AP Physics 2 Practice Test 18 – 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 reservoirs at 700 K and 300 K. What is its maximum possible efficiency?

  • (A) 114.29 %
  • (B) 57.14 %
  • (C) 28.57 %
  • (D) 85.71 %
Show Answer

Answer: B – Carnot efficiency = 1 – TC ÷ TH = 1 – 300 ÷ 700 = 0.571 = 57.14%.

Q2. A heat engine takes in 1000 J of heat from a hot reservoir and rejects 800 J to a cold reservoir in one cycle. How much useful work does it do per cycle?

  • (A) 400 J
  • (B) 200 J
  • (C) 100 J
  • (D) 300 J
Show Answer

Answer: B – W = QH – QC = 1000 – 800 = 200 J.

Q3. The absolute temperature of a sample of neon gas rises from 350 K to 175 K, i.e. it is halved. What happens to the average kinetic energy of its molecules?

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

Answer: D – Average kinetic energy is proportional to absolute temperature, so going from 350 K to 175 K halves the average kinetic energy.

Q4. A sample of air expands slowly in a water bath so its temperature stays constant. Which name describes this process?

  • (A) isobaric
  • (B) isothermal
  • (C) adiabatic
  • (D) isovolumetric
Show Answer

Answer: B – The process is isothermal: temperature is constant.

Q5. How much energy is needed to melt 1.2 kg of ice at its melting point without changing its temperature? (Latent heat = 334000 J/kg.)

  • (A) 400800 J
  • (B) 801600 J
  • (C) 200400 J
  • (D) 601200 J
Show Answer

Answer: A – Q = mL = 1.2 × 334000 = 400800 J.

Q6. 9000 J of heat raises the temperature of a 1.5 kg sample by 20 K. What is the specific heat capacity of the sample material?

  • (A) 600 J/kg·K
  • (B) 150 J/kg·K
  • (C) 450 J/kg·K
  • (D) 300 J/kg·K
Show Answer

Answer: D – c = Q ÷ (mΔT) = 9000 ÷ (1.5 × 20) = 300 J/kg·K.

Q7. What is the internal energy of 1 mol of a monatomic ideal gas at 500 K? (Use U = (3/2)nRT and R = 8.31 J/mol·K.)

  • (A) 12465 J
  • (B) 3116.25 J
  • (C) 9348.75 J
  • (D) 6232.5 J
Show Answer

Answer: D – U = (3/2)nRT = 1.5 × 1 × 8.31 × 500 = 6232.5 J.

Q8. A gas expands adiabatically and does 200 J of work on its surroundings. What is the change in its internal energy?

  • (A) -400 J
  • (B) -100 J
  • (C) -300 J
  • (D) -200 J
Show Answer

Answer: D – In an adiabatic process Q = 0, so ΔU = Q – Wby = 0 – 200 = -200 J. The internal energy decreases.

Q9. A storage cylinder holds 4.5 mol of an ideal gas at 450 K in a volume of 0.01 m3. What is the pressure of the gas, in kPa? (Use R = 8.31 J/mol·K.)

  • (A) 1121.85 kPa
  • (B) 2243.7 kPa
  • (C) 560.92 kPa
  • (D) 1682.77 kPa
Show Answer

Answer: A – P = nRT ÷ V = (4.5 × 8.31 × 450) ÷ 0.01 = 1121850 Pa = 1121.85 kPa.

Unit 10: Electric Force, Field, and Potential

Q10. A charge is moved slowly along an equipotential surface around a charged sphere. How much work is done by the electric field?

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

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

Q11. The distance between two point charges is doubled 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 four times as large
  • (D) It becomes one-quarter as large
Show Answer

Answer: D – Coulomb’s law has force proportional to 1/r2, so doubling r gives a factor 1/22 = 1/4.

Q12. Parallel plates 8 mm apart have a potential difference of 300 V. A 3 μC charge between them experiences what force magnitude?

  • (A) 0.23 N
  • (B) 0.06 N
  • (C) 0.11 N
  • (D) 0.17 N
Show Answer

Answer: C – E = V ÷ d = 300 ÷ 0.01 = 37500 N/C. F = qE = (3 × 10-6) × 37500 = 0.11 N.

