AP Physics 2 Practice Test 16 – 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 pressure chamber holds 3 mol of an ideal gas at 350 K in a volume of 0.02 m3. What is the pressure of the gas, in kPa? (Use R = 8.31 J/mol·K.)

  • (A) 872.55 kPa
  • (B) 436.27 kPa
  • (C) 218.14 kPa
  • (D) 654.41 kPa
Show Answer

Answer: B – P = nRT ÷ V = (3 × 8.31 × 350) ÷ 0.02 = 436275 Pa = 436.27 kPa.

Q2. 8 mol of an ideal gas at 350 K exerts a pressure of 200 kPa. What volume does the gas occupy? (Use R = 8.31 J/mol·K.)

  • (A) 0.23 m3
  • (B) 0.12 m3
  • (C) 0.06 m3
  • (D) 0.17 m3
Show Answer

Answer: B – V = nRT ÷ P = (8 × 8.31 × 350) ÷ 200000 = 0.12 m3.

Q3. An ideal gas sample of 3 mol in a 0.02 m3 container exerts a pressure of 120 kPa. What is its temperature? (Use R = 8.31 J/mol·K.)

  • (A) 96.27 K
  • (B) 192.54 K
  • (C) 48.13 K
  • (D) 144.4 K
Show Answer

Answer: A – T = PV ÷ nR = (120000 × 0.02) ÷ (3 × 8.31) = 96.27 K.

Q4. A 1.2 kg brass disc with specific heat capacity 900 J/kg·K is heated through 40 K. How much thermal energy is transferred to it?

  • (A) 86400 J
  • (B) 21600 J
  • (C) 43200 J
  • (D) 64800 J
Show Answer

Answer: C – Q = mcΔT = 1.2 × 900 × 40 = 43200 J.

Q5. 300 J of heat is added to a gas in a piston while it does 200 J of work on its surroundings. What is the change in its internal energy?

  • (A) 200 J
  • (B) 100 J
  • (C) 50 J
  • (D) 150 J
Show Answer

Answer: B – First law: ΔU = Q – Wby = 300 – 200 = 100 J. A positive value means the internal energy increases.

Q6. A gas expands at a constant pressure of 80 kPa, increasing its volume by 0.02 m3. How much work does the gas do?

  • (A) 3200 J
  • (B) 1600 J
  • (C) 800 J
  • (D) 2400 J
Show Answer

Answer: B – At constant pressure, W = PΔV = 80000 × 0.02 = 1600 J.

Q7. A reversible heat engine operates between reservoirs at 600 K and 300 K. What is its maximum possible efficiency?

  • (A) 100 %
  • (B) 50 %
  • (C) 25 %
  • (D) 75 %
Show Answer

Answer: B – Carnot efficiency = 1 – TC ÷ TH = 1 – 300 ÷ 600 = 0.5 = 50%.

Q8. A heat engine takes in 1500 J of heat from a hot reservoir and rejects 300 J to a cold reservoir in one cycle. How much useful work does it do per cycle?

  • (A) 2400 J
  • (B) 600 J
  • (C) 1800 J
  • (D) 1200 J
Show Answer

Answer: D – W = QH – QC = 1500 – 300 = 1200 J.

Q9. The absolute temperature of a sample of nitrogen gas rises from 400 K to 1200 K, i.e. it is tripled. What happens to the average kinetic energy of its molecules?

  • (A) doubles
  • (B) halves
  • (C) triples
  • (D) quadruples
Show Answer

Answer: C – Average kinetic energy is proportional to absolute temperature, so going from 400 K to 1200 K triples the average kinetic energy.

Unit 10: Electric Force, Field, and Potential

Q10. A charge of 4 μC is placed in a uniform electric field of 200 N/C. What is the magnitude of the force on the charge, in mN?

  • (A) 1.6 mN
  • (B) 0.4 mN
  • (C) 1.2 mN
  • (D) 0.8 mN
Show Answer

Answer: D – F = qE = (4 × 10-6) × 200 = 0.0008 N = 0.8 mN.

