AP Physics 2 Practice Test 10 – 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). Take g = 9.8 m/s2 unless stated otherwise.

Unit 9: Thermodynamics

Q1. A sample of helium in a weather-balloon test chamber is initially at 240 kPa, occupies 4 L and is at 300 K. It is then compressed or expanded to 8 L and its temperature changes to 150 K. What is the final pressure of the gas?

  • (A) 120 kPa
  • (B) 30 kPa
  • (C) 60 kPa
  • (D) 90 kPa
Show Answer

Answer: C – P1V1 ÷ T1 = P2V2 ÷ T2, so P2 = P1 × (V1 ÷ V2) × (T2 ÷ T1) = 240 × (4 ÷ 8) × (150 ÷ 300) = 60 kPa.

Q2. 650 J of heat is added to a gas in a compressor while the gas has 250 J of work done on it. What is the change in internal energy of the gas?

  • (A) 400 J
  • (B) -900 J
  • (C) 650 J
  • (D) 900 J
Show Answer

Answer: D – First law: ΔU = Q – Wby = 650 – (-250) = 900 J.

Q3. A gas in a piston expands at a constant pressure of 250 kPa from 1 L to 5 L. How much work is done by the gas?

  • (A) 1000 J
  • (B) 2000 J
  • (C) 500 J
  • (D) 1500 J
Show Answer

Answer: A – W = PΔV = 250 kPa × (5 – 1) L = 250 × 4 = 1000 J, since 1 kPa × 1 L = 1 J.

Q4. How much heat must be supplied to a 2.5 kg jug of water of specific heat capacity 4200 J/kg·K to raise its temperature by 10 K? (Assume no heat is lost to the surroundings.)

  • (A) 210,000 J
  • (B) 52,500 J
  • (C) 10,500 J
  • (D) 105,000 J
Show Answer

Answer: D – Q = mcΔT = 2.5 × 4200 × 10 = 105,000 J.

Q5. A heat engine does 240 J of useful work in each cycle while absorbing 1200 J of heat from a hot reservoir. What is the efficiency of the engine?

  • (A) 80%
  • (B) 40%
  • (C) 10%
  • (D) 20%
Show Answer

Answer: D – Efficiency = W ÷ Qh = 240 ÷ 1200 = 0.2 = 20%.

Q6. An ideal Carnot engine operates between reservoirs at 1200 K and 300 K. What is its maximum possible efficiency?

  • (A) 25%
  • (B) 37.5%
  • (C) 75%
  • (D) None of the other options
Show Answer

Answer: C – Carnot efficiency = 1 – Tc ÷ Th = 1 – 300 ÷ 1200 = 1 – 0.25 = 0.75 = 75%.

Q7. The root-mean-square speed of the molecules in a gas sample is 350 m/s. If the absolute temperature of the gas is multiplied by 9 while the gas stays the same, what is the new root-mean-square speed?

  • (A) 3150 m/s
  • (B) 1050 m/s
  • (C) 525 m/s
  • (D) 350 m/s
Show Answer

Answer: B – vrms is proportional to √T, so the speed is multiplied by √9 = 3. New vrms = 350 × 3 = 1050 m/s.

Q8. A hot metal block is placed in cold water inside an insulated box and the two reach a common temperature. What happens to the total entropy of the block and water?

  • (A) It decreases
  • (B) It increases
  • (C) It stays the same
  • (D) It becomes zero
Show Answer

Answer: B – Entropy measures disorder. Heat flows irreversibly from hot to cold in an isolated system, so total entropy rises.

Q9. What is the internal energy of 3 mol of a monatomic ideal gas at 400 K? (Use R = 8.31 J/mol·K.)

  • (A) 14958 J
  • (B) 29916 J
  • (C) 7479 J
  • (D) 9972 J
Show Answer

Answer: A – For a monatomic ideal gas, U = (3 ÷ 2)nRT = 1.5 × 3 × 8.31 × 400 = 14958 J.

