AP Physics 2 Practice Test 14 – 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 ideal gas in an insulated chamber absorbs 800 J of heat from its surroundings and does 200 J of work on its surroundings. What is the change in its internal energy?

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

Answer: B – First law: ΔU = Q – W = 800 – 200 = 600 J.

Q2. A gas expands at a constant pressure of 400 kPa, increasing its volume by 0.1 m3. How much work does the gas do?

  • (A) 80000 J
  • (B) 20000 J
  • (C) 40000 J
  • (D) 120000 J
Show Answer

Answer: C – At constant pressure, W = pΔV = 400000 × 0.1 = 40000 J.

Q3. A 0.25 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) 6000 J
  • (B) 3000 J
  • (C) 1500 J
  • (D) 9000 J
Show Answer

Answer: B – Q = mcΔT = 0.25 × 240 × 50 = 3000 J.

Q4. How much heat is needed to freeze water at 0°C (specific latent heat of fusion 334,000 J/kg) if the mass involved is 0.8 kg?

  • (A) 534400 J
  • (B) 133600 J
  • (C) 267200 J
  • (D) 801600 J
Show Answer

Answer: C – Q = mL = 0.8 × 334000 = 267200 J.

Q5. A reversible heat engine operates between a hot reservoir at 800 K and a cold reservoir at 200 K. What is its maximum possible efficiency?

  • (A) 37.5 %
  • (B) 25 %
  • (C) 85 %
  • (D) 75 %
Show Answer

Answer: D – Carnot efficiency = 1 – Tc ÷ Th = 1 – 200 ÷ 800 = 75%.

Q6. A heat engine takes in 1000 J of heat in one cycle and does 200 J of useful work. How much heat does it reject to the cold reservoir?

  • (A) 800 J
  • (B) 1600 J
  • (C) 400 J
  • (D) 2400 J
Show Answer

Answer: A – Energy conservation: Qh = W + Qc, so Qc = 1000 – 200 = 800 J.

Q7. A gas at 150 kPa and 300 K occupies 4 L. Its volume changes to 8 L and its temperature changes to 150 K. What is its new pressure?

  • (A) 75 kPa
  • (B) 18.75 kPa
  • (C) 37.5 kPa
  • (D) 112.5 kPa
Show Answer

Answer: C – p1V1 ÷ T1 = p2V2 ÷ T2, so p2 = 150 × 4 × 150 ÷ (8 × 300) = 37.5 kPa.

Q8. An ideal gas is compressed isothermally. What is the change in its internal energy?

  • (A) Positive and equal to the work done on it
  • (B) Negative and equal to the work done by it
  • (C) Zero
  • (D) Positive and equal to the heat transferred
Show Answer

Answer: C – For an ideal gas internal energy depends only on temperature, which is constant in an isothermal process, so ΔU = 0.

Q9. A sealed copper chamber holds 1 mol of an ideal gas at 600 K in a volume of 0.02 m3. What is the pressure of the gas? (Use R = 8.31 J/(mol·K).)

  • (A) 498600 Pa
  • (B) 124650 Pa
  • (C) 249300 Pa
  • (D) 747900 Pa
Show Answer

Answer: C – pV = nRT gives p = nRT ÷ V = (1 × 8.31 × 600) ÷ 0.02 = 249300 Pa.

Unit 10: Electric Force, Field, and Potential

Q10. What is the electric potential 1 m from a point charge of 8 µC in a vacuum? (Use k = 9.0 × 109 N·m2/C2.)

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

Answer: C – V = kQ ÷ r = (9.0 × 109 × 8 × 10-6) ÷ 1 = 72000 V.

Q11. Two point charges of 4 µC and 4 µC are 0.1 m apart in a vacuum. What is their electric potential energy? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 2.88 J
  • (B) 0.72 J
  • (C) 1.44 J
  • (D) 4.32 J
Show Answer

Answer: C – U = kq1q2 ÷ r = (9.0 × 109 × 4 × 10-6 × 4 × 10-6) ÷ 0.1 = 1.44 J.