Q11. What is the electric potential 20 cm from a point charge of 4 μC in a vacuum, taking potential at infinity as zero? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 360000 V
  • (B) 90000 V
  • (C) 180000 V
  • (D) 270000 V
Show Answer

Answer: C – V = kQ ÷ r = (9.0 × 109 × 4 × 10-6) ÷ 0.2 = 180000 V.

Q12. How much work is done by an external agent moving a 5 μC charge through a potential difference of 200 V at constant speed? Give the answer in mJ.

  • (A) 10 mJ
  • (B) 0.1 mJ
  • (C) 2 mJ
  • (D) 1 mJ
Show Answer

Answer: D – W = qΔV = (5 × 10-6) × 200 = 1000 μJ = 1 mJ.

Q13. The potential difference between two parallel plates 2 mm apart is 300 V. What is the magnitude of the uniform electric field between the plates?

  • (A) 300000 N/C
  • (B) 150000 N/C
  • (C) 75000 N/C
  • (D) 225000 N/C
Show Answer

Answer: B – E = V ÷ d = 300 ÷ 0 = 150000 N/C.

Q14. Two point charges of 5 μC and 3 μC are held 50 cm apart. What is their electric potential energy? (Use k = 9.0 × 109 N·m2/C2, and take zero at infinity.)

  • (A) 0.27 J
  • (B) 0.54 J
  • (C) 0.13 J
  • (D) 0.4 J
Show Answer

Answer: A – U = kq1q2 ÷ r = (9.0 × 109 × 5 × 10-6 × 3 × 10-6) ÷ 0.5 = 0.27 J.

Q15. A positively charged oil drop is placed in an electric field. In which direction does the electric force on it act?

  • (A) Opposite to the field
  • (B) In the direction of the field
  • (C) Perpendicular to the field
  • (D) In the direction of decreasing potential only
Show Answer

Answer: B – A positive charge experiences a force in the direction of the electric field, by definition of field direction.

Q16. A alpha particle is moved slowly along an equipotential surface around a point charge. How much work is done by the electric field?

  • (A) Positive
  • (B) Zero
  • (C) Negative
  • (D) Cannot be determined without the charge
Show Answer

Answer: B – An equipotential surface has the same potential everywhere, so the potential difference along it is zero and W = qΔV = 0.

Q17. The distance between two point charges is halved while both charges stay the same. By what factor does the electrostatic force between them change?

  • (A) It doubles
  • (B) It halves
  • (C) It becomes one-quarter as large
  • (D) It becomes four times as large
Show Answer

Answer: D – Coulomb’s law has force proportional to 1/r2, so halving r gives a factor 1/(1/2)2 = 4.

Q18. Parallel plates 2 mm apart have a potential difference of 200 V. A 2 μC charge between them experiences what force magnitude?

  • (A) 0.4 N
  • (B) 0.1 N
  • (C) 0.3 N
  • (D) 0.2 N
Show Answer

Answer: D – E = V ÷ d = 200 ÷ 0 = 100000 N/C. F = qE = (2 × 10-6) × 100000 = 0.2 N.

Unit 11: Electric Circuits

Q19. A current of 3 A flows through a 25 Ω resistor. How much power is dissipated as heat in the resistor?

  • (A) 450 W
  • (B) 112.5 W
  • (C) 337.5 W
  • (D) 225 W
Show Answer

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

Q20. A wire of length 20 m and cross-sectional area 2 mm2 is made of a material of resistivity 5 × 10-8 Ω·m. What is the resistance of the wire?

  • (A) 1 Ω
  • (B) 0.25 Ω
  • (C) 0.75 Ω
  • (D) 0.5 Ω
Show Answer

Answer: D – R = μL ÷ A = (5 × 10-8 × 20) ÷ (2 × 10-6) = 0.5 Ω.