Unit 10: Electric Force, Field, and Potential

Q10. Two point charges of 4 μC and 4 μC are placed 0.2 m apart. What is the magnitude of the electrostatic force between them? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 3.6 N
  • (B) 7.2 N
  • (C) 1.8 N
  • (D) 14.4 N
Show Answer

Answer: A – F = kq1q2 ÷ r2 = 9.0 × 109 × (4 × 10-6) × (4 × 10-6) ÷ 0.22 = 3.6 N.

Q11. What is the magnitude of the electric field at a point 0.4 m from a point charge of 8 μC? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 450,000 N/C
  • (B) 900,000 N/C
  • (C) 225,000 N/C
  • (D) 1,800,000 N/C
Show Answer

Answer: A – E = kQ ÷ r2 = 9.0 × 109 × (8 × 10-6) ÷ 0.42 = 450,000 N/C.

Q12. What is the electric potential at a point 0.3 m from a point charge of 9 μC, taking potential at infinity to be zero? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 540,000 V
  • (B) 135,000 V
  • (C) 270,000 V
  • (D) 67,500 V
Show Answer

Answer: C – V = kQ ÷ r = 9.0 × 109 × (9 × 10-6) ÷ 0.3 = 270,000 V.

Q13. A charge of 7 μC is placed in a uniform electric field of 2000 N/C. What is the magnitude of the electric force on the charge?

  • (A) 0.14 N
  • (B) 0 N
  • (C) 0.01 N
  • (D) 0.03 N
Show Answer

Answer: C – F = qE = (7 × 10-6) × 2000 = 0.01 N.

Q14. Two parallel metal plates are 0.04 m apart and have a potential difference of 96 V between them. What is the magnitude of the uniform electric field between the plates?

  • (A) 4,800 V/m
  • (B) 1,200 V/m
  • (C) 2,400 V/m
  • (D) 4 V/m
Show Answer

Answer: C – E = V ÷ d = 96 ÷ 0.04 = 2,400 V/m.

Q15. How much work is done by the electric field when a charge of 8 μC moves through a potential drop of 150 V?

  • (A) 0.01 J
  • (B) None of the other options
  • (C) It cannot be determined from the data given
  • (D) 0 J
Show Answer

Answer: D – W = qΔV = (8 × 10-6) × 150 = 0 J.

Q16. An electron is released from rest in a uniform electric field pointing to the right. In which direction does it begin to accelerate?

  • (A) To the right
  • (B) To the left
  • (C) Upward
  • (D) It does not accelerate
Show Answer

Answer: B – The electric field between parallel plates points from the positive plate toward the negative plate, and a negative charge accelerates opposite to the field direction.

Q17. Both point charges are doubled and the distance between them is also doubled. How does the electrostatic force between them change?

  • (A) It becomes 2 times the original force
  • (B) It becomes 4 times the original force
  • (C) It becomes 1/4 of the original force
  • (D) It stays the same
Show Answer

Answer: D – F is proportional to q1q2 ÷ r2: the numerator gains a factor 4 and the denominator gains a factor 4, so F is unchanged.

Q18. A electron starting from rest is accelerated through a potential difference of 600 V. What kinetic energy does it gain, in electronvolts?

  • (A) 1200 eV
  • (B) 300 eV
  • (C) 600 eV
  • (D) 2400 eV
Show Answer

Answer: C – Kinetic energy gained = qV. A proton or electron carries one elementary charge, so KE = 600 eV.

Unit 11: Electric Circuits

Q19. A current of 0.5 A flows through a 40 Ω resistor. What is the potential difference across the resistor?

  • (A) 80 V
  • (B) 40.5 V
  • (C) 20 V
  • (D) 0.0125 V
Show Answer

Answer: C – V = IR = 0.5 × 40 = 20 V.

Q20. Three resistors of 10 Ω, 10 Ω and 5 Ω are connected in series. What is their equivalent resistance?

  • (A) 12.5 Ω
  • (B) 20 Ω
  • (C) 35 Ω
  • (D) 25 Ω
Show Answer

Answer: D – For series resistors, Req = R1 + R2 + R3 = 10 + 10 + 5 = 25 Ω.

Q21. Two resistors of 20 Ω and 5 Ω are connected in parallel. What is their equivalent resistance?