Q13. In a uniform field of 800 N/C, a charge moves 5 cm parallel to the field. What is the magnitude of the potential difference between its start and end points?

  • (A) 40 V
  • (B) 80 V
  • (C) 20 V
  • (D) 60 V
Show Answer

Answer: A – |ΔV| = Ed = 800 × 0.05 = 40 V.

Q14. Two point charges of 8 μC and 8 μC are 20 cm apart in a vacuum. What is the magnitude of the electrostatic force between them? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 28.8 N
  • (B) 7.2 N
  • (C) 14.4 N
  • (D) 21.6 N
Show Answer

Answer: C – F = kq1q2 ÷ r2 = (9.0 × 109 × 8 × 10-6 × 8 × 10-6) ÷ (0.2)2 = 14.4 N.

Q15. What is the magnitude of the electric field 10 cm from a point charge of 2 μC in a vacuum? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 3600000 N/C
  • (B) 1800000 N/C
  • (C) 900000 N/C
  • (D) 2700000 N/C
Show Answer

Answer: B – E = kQ ÷ r2 = (9.0 × 109 × 2 × 10-6) ÷ (0.1)2 = 1800000 N/C.

Q16. A charge of 5 μC is placed in a uniform electric field of 500 N/C. What is the magnitude of the force on the charge, in mN?

  • (A) 5 mN
  • (B) 1.25 mN
  • (C) 3.75 mN
  • (D) 2.5 mN
Show Answer

Answer: D – F = qE = (5 × 10-6) × 500 = 0.0025 N = 2.5 mN.

Q17. What is the electric potential 10 cm from a point charge of 8 μC in a vacuum, taking potential at infinity as zero? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 720000 V
  • (B) 1440000 V
  • (C) 360000 V
  • (D) 1080000 V
Show Answer

Answer: A – V = kQ ÷ r = (9.0 × 109 × 8 × 10-6) ÷ 0.1 = 720000 V.

Q18. How much work is done by an external agent moving a 5 μC charge through a potential difference of 250 V at constant speed? Give the answer in mJ.

  • (A) 12.5 mJ
  • (B) 0.12 mJ
  • (C) 1.25 mJ
  • (D) 2.5 mJ
Show Answer

Answer: C – W = qΔV = (5 × 10-6) × 250 = 1250 μJ = 1.25 mJ.

Unit 11: Electric Circuits

Q19. A circuit contains a 100 kΩ resistor in series with a 50 μF capacitor. What is the time constant of the circuit?

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

Answer: B – μ = RC = (100 × 103) × (50 × 10-6) = 5 s.

Q20. A battery of emf 12 V and internal resistance 2 Ω delivers a current of 3 A. What is the terminal potential difference of the battery?

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

Answer: A – V = emf – Ir = 12 – (3 × 2) = 6 V.

Q21. Two identical lamps are connected in series with a 12 V battery, then reconnected in parallel with the same battery. Compared with the series arrangement, each lamp in parallel is:

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

Answer: C – 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.

Q22. A total current of 6 A splits between parallel resistors of 4 Ω and 4 Ω. What current flows through the 4 Ω resistor?

  • (A) 6 A
  • (B) 1.5 A
  • (C) 3 A
  • (D) 4.5 A
Show Answer

Answer: C – Current divides in inverse proportion to resistance: I4 = Itotal × 4 ÷ (4 + 4) = 6 × 4 ÷ 8 = 3 A.

Q23. A resistor of 6 Ω is connected across a 12 V battery. What current flows through the resistor?

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

Answer: C – I = V ÷ R = 12 ÷ 6 = 2 A.

Q24. Resistors of 12 Ω, 2 Ω and 6 Ω are connected in series across a 30 V battery. What is the current in the circuit?

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

Answer: A – Req = 12 + 2 + 6 = 20 Ω. I = V ÷ Req = 30 ÷ 20 = 1.5 A.

Q25. Resistors of 12 Ω and 6 Ω are connected in parallel across a 36 V battery. What is the total current drawn from the battery?

  • (A) 18 A
  • (B) 4.5 A
  • (C) 13.5 A
  • (D) 9 A
Show Answer

Answer: D – 1/Req = 1/12 + 1/6, so Req = 4 Ω. I = V ÷ Req = 36 ÷ 4 = 9 A.