Q12. A charge of 3 µC is moved through a potential difference of 100 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) 6 × 10-4 J
  • (B) 1.5 × 10-4 J
  • (C) 3 × 10-4 J
  • (D) 9 × 10-4 J
Show Answer

Answer: C – W = qΔV = (3 × 10-6) × 100 = 3 × 10-4 J.

Q13. Two parallel plates are 0.03 m apart and have a potential difference of 500 V between them. What is the magnitude of the uniform electric field between the plates?

  • (A) 40000 N/C
  • (B) 20000 N/C
  • (C) 10000 N/C
  • (D) 60000 N/C
Show Answer

Answer: B – E = V ÷ d = 500 ÷ 0.03 = 20000 N/C.

Q14. A positive test charge is placed in an electric field. In which direction does the electric force on it act?

  • (A) Opposite to the direction of the electric field
  • (B) Perpendicular to the electric field
  • (C) In the direction of the electric field
  • (D) Toward the nearest negative charge only
Show Answer

Answer: C – By definition, the electric field direction is the direction of the force on a positive test charge.

Q15. A charge of 5 µC sits at a point where the electric potential is 200 V. What is its electric potential energy?

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

Answer: A – U = qV = (5 × 10-6) × 200 = 1 × 10-3 J.

Q16. Two small charged spheres carry charges of 2 µC and 3 µC and are 0.1 m apart in a vacuum. What is the magnitude of the electrostatic force between them? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 10.8 N
  • (B) 2.7 N
  • (C) 5.4 N
  • (D) 16.2 N
Show Answer

Answer: C – F = kq1q2 ÷ r2 = (9.0 × 109 × 2 × 10-6 × 3 × 10-6) ÷ 0.12 = 5.4 N.

Q17. The distance between two point charges is halved, with both charges unchanged. By what factor does the electrostatic force change?

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

Answer: D – Force is proportional to 1 ÷ r2, so halving r gives a factor of 1 ÷ (0.5)2 = 4.

Q18. What is the magnitude of the electric field 0.1 m from a point charge of 4 µC in a vacuum? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 3.6 × 106 N/C
  • (B) 7.2 × 106 N/C
  • (C) 1.8 × 106 N/C
  • (D) 1.08 × 107 N/C
Show Answer

Answer: A – E = kQ ÷ r2 = (9.0 × 109 × 4 × 10-6) ÷ 0.12 = 3.6 × 106 N/C.

Unit 11: Electric Circuits

Q19. A 4 µF capacitor is charged to 50 V. How much energy is stored in it, in millijoules?

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

Answer: D – U = 0.5CV2 = 0.5 × 4 × 10-6 × 502 = 5 × 10-3 J = 5 mJ.

Q20. Capacitors of 10 µF and 2 µF are connected in parallel. What is their equivalent capacitance?

  • (A) 12 µF
  • (B) 24 µF
  • (C) 6 µF
  • (D) 36 µF
Show Answer

Answer: A – In parallel, Ceq = C1 + C2 = 10 + 2 = 12 µF.

Q21. A 36 V battery is connected across resistors of 10 Ω and 12 Ω in series. What is the potential difference across the 10 Ω resistor?

  • (A) 32.73 V
  • (B) 16.36 V
  • (C) 8.18 V
  • (D) 49.09 V
Show Answer

Answer: B – Total resistance = 10 + 12 = 22 Ω. Current = 36 ÷ 22 = 1.64 A. Voltage across the first resistor = 1.64 × 10 = 16.36 V.

Q22. The resistance of a wire is doubled while the voltage across it stays fixed. What happens to the power dissipated?

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

Answer: D – P = V2 ÷ R, so doubling R with fixed V halves the power.

Q23. A current of 4 A flows through a resistor of 30 Ω. What is the potential difference across the resistor?

  • (A) 240 V
  • (B) 120 V
  • (C) 60 V
  • (D) 360 V
Show Answer

Answer: B – V = IR = 4 × 30 = 120 V.

Q24. Resistors of 5 Ω, 20 Ω, and 12 Ω are connected in series. What is their equivalent resistance?

  • (A) 74 Ω
  • (B) 37 Ω
  • (C) 18.5 Ω
  • (D) 111 Ω
Show Answer

Answer: B – In series, Req = R1 + R2 + R3 = 5 + 20 + 12 = 37 Ω.

