AP Physics 2 Practice Test 19 – 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 sample of helium expands quickly in an insulated cylinder so no heat enters or leaves. Which name describes this process?

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

Answer: A – The process is adiabatic: no heat is transferred.

Q2. How much energy is needed to melt 1.2 kg of wax at its melting point without changing its temperature? (Latent heat = 250000 J/kg.)

  • (A) 600000 J
  • (B) 150000 J
  • (C) 300000 J
  • (D) 450000 J
Show Answer

Answer: C – Q = mL = 1.2 × 250000 = 300000 J.

Q3. 13500 J of heat raises the temperature of a 2 kg sample by 10 K. What is the specific heat capacity of the sample material?

  • (A) 675 J/kg·K
  • (B) 1350 J/kg·K
  • (C) 337.5 J/kg·K
  • (D) 1012.5 J/kg·K
Show Answer

Answer: A – c = Q ÷ (mΔT) = 13500 ÷ (2 × 10) = 675 J/kg·K.

Q4. What is the internal energy of 1 mol of a monatomic ideal gas at 600 K? (Use U = (3/2)nRT and R = 8.31 J/mol·K.)

  • (A) 14958 J
  • (B) 7479 J
  • (C) 3739.5 J
  • (D) 11218.5 J
Show Answer

Answer: B – U = (3/2)nRT = 1.5 × 1 × 8.31 × 600 = 7479 J.

Q5. A gas expands adiabatically and does 800 J of work on its surroundings. What is the change in its internal energy?

  • (A) -800 J
  • (B) -1600 J
  • (C) -400 J
  • (D) -1200 J
Show Answer

Answer: A – In an adiabatic process Q = 0, so ΔU = Q – Wby = 0 – 800 = -800 J. The internal energy decreases.

Q6. A laboratory flask holds 2 mol of an ideal gas at 300 K in a volume of 0.04 m3. What is the pressure of the gas, in kPa? (Use R = 8.31 J/mol·K.)

  • (A) 249.3 kPa
  • (B) 124.65 kPa
  • (C) 62.33 kPa
  • (D) 186.98 kPa
Show Answer

Answer: B – P = nRT ÷ V = (2 × 8.31 × 300) ÷ 0.04 = 124650 Pa = 124.65 kPa.

Q7. 4 mol of an ideal gas at 400 K exerts a pressure of 200 kPa. What volume does the gas occupy? (Use R = 8.31 J/mol·K.)

  • (A) 0.13 m3
  • (B) 0.03 m3
  • (C) 0.1 m3
  • (D) 0.07 m3
Show Answer

Answer: D – V = nRT ÷ P = (4 × 8.31 × 400) ÷ 200000 = 0.07 m3.

Q8. An ideal gas sample of 2 mol in a 0.04 m3 container exerts a pressure of 120 kPa. What is its temperature? (Use R = 8.31 J/mol·K.)

  • (A) 577.62 K
  • (B) 288.81 K
  • (C) 144.4 K
  • (D) 433.21 K
Show Answer

Answer: B – T = PV ÷ nR = (120000 × 0.04) ÷ (2 × 8.31) = 288.81 K.

Q9. A 0.6 kg copper rod with specific heat capacity 500 J/kg·K is heated through 25 K. How much thermal energy is transferred to it?

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

Answer: D – Q = mcΔT = 0.6 × 500 × 25 = 7500 J.

Unit 10: Electric Force, Field, and Potential

Q10. In a uniform field of 800 N/C, a charge moves 4 cm parallel to the field. What is the magnitude of the potential difference between its start and end points?

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

Answer: C – |ΔV| = Ed = 800 × 0.04 = 32 V.

Q11. Two point charges of 3 μC and 6 μC are 25 cm apart in a vacuum. What is the magnitude of the electrostatic force between them? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 5.18 N
  • (B) 2.59 N
  • (C) 1.3 N
  • (D) 3.89 N
Show Answer

Answer: B – F = kq1q2 ÷ r2 = (9.0 × 109 × 3 × 10-6 × 6 × 10-6) ÷ (0.25)2 = 2.59 N.

Q12. What is the magnitude of the electric field 20 cm from a point charge of 5 μC in a vacuum? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 2250000 N/C
  • (B) 562500 N/C
  • (C) 1687500 N/C
  • (D) 1125000 N/C
Show Answer

Answer: D – E = kQ ÷ r2 = (9.0 × 109 × 5 × 10-6) ÷ (0.2)2 = 1125000 N/C.