  • (A) 25 Ω
  • (B) 4 Ω
  • (C) 8 Ω
  • (D) 6.25 Ω
Show Answer

Answer: B – For two parallel resistors, Req = (R1R2) ÷ (R1 + R2) = (20 × 5) ÷ (20 + 5) = 100 ÷ 25 = 4 Ω.

Q22. A current of 4 A flows through a 5 Ω resistor. What power is dissipated?

  • (A) 80 W
  • (B) 20 W
  • (C) 100 W
  • (D) 0.8 W
Show Answer

Answer: A – P = I2R = 42 × 5 = 16 × 5 = 80 W.

Q23. A battery of 48 V is connected across two resistors of 10 Ω and 30 Ω in series. What is the potential difference across the 10 Ω resistor?

  • (A) 12 V
  • (B) 36 V
  • (C) 24 V
  • (D) 6 V
Show Answer

Answer: A – The same current flows through both resistors, so V1 = V × R1 ÷ (R1 + R2) = 48 × 10 ÷ 40 = 12 V.

Q24. A 36 V battery is connected across two resistors in parallel. One branch has resistance 12 Ω. What current flows in that branch?

  • (A) 6 A
  • (B) 1.5 A
  • (C) 48 A
  • (D) 3 A
Show Answer

Answer: D – Each parallel branch has the full battery voltage, so I = V ÷ R = 36 ÷ 12 = 3 A.

Q25. A capacitor of capacitance 100 μF is charged to 6 V. How much charge is stored on it?

  • (A) 106 μC
  • (B) 600 μC
  • (C) 16.7 μC
  • (D) 1200 μC
Show Answer

Answer: B – Q = CV = 100 × 6 = 600 μC.

Q26. Two capacitors of 12 μF and 4 μF are connected in series. What is their equivalent capacitance?

  • (A) 16 μF
  • (B) 48 μF
  • (C) 8 μF
  • (D) 3 μF
Show Answer

Answer: D – Series: Ceq = (12 × 4) ÷ (12 + 4) = 48 ÷ 16 = 3 μF.

Q27. A cell of emf 24 V and internal resistance 2 Ω supplies a current of 4 A. What is the terminal potential difference of the cell?

  • (A) 24 V
  • (B) 16 V
  • (C) 32 V
  • (D) 8 V
Show Answer

Answer: B – Terminal voltage V = emf – Ir = 24 – (4 × 2) = 24 – 8 = 16 V.

Unit 12: Magnetism and Electromagnetism

Q28. A particle with charge 5 μC moves at 6,000 m/s perpendicular to a uniform magnetic field of 0.5 T. What is the magnitude of the magnetic force on the particle?

  • (A) 0.15 N
  • (B) 0 N
  • (C) 0.03 N
  • (D) 0.01 N
Show Answer

Answer: D – F = qvB = (5 × 10-6) × 6,000 × 0.5 = 0.01 N.

Q29. A straight wire of length 0.25 m carries a current of 8 A perpendicular to a uniform magnetic field of 0.6 T. What is the magnitude of the magnetic force on the wire?

  • (A) 2.4 N
  • (B) 0.6 N
  • (C) 1.2 N
  • (D) 12 N
Show Answer

Answer: C – F = ILB = 8 × 0.25 × 0.6 = 1.2 N.

Q30. What is the magnitude of the magnetic field at a distance of 0.25 m from a long straight wire carrying a current of 15 A?

  • (A) 24 μT
  • (B) 12 μT
  • (C) 6 μT
  • (D) 120 μT
Show Answer

Answer: B – B = μ0I ÷ (2πr) = (2 × 10-7 × 15) ÷ 0.25 = 12 × 10-6 T = 12 μT.


Q31. A long solenoid has 900 turns per metre and carries a current of 4 A. What is the magnitude of the magnetic field inside the solenoid? (Use μ0 = 4π × 10-7 T·m/A and π = 3.14.)

  • (A) 9.04 mT
  • (B) 2.26 mT
  • (C) 4.52 mT
  • (D) 45.22 mT
Show Answer

Answer: C – B = μ0nI = 4π × 10-7 × 900 × 4 = 4.52 × 10-3 T = 4.52 mT.

Q32. A metal rod of length 1.5 m moves at 5 m/s perpendicular to a uniform magnetic field of 0.4 T, with the rod perpendicular to both the field and its motion. What emf is induced between the ends of the rod?

