AP Physics 2 Practice Test 12 – 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 reversible heat engine operates between a hot reservoir at 1000 K and a cold reservoir at 250 K. What is its maximum possible efficiency?

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

Answer: A – Carnot efficiency = 1 – Tc ÷ Th = 1 – 250 ÷ 1000 = 75%.

Q2. A heat engine takes in 1000 J of heat in one cycle and does 400 J of useful work. How much heat does it reject to the cold reservoir?

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

Answer: C – Energy conservation: Qh = W + Qc, so Qc = 1000 – 400 = 600 J.

Q3. A gas at 200 kPa occupies 5 L at 600 K. Its pressure changes to 400 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 = 400 × 2.5 × 600 ÷ (200 × 5) = 600 K.

Q4. The average kinetic energy of the molecules of an ideal gas is doubled. By what factor does its absolute temperature change?

  • (A) It halves
  • (B) It increases by a factor of 4
  • (C) It stays the same
  • (D) It doubles
Show Answer

Answer: D – Average molecular kinetic energy is proportional to absolute temperature, so doubling the energy doubles the temperature.

Q5. A closed laboratory flask holds 3 mol of an ideal gas at 400 K in a volume of 0.08 m3. What is the pressure of the gas? (Use R = 8.31 J/(mol·K).)

  • (A) 124650 Pa
  • (B) 249300 Pa
  • (C) 62325 Pa
  • (D) 373950 Pa
Show Answer

Answer: A – pV = nRT gives p = nRT ÷ V = (3 × 8.31 × 400) ÷ 0.08 = 124650 Pa.

Q6. An ideal gas sample contains 4 mol at 273 K and a pressure of 150 kPa. What is its volume? (Use R = 8.31 J/(mol·K).)

  • (A) 0.06 m3
  • (B) 0.12 m3
  • (C) 0.03 m3
  • (D) 0.18 m3
Show Answer

Answer: A – pV = nRT gives V = nRT ÷ p = (4 × 8.31 × 273) ÷ 150000 = 0.06 m3.

Q7. A steam sample absorbs 1000 J of heat from its surroundings and does 250 J of work on its surroundings. What is the change in its internal energy?

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

Answer: D – First law: ΔU = Q – W = 1000 – 250 = 750 J.

Q8. A gas expands at a constant pressure of 400 kPa, increasing its volume by 0.03 m3. How much work does the gas do?

  • (A) 24000 J
  • (B) 12000 J
  • (C) 6000 J
  • (D) 36000 J
Show Answer

Answer: B – At constant pressure, W = pΔV = 400000 × 0.03 = 12000 J.

Q9. A 4 kg block of copper (specific heat capacity 390 J/(kg·K)) is heated so its temperature rises by 40 K. How much heat is supplied?

  • (A) 124800 J
  • (B) 31200 J
  • (C) 187200 J
  • (D) 62400 J
Show Answer

Answer: D – Q = mcΔT = 4 × 390 × 40 = 62400 J.

Unit 10: Electric Force, Field, and Potential

Q10. Two equal positive charges are fixed in place. At the midpoint between them, what is the electric field?

  • (A) Zero
  • (B) Twice the field of one charge
  • (C) Half the field of one charge
  • (D) Directed toward one of the charges
Show Answer

Answer: A – The two fields are equal in magnitude and opposite in direction at the midpoint, so they cancel.

Q11. A charge of 5 µC sits at a point where the electric potential is 300 V. What is its electric potential energy?

  • (A) 1.5 × 10-3 J
  • (B) 3 × 10-3 J
  • (C) 7.5 × 10-4 J
  • (D) 4.5 × 10-3 J
Show Answer

Answer: A – U = qV = (5 × 10-6) × 300 = 1.5 × 10-3 J.

Q12. Two small charged spheres carry charges of 6 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) 21.6 N
  • (B) 10.8 N
  • (C) 5.4 N
  • (D) 32.4 N
Show Answer

Answer: B – F = kq1q2 ÷ r2 = (9.0 × 109 × 6 × 10-6 × 8 × 10-6) ÷ 0.22 = 10.8 N.

Q13. Both point charges are doubled, with their separation unchanged. By what factor does the electrostatic force change?

  • (A) It becomes 2 times its original value
  • (B) It becomes 8 times its original value
  • (C) It becomes 4 times its original value
  • (D) It stays the same
Show Answer

Answer: C – Force is proportional to q1q2, so doubling both gives a factor of 2 × 2 = 4.

Q14. What is the magnitude of the electric field 0.1 m from a point charge of 20 µC in a vacuum? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 3.6 × 107 N/C
  • (B) 1.8 × 107 N/C
  • (C) 9 × 106 N/C
  • (D) 5.4 × 107 N/C
Show Answer

Answer: B – E = kQ ÷ r2 = (9.0 × 109 × 20 × 10-6) ÷ 0.12 = 1.8 × 107 N/C.

