AP Physics 2 Practice Test 17 – 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 0.6 kg lead weight with specific heat capacity 900 J/kg·K is heated through 15 K. How much thermal energy is transferred to it?

  • (A) 16200 J
  • (B) 8100 J
  • (C) 4050 J
  • (D) 12150 J
Show Answer

Answer: B – Q = mcΔT = 0.6 × 900 × 15 = 8100 J.

Q2. 1500 J of heat is added to a air in a cylinder while it does 100 J of work on its surroundings. What is the change in its internal energy?

  • (A) 2800 J
  • (B) 700 J
  • (C) 2100 J
  • (D) 1400 J
Show Answer

Answer: D – First law: ΔU = Q – Wby = 1500 – 100 = 1400 J. A positive value means the internal energy increases.

Q3. A gas expands at a constant pressure of 200 kPa, increasing its volume by 0.03 m3. How much work does the gas do?

  • (A) 12000 J
  • (B) 3000 J
  • (C) 9000 J
  • (D) 6000 J
Show Answer

Answer: D – At constant pressure, W = PΔV = 200000 × 0.03 = 6000 J.

Q4. A reversible heat engine operates between reservoirs at 1000 K and 320 K. What is its maximum possible efficiency?

  • (A) 136 %
  • (B) 34 %
  • (C) 102 %
  • (D) 68 %
Show Answer

Answer: D – Carnot efficiency = 1 – TC ÷ TH = 1 – 320 ÷ 1000 = 0.68 = 68%.

Q5. A heat engine takes in 800 J of heat from a hot reservoir and rejects 400 J to a cold reservoir in one cycle. How much useful work does it do per cycle?

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

Answer: B – W = QH – QC = 800 – 400 = 400 J.

Q6. The absolute temperature of a sample of nitrogen gas rises from 250 K to 750 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 250 K to 750 K triples the average kinetic energy.

Q7. A sample of helium expands in a piston while the pressure stays constant. Which name describes this process?

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

Answer: B – The process is isobaric: pressure is constant.

Q8. How much energy is needed to melt 3 kg of water at its melting point without changing its temperature? (Latent heat = 2260000 J/kg.)

  • (A) 13560000 J
  • (B) 3390000 J
  • (C) 10170000 J
  • (D) 6780000 J
Show Answer

Answer: D – Q = mL = 3 × 2260000 = 6780000 J.

Q9. 13500 J of heat raises the temperature of a 3 kg sample by 15 K. What is the specific heat capacity of the sample material?

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

Answer: A – c = Q ÷ (mΔT) = 13500 ÷ (3 × 15) = 300 J/kg·K.

Unit 10: Electric Force, Field, and Potential

Q10. The potential difference between two parallel plates 10 mm apart is 200 V. What is the magnitude of the uniform electric field between the plates?

  • (A) 40000 N/C
  • (B) 20000 N/C
  • (C) 10000 N/C
  • (D) 30000 N/C
Show Answer

Answer: B – E = V ÷ d = 200 ÷ 0.01 = 20000 N/C.

Q11. Two point charges of 5 μC and 4 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) 1.44 J
  • (B) 0.36 J
  • (C) 1.08 J
  • (D) 0.72 J
Show Answer

Answer: D – U = kq1q2 ÷ r = (9.0 × 109 × 5 × 10-6 × 4 × 10-6) ÷ 0.25 = 0.72 J.

Q12. A alpha particle is placed in an electric field. In which direction does the electric force on it act?

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

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

Q13. A charge 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.

Q14. The distance between two point charges is tripled 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-ninth as large
Show Answer

Answer: D – Coulomb’s law has force proportional to 1/r2, so tripling r gives a factor 1/32 = 1/9.

Q15. Parallel plates 4 mm apart have a potential difference of 360 V. A 10 μC charge between them experiences what force magnitude?

  • (A) 0.9 N
  • (B) 1.8 N
  • (C) 0.45 N
  • (D) 1.35 N
Show Answer

Answer: A – E = V ÷ d = 360 ÷ 0 = 90000 N/C. F = qE = (10 × 10-6) × 90000 = 0.9 N.

Q16. In a uniform field of 1000 N/C, a charge moves 40 cm parallel to the field. What is the magnitude of the potential difference between its start and end points?

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

Answer: C – |ΔV| = Ed = 1000 × 0.4 = 400 V.