Q25. Resistors of 8 Ω and 8 Ω are connected in parallel. What is their equivalent resistance?

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

Answer: A – 1 ÷ Req = 1 ÷ 8 + 1 ÷ 8, so Req = (8 × 8) ÷ (8 + 8) = 4 Ω.

Q26. A current of 5 A flows through a 10 Ω resistor. What power is dissipated in the resistor?

  • (A) 500 W
  • (B) 125 W
  • (C) 750 W
  • (D) 250 W
Show Answer

Answer: D – P = I2R = 52 × 10 = 250 W.

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 equivalent resistance of the whole combination?

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

Answer: B – Parallel part: (20 × 5) ÷ (20 + 5) = 4 Ω. Adding the series resistor: 4 + 6 = 10 Ω.

Unit 12: Magnetism and Electromagnetism

Q28. The magnetic flux through a coil is decreasing. According to Lenz’s law, what does the induced current do?

  • (A) It creates a magnetic field that tries to increase the flux
  • (B) It creates a magnetic field that decreases the flux further
  • (C) It flows in the same direction as the original current only
  • (D) It stops immediately
Show Answer

Answer: A – Lenz’s law states that an induced current opposes the change in flux, so it tries to replace the lost flux.

Q29. A particle with charge 5 µC moves at 4000 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) 40 mN
  • (B) 10 mN
  • (C) 60 mN
  • (D) 20 mN
Show Answer

Answer: D – F = qvB = (5 × 10-6) × 4000 × 1 = 0.02 N = 20 mN.

Q30. A particle with charge 4 µC moves at 2000 m/s perpendicular to a uniform magnetic field of 0.2 T. What is the magnitude of the magnetic force on the particle, in millinewtons?

  • (A) 3.2 mN
  • (B) 0.8 mN
  • (C) 4.8 mN
  • (D) 1.6 mN
Show Answer

Answer: D – F = qvB = (4 × 10-6) × 2000 × 0.2 = 1.6 × 10-3 N = 1.6 mN.


Q31. A straight wire of length 0.5 m carries a current of 10 A perpendicular to a uniform magnetic field of 0.1 T. What is the magnitude of the magnetic force on the wire?

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

Answer: B – F = ILB = 10 × 0.5 × 0.1 = 0.5 N.

Q32. A long solenoid has 800 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) 4.02 mT
  • (B) 1.01 mT
  • (C) 2.01 mT
  • (D) 6.03 mT
Show Answer

Answer: C – B = μ0nI = (4π × 10-7) × 800 × 2 = 2.01 mT.

Q33. A long straight wire carries a current of 20 A. What is the magnetic field strength at a distance of 0.1 m from the wire, in microtesla? (Use μ0 = 4π × 10-7 T·m/A.)

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

Answer: C – B = μ0I ÷ (2πr) = (4π × 10-7 × 20) ÷ (2π × 0.1) = 40 µT.

Q34. A flat coil of area 0.01 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) 0.04 Wb
  • (B) 0.01 Wb
  • (C) 0.06 Wb
  • (D) 0.02 Wb
Show Answer

Answer: D – Φ = BA cos θ = 2 × 0.01 × 1 = 0.02 Wb.

Q35. A coil with 50 turns experiences a change in magnetic flux of 0.04 Wb in 0.01 s. What is the magnitude of the induced emf in the coil?

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

Answer: A – Magnitude of emf = N ΔΦ ÷ Δt = 50 × 0.04 ÷ 0.01 = 200 V.

Unit 13: Geometric Optics

Q36. An object is placed 30 cm from a concave mirror of focal length 15 cm. What is the image distance?

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

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

Q37. An object is 60 cm from a mirror and its image forms 10 cm from the mirror. What is the lateral magnification?

  • (A) -0.33
  • (B) -0.08
  • (C) -0.5
  • (D) -0.17
Show Answer

Answer: D – m = -di ÷ do = -10 ÷ 60 = -0.17.

Q38. An object is placed 24 cm from a converging lens of focal length 12 cm. What is the image distance?

