AP Physics 2 Practice Test 20 – 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. What is the internal energy of 4 mol of a monatomic ideal gas at 300 K? (Use U = (3/2)nRT and R = 8.31 J/mol·K.)

  • (A) 29916 J
  • (B) 14958 J
  • (C) 7479 J
  • (D) 22437 J
Show Answer

Answer: B – U = (3/2)nRT = 1.5 × 4 × 8.31 × 300 = 14958 J.

Q2. A gas expands adiabatically and does 600 J of work on its surroundings. What is the change in its internal energy?

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

Answer: B – In an adiabatic process Q = 0, so ΔU = Q – Wby = 0 – 600 = -600 J. The internal energy decreases.

Q3. A laboratory flask holds 3.5 mol of an ideal gas at 320 K in a volume of 0.03 m3. What is the pressure of the gas, in kPa? (Use R = 8.31 J/mol·K.)

  • (A) 620.48 kPa
  • (B) 310.24 kPa
  • (C) 155.12 kPa
  • (D) 465.36 kPa
Show Answer

Answer: B – P = nRT ÷ V = (3.5 × 8.31 × 320) ÷ 0.03 = 310240 Pa = 310.24 kPa.

Q4. 6 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) 0.09 m3
  • (B) 0.17 m3
  • (C) 0.04 m3
  • (D) 0.13 m3
Show Answer

Answer: A – V = nRT ÷ P = (6 × 8.31 × 350) ÷ 200000 = 0.09 m3.

Q5. An ideal gas sample of 5 mol in a 0.02 m3 container exerts a pressure of 200 kPa. What is its temperature? (Use R = 8.31 J/mol·K.)

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

Answer: D – T = PV ÷ nR = (200000 × 0.02) ÷ (5 × 8.31) = 96.27 K.

Q6. A 0.8 kg iron plate with specific heat capacity 385 J/kg·K is heated through 15 K. How much thermal energy is transferred to it?

  • (A) 9240 J
  • (B) 4620 J
  • (C) 2310 J
  • (D) 6930 J
Show Answer

Answer: B – Q = mcΔT = 0.8 × 385 × 15 = 4620 J.

Q7. 300 J of heat is added to a gas sample while it does 150 J of work on its surroundings. What is the change in its internal energy?

  • (A) 150 J
  • (B) 300 J
  • (C) 75 J
  • (D) 225 J
Show Answer

Answer: A – First law: ΔU = Q – Wby = 300 – 150 = 150 J. A positive value means the internal energy increases.

Q8. A gas expands at a constant pressure of 150 kPa, increasing its volume by 0.05 m3. How much work does the gas do?

  • (A) 15000 J
  • (B) 3750 J
  • (C) 7500 J
  • (D) 11250 J
Show Answer

Answer: C – At constant pressure, W = PΔV = 150000 × 0.05 = 7500 J.

Q9. A reversible heat engine operates between reservoirs at 800 K and 350 K. What is its maximum possible efficiency?

  • (A) 112.5 %
  • (B) 28.12 %
  • (C) 56.25 %
  • (D) 84.38 %
Show Answer

Answer: C – Carnot efficiency = 1 – TC ÷ TH = 1 – 350 ÷ 800 = 0.562 = 56.25%.

Unit 10: Electric Force, Field, and Potential

Q10. A charge of 2 μC is placed in a uniform electric field of 1000 N/C. What is the magnitude of the force on the charge, in mN?

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

Answer: B – F = qE = (2 × 10-6) × 1000 = 0.002 N = 2 mN.

Q11. What is the electric potential 40 cm from a point charge of 6 μC in a vacuum, taking potential at infinity as zero? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 270000 V
  • (B) 67500 V
  • (C) 202500 V
  • (D) 135000 V
Show Answer

Answer: D – V = kQ ÷ r = (9.0 × 109 × 6 × 10-6) ÷ 0.4 = 135000 V.

Q12. How much work is done by an external agent moving a 4 μC charge through a potential difference of 400 V at constant speed? Give the answer in mJ.

  • (A) 16 mJ
  • (B) 0.16 mJ
  • (C) 1.6 mJ
  • (D) 3.2 mJ
Show Answer

Answer: C – W = qΔV = (4 × 10-6) × 400 = 1600 μJ = 1.6 mJ.

Q13. The potential difference between two parallel plates 5 mm apart is 600 V. What is the magnitude of the uniform electric field between the plates?

  • (A) 120000 N/C
  • (B) 240000 N/C
  • (C) 60000 N/C
  • (D) 180000 N/C
Show Answer

Answer: A – E = V ÷ d = 600 ÷ 0.01 = 120000 N/C.