Q21. A 50 μF capacitor is connected across a 12 V battery. What charge is stored on each plate, in μC?

  • (A) 1200 μC
  • (B) 300 μC
  • (C) 600 μC
  • (D) 900 μC
Show Answer

Answer: C – Q = CV = 50 × 12 = 600 μC.

Q22. A 50 μF capacitor charged to 30 V stores how much energy, in mJ?

  • (A) 45 mJ
  • (B) 11.25 mJ
  • (C) 22.5 mJ
  • (D) 33.75 mJ
Show Answer

Answer: C – U = (1/2)CV2 = 0.5 × (50 × 10-6) × 302 = 0.0225 J = 22.5 mJ.

Q23. Capacitors of 10 μF and 40 μF are connected in parallel across a 24 V battery. What total charge is stored, in μC?

  • (A) 2400 μC
  • (B) 600 μC
  • (C) 1800 μC
  • (D) 1200 μC
Show Answer

Answer: D – Ceq = 10 + 40 = 50 μF. Q = CeqV = 50 × 24 = 1200 μC.

Q24. Capacitors of 20 μF and 20 μF are connected in series across a 12 V battery. What charge is stored on each capacitor, in μC?

  • (A) 240 μC
  • (B) 120 μC
  • (C) 60 μC
  • (D) 180 μC
Show Answer

Answer: B – 1/Ceq = 1/20 + 1/20, so Ceq = 10 μF. Each series capacitor carries Q = CeqV = 10 × 12 = 120 μC.

Q25. A circuit contains a 20 kΩ resistor in series with a 20 μF capacitor. What is the time constant of the circuit?

  • (A) 0.8 s
  • (B) 0.2 s
  • (C) 0.6 s
  • (D) 0.4 s
Show Answer

Answer: D – μ = RC = (20 × 103) × (20 × 10-6) = 0.4 s.

Q26. A battery of emf 24 V and internal resistance 1 Ω delivers a current of 2 A. What is the terminal potential difference of the battery?

  • (A) 44 V
  • (B) 11 V
  • (C) 22 V
  • (D) 33 V
Show Answer

Answer: C – V = emf – Ir = 24 – (2 × 1) = 22 V.

Q27. Two identical bulbs are connected in series with a 24 V battery, then reconnected in parallel with the same battery. Compared with the series arrangement, each bulb in parallel is:

  • (A) dimmer
  • (B) the same brightness
  • (C) off
  • (D) brighter
Show Answer

Answer: D – In parallel each bulb gets the full battery voltage, whereas in series the voltage is shared, so each parallel bulb dissipates more power and glows brighter.

Unit 12: Magnetism and Electromagnetism

Q28. A metal rod of length 0.5 m moves at 2 m/s perpendicular to a uniform magnetic field of 0.5 T, with its length perpendicular to both motion and field. What motional emf is induced between its ends?

  • (A) 0.5 V
  • (B) 1 V
  • (C) 0.25 V
  • (D) 0.75 V
Show Answer

Answer: A – emf = BLv = 0.5 × 0.5 × 2 = 0.5 V.

Q29. A proton (mass 1.67 × 10-27 kg, charge 1.6 × 10-19 C) moves at 3 × 105 m/s perpendicular to a uniform magnetic field of 0.5 T. What is the radius of its circular path, in mm?

  • (A) 6.26 mm
  • (B) 12.53 mm
  • (C) 3.13 mm
  • (D) 9.39 mm
Show Answer

Answer: A – r = mv ÷ qB = (1.67 × 10-27 × 3 × 105) ÷ (1.6 × 10-19 × 0.5) = 0.0063 m = 6.26 mm.

Q30. The magnetic flux through a closed aluminium loop increases. According to Lenz’s law, the induced current in the loop creates a magnetic field that:

  • (A) opposes the increase in flux
  • (B) opposes the decrease in flux
  • (C) reinforces the change in flux
  • (D) is zero
Show Answer

Answer: A – Lenz’s law states that an induced current opposes the change in flux that produced it, so its field opposes the increase.