Q26. A device operates at 12 V and draws 3 A. At what rate does it dissipate energy?

  • (A) 36 W
  • (B) 72 W
  • (C) 18 W
  • (D) 54 W
Show Answer

Answer: A – P = IV = 3 × 12 = 36 W.

Q27. A current of 5 A flows through a 5 Ω resistor. How much power is dissipated as heat in the resistor?

  • (A) 250 W
  • (B) 62.5 W
  • (C) 187.5 W
  • (D) 125 W
Show Answer

Answer: D – P = I2R = 52 × 5 = 125 W.

Unit 12: Magnetism and Electromagnetism

Q28. A straight wire of length 0.2 m carries a current of 6 A perpendicular to a uniform magnetic field of 0.2 T. What is the magnitude of the force on the wire?

  • (A) 0.24 N
  • (B) 0.48 N
  • (C) 0.12 N
  • (D) 0.36 N
Show Answer

Answer: A – F = ILB = 6 × 0.2 × 0.2 = 0.24 N.

Q29. What is the magnitude of the magnetic field 5 cm from a long straight wire carrying a current of 8 A?

  • (A) 32 μT
  • (B) 64 μT
  • (C) 16 μT
  • (D) 48 μT
Show Answer

Answer: A – B = μ0I ÷ (2πr) = (2 × 10-7 × 8) ÷ 0.05 = 32 μT.

Q30. A solenoid has 200 turns per metre and carries a current of 5 A. What is the magnetic field inside the solenoid, in mT? (Use μ0 = 4π × 10-7 T·m/A.)

  • (A) 2.51 mT
  • (B) 0.63 mT
  • (C) 1.88 mT
  • (D) 1.26 mT
Show Answer

Answer: D – B = μ0nI = (4π × 10-7) × 200 × 5 = 0.00126 T = 1.26 mT.


Q31. A uniform magnetic field of 0.5 T passes through a flat loop of area 0.1 m2. The field makes an angle of 60° with the normal to the loop. What is the magnetic flux through the loop, in mWb?

  • (A) 25 mWb
  • (B) 50 mWb
  • (C) 12.5 mWb
  • (D) 37.5 mWb
Show Answer

Answer: A – Φ = BA cosθ = 0.5 × 0.1 × 0.5 = 0.03 Wb = 25 mWb.

Q32. The magnetic flux through a single loop changes by 0.05 Wb in 0.1 s. What is the magnitude of the average induced emf in the loop?

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

Answer: D – emf = |ΔΦ ÷ Δt| = 0.05 ÷ 0.1 = 0.5 V.

Q33. A metal rod of length 0.4 m moves at 2 m/s perpendicular to a uniform magnetic field of 0.2 T, with its length perpendicular to both motion and field. What motional emf is induced between its ends?

  • (A) 0.32 V
  • (B) 0.08 V
  • (C) 0.16 V
  • (D) 0.24 V
Show Answer

Answer: C – emf = BLv = 0.2 × 0.4 × 2 = 0.16 V.

Q34. A proton (mass 1.67 × 10-27 kg, charge 1.6 × 10-19 C) moves at 2 × 105 m/s perpendicular to a uniform magnetic field of 0.8 T. What is the radius of its circular path, in mm?

  • (A) 2.61 mm
  • (B) 5.22 mm
  • (C) 1.3 mm
  • (D) 3.91 mm
Show Answer

Answer: A – r = mv ÷ qB = (1.67 × 10-27 × 2 × 105) ÷ (1.6 × 10-19 × 0.8) = 0.0026 m = 2.61 mm.

Q35. The magnetic flux through a closed copper loop decreases. According to Lenz’s law, the induced current in the loop creates a magnetic field that:

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

Answer: C – Lenz’s law states that an induced current opposes the change in flux that produced it, so its field tries to maintain the flux and opposes the decrease.

Unit 13: Geometric Optics

Q36. An object is placed 62.5 cm in front of a converging lens of focal length 25 cm. How far from the lens is the image formed?