Q17. Two point charges of 2 μC and 8 μC are 10 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 × 2 × 10-6 × 8 × 10-6) ÷ (0.1)2 = 14.4 N.

Q18. What is the magnitude of the electric field 10 cm from a point charge of 10 μC in a vacuum? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 18000000 N/C
  • (B) 4500000 N/C
  • (C) 13500000 N/C
  • (D) 9000000 N/C
Show Answer

Answer: D – E = kQ ÷ r2 = (9.0 × 109 × 10 × 10-6) ÷ (0.1)2 = 9000000 N/C.

Unit 11: Electric Circuits

Q19. A 50 μF capacitor charged to 10 V stores how much energy, in mJ?

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

Answer: A – U = (1/2)CV2 = 0.5 × (50 × 10-6) × 102 = 0.0025 J = 2.5 mJ.

Q20. Capacitors of 50 μF and 20 μF are connected in parallel across a 24 V battery. What total charge is stored, in μC?

  • (A) 1680 μC
  • (B) 3360 μC
  • (C) 840 μC
  • (D) 2520 μC
Show Answer

Answer: A – Ceq = 50 + 20 = 70 μF. Q = CeqV = 70 × 24 = 1680 μC.

Q21. Capacitors of 10 μF and 30 μF are connected in series across a 12 V battery. What charge is stored on each capacitor, in μC?

  • (A) 180 μC
  • (B) 45 μC
  • (C) 135 μC
  • (D) 90 μC
Show Answer

Answer: D – 1/Ceq = 1/10 + 1/30, so Ceq = 7.5 μF. Each series capacitor carries Q = CeqV = 7.5 × 12 = 90 μC.

Q22. A circuit contains a 100 kΩ resistor in series with a 100 μF capacitor. What is the time constant of the circuit?

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

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

Q23. A battery of emf 36 V and internal resistance 1 Ω delivers a current of 3 A. What is the terminal potential difference of the battery?

  • (A) 66 V
  • (B) 16.5 V
  • (C) 49.5 V
  • (D) 33 V
Show Answer

Answer: D – V = emf – Ir = 36 – (3 × 1) = 33 V.

Q24. Two identical lamps are connected in series with a 24 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.

Q25. A total current of 15 A splits between parallel resistors of 4 Ω and 4 Ω. What current flows through the 4 Ω resistor?

  • (A) 15 A
  • (B) 7.5 A
  • (C) 3.75 A
  • (D) 11.25 A
Show Answer

Answer: B – Current divides in inverse proportion to resistance: I4 = Itotal × 4 ÷ (4 + 4) = 15 × 4 ÷ 8 = 7.5 A.

Q26. A resistor of 6 Ω is connected across a 24 V battery. What current flows through the resistor?

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

Answer: A – I = V ÷ R = 24 ÷ 6 = 4 A.

Q27. Resistors of 10 Ω, 8 Ω and 15 Ω are connected in series across a 60 V battery. What is the current in the circuit?

  • (A) 3.64 A
  • (B) 0.91 A
  • (C) 2.73 A
  • (D) 1.82 A
Show Answer

Answer: D – Req = 10 + 8 + 15 = 33 Ω. I = V ÷ Req = 60 ÷ 33 = 1.82 A.

Unit 12: Magnetism and Electromagnetism

Q28. What is the magnetic field at the centre of a circular loop of radius 4 cm carrying a current of 4 A?

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

Answer: B – B = μ0I ÷ (2R) = (4π × 10-7 × 4) ÷ (2 × 0.04) = 62.83 μT.

Q29. Two long parallel wires carry currents in the opposite direction. The magnetic force between them is:

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

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

Q30. A 8 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) 960 mN
  • (B) 480 mN
  • (C) 240 mN
  • (D) 720 mN
Show Answer

Answer: B – F = qvB = (8 × 10-6) × (3 × 104) × 2 = 0.48 N = 480 mN.


Q31. A straight wire of length 1 m carries a current of 2 A perpendicular to a uniform magnetic field of 0.5 T. What is the magnitude of the force on the wire?

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

Answer: D – F = ILB = 2 × 1 × 0.5 = 1 N.

Q32. What is the magnitude of the magnetic field 10 cm from a long straight wire carrying a current of 1 A?

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

Answer: D – B = μ0I ÷ (2πr) = (2 × 10-7 × 1) ÷ 0.1 = 2 μT.