Q31. What is the magnetic field at the centre of a circular loop of radius 8 cm carrying a current of 8 A?

  • (A) 62.83 μT
  • (B) 125.66 μT
  • (C) 31.42 μT
  • (D) 94.25 μT
Show Answer

Answer: A – B = μ0I ÷ (2R) = (4π × 10-7 × 8) ÷ (2 × 0.08) = 62.83 μT.

Q32. Two long parallel metal rods carry currents in the opposite direction. The magnetic force between them is:

  • (A) attractive
  • (B) zero
  • (C) repulsive
  • (D) perpendicular to both wires
Show Answer

Answer: C – Parallel currents in the same direction attract each other; opposite currents repel.

Q33. A 4 μC particle moves at 4 × 104 m/s perpendicular to a magnetic field of 0.5 T. What is the magnitude of the magnetic force on it, in mN?

  • (A) 80 mN
  • (B) 160 mN
  • (C) 40 mN
  • (D) 120 mN
Show Answer

Answer: A – F = qvB = (4 × 10-6) × (4 × 104) × 0.5 = 0.08 N = 80 mN.

Q34. A straight wire of length 0.4 m carries a current of 2 A perpendicular to a uniform magnetic field of 1.5 T. What is the magnitude of the force on the wire?

  • (A) 2.4 N
  • (B) 0.6 N
  • (C) 1.2 N
  • (D) 1.8 N
Show Answer

Answer: C – F = ILB = 2 × 0.4 × 1.5 = 1.2 N.

Q35. What is the magnitude of the magnetic field 40 cm from a long straight wire carrying a current of 10 A?

  • (A) 5 μT
  • (B) 10 μT
  • (C) 2.5 μT
  • (D) 7.5 μT
Show Answer

Answer: A – B = μ0I ÷ (2πr) = (2 × 10-7 × 10) ÷ 0.4 = 5 μT.

Unit 13: Geometric Optics

Q36. Total internal reflection between water and air can occur only when light travels:

  • (A) from water to air at an angle above the critical angle
  • (B) from air to water
  • (C) at normal incidence
  • (D) from water to air at any angle
Show Answer

Answer: A – Total internal reflection needs light going from higher to lower refractive index, here water to air, with incidence above the critical angle.

Q37. An candle stands 6 m in front of a plane mirror. How far behind the mirror does its image appear to be?

  • (A) 12 m
  • (B) 6 m
  • (C) 3 m
  • (D) 9 m
Show Answer

Answer: B – A plane mirror forms an image as far behind the mirror as the object is in front, so the image distance is 6 m.

Q38. A concave lens of focal length -30 cm always forms an image that is:

  • (A) virtual, upright and diminished
  • (B) real, inverted and magnified
  • (C) virtual, inverted and magnified
  • (D) real, upright and diminished
Show Answer

Answer: A – A diverging lens spreads rays apart, so the image is always virtual, upright and smaller than the object.

Q39. A ray strikes a plane mirror at 70° to the normal. At what angle to the normal is the reflected ray?

  • (A) 140 degrees
  • (B) 35 degrees
  • (C) 70 degrees
  • (D) 105 degrees
Show Answer

Answer: C – By the law of reflection, angle of reflection equals angle of incidence, so it is 70° to the normal.

Q40. An object is placed 54 cm in front of a concave mirror of focal length 18 cm. How far from the mirror is the image formed?

  • (A) 27 cm
  • (B) 54 cm
  • (C) 13.5 cm
  • (D) 40.5 cm
Show Answer

Answer: A – 1/f = 1/do + 1/di, so 1/di = 1/18 – 1/54, giving di = 27 cm.

Q41. An object 27 cm in front of a concave mirror forms an image 54 cm from the mirror. What is the lateral magnification?