Q15. A small charge of 4 µC is placed in a uniform electric field of 400 N/C. What is the magnitude of the electric force on the charge?

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

Answer: B – F = qE = (4 × 10-6) × 400 = 1.6 × 10-3 N.

Q16. What is the electric potential 0.2 m from a point charge of 10 µC in a vacuum? (Use k = 9.0 × 109 N·m2/C2.)

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

Answer: C – V = kQ ÷ r = (9.0 × 109 × 10 × 10-6) ÷ 0.2 = 450000 V.

Q17. Two point charges of 5 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) 0.72 J
  • (B) 0.36 J
  • (C) 0.18 J
  • (D) 1.08 J
Show Answer

Answer: B – U = kq1q2 ÷ r = (9.0 × 109 × 5 × 10-6 × 4 × 10-6) ÷ 0.5 = 0.36 J.

Q18. A charge of 2 µC is moved through a potential difference of 500 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 = (2 × 10-6) × 500 = 1 × 10-3 J.

Unit 11: Electric Circuits

Q19. A potential difference of 60 V is applied across a resistor of 8 Ω. What current flows through it?

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

Answer: A – I = V ÷ R = 60 ÷ 8 = 7.5 A.

Q20. Resistors of 5 Ω, 2 Ω, and 8 Ω are connected in series. What is their equivalent resistance?

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

Answer: B – In series, Req = R1 + R2 + R3 = 5 + 2 + 8 = 15 Ω.

Q21. Resistors of 12 Ω and 4 Ω are connected in parallel. What is their equivalent resistance?

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

Answer: A – 1 ÷ Req = 1 ÷ 12 + 1 ÷ 4, so Req = (12 × 4) ÷ (12 + 4) = 3 Ω.

Q22. A 18 Ω resistor has a potential difference of 60 V across it. What power is dissipated in it?

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

Answer: C – P = V2 ÷ R = 602 ÷ 18 = 200 W.

Q23. Resistors of 6 Ω and 3 Ω are connected in parallel, and this combination is connected in series with a 4 Ω resistor. What is the equivalent resistance of the whole combination?

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

Answer: A – Parallel part: (6 × 3) ÷ (6 + 3) = 2 Ω. Adding the series resistor: 2 + 4 = 6 Ω.

Q24. A capacitor of capacitance 10 µF is connected across a 12 V battery. What charge is stored on the capacitor?

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

Answer: D – Q = CV = 10 × 12 = 120 µC.

Q25. A 5 µF capacitor is charged to 50 V. How much energy is stored in it, in millijoules?

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

Answer: C – U = 0.5CV2 = 0.5 × 5 × 10-6 × 502 = 6.25 × 10-3 J = 6.25 mJ.

Q26. Capacitors of 10 µF and 10 µF are connected in series. What is their equivalent capacitance?

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

Answer: A – 1 ÷ Ceq = 1 ÷ 10 + 1 ÷ 10, so Ceq = (10 × 10) ÷ (10 + 10) = 5 µF.

Q27. A 60 V battery is connected across resistors of 10 Ω and 4 Ω in series. What is the potential difference across the 10 Ω resistor?

  • (A) 85.71 V
  • (B) 21.43 V
  • (C) 42.86 V
  • (D) 128.57 V
Show Answer

Answer: C – Total resistance = 10 + 4 = 14 Ω. Current = 60 ÷ 14 = 4.29 A. Voltage across the first resistor = 4.29 × 10 = 42.86 V.

Unit 12: Magnetism and Electromagnetism

Q28. A long solenoid has 800 turns per metre and carries a current of 5 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) × 800 × 5 = 5.03 mT.

Q29. A long straight wire carries a current of 10 A. What is the magnetic field strength at a distance of 0.2 m from the wire, in microtesla? (Use μ0 = 4π × 10-7 T·m/A.)

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

Answer: A – B = μ0I ÷ (2πr) = (4π × 10-7 × 10) ÷ (2π × 0.2) = 10 µT.

Q30. A flat coil of area 0.04 m2 sits in a uniform magnetic field of 0.2 T. The field makes an angle of 60° with the normal to the coil. What is the magnetic flux through the coil?

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

Answer: B – Φ = BA cos θ = 0.2 × 0.04 × 0.5 = 4 × 10-3 Wb.


Q31. A coil with 10 turns experiences a change in magnetic flux of 0.2 Wb in 0.1 s. What is the magnitude of the induced emf in the coil?

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

Answer: C – Magnitude of emf = N ΔΦ ÷ Δt = 10 × 0.2 ÷ 0.1 = 20 V.