Q13. A charge of 3 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) 3 mN
  • (B) 0.75 mN
  • (C) 2.25 mN
  • (D) 1.5 mN
Show Answer

Answer: D – F = qE = (3 × 10-6) × 500 = 0.0015 N = 1.5 mN.

Q14. What is the electric potential 30 cm from a point charge of 12 μC in a vacuum, taking potential at infinity as zero? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 360000 V
  • (B) 720000 V
  • (C) 180000 V
  • (D) 540000 V
Show Answer

Answer: A – V = kQ ÷ r = (9.0 × 109 × 12 × 10-6) ÷ 0.3 = 360000 V.

Q15. How much work is done by an external agent moving a 5 μC charge through a potential difference of 50 V at constant speed? Give the answer in mJ.

  • (A) 2.5 mJ
  • (B) 0.03 mJ
  • (C) 0.25 mJ
  • (D) 0.5 mJ
Show Answer

Answer: C – W = qΔV = (5 × 10-6) × 50 = 250 μJ = 0.25 mJ.

Q16. The potential difference between two parallel plates 2 mm apart is 100 V. What is the magnitude of the uniform electric field between the plates?

  • (A) 100000 N/C
  • (B) 25000 N/C
  • (C) 50000 N/C
  • (D) 75000 N/C
Show Answer

Answer: C – E = V ÷ d = 100 ÷ 0 = 50000 N/C.

Q17. Two point charges of 2 μC and 6 μC are held 25 cm apart. What is their electric potential energy? (Use k = 9.0 × 109 N·m2/C2, and take zero at infinity.)

  • (A) 0.86 J
  • (B) 0.22 J
  • (C) 0.65 J
  • (D) 0.43 J
Show Answer

Answer: D – U = kq1q2 ÷ r = (9.0 × 109 × 2 × 10-6 × 6 × 10-6) ÷ 0.25 = 0.43 J.

Q18. A proton 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.

Unit 11: Electric Circuits

Q19. A total current of 6 A splits between parallel resistors of 3 Ω and 6 Ω. What current flows through the 3 Ω resistor?

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

Answer: C – Current divides in inverse proportion to resistance: I3 = Itotal × 6 ÷ (3 + 6) = 6 × 6 ÷ 9 = 4 A.

Q20. A resistor of 4 Ω is connected across a 60 V battery. What current flows through the resistor?

  • (A) 30 A
  • (B) 7.5 A
  • (C) 22.5 A
  • (D) 15 A
Show Answer

Answer: D – I = V ÷ R = 60 ÷ 4 = 15 A.

Q21. Resistors of 12 Ω, 5 Ω and 8 Ω are connected in series across a 30 V battery. What is the current in the circuit?

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

Answer: D – Req = 12 + 5 + 8 = 25 Ω. I = V ÷ Req = 30 ÷ 25 = 1.2 A.

Q22. Resistors of 12 Ω and 6 Ω are connected in parallel across a 24 V battery. What is the total current drawn from the battery?

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

Answer: C – 1/Req = 1/12 + 1/6, so Req = 4 Ω. I = V ÷ Req = 24 ÷ 4 = 6 A.

Q23. A device operates at 12 V and draws 6 A. At what rate does it dissipate energy?

  • (A) 144 W
  • (B) 36 W
  • (C) 72 W
  • (D) 108 W
Show Answer

Answer: C – P = IV = 6 × 12 = 72 W.

Q24. A current of 4 A flows through a 5 Ω resistor. How much power is dissipated as heat in the resistor?

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

Answer: B – P = I2R = 42 × 5 = 80 W.

Q25. A wire of length 20 m and cross-sectional area 1 mm2 is made of a material of resistivity 1.7 × 10-8 Ω·m. What is the resistance of the wire?

  • (A) 0.68 Ω
  • (B) 0.17 Ω
  • (C) 0.51 Ω
  • (D) 0.34 Ω
Show Answer

Answer: D – R = μL ÷ A = (1.7 × 10-8 × 20) ÷ (1 × 10-6) = 0.34 Ω.

Q26. A 100 μF capacitor is connected across a 30 V battery. What charge is stored on each plate, in μC?

  • (A) 3000 μC
  • (B) 6000 μC
  • (C) 1500 μC
  • (D) 4500 μC
Show Answer

Answer: A – Q = CV = 100 × 30 = 3000 μC.

Q27. A 50 μF capacitor charged to 40 V stores how much energy, in mJ?

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

Answer: C – U = (1/2)CV2 = 0.5 × (50 × 10-6) × 402 = 0.04 J = 40 mJ.