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

Answer: B – m = -di ÷ do = -54 ÷ 27 = -2. The negative sign shows the image is inverted.

Q42. An object is placed 27 cm in front of a converging lens of focal length 18 cm. How far from the lens is the image formed?

  • (A) 54 cm
  • (B) 108 cm
  • (C) 27 cm
  • (D) 81 cm
Show Answer

Answer: A – 1/f = 1/do + 1/di, so 1/di = 1/18 – 1/27, giving di = 54 cm.

Q43. A concave mirror has a radius of curvature of 22 cm. What is its focal length?

  • (A) 22 cm
  • (B) 5.5 cm
  • (C) 11 cm
  • (D) 16.5 cm
Show Answer

Answer: C – f = R ÷ 2 = 22 ÷ 2 = 11 cm.

Unit 14: Waves, Sound, and Physical Optics

Q44. An oil film in air appears dark for a certain wavelength in reflected light. Which condition on its thickness t best describes this, for the thinnest such film case?

  • (A) 2t = (m + 1/2) times the wavelength in the film
  • (B) t = the wavelength in the film
  • (C) 2t = one-quarter times the wavelength in the film
  • (D) 2t = m times the wavelength in the film
Show Answer

Answer: D – With one phase reversal at the first surface, destructive interference in reflection needs a path difference of a whole number of wavelengths, so 2t is a whole-number multiple of the wavelength in the film.

Q45. A wave has frequency 200 Hz and wavelength 6.8 m. What is its speed?

  • (A) 2720 m/s
  • (B) 680 m/s
  • (C) 2040 m/s
  • (D) 1360 m/s
Show Answer

Answer: D – v = fλ = 200 × 6.8 = 1360 m/s.

Q46. A oscillator completes one oscillation in 2 s. What is its frequency?

  • (A) 1 Hz
  • (B) 0.25 Hz
  • (C) 0.75 Hz
  • (D) 0.5 Hz
Show Answer

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

Q47. A string of length 1.2 m fixed at both ends supports waves travelling at 120 m/s. What is its fundamental frequency?

  • (A) 100 Hz
  • (B) 25 Hz
  • (C) 50 Hz
  • (D) 75 Hz
Show Answer

Answer: C – For a string fixed at both ends, fn = nv ÷ (2L) = 1 × 120 ÷ (2 × 1.2) = 50 Hz.

Q48. An open pipe of length 1.1 m resonates with sound waves travelling at 340 m/s. What is its fundamental frequency?

  • (A) 154.55 Hz
  • (B) 309.09 Hz
  • (C) 77.27 Hz
  • (D) 231.82 Hz
Show Answer

Answer: A – For an open pipe, f1 = v ÷ (2L) = 340 ÷ (2 × 1.1) = 154.55 Hz.

Q49. A pipe closed at one end has length 0.5 m. Sound travels at 340 m/s. What is its fundamental frequency?

  • (A) 340 Hz
  • (B) 85 Hz
  • (C) 255 Hz
  • (D) 170 Hz
Show Answer

Answer: D – For a pipe closed at one end, f1 = v ÷ (4L) = 340 ÷ (4 × 0.5) = 170 Hz.

Q50. Two tuning forks of frequencies 480 Hz and 490 Hz sound together. What is the beat frequency heard?

  • (A) 10 Hz
  • (B) 20 Hz
  • (C) 5 Hz
  • (D) 15 Hz
Show Answer

Answer: A – Beat frequency = |f1 – f2| = |480 – 490| = 10 Hz.

Q51. A source emitting 850 Hz moves towards a stationary observer at 68 m/s. Sound travels at 340 m/s. What frequency does the observer hear?

  • (A) 2125 Hz
  • (B) 1062.5 Hz
  • (C) 531.25 Hz
  • (D) 1593.75 Hz
Show Answer

Answer: B – f’ = f × v ÷ (v – vs) = 850 × 340 ÷ (340 – 68) = 1062.5 Hz.