  • (A) 6 V
  • (B) 1.5 V
  • (C) 30 V
  • (D) 3 V
Show Answer

Answer: D – Motional emf = BLv = 0.4 × 1.5 × 5 = 3 V.

Q33. The magnetic flux through a coil of 120 turns changes by 0.03 Wb in 0.6 s. What is the magnitude of the average induced emf in the coil?

  • (A) 12 V
  • (B) 3 V
  • (C) 6 V
  • (D) 0.6 V
Show Answer

Answer: C – Faraday’s law: emf = N × (Δflux ÷ time) = 120 × (0.03 ÷ 0.6) = 120 × 0.05 = 6 V.

Q34. A flat loop of area 1 m2 is placed in a uniform magnetic field of 0.3 T. The field makes an angle of 0° with the normal to the loop. What is the magnetic flux through the loop?

  • (A) 0.6 Wb
  • (B) 0.3 Wb
  • (C) 0.15 Wb
  • (D) None of the other options
Show Answer

Answer: B – Flux = BA cos θ = 0.3 × 1 × cos 0° = 0.3 × 1 = 0.3 Wb.

Q35. A magnet is dropped through a vertical copper tube. Compared with free fall, what happens to the magnet’s fall?

  • (A) It falls more slowly, because induced currents oppose its motion
  • (B) It falls faster, because induced currents assist its motion
  • (C) It falls at exactly the same rate as in free fall
  • (D) It stops and hangs in the tube
Show Answer

Answer: A – Lenz’s law: an induced current always opposes the change that produced it. An approaching pole is repelled, a receding pole is attracted, and a falling magnet is slowed by the opposing induced currents.

Unit 13: Geometric Optics

Q36. An object is placed 24 cm in front of a concave headlight reflector of focal length 8 cm. How far from the mirror is the image formed?

  • (A) 24 cm
  • (B) 6 cm
  • (C) 12 cm
  • (D) 16 cm
Show Answer

Answer: C – Mirror equation: 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ di = 1 ÷ 8 – 1 ÷ 24 = 0.08 cm-1, giving di = 12 cm.

Q37. An object is placed 18 cm in front of a converging lens of focal length 6 cm. How far from the lens is the image formed?

  • (A) 18 cm
  • (B) 4.5 cm
  • (C) 24 cm
  • (D) 9 cm
Show Answer

Answer: D – Lens equation: 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ di = 1 ÷ 6 – 1 ÷ 18, giving di = 9 cm.

Q38. An object is 18 cm from a mirror and its image is formed 54 cm from the mirror. What is the lateral magnification of the image?

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

Answer: C – Magnification m = -di ÷ do = -54 ÷ 18 = -3. The negative sign shows the image is inverted.

Q39. A ray of light travels from a medium of refractive index 1 into a medium of refractive index 2.42 at an angle of incidence of 45°. What is the angle of refraction, to the nearest 0.1°?

  • (A) 27.0°
  • (B) 17.0°
  • (C) 7.0°
  • (D) 73.0°
Show Answer

Answer: B – Snell law: n1 sin θ1 = n2 sin θ2, so sin θ2 = (1 × sin 45°) ÷ 2.42 = 0.29. Hence θ2 = 17.0°.

Q40. What is the critical angle for light travelling from diamond of refractive index 2.42 into air, to the nearest 0.1°?

  • (A) 34.4°
  • (B) 14.4°
  • (C) 65.6°
  • (D) 24.4°
Show Answer

Answer: D – At the critical angle, sin θc = n2 ÷ n1 = 1 ÷ 2.42 = 0.41, so θc = 24.4°.

Q41. Light travels at 1.2 × 108 m/s in a transparent material. What is the refractive index of the material? (Take c = 3.00 × 108 m/s.)

  • (A) 2.5
  • (B) 5
  • (C) 1.25
  • (D) 1.8
Show Answer

Answer: A – n = c ÷ v = (3.00 × 108) ÷ (1.2 × 108) = 2.5.

Q42. Light of wavelength 720 nm in air enters a medium of refractive index 1.2. What is its wavelength in the medium?