Q32. A conducting rod of length 0.5 m moves at 2 m/s perpendicular to a uniform magnetic field of 1 T. What is the motional emf induced between its ends?

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

Answer: C – Motional emf = BLv = 1 × 0.5 × 2 = 1 V.

Q33. A rectangular coil with 50 turns, area 0.02 m2, carries 5 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) 2.5 N·m
  • (B) 0.62 N·m
  • (C) 1.25 N·m
  • (D) 3.75 N·m
Show Answer

Answer: C – τ = NIAB sin θ = 50 × 5 × 0.02 × 0.5 × 0.5 = 1.25 N·m.

Q34. A positively charged particle moves into a magnetic field directed into the page, travelling to the right. In which direction does the magnetic force initially act?

  • (A) Downward
  • (B) Into the page
  • (C) Out of the page
  • (D) Upward
Show Answer

Answer: D – Using the right-hand rule for a positive charge, v to the right and B into the page give a force upward.

Q35. The speed of a charged particle moving perpendicular to a magnetic field is doubled and the field strength is halved. By what factor does the magnetic force change?

  • (A) It doubles
  • (B) It stays the same
  • (C) It halves
  • (D) It becomes 4 times larger
Show Answer

Answer: B – F is proportional to vB, so the factor is 2 × 0.5 = 1.

Unit 13: Geometric Optics

Q36. Light travels at 2.25 × 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) 2.67
  • (B) 0.67
  • (C) 1.33
  • (D) 4
Show Answer

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

Q37. A transparent material has refractive index 1.25. At what speed does light travel in it, in units of 108 m/s? (Use c = 3.00 × 108 m/s.)

  • (A) 4.8 × 108 m/s
  • (B) 1.2 × 108 m/s
  • (C) 7.2 × 108 m/s
  • (D) 2.4 × 108 m/s
Show Answer

Answer: D – v = c ÷ n = 3.00 ÷ 1.25 = 2.4, so the speed is 2.4 × 108 m/s.

Q38. Light in a medium of refractive index 1.5 strikes the boundary with a medium of refractive index 1.33. What is sin of the critical angle for total internal reflection?

  • (A) 0.44
  • (B) 0.89
  • (C) 0.11
  • (D) 0.69
Show Answer

Answer: B – sin θc = n2 ÷ n1 = 1.33 ÷ 1.5 = 0.89.

Q39. A spherical mirror has a radius of curvature of 64 cm. What is its focal length?

  • (A) 64 cm
  • (B) 32 cm
  • (C) 16 cm
  • (D) 96 cm
Show Answer

Answer: B – f = R ÷ 2 = 64 ÷ 2 = 32 cm.

Q40. A converging lens of focal length 25 cm forms an image 50 cm from the lens. How far from the lens is the object?

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

Answer: B – 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ do = 1 ÷ 25 – 1 ÷ 50, giving do = 50 cm.

Q41. A ray of light passes from air into water. What happens to its speed and wavelength?

  • (A) Both increase
  • (B) Speed decreases and wavelength increases
  • (C) Speed increases and wavelength decreases
  • (D) Both decrease
Show Answer

Answer: D – In a denser medium light slows down, and with frequency fixed, wavelength = speed ÷ frequency also decreases.

Q42. An object is placed 30 cm from a concave mirror of focal length 10 cm. What is the image distance?

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

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

Q43. An object is 20 cm from a mirror and its image forms 30 cm from the mirror. What is the lateral magnification?

  • (A) -1.5
  • (B) -3
  • (C) -0.75
  • (D) -4.5
Show Answer

Answer: A – m = -di ÷ do = -30 ÷ 20 = -1.5.

Unit 14: Waves, Sound, and Physical Optics

Q44. A source emitting 600 Hz moves away from a stationary observer at 170 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 away, f′ = f × v ÷ (v + vs) = 600 × 340 ÷ (340 + 170) = 400 Hz.

Q45. A wave completes one full oscillation in 0.05 s. What is its frequency?

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

Answer: A – f = 1 ÷ T = 1 ÷ 0.05 = 20 Hz.

Q46. Light passes through a narrow single slit. What happens to the width of the central diffraction maximum if the slit is made narrower?

  • (A) It becomes narrower
  • (B) It stays the same
  • (C) It becomes wider
  • (D) It disappears
Show Answer

Answer: C – A narrower slit spreads light more, so the central maximum widens.

Q47. A pipe closed at one end has length 1.7 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) 300 Hz
  • (B) 75 Hz
  • (C) 150 Hz
  • (D) 450 Hz
Show Answer

Answer: C – For a closed pipe, fn = nv ÷ (4L) = 3 × 340 ÷ (4 × 1.7) = 150 Hz.

Q48. A wave has frequency 500 Hz and wavelength 1 m. What is its speed?