Q14. Two point charges of 2 μC and 4 μC are held 20 cm apart. What is their electric potential energy? (Use k = 9.0 × 109 N·m2/C2, and take zero at infinity.)

  • (A) 0.72 J
  • (B) 0.18 J
  • (C) 0.36 J
  • (D) 0.54 J
Show Answer

Answer: C – U = kq1q2 ÷ r = (9.0 × 109 × 2 × 10-6 × 4 × 10-6) ÷ 0.2 = 0.36 J.

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

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

Answer: D – 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 a flat equipotential plane between parallel plates. 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 charged particles 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 10 mm apart have a potential difference of 120 V. A 5 μC charge between them experiences what force magnitude?

  • (A) 0.12 N
  • (B) 0.03 N
  • (C) 0.09 N
  • (D) 0.06 N
Show Answer

Answer: D – E = V ÷ d = 120 ÷ 0.01 = 12000 N/C. F = qE = (5 × 10-6) × 12000 = 0.06 N.

Unit 11: Electric Circuits

Q19. Resistors of 12 Ω and 6 Ω are connected in parallel across a 12 V battery. What is the total current drawn from the battery?

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

Answer: C – 1/Req = 1/12 + 1/6, so Req = 4 Ω. I = V ÷ Req = 12 ÷ 4 = 3 A.

Q20. A device operates at 48 V and draws 5 A. At what rate does it dissipate energy?

  • (A) 240 W
  • (B) 480 W
  • (C) 120 W
  • (D) 360 W
Show Answer

Answer: A – P = IV = 5 × 48 = 240 W.

Q21. A current of 3 A flows through a 20 Ω resistor. How much power is dissipated as heat in the resistor?

  • (A) 360 W
  • (B) 180 W
  • (C) 90 W
  • (D) 270 W
Show Answer

Answer: B – P = I2R = 32 × 20 = 180 W.

Q22. A wire of length 25 m and cross-sectional area 2 mm2 is made of a material of resistivity 2.8 × 10-8 Ω·m. What is the resistance of the wire?

  • (A) 0.7 Ω
  • (B) 0.18 Ω
  • (C) 0.35 Ω
  • (D) 0.53 Ω
Show Answer

Answer: C – R = μL ÷ A = (2.8 × 10-8 × 25) ÷ (2 × 10-6) = 0.35 Ω.

Q23. A 20 μF capacitor is connected across a 60 V battery. What charge is stored on each plate, in μC?

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

Answer: D – Q = CV = 20 × 60 = 1200 μC.

Q24. A 40 μF capacitor charged to 20 V stores how much energy, in mJ?

  • (A) 16 mJ
  • (B) 8 mJ
  • (C) 4 mJ
  • (D) 12 mJ
Show Answer

Answer: B – U = (1/2)CV2 = 0.5 × (40 × 10-6) × 202 = 0.008 J = 8 mJ.

Q25. Capacitors of 50 μF and 30 μF are connected in parallel across a 12 V battery. What total charge is stored, in μC?

  • (A) 1920 μC
  • (B) 480 μC
  • (C) 960 μC
  • (D) 1440 μC
Show Answer

Answer: C – Ceq = 50 + 30 = 80 μF. Q = CeqV = 80 × 12 = 960 μC.

Q26. Capacitors of 10 μF and 30 μF are connected in series across a 36 V battery. What charge is stored on each capacitor, in μC?

  • (A) 540 μC
  • (B) 135 μC
  • (C) 270 μC
  • (D) 405 μC
Show Answer

Answer: C – 1/Ceq = 1/10 + 1/30, so Ceq = 7.5 μF. Each series capacitor carries Q = CeqV = 7.5 × 36 = 270 μC.

Q27. A circuit contains a 100 kΩ resistor in series with a 10 μF capacitor. What is the time constant of the circuit?

  • (A) 1 s
  • (B) 2 s
  • (C) 0.5 s
  • (D) 1.5 s
Show Answer

Answer: A – μ = RC = (100 × 103) × (10 × 10-6) = 1 s.

Unit 12: Magnetism and Electromagnetism

Q28. A proton (mass 1.67 × 10-27 kg, charge 1.6 × 10-19 C) moves at 8 × 105 m/s perpendicular to a uniform magnetic field of 0.25 T. What is the radius of its circular path, in mm?