Unit 12: Magnetism and Electromagnetism

Q28. A uniform magnetic field of 0.8 T passes through a flat loop of area 0.05 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) 40 mWb
  • (B) 10 mWb
  • (C) 30 mWb
  • (D) 20 mWb
Show Answer

Answer: D – Φ = BA cosθ = 0.8 × 0.05 × 0.5 = 0.02 Wb = 20 mWb.

Q29. The magnetic flux through a single loop changes by 0.02 Wb in 0.5 s. What is the magnitude of the average induced emf in the loop?

  • (A) 0.08 V
  • (B) 0.02 V
  • (C) 0.06 V
  • (D) 0.04 V
Show Answer

Answer: D – emf = |ΔΦ ÷ Δt| = 0.02 ÷ 0.5 = 0.04 V.

Q30. A metal rod of length 0.5 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) 0.2 V
  • (B) 0.4 V
  • (C) 0.1 V
  • (D) 0.3 V
Show Answer

Answer: A – emf = BLv = 0.2 × 0.5 × 2 = 0.2 V.


Q31. 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 0.5 T. What is the radius of its circular path, in mm?

  • (A) 8.35 mm
  • (B) 2.09 mm
  • (C) 6.26 mm
  • (D) 4.18 mm
Show Answer

Answer: D – r = mv ÷ qB = (1.67 × 10-27 × 2 × 105) ÷ (1.6 × 10-19 × 0.5) = 0.0042 m = 4.18 mm.

Q32. The magnetic flux through a closed metal coil 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.

Q33. What is the magnetic field at the centre of a circular loop of radius 4 cm carrying a current of 5 A?

  • (A) 78.54 μT
  • (B) 157.08 μT
  • (C) 39.27 μT
  • (D) 117.81 μT
Show Answer

Answer: A – B = μ0I ÷ (2R) = (4π × 10-7 × 5) ÷ (2 × 0.04) = 78.54 μT.

Q34. Two long parallel conductors 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.

Q35. A 8 μC particle moves at 6 × 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) 240 mN
  • (B) 480 mN
  • (C) 120 mN
  • (D) 360 mN
Show Answer

Answer: A – F = qvB = (8 × 10-6) × (6 × 104) × 0.5 = 0.24 N = 240 mN.

Unit 13: Geometric Optics

Q36. What is the critical angle for light travelling from a medium of refractive index 1.5 into air? (Use sinθc = 1 ÷ n.)

  • (A) 83.6 degrees
  • (B) 41.8 degrees
  • (C) 20.9 degrees
  • (D) 62.7 degrees
Show Answer

Answer: B – sinθc = 1 ÷ 1.5 = 0.667, which gives θc = 41.8°.

Q37. Light travels at 2.25 × 108 m/s in a plastic sample. What is the refractive index of the material? (Use c = 3.0 × 108 m/s.)

  • (A) 1.33
  • (B) 2.67
  • (C) 0.67
  • (D) 2
Show Answer

Answer: A – n = c ÷ v = (3.0 × 108) ÷ (2.25 × 108) = 1.33.

Q38. An object is placed 15 cm from a converging lens of focal length 30 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.

Q39. Total internal reflection at a glass-air boundary can occur only when light travels:

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

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

Q40. An candle stands 1.4 m in front of a plane mirror. How far behind the mirror does its image appear to be?

  • (A) 1.4 m
  • (B) 2.8 m
  • (C) 0.7 m
  • (D) 2.1 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 1.4 m.

Q41. A diverging lens of focal length -20 cm always forms an image that is:

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

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

Q42. A ray strikes a plane mirror at 30° to the normal. At what angle to the normal is the reflected ray?

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

Answer: D – By the law of reflection, angle of reflection equals angle of incidence, so it is 30° to the normal.

Q43. An object is placed 36 cm in front of a concave mirror of focal length 18 cm. How far from the mirror is the image formed?

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

Answer: D – 1/f = 1/do + 1/di, so 1/di = 1/18 – 1/36, giving di = 36 cm.

Unit 14: Waves, Sound, and Physical Optics

Q44. A point source radiates 120 W of sound uniformly in all directions. What is the intensity at a distance of 2 m? (Use π = 3.14.)

  • (A) 4.77 W/m2
  • (B) 2.39 W/m2
  • (C) 1.19 W/m2
  • (D) 3.58 W/m2
Show Answer

Answer: B – I = P ÷ (4πr2) = 120 ÷ (4 × 3.14 × 22) = 2.3873 W/m2.