Q33. A solenoid has 500 turns per metre and carries a current of 1 A. What is the magnetic field inside the solenoid, in mT? (Use μ0 = 4π × 10-7 T·m/A.)

  • (A) 1.26 mT
  • (B) 0.63 mT
  • (C) 0.31 mT
  • (D) 0.94 mT
Show Answer

Answer: B – B = μ0nI = (4π × 10-7) × 500 × 1 = 0.00063 T = 0.63 mT.

Q34. A uniform magnetic field of 0.5 T passes through a flat loop of area 0.05 m2. The field makes an angle of 0° with the normal to the loop. What is the magnetic flux through the loop, in mWb?

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

Answer: B – Φ = BA cosθ = 0.5 × 0.05 × 1 = 0.03 Wb = 25 mWb.

Q35. The magnetic flux through a single loop changes by 0.05 Wb in 0.2 s. What is the magnitude of the average induced emf in the loop?

  • (A) 0.5 V
  • (B) 0.25 V
  • (C) 0.12 V
  • (D) 0.38 V
Show Answer

Answer: B – emf = |ΔΦ ÷ Δt| = 0.05 ÷ 0.2 = 0.25 V.

Unit 13: Geometric Optics

Q36. A ray strikes a plane mirror at 50° to the normal. At what angle to the normal is the reflected ray?

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

Answer: C – By the law of reflection, angle of reflection equals angle of incidence, so it is 50° to the normal.

Q37. An object is placed 40 cm in front of a concave mirror of focal length 20 cm. How far from the mirror is the image formed?

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

Answer: A – 1/f = 1/do + 1/di, so 1/di = 1/20 – 1/40, giving di = 40 cm.

Q38. An object 60 cm in front of a concave mirror forms an image 120 cm from the mirror. What is the lateral magnification?

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

Answer: C – m = -di ÷ do = -120 ÷ 60 = -2. The negative sign shows the image is inverted.

Q39. An object is placed 75 cm in front of a converging lens of focal length 25 cm. How far from the lens is the image formed?

  • (A) 75 cm
  • (B) 18.75 cm
  • (C) 37.5 cm
  • (D) 56.25 cm
Show Answer

Answer: C – 1/f = 1/do + 1/di, so 1/di = 1/25 – 1/75, giving di = 37.5 cm.

Q40. A concave mirror has a radius of curvature of 80 cm. What is its focal length?

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

Answer: C – f = R ÷ 2 = 80 ÷ 2 = 40 cm.

Q41. A ray in air enters a transparent block at an angle of incidence of 50° and refracts at 30° to the normal. Using sin 50° = 0.766 and sin 30° = 0.5, what is the refractive index of the block?

  • (A) 3.06
  • (B) 0.77
  • (C) 1.53
  • (D) 2.29
Show Answer

Answer: C – Snell: n1 sinθ1 = n2 sinθ2, so n2 = (1 × 0.766) ÷ 0.5 = 1.53.

Q42. What is the critical angle for light travelling from a medium of refractive index 1.33 into air? (Use sinθc = 1 ÷ n.)

  • (A) 97.6 degrees
  • (B) 24.4 degrees
  • (C) 48.8 degrees
  • (D) 73.2 degrees
Show Answer

Answer: C – sinθc = 1 ÷ 1.33 = 0.752, which gives θc = 48.8°.

Q43. Light travels at 2.4 × 108 m/s in a oil sample. What is the refractive index of the material? (Use c = 3.0 × 108 m/s.)

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

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

Unit 14: Waves, Sound, and Physical Optics

Q44. A string of length 0.4 m fixed at both ends supports waves travelling at 200 m/s. What is the frequency of its 2th harmonic?

  • (A) 1000 Hz
  • (B) 500 Hz
  • (C) 250 Hz
  • (D) 750 Hz
Show Answer

Answer: B – For a string fixed at both ends, fn = nv ÷ (2L) = 2 × 200 ÷ (2 × 0.4) = 500 Hz.

Q45. An open pipe of length 2.12 m resonates with sound waves travelling at 340 m/s. What is its fundamental frequency?

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

Answer: A – For an open pipe, f1 = v ÷ (2L) = 340 ÷ (2 × 2.12) = 80 Hz.

Q46. A pipe closed at one end has length 0.85 m. Sound travels at 340 m/s. What is its fundamental frequency?