  • (A) 83.33 cm
  • (B) 20.83 cm
  • (C) 41.67 cm
  • (D) 62.5 cm
Show Answer

Answer: C – 1/f = 1/do + 1/di, so 1/di = 1/25 – 1/62.5, giving di = 41.67 cm.

Q37. A concave mirror has a radius of curvature of 18 cm. What is its focal length?

  • (A) 18 cm
  • (B) 4.5 cm
  • (C) 9 cm
  • (D) 13.5 cm
Show Answer

Answer: C – f = R ÷ 2 = 18 ÷ 2 = 9 cm.

Q38. A ray in air enters a transparent block at an angle of incidence of 60° and refracts at 45° to the normal. Using sin 60° = 0.866 and sin 45° = 0.707, what is the refractive index of the block?

  • (A) 2.44
  • (B) 1.22
  • (C) 0.61
  • (D) 1.83
Show Answer

Answer: B – Snell: n1 sinθ1 = n2 sinθ2, so n2 = (1 × 0.866) ÷ 0.707 = 1.22.

Q39. What is the critical angle for light travelling from a medium of refractive index 1.52 into air? (Use sinθc = 1 ÷ n.)

  • (A) 82.2 degrees
  • (B) 41.1 degrees
  • (C) 20.55 degrees
  • (D) 61.65 degrees
Show Answer

Answer: B – sinθc = 1 ÷ 1.52 = 0.658, which gives θc = 41.1°.

Q40. Light travels at 2.4 × 108 m/s in a plastic sample. What is the refractive index of the material? (Use c = 3.0 × 108 m/s.)

  • (A) 2.5
  • (B) 1.25
  • (C) 0.62
  • (D) 1.88
Show Answer

Answer: B – n = c ÷ v = (3.0 × 108) ÷ (2.4 × 108) = 1.25.

Q41. An object is placed 10 cm from a converging lens of focal length 20 cm, so it is inside the focal length. The image formed is:

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

Answer: B – For a converging lens with the object inside f, the rays diverge after the lens and appear to come from a virtual, upright, magnified image on the same side as the object.

Q42. Total internal reflection between diamond and air can occur only when light travels:

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

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

Q43. An candle stands 1.2 m in front of a plane mirror. How far behind the mirror does its image appear to be?

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

Answer: C – A plane mirror forms an image as far behind the mirror as the object is in front, so the image distance is 1.2 m.

Unit 14: Waves, Sound, and Physical Optics

Q44. Two tuning forks of frequencies 320 Hz and 330 Hz sound together. What is the beat frequency heard?

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

Answer: C – Beat frequency = |f1 – f2| = |320 – 330| = 10 Hz.

Q45. A source emitting 500 Hz moves towards a stationary observer at 68 m/s. Sound travels at 340 m/s. What frequency does the observer hear?

  • (A) 625 Hz
  • (B) 1250 Hz
  • (C) 312.5 Hz
  • (D) 937.5 Hz
Show Answer

Answer: A – f’ = f × v ÷ (v – vs) = 500 × 340 ÷ (340 – 68) = 625 Hz.

Q46. In a double-slit experiment, light of wavelength 600 nm passes through slits 0.5 mm apart, with a screen 2 m away. How far from the central maximum is the 1th bright fringe, in mm?

  • (A) 4.8 mm
  • (B) 1.2 mm
  • (C) 3.6 mm
  • (D) 2.4 mm
Show Answer

Answer: D – y = mλL ÷ d = 1 × (600 × 10-9) × 2 ÷ (0.5 × 10-3) = 0.0024 m = 2.4 mm.

Q47. A point source radiates 80 W of sound uniformly in all directions. What is the intensity at a distance of 4 m? (Use π = 3.14.)

  • (A) 0.8 W/m2
  • (B) 0.2 W/m2
  • (C) 0.6 W/m2
  • (D) 0.4 W/m2
Show Answer

Answer: D – I = P ÷ (4πr2) = 80 ÷ (4 × 3.14 × 42) = 0.3979 W/m2.

Q48. In a single-slit diffraction pattern, using a shorter wavelength while keeping the other factors fixed causes the central maximum to become:

  • (A) wider
  • (B) narrower
  • (C) unchanged in width
  • (D) brighter but the same width
Show Answer

Answer: B – The angular width of the central maximum is proportional to wavelength divided by slit width, so a shorter wavelength gives a narrower central maximum.

