × (3.00 × 108)2 = 4.5 × 1016 J.

Q59. A photon has energy 3.31 × 10-19 J. What is its frequency, in units of 1014 Hz? (Use h = 6.63 × 10-34 J·s.)

  • (A) 10 × 1014 Hz
  • (B) 2.5 × 1014 Hz
  • (C) 15 × 1014 Hz
  • (D) 5 × 1014 Hz
Show Answer

Answer: D – f = E ÷ h = (3.31 × 10-19) ÷ (6.63 × 10-34) = 5 × 1014 Hz, so the value is 5 in units of 1014 Hz.

Q60. Increasing the intensity of light falling on a metal surface, without changing its frequency, increases which quantity in the photoelectric effect?

  • (A) The maximum kinetic energy of each photoelectron
  • (B) The number of photoelectrons emitted per second
  • (C) The work function of the metal
  • (D) The threshold frequency of the metal
Show Answer

Answer: B – Higher intensity means more photons per second, so more electrons are emitted per second, but each photon still has the same energy.

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