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

Answer: A – For a pipe closed at one end, f1 = v ÷ (4L) = 340 ÷ (4 × 0.85) = 100 Hz.

Q47. Two tuning forks of frequencies 660 Hz and 665 Hz sound together. What is the beat frequency heard?

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

Answer: A – Beat frequency = |f1 – f2| = |660 – 665| = 5 Hz.

Q48. A source emitting 680 Hz moves towards a stationary observer at 68 m/s. Sound travels at 340 m/s. What frequency does the observer hear?

  • (A) 1700 Hz
  • (B) 425 Hz
  • (C) 850 Hz
  • (D) 1275 Hz
Show Answer

Answer: C – f’ = f × v ÷ (v – vs) = 680 × 340 ÷ (340 – 68) = 850 Hz.

Q49. In a double-slit experiment, light of wavelength 650 nm passes through slits 1 mm apart, with a screen 1 m away. How far from the central maximum is the 1th bright fringe, in mm?

  • (A) 1.3 mm
  • (B) 0.33 mm
  • (C) 0.65 mm
  • (D) 0.98 mm
Show Answer

Answer: C – y = mλL ÷ d = 1 × (650 × 10-9) × 1 ÷ (1 × 10-3) = 0.00065 m = 0.65 mm.

Q50. A point source radiates 120 W of sound uniformly in all directions. What is the intensity at a distance of 4 m? (Use π = 3.14.)

  • (A) 1.19 W/m2
  • (B) 0.6 W/m2
  • (C) 0.3 W/m2
  • (D) 0.9 W/m2
Show Answer

Answer: B – I = P ÷ (4πr2) = 120 ÷ (4 × 3.14 × 42) = 0.5968 W/m2.

Q51. In a single-slit diffraction pattern, using a wider slit while keeping the other factors fixed causes the central maximum to become:

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

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

Unit 15: Modern Physics

Q52. A hydrogen atom transitions from energy level n = 4 to n = 1. What is the energy of the photon emitted, in eV? (Levels: En = -13.6 ÷ n2 eV.)

  • (A) 25.5 eV
  • (B) 6.38 eV
  • (C) 19.12 eV
  • (D) 12.75 eV
Show Answer

Answer: D – ΔE = 13.6 × (1/12 – 1/42) = 12.75 eV.

Q53. What is 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) -2.42 ×10-19 J
  • (D) 4.84 ×10-19 J
Show Answer

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

Q54. A radioactive sample of mass 80 g has a half-life of one hour. How much of it remains undecayed after 4 hours?

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

Answer: B – After 4 half-lives the fraction remaining is (1/2)4 = 0.062, so remaining mass = 80 × 0.062 = 5 g.

Q55. A radioactive decay has a mass defect of 0.01 u. How much energy is released? (Use 1 u = 931.5 MeV/c2.)

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

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

Q56. A copper surface has work function 1.9 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) 4.59 ×1014 Hz
  • (B) 9.17 ×1014 Hz
  • (C) 2.29 ×1014 Hz
  • (D) 6.88 ×1014 Hz
Show Answer

Answer: A – f0 = Φ ÷ h = (1.9 × 1.6 × 10-19) ÷ (6.63 × 10-34) = 4.59 × 1014 Hz.

Q57. The photoelectric effect provides evidence that light behaves as:

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

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

Q58. According to the Heisenberg uncertainty principle, measuring a electron’s position 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: B – Position and momentum uncertainties obey a joint limit, so reducing position uncertainty increases the minimum possible momentum uncertainty.

Q59. What is the energy, in units of 10-19 J, of a photon of frequency 8 × 1014 Hz? (Use h = 6.63 × 10-34 J·s.)

  • (A) 5.3 ×10-19 J
  • (B) 10.61 ×10-19 J
  • (C) 2.65 ×10-19 J
  • (D) 7.96 ×10-19 J
Show Answer

Answer: A – E = hf = (6.63 × 10-34) × (8 × 1014) = 5.3 × 10-19 J.

Q60. Light photons of energy 5.5 eV strike a metal with work function 2 eV. What is the maximum kinetic energy of the emitted photoelectrons, in eV?

  • (A) 7 eV
  • (B) 1.75 eV
  • (C) 3.5 eV
  • (D) 5.25 eV
Show Answer

Answer: C – Kmax = hf – Φ = 5.5 – 2 = 3.5 eV.

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