  • (A) 500 m/s
  • (B) 1000 m/s
  • (C) 250 m/s
  • (D) 1500 m/s
Show Answer

Answer: A – v = fλ = 500 × 1 = 500 m/s.

Q49. A string of length 0.75 m is fixed at both ends. Waves travel along it at 240 m/s. What is the frequency of its fourth harmonic?

  • (A) 1280 Hz
  • (B) 640 Hz
  • (C) 320 Hz
  • (D) 1920 Hz
Show Answer

Answer: B – For a string fixed at both ends, fn = nv ÷ (2L) = 4 × 240 ÷ (2 × 0.75) = 640 Hz.

Q50. An open pipe of length 0.85 m is filled with air in which sound travels at 340 m/s. What is its fundamental frequency?

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

Answer: C – For an open pipe, f1 = v ÷ (2L) = 340 ÷ (2 × 0.85) = 200 Hz.

Q51. Two tuning forks of frequencies 512 Hz and 516 Hz sound together. What is the beat frequency heard?

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

Answer: D – Beat frequency = |f1 – f2| = |512 – 516| = 4 Hz.

Unit 15: Modern Physics

Q52. A photon has energy 1.99 × 10-19 J. What is its frequency, in units of 1014 Hz? (Use h = 6.63 × 10-34 J·s.)

  • (A) 6 × 1014 Hz
  • (B) 3 × 1014 Hz
  • (C) 1.5 × 1014 Hz
  • (D) 9 × 1014 Hz
Show Answer

Answer: B – f = E ÷ h = (1.99 × 10-19) ÷ (6.63 × 10-34) = 3 × 1014 Hz, so the value is 3 in units of 1014 Hz.

Q53. Light of frequency below the threshold frequency falls on a metal, even at very high intensity. How many photoelectrons are emitted?

  • (A) A few with low kinetic energy
  • (B) Many with high kinetic energy
  • (C) One per photon regardless of frequency
  • (D) None
Show Answer

Answer: D – Below the threshold frequency no single photon has enough energy to overcome the work function, so no electrons are emitted.

Q54. A photon has frequency 4 × 1014 Hz. What is its energy? (Use h = 6.63 × 10-34 J·s.)

  • (A) 5.3 × 10-19 J
  • (B) 1.33 × 10-19 J
  • (C) 7.96 × 10-19 J
  • (D) 2.65 × 10-19 J
Show Answer

Answer: D – E = hf = (6.63 × 10-34) × (4 × 1014) = 2.65 × 10-19 J.

Q55. What is the energy of a photon of wavelength 600 nm? (Use h = 6.63 × 10-34 J·s and c = 3.00 × 108 m/s.)

  • (A) 6.63 × 10-19 J
  • (B) 1.66 × 10-19 J
  • (C) 9.94 × 10-19 J
  • (D) 3.31 × 10-19 J
Show Answer

Answer: D – E = hc ÷ λ = (6.63 × 10-34 × 3.00 × 108) ÷ (600 × 10-9) = 3.31 × 10-19 J.

Q56. Light of photon energy 4.2 eV falls on a metal with work function 1.9 eV. What is the maximum kinetic energy of the emitted photoelectrons?

  • (A) 2.3 eV
  • (B) 4.6 eV
  • (C) 1.15 eV
  • (D) 6.9 eV
Show Answer

Answer: A – Kmax = hf – Φ = 4.2 – 1.9 = 2.3 eV.

Q57. A particle has momentum 33.15 × 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) 0.4
  • (B) 0.2
  • (C) 0.1
  • (D) 0.6
Show Answer

Answer: B – λ = h ÷ p = (6.63 × 10-34) ÷ (33.15 × 10-24) = 0.2 × 10-10 m, so the value is 0.2 in units of 10-10 m.

Q58. In the Bohr model, the energy of level n is En = -13.6 ÷ n2 eV. What is the energy of level n = 4?

  • (A) -1.7 eV
  • (B) -0.42 eV
  • (C) -2.55 eV
  • (D) -0.85 eV
Show Answer

Answer: D – E4 = -13.6 ÷ 42 = -13.6 ÷ 16 = -0.85 eV.

Q59. An electron in a hydrogen atom drops from level n = 2 to level n = 1. What is the energy of the photon emitted?

  • (A) 20.4 eV
  • (B) 5.1 eV
  • (C) 30.6 eV
  • (D) 10.2 eV
Show Answer

Answer: D – Photon energy = Ehigh – Elow = (-13.6 ÷ 22) – (-13.6 ÷ 12) = 10.2 eV.

Q60. A radioactive sample initially has mass 160 g. After 3 half-lives, what mass remains undecayed?

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

Answer: C – After 3 half-lives the fraction remaining is (1/2)3 = 0.12, so the mass remaining is 160 × 0.12 = 20 g.

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