  • (A) 66.8 mm
  • (B) 33.4 mm
  • (C) 16.7 mm
  • (D) 50.1 mm
Show Answer

Answer: B – r = mv ÷ qB = (1.67 × 10-27 × 8 × 105) ÷ (1.6 × 10-19 × 0.25) = 0.0334 m = 33.4 mm.

Q29. The magnetic flux through a closed wire 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.

Q30. What is the magnetic field at the centre of a circular loop of radius 25 cm carrying a current of 4 A?

  • (A) 20.11 μT
  • (B) 10.05 μT
  • (C) 5.03 μT
  • (D) 15.08 μT
Show Answer

Answer: B – B = μ0I ÷ (2R) = (4π × 10-7 × 4) ÷ (2 × 0.25) = 10.05 μT.


Q31. Two long parallel cables carry currents in the opposite direction. The magnetic force between them is:

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

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

Q32. A 4 μC particle moves at 3 × 104 m/s perpendicular to a magnetic field of 2.0 T. What is the magnitude of the magnetic force on it, in mN?

  • (A) 480 mN
  • (B) 240 mN
  • (C) 120 mN
  • (D) 360 mN
Show Answer

Answer: B – F = qvB = (4 × 10-6) × (3 × 104) × 2 = 0.24 N = 240 mN.

Q33. A straight wire of length 0.5 m carries a current of 5 A perpendicular to a uniform magnetic field of 0.5 T. What is the magnitude of the force on the wire?

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

Answer: C – F = ILB = 5 × 0.5 × 0.5 = 1.25 N.

Q34. What is the magnitude of the magnetic field 10 cm from a long straight wire carrying a current of 2 A?

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

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

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

  • (A) 10.05 mT
  • (B) 2.51 mT
  • (C) 7.54 mT
  • (D) 5.03 mT
Show Answer

Answer: D – B = μ0nI = (4π × 10-7) × 1000 × 4 = 0.00503 T = 5.03 mT.

Unit 13: Geometric Optics

Q36. Total internal reflection at a diamond-air boundary can occur only when light travels:

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

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

Q37. An book stands 1.6 m in front of a plane mirror. How far behind the mirror does its image appear to be?

  • (A) 1.6 m
  • (B) 3.2 m
  • (C) 0.8 m
  • (D) 2.4 m
Show Answer

Answer: A – A plane mirror forms an image as far behind the mirror as the object is in front, so the image distance is 1.6 m.

Q38. A concave lens of focal length -20 cm always forms an image that is:

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

Answer: D – 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 20° to the normal. At what angle to the normal is the reflected ray?

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

Answer: A – By the law of reflection, angle of reflection equals angle of incidence, so it is 20° to the normal.

Q40. An object is placed 45 cm in front of a concave mirror of focal length 18 cm. How far from the mirror is the image formed?

  • (A) 60 cm
  • (B) 15 cm
  • (C) 45 cm
  • (D) 30 cm
Show Answer

Answer: D – 1/f = 1/do + 1/di, so 1/di = 1/18 – 1/45, giving di = 30 cm.

Q41. An object 15 cm in front of a concave mirror forms an image 30 cm from the mirror. What is the lateral magnification?

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

Answer: C – m = -di ÷ do = -30 ÷ 15 = -2. The negative sign shows the image is inverted.

Q42. An object is placed 108 cm in front of a converging lens of focal length 36 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 = 1/36 – 1/108, giving di = 54 cm.

Q43. A concave mirror has a radius of curvature of 36 cm. What is its focal length?

  • (A) 36 cm
  • (B) 9 cm
  • (C) 27 cm
  • (D) 18 cm
Show Answer

Answer: D – f = R ÷ 2 = 36 ÷ 2 = 18 cm.

Unit 14: Waves, Sound, and Physical Optics

Q44. A wave has frequency 20 Hz and wavelength 1.2 m. What is its speed?

  • (A) 48 m/s
  • (B) 12 m/s
  • (C) 24 m/s
  • (D) 36 m/s
Show Answer

Answer: C – v = fλ = 20 × 1.2 = 24 m/s.

Q45. A oscillator completes one oscillation in 0.8 s. What is its frequency?

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

Answer: C – f = 1 ÷ T = 1 ÷ 0.8 = 1.25 Hz.

Q46. A string of length 0.5 m fixed at both ends supports waves travelling at 480 m/s. What is the frequency of its 3th harmonic?

  • (A) 1440 Hz
  • (B) 2880 Hz
  • (C) 720 Hz
  • (D) 2160 Hz
Show Answer

Answer: A – For a string fixed at both ends, fn = nv ÷ (2L) = 3 × 480 ÷ (2 × 0.5) = 1440 Hz.