Q49. An oil film in air appears bright 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 times the wavelength in the film
  • (B) 2t = (m + 1/2) times the wavelength in the film
  • (C) t = the wavelength in the film
  • (D) 2t = one-quarter times the wavelength in the film
Show Answer

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

Q50. A wave has frequency 100 Hz and wavelength 4 m. What is its speed?

  • (A) 800 m/s
  • (B) 400 m/s
  • (C) 200 m/s
  • (D) 600 m/s
Show Answer

Answer: B – v = fλ = 100 × 4 = 400 m/s.

Q51. A oscillator completes one oscillation in 5 s. What is its frequency?

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

Answer: D – f = 1 ÷ T = 1 ÷ 5 = 0.2 Hz.

Unit 15: Modern Physics

Q52. A radioactive decay has a mass defect of 0.02 u. How much energy is released? (Use 1 u = 931.5 MeV/c2.)

  • (A) 37.26 MeV
  • (B) 9.31 MeV
  • (C) 27.95 MeV
  • (D) 18.63 MeV
Show Answer

Answer: D – E = Δm c2 = 0.02 × 931.5 = 18.63 MeV.

Q53. A zinc surface has work function 4.5 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) 21.72 ×1014 Hz
  • (B) 10.86 ×1014 Hz
  • (C) 5.43 ×1014 Hz
  • (D) 16.29 ×1014 Hz
Show Answer

Answer: B – f0 = Φ ÷ h = (4.5 × 1.6 × 10-19) ÷ (6.63 × 10-34) = 10.86 × 1014 Hz.

Q54. The Compton scattering is best explained if light behaves as:

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

Answer: A – The Compton scattering depends on photon energy and momentum in a way explained by light arriving as discrete photons.

Q55. According to the Heisenberg uncertainty principle, measuring a electron’s momentum more precisely necessarily:

  • (A) increases the uncertainty in its momentum
  • (B) increases the uncertainty in its position
  • (C) decreases the uncertainty in the other quantity
  • (D) leaves the other uncertainty unchanged
Show Answer

Answer: B – Position and momentum uncertainties obey a joint limit, so reducing momentum uncertainty increases the minimum possible position uncertainty.

Q56. What is the energy, in units of 10-19 J, of a photon of frequency 4 × 1014 Hz? (Use h = 6.63 × 10-34 J·s.)

  • (A) 5.3 ×10-19 J
  • (B) 1.33 ×10-19 J
  • (C) 2.65 ×10-19 J
  • (D) 3.98 ×10-19 J
Show Answer

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

Q57. Light photons of energy 5 eV strike a metal with work function 2.5 eV. What is the maximum kinetic energy of the emitted photoelectrons, in eV?

  • (A) 5 eV
  • (B) 1.25 eV
  • (C) 3.75 eV
  • (D) 2.5 eV
Show Answer

Answer: D – Kmax = hf – Φ = 5 – 2.5 = 2.5 eV.

Q58. Photoelectrons are emitted with maximum kinetic energy 1.2 eV. What stopping potential is needed to bring the fastest electrons to rest?

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

Answer: B – eVs = Kmax, so Vs = 1.2 eV ÷ e = 1.2 V.

Q59. An electron (mass 9.11 × 10-31 kg) moves at 1 × 106 m/s. What is its de Broglie wavelength, in pm? (Use h = 6.63 × 10-34 J·s.)

  • (A) 1455.54 pm
  • (B) 363.89 pm
  • (C) 727.77 pm
  • (D) 1091.66 pm
Show Answer

Answer: C – λ = h ÷ p = h ÷ (mv) = (6.63 × 10-34) ÷ (9.11 × 10-31 × 1 × 106) = 0.728 nm = 727.77 pm.

Q60. A hydrogen atom transitions from energy level n = 5 to n = 4. What is the energy of the photon emitted, in eV? (Levels: En = -13.6 ÷ n2 eV.)

  • (A) 0.31 eV
  • (B) 0.62 eV
  • (C) 0.15 eV
  • (D) 0.46 eV
Show Answer

Answer: A – ΔE = 13.6 × (1/42 – 1/52) = 0.31 eV.

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