Q45. In a single-slit diffraction pattern, using a longer wavelength while keeping the other factors fixed causes the central maximum to become:

  • (A) narrower
  • (B) wider
  • (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 longer wavelength gives a wider central maximum.

Q46. A soap 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) t = the wavelength in the film
  • (C) 2t = (m + 1/2) times the wavelength in the film
  • (D) 2t = one-quarter times the wavelength in the film
Show Answer

Answer: C – 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.

Q47. A wave has frequency 50 Hz and wavelength 6.8 m. What is its speed?

  • (A) 680 m/s
  • (B) 170 m/s
  • (C) 340 m/s
  • (D) 510 m/s
Show Answer

Answer: C – v = fλ = 50 × 6.8 = 340 m/s.

Q48. A wave completes one oscillation in 2.5 s. What is its frequency?

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

Answer: A – f = 1 ÷ T = 1 ÷ 2.5 = 0.4 Hz.

Q49. A string of length 1 m fixed at both ends supports waves travelling at 100 m/s. What is its fundamental frequency?

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

Answer: A – For a string fixed at both ends, fn = nv ÷ (2L) = 1 × 100 ÷ (2 × 1) = 50 Hz.

Q50. An open pipe of length 0.425 m resonates with sound waves travelling at 340 m/s. What is its fundamental frequency?

  • (A) 800 Hz
  • (B) 400 Hz
  • (C) 200 Hz
  • (D) 600 Hz
Show Answer

Answer: B – For an open pipe, f1 = v ÷ (2L) = 340 ÷ (2 × 0.425) = 400 Hz.

Q51. A pipe closed at one end has length 0.75 m. Sound travels at 340 m/s. What is its fundamental frequency?

  • (A) 113.33 Hz
  • (B) 226.67 Hz
  • (C) 56.67 Hz
  • (D) 170 Hz
Show Answer

Answer: A – For a pipe closed at one end, f1 = v ÷ (4L) = 340 ÷ (4 × 0.75) = 113.33 Hz.

Unit 15: Modern Physics

Q52. According to the Heisenberg uncertainty principle, measuring a particle’s momentum more precisely necessarily:

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

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

Q53. What is the energy, in units of 10-19 J, of a photon of frequency 7 × 1014 Hz? (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) 6.96 ×10-19 J
Show Answer

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

Q54. Light photons of energy 6.5 eV strike a metal with work function 4 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 = 6.5 – 4 = 2.5 eV.

Q55. Photoelectrons are emitted with maximum kinetic energy 2 eV. What stopping potential is needed to bring the fastest electrons to rest?

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

Answer: C – eVs = Kmax, so Vs = 2 eV ÷ e = 2 V.

Q56. An electron (mass 9.11 × 10-31 kg) moves at 2 × 106 m/s. What is its de Broglie wavelength, in pm? (Use h = 6.63 × 10-34 J·s.)

  • (A) 727.77 pm
  • (B) 363.89 pm
  • (C) 181.94 pm
  • (D) 545.83 pm
Show Answer

Answer: B – λ = h ÷ p = h ÷ (mv) = (6.63 × 10-34) ÷ (9.11 × 10-31 × 2 × 106) = 0.364 nm = 363.89 pm.

Q57. A hydrogen atom transitions from energy level n = 5 to n = 2. What is the energy of the photon emitted, in eV? (Levels: En = -13.6 ÷ n2 eV.)

  • (A) 5.72 eV
  • (B) 2.86 eV
  • (C) 1.43 eV
  • (D) 4.29 eV
Show Answer

Answer: B – ΔE = 13.6 × (1/22 – 1/52) = 2.86 eV.

Q58. What is the energy of a hydrogen atom in level n = 2, in eV? (Use En = -13.6 ÷ n2 eV.)

  • (A) -6.8 eV
  • (B) -3.4 eV
  • (C) -1.7 eV
  • (D) 6.8 eV
Show Answer

Answer: B – E2 = -13.6 ÷ 22 = -3.4 eV.

Q59. A radioactive sample of mass 80 g has a half-life of one hour. How much of it remains undecayed after 3 hours?

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

Answer: B – After 3 half-lives the fraction remaining is (1/2)3 = 0.125, so remaining mass = 80 × 0.125 = 10 g.

Q60. A fusion reaction has a mass defect of 0.2 u. How much energy is released? (Use 1 u = 931.5 MeV/c2.)

  • (A) 372.6 MeV
  • (B) 186.3 MeV
  • (C) 93.15 MeV
  • (D) 279.45 MeV
Show Answer

Answer: B – E = Δm c2 = 0.2 × 931.5 = 186.3 MeV.

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