  • (A) 600 nm
  • (B) 720 nm
  • (C) 1200 nm
  • (D) 300 nm
Show Answer

Answer: A – Wavelength in a medium: λ = λ0 ÷ n = 720 ÷ 1.2 = 600 nm. The frequency stays the same, so the wavelength shortens.

Q43. An object is placed in front of a diverging lens. Which description always fits the image formed?

  • (A) Real, inverted and magnified
  • (B) Virtual, upright and diminished
  • (C) Virtual, upright and magnified
  • (D) Real, inverted and diminished
Show Answer

Answer: B – Ray diagrams show: a concave mirror gives a real, inverted, diminished image beyond the centre of curvature and a virtual, upright, magnified image inside the focal point; convex mirrors and diverging lenses always give virtual, upright, diminished images; a converging lens with the object beyond 2F gives a real, inverted, diminished image.

Unit 14: Waves, Sound, and Physical Optics

Q44. A wave has frequency 8 Hz and wavelength 12.5 m. What is its speed?

  • (A) 200 m/s
  • (B) 50 m/s
  • (C) 20.5 m/s
  • (D) 100 m/s
Show Answer

Answer: D – v = fλ = 8 × 12.5 = 100 m/s.

Q45. A string of length 1.8 m is fixed at both ends. Waves travel along it at 360 m/s. What is the frequency of its second harmonic?

  • (A) 200 Hz
  • (B) 400 Hz
  • (C) 100 Hz
  • (D) None of the other options
Show Answer

Answer: A – For a string fixed at both ends, fn = nv ÷ (2L) = 2 × 360 ÷ (2 × 1.8) = 200 Hz.

Q46. An air column of length 0.68 m is open at both ends. Taking the speed of sound as 340 m/s, what is its fundamental frequency?

  • (A) 250 Hz
  • (B) 500 Hz
  • (C) 125 Hz
  • (D) 1000 Hz
Show Answer

Answer: A – For a pipe open at both ends, f = v ÷ (2L) = 340 ÷ (2 × 0.68) = 250 Hz.

Q47. Two tuning forks of frequencies 1024 Hz and 1015 Hz are sounded together. What is the beat frequency heard?

  • (A) 18 Hz
  • (B) 9 Hz
  • (C) 11 Hz
  • (D) 2039 Hz
Show Answer

Answer: B – Beat frequency = |f1 – f2| = |1024 – 1015| = 9 Hz.

Q48. In a double-slit experiment, light of wavelength 700 nm passes through slits 0.2 mm apart and forms fringes on a screen 1 m away. How far from the central bright fringe is the first bright fringe, in millimetres?

  • (A) 7 mm
  • (B) 3.5 mm
  • (C) 1.75 mm
  • (D) 35 mm
Show Answer

Answer: B – Fringe position y = mλL ÷ d = 1 × (700 × 10-9) × 1 ÷ (0.2 × 10-3) = 3.5 × 10-3 m = 3.5 mm.

Q49. A sound source emitting 700 Hz moves at 51 m/s toward a stationary observer. Taking the speed of sound as 340 m/s, what frequency does the observer hear, to the nearest hertz?

  • (A) 864 Hz
  • (B) 784 Hz
  • (C) 700 Hz
  • (D) 824 Hz
Show Answer

Answer: D – Doppler shift for a moving source: f’ = f × v ÷ (v – vs) = 700 × 340 ÷ 289 = 824 Hz to the nearest hertz.

Q50. The power output of a point source of sound is doubled, with the listener staying at the same distance. How does the sound intensity at the listener change?

  • (A) It becomes 2 times the original intensity
  • (B) It becomes 4 times the original intensity
  • (C) It becomes 1/2 of the original intensity
  • (D) It stays the same
Show Answer

Answer: A – At fixed distance, intensity is proportional to source power, so doubling the power doubles I.

Q51. A diffraction grating has 500 lines per millimetre. Light of wavelength 700 nm is incident normally on it. What is the angle of the first-order maximum, to the nearest 0.1°?

  • (A) 30.5°
  • (B) 10.5°
  • (C) 20.5°
  • (D) 69.5°
Show Answer

Answer: C – Grating spacing d = 1 ÷ 500 mm = 2000 nm. For the first order, d sin θ = λ, so sin θ = 700 ÷ 2000 = 0.35, giving θ = 20.5°.