  • (A) 48 cm
  • (B) 12 cm
  • (C) 72 cm
  • (D) 24 cm
Show Answer

Answer: D – 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ di = 1 ÷ 12 – 1 ÷ 24, giving di = 24 cm.

Q39. A ray travels from glass (refractive index 1.5) into air (refractive index 1) with sin of the angle of incidence equal to 0.4. What is sin of the angle of refraction?

  • (A) 0.6
  • (B) 0.4
  • (C) 0.3
  • (D) 0.7
Show Answer

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

Q40. Light travels at 1.5 × 108 m/s in a sample of clear plastic. What is the refractive index of the sample? (Use c = 3.00 × 108 m/s.)

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

Answer: B – n = c ÷ v = (3.00 × 108) ÷ (1.5 × 108) = 2.

Q41. A transparent material has refractive index 2.42. At what speed does light travel in it, in units of 108 m/s? (Use c = 3.00 × 108 m/s.)

  • (A) 2.48 × 108 m/s
  • (B) 1.24 × 108 m/s
  • (C) 0.62 × 108 m/s
  • (D) 3.72 × 108 m/s
Show Answer

Answer: B – v = c ÷ n = 3.00 ÷ 2.42 = 1.24, so the speed is 1.24 × 108 m/s.

Q42. Light in a medium of refractive index 2.42 strikes the boundary with a medium of refractive index 1. What is sin of the critical angle for total internal reflection?

  • (A) 0.21
  • (B) 0.41
  • (C) 0.59
  • (D) 0.61
Show Answer

Answer: B – sin θc = n2 ÷ n1 = 1 ÷ 2.42 = 0.41.

Q43. A spherical mirror has a radius of curvature of 84 cm. What is its focal length?

  • (A) 84 cm
  • (B) 21 cm
  • (C) 126 cm
  • (D) 42 cm
Show Answer

Answer: D – f = R ÷ 2 = 84 ÷ 2 = 42 cm.

Unit 14: Waves, Sound, and Physical Optics

Q44. An open pipe of length 1 m is filled with air in which sound travels at 340 m/s. What is its fundamental frequency?

  • (A) 170 Hz
  • (B) 340 Hz
  • (C) 85 Hz
  • (D) 510 Hz
Show Answer

Answer: A – For an open pipe, f1 = v ÷ (2L) = 340 ÷ (2 × 1) = 170 Hz.

Q45. Two tuning forks of frequencies 512 Hz and 518 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| = |512 – 518| = 6 Hz.

Q46. In a double-slit experiment, light of wavelength 600 nm passes through slits 0.2 mm apart and falls on a screen 1 m away. How far from the central maximum is the third bright fringe, in millimetres?

  • (A) 18 mm
  • (B) 4.5 mm
  • (C) 27 mm
  • (D) 9 mm
Show Answer

Answer: D – y = mλL ÷ d = 3 × 600 × 10-9 × 1 ÷ (0.2 × 10-3) = 9 mm.

Q47. Light of wavelength 550 nm passes through double slits 0.15 mm apart onto a screen 1.5 m away. What is the spacing between adjacent bright fringes, in millimetres?

  • (A) 11 mm
  • (B) 2.75 mm
  • (C) 16.5 mm
  • (D) 5.5 mm
Show Answer

Answer: D – Fringe spacing Δy = λL ÷ d = 550 × 10-9 × 1.5 ÷ (0.15 × 10-3) = 5.5 mm.

Q48. A source emitting 450 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) 750 Hz
  • (B) 375 Hz
  • (C) 187.5 Hz
  • (D) 1125 Hz
Show Answer

Answer: B – For a source moving away, f′ = f × v ÷ (v + vs) = 450 × 340 ÷ (340 + 68) = 375 Hz.

Q49. A wave completes one full oscillation in 0.02 s. What is its frequency?

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

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

Q50. Two waves of the same frequency meet in phase (crest on crest). What kind of interference occurs and what happens to the amplitude?

  • (A) Destructive interference; amplitude increases
  • (B) Constructive interference; amplitude decreases
  • (C) Destructive interference; amplitude stays the same
  • (D) Constructive interference; amplitude increases
Show Answer

Answer: D – In-phase superposition is constructive, so the amplitudes add and the resultant amplitude is larger.