Unit 15: Modern Physics

Q52. Light photons of energy 6 eV strike a metal with work function 2 eV. What is the maximum kinetic energy of the emitted photoelectrons, in eV?

  • (A) 8 eV
  • (B) 2 eV
  • (C) 4 eV
  • (D) 6 eV
Show Answer

Answer: C – Kmax = hf – Φ = 6 – 2 = 4 eV.

Q53. Photoelectrons are emitted with maximum kinetic energy 1.5 eV. What stopping potential is needed to bring the fastest electrons to rest?

  • (A) 3 V
  • (B) 0.75 V
  • (C) 1.5 V
  • (D) 2.25 V
Show Answer

Answer: C – eVs = Kmax, so Vs = 1.5 eV ÷ e = 1.5 V.

Q54. An electron (mass 9.11 × 10-31 kg) moves at 3 × 106 m/s. What is its de Broglie wavelength, in pm? (Use h = 6.63 × 10-34 J·s.)

  • (A) 485.18 pm
  • (B) 242.59 pm
  • (C) 121.3 pm
  • (D) 363.89 pm
Show Answer

Answer: B – λ = h ÷ p = h ÷ (mv) = (6.63 × 10-34) ÷ (9.11 × 10-31 × 3 × 106) = 0.243 nm = 242.59 pm.

Q55. A hydrogen atom drops from energy level n = 4 to n = 3. What is the energy of the photon emitted, in eV? (Levels: En = -13.6 ÷ n2 eV.)

  • (A) 1.32 eV
  • (B) 0.66 eV
  • (C) 0.33 eV
  • (D) 0.99 eV
Show Answer

Answer: B – ΔE = 13.6 × (1/32 – 1/42) = 0.66 eV.

Q56. Calculate the energy of a hydrogen atom in level n = 3, in units of 10-19 J? (Use En = -13.6 ÷ n2 eV and 1 eV = 1.6 × 10-19 J.)

  • (A) -4.84 ×10-19 J
  • (B) -1.21 ×10-19 J
  • (C) 4.84 ×10-19 J
  • (D) -2.42 ×10-19 J
Show Answer

Answer: D – E3 = -13.6 ÷ 32 = -1.51 eV = -2.42 × 10-19 J.

Q57. A radioactive sample of mass 80 g has a half-life of one hour. How much of it remains undecayed after 2 hours?

  • (A) 20 g
  • (B) 40 g
  • (C) 10 g
  • (D) 30 g
Show Answer

Answer: A – After 2 half-lives the fraction remaining is (1/2)2 = 0.25, so remaining mass = 80 × 0.25 = 20 g.

Q58. A nuclear reaction has a mass defect of 0.05 u. How much energy is released? (Use 1 u = 931.5 MeV/c2.)

  • (A) 93.15 MeV
  • (B) 23.29 MeV
  • (C) 46.58 MeV
  • (D) 69.86 MeV
Show Answer

Answer: C – E = Δm c2 = 0.05 × 931.5 = 46.58 MeV.

Q59. A caesium surface has work function 2 eV. What is its threshold frequency, in units of 1014 Hz? (Use h = 6.63 × 10-34 J·s and 1 eV = 1.6 × 10-19 J.)

  • (A) 9.65 ×1014 Hz
  • (B) 2.41 ×1014 Hz
  • (C) 4.83 ×1014 Hz
  • (D) 7.24 ×1014 Hz
Show Answer

Answer: C – f0 = Φ ÷ h = (2 × 1.6 × 10-19) ÷ (6.63 × 10-34) = 4.83 × 1014 Hz.

Q60. The photoelectric effect shows that light behaves as:

  • (A) waves only
  • (B) particles (photons)
  • (C) a continuous fluid
  • (D) a magnetic field only
Show Answer

Answer: B – The photoelectric effect depends on photon energy and momentum in a way explained by light arriving as discrete photons.

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