Q47. An open pipe of length 0.75 m resonates with sound waves travelling at 340 m/s. What is its fundamental frequency?

  • (A) 226.67 Hz
  • (B) 453.33 Hz
  • (C) 113.33 Hz
  • (D) 340 Hz
Show Answer

Answer: A – For an open pipe, f1 = v ÷ (2L) = 340 ÷ (2 × 0.75) = 226.67 Hz.

Q48. A pipe closed at one end has length 0.42 m. Sound travels at 340 m/s. What is its fundamental frequency?

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

Answer: B – For a pipe closed at one end, f1 = v ÷ (4L) = 340 ÷ (4 × 0.42) = 200 Hz.

Q49. Two tuning forks of frequencies 440 Hz and 448 Hz sound together. What is the beat frequency heard?

  • (A) 16 Hz
  • (B) 4 Hz
  • (C) 12 Hz
  • (D) 8 Hz
Show Answer

Answer: D – Beat frequency = |f1 – f2| = |440 – 448| = 8 Hz.

Q50. A source emitting 850 Hz moves towards a stationary observer at 17 m/s. Sound travels at 340 m/s. What frequency does the observer hear?

  • (A) 894.74 Hz
  • (B) 1789.47 Hz
  • (C) 447.37 Hz
  • (D) 1342.11 Hz
Show Answer

Answer: A – f’ = f × v ÷ (v – vs) = 850 × 340 ÷ (340 – 17) = 894.74 Hz.

Q51. In a double-slit experiment, light of wavelength 500 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 mm
  • (B) 2 mm
  • (C) 1 mm
  • (D) 3 mm
Show Answer

Answer: B – y = mλL ÷ d = 1 × (500 × 10-9) × 2 ÷ (0.5 × 10-3) = 0.002 m = 2 mm.

Unit 15: Modern Physics

Q52. Photoelectrons are emitted with maximum kinetic energy 2.5 eV. What stopping potential is needed to bring the fastest electrons to rest?

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

Answer: A – eVs = Kmax, so Vs = 2.5 eV ÷ e = 2.5 V.

Q53. An electron (mass 9.11 × 10-31 kg) moves at 4 × 106 m/s. What is its de Broglie wavelength, in pm? (Use h = 6.63 × 10-34 J·s.)

  • (A) 363.89 pm
  • (B) 90.97 pm
  • (C) 272.91 pm
  • (D) 181.94 pm
Show Answer

Answer: D – λ = h ÷ p = h ÷ (mv) = (6.63 × 10-34) ÷ (9.11 × 10-31 × 4 × 106) = 0.182 nm = 181.94 pm.

Q54. A hydrogen atom transitions from energy level n = 3 to n = 2. What is the energy of the photon emitted, in eV? (Levels: En = -13.6 ÷ n2 eV.)

  • (A) 3.78 eV
  • (B) 1.89 eV
  • (C) 0.94 eV
  • (D) 2.83 eV
Show Answer

Answer: B – ΔE = 13.6 × (1/22 – 1/32) = 1.89 eV.

Q55. Calculate the energy of a hydrogen atom in level n = 5, in eV? (Use En = -13.6 ÷ n2 eV.)

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

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

Q56. A radioactive sample of mass 320 g has a half-life of one hour. How much of it remains undecayed after 4 hours?

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

Answer: B – After 4 half-lives the fraction remaining is (1/2)4 = 0.062, so remaining mass = 320 × 0.062 = 20 g.

Q57. A fission reaction has a mass defect of 0.01 u. How much energy is released? (Use 1 u = 931.5 MeV/c2.)

  • (A) 18.63 MeV
  • (B) 4.66 MeV
  • (C) 13.97 MeV
  • (D) 9.31 MeV
Show Answer

Answer: D – E = Δm c2 = 0.01 × 931.5 = 9.31 MeV.

Q58. A caesium surface has work function 3.5 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) 16.89 ×1014 Hz
  • (B) 4.22 ×1014 Hz
  • (C) 8.45 ×1014 Hz
  • (D) 12.67 ×1014 Hz
Show Answer

Answer: C – f0 = Φ ÷ h = (3.5 × 1.6 × 10-19) ÷ (6.63 × 10-34) = 8.45 × 1014 Hz.

Q59. The photoelectric effect 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 photoelectric effect depends on photon energy and momentum in a way explained by light arriving as discrete photons.

Q60. According to the Heisenberg uncertainty principle, measuring a neutron’s momentum more precisely necessarily:

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

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

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