Q51. A pipe closed at one end has length 1 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) 127.5 Hz
  • (C) 255 Hz
  • (D) 765 Hz
Show Answer

Answer: C – For a closed pipe, fn = nv ÷ (4L) = 3 × 340 ÷ (4 × 1) = 255 Hz.

Unit 15: Modern Physics

Q52. In the Bohr model, the energy of level n is En = -13.6 ÷ n2 eV. What is the energy of level n = 5?

  • (A) -0.54 eV
  • (B) -1.09 eV
  • (C) -0.27 eV
  • (D) -1.63 eV
Show Answer

Answer: A – E5 = -13.6 ÷ 52 = -13.6 ÷ 25 = -0.54 eV.

Q53. An electron in a hydrogen atom drops from level n = 4 to level n = 3. What is the energy of the photon emitted?

  • (A) 0.66 eV
  • (B) 1.32 eV
  • (C) 0.33 eV
  • (D) 1.98 eV
Show Answer

Answer: A – Photon energy = Ehigh – Elow = (-13.6 ÷ 42) – (-13.6 ÷ 32) = 0.66 eV.

Q54. A radioactive sample initially has mass 100 g. After 4 half-lives, what mass remains undecayed?

  • (A) 12.5 g
  • (B) 3.12 g
  • (C) 6.25 g
  • (D) 18.75 g
Show Answer

Answer: C – After 4 half-lives the fraction remaining is (1/2)4 = 0.06, so the mass remaining is 100 × 0.06 = 6.25 g.

Q55. How much energy is equivalent to a mass of 1 kg? (Use c = 3.00 × 108 m/s.)

  • (A) 1.8 × 1017 J
  • (B) 4.5 × 1016 J
  • (C) 9 × 1016 J
  • (D) 2.7 × 1017 J
Show Answer

Answer: C – E = mc2 = 1 × (3.00 × 108)2 = 9 × 1016 J.

Q56. A photon has energy 6.63 × 10-19 J. What is its frequency, in units of 1014 Hz? (Use h = 6.63 × 10-34 J·s.)

  • (A) 20 × 1014 Hz
  • (B) 10 × 1014 Hz
  • (C) 5 × 1014 Hz
  • (D) 30 × 1014 Hz
Show Answer

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

Q57. An electron and a proton have the same momentum. Which has the longer de Broglie wavelength?

  • (A) The electron, because it is lighter
  • (B) They have the same wavelength
  • (C) The proton, because it is heavier
  • (D) Neither has a wavelength
Show Answer

Answer: B – De Broglie wavelength depends only on momentum, λ = h ÷ p, so equal momentum gives equal wavelength.

Q58. A photon has frequency 7 × 1014 Hz. What is its energy? (Use h = 6.63 × 10-34 J·s.)

  • (A) 4.64 × 10-19 J
  • (B) 9.28 × 10-19 J
  • (C) 2.32 × 10-19 J
  • (D) 1.39 × 10-18 J
Show Answer

Answer: A – E = hf = (6.63 × 10-34) × (7 × 1014) = 4.64 × 10-19 J.

Q59. What is the energy of a photon of wavelength 450 nm? (Use h = 6.63 × 10-34 J·s and c = 3.00 × 108 m/s.)

  • (A) 4.42 × 10-19 J
  • (B) 8.84 × 10-19 J
  • (C) 2.21 × 10-19 J
  • (D) 1.33 × 10-18 J
Show Answer

Answer: A – E = hc ÷ λ = (6.63 × 10-34 × 3.00 × 108) ÷ (450 × 10-9) = 4.42 × 10-19 J.

Q60. Light of photon energy 5.6 eV falls on a metal with work function 2.1 eV. What is the maximum kinetic energy of the emitted photoelectrons?

  • (A) 7 eV
  • (B) 3.5 eV
  • (C) 1.75 eV
  • (D) 10.5 eV
Show Answer

Answer: B – Kmax = hf – Φ = 5.6 – 2.1 = 3.5 eV.

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