Unit 15: Modern Physics

Q52. What is the energy, in electronvolts, of a photon of wavelength 775 nm? (Use hc = 1240 eV·nm.)

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

Answer: C – E = hc ÷ λ = 1240 ÷ 775 = 1.6 eV.

Q53. Light of photon energy 7 eV falls on a metal surface whose work function is 4.2 eV. What is the maximum kinetic energy of the emitted photoelectrons?

  • (A) 2.8 eV
  • (B) 11.2 eV
  • (C) 7 eV
  • (D) 5.6 eV
Show Answer

Answer: A – Einstein photoelectric equation: Kmax = hf – work function = 7 – 4.2 = 2.8 eV.

Q54. A metal has a work function of 4 eV. What is its threshold frequency, to two decimal places in units of 1014 Hz? (Use h = 6.63 × 10-34 J·s and 1 eV = 1.60 × 10-19 J.)

  • (A) 9.65 × 1014 Hz
  • (B) 19.31 × 1014 Hz
  • (C) 4.83 × 1014 Hz
  • (D) 14.48 × 1014 Hz
Show Answer

Answer: A – Threshold: hf0 = work function = 4 × 1.60 × 10-19 J = 6.4 × 10-19 J. So f0 = (6.4 × 10-19) ÷ (6.63 × 10-34) = 9.65 × 1014 Hz.

Q55. An electron moves at 2.5 × 106 m/s. What is its de Broglie wavelength, in nanometres? (Use h = 6.63 × 10-34 J·s and electron mass 9.11 × 10-31 kg.)

  • (A) 0.58 nm
  • (B) 0.15 nm
  • (C) 2.91 nm
  • (D) 0.29 nm
Show Answer

Answer: D – λ = h ÷ p = h ÷ (mv) = (6.63 × 10-34) ÷ (9.11 × 10-31 × 2.5 × 106) = 0.29 × 10-9 m = 0.29 nm.

Q56. A hydrogen atom makes a transition from energy level n = 4 to n = 1. How much energy is released as a photon, in electronvolts? (Use En = -13.6 ÷ n2 eV.)

  • (A) 12.75 eV
  • (B) 25.5 eV
  • (C) 6.38 eV
  • (D) 0.85 eV
Show Answer

Answer: A – Energy released = E4 – E1 in magnitude = 13.6 × (1 ÷ 12 – 1 ÷ 42) = 13.6 × 0.94 = 12.75 eV.

Q57. A radioactive sample initially contains 96 g of an isotope. After 5 half-lives, how much of the isotope remains?

  • (A) 6 g
  • (B) 3 g
  • (C) 1.5 g
  • (D) 93 g
Show Answer

Answer: B – After each half-life half remains, so after 5 half-lives the fraction left is (1 ÷ 2)5 = 1 ÷ 32. Remaining mass = 96 ÷ 32 = 3 g.

Q58. In a nuclear reaction the total mass decreases by 0.08 u. How much energy is released? (Use 1 u = 931.5 MeV/c2.)

  • (A) 149.04 MeV
  • (B) 74.52 MeV
  • (C) 37.26 MeV
  • (D) 745.2 MeV
Show Answer

Answer: B – E = Δmc2 = 0.08 × 931.5 = 74.52 MeV.

Q59. A clock on a spacecraft moving at 0.6c ticks 4 s as measured in the spacecraft frame. How much time passes on Earth while the spacecraft clock ticks 4 s?

  • (A) 4 s
  • (B) 5 s
  • (C) 10 s
  • (D) 8 s
Show Answer

Answer: B – For v = 0.6c, the Lorentz factor is 1.25. Dilated time = 1.25 × 4 = 5 s.

Q60. A spacecraft of proper length 60 m travels past Earth at 0.6c. What length does an observer on Earth measure for the spacecraft?

  • (A) 60 m
  • (B) 96 m
  • (C) 48 m
  • (D) 30 m
Show Answer

Answer: C – Length contraction: L = L0 ÷ Lorentz factor = 60 ÷ 1.25 = 48 m.

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