AP Physics 2 Practice Test 7 – 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 gas sealed inside a laboratory syringe is initially at 200 kPa, occupies 5 L and is at 300 K. It is then compressed or expanded to 5 L and its temperature changes to 600 K. What is the final pressure of the gas?

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

Answer: C – P1V1 ÷ T1 = P2V2 ÷ T2, so P2 = P1 × (V1 ÷ V2) × (T2 ÷ T1) = 200 × (5 ÷ 5) × (600 ÷ 300) = 400 kPa.

Q2. 800 J of heat is added to steam in a turbine chamber while the gas does 350 J of work on its surroundings. What is the change in internal energy of the gas?

  • (A) 450 J
  • (B) 1150 J
  • (C) -450 J
  • (D) 800 J
Show Answer

Answer: A – First law: ΔU = Q – Wby = 800 – (350) = 450 J.

Q3. A gas in a syringe expands at a constant pressure of 150 kPa from 2 L to 9 L. How much work is done by the gas?

  • (A) 2100 J
  • (B) 525 J
  • (C) 1050 J
  • (D) 1650 J
Show Answer

Answer: C – W = PΔV = 150 kPa × (9 – 2) L = 150 × 7 = 1050 J, since 1 kPa × 1 L = 1 J.

Q4. How much heat must be supplied to a 2 kg copper rod of specific heat capacity 385 J/kg·K to raise its temperature by 40 K? (Assume no heat is lost to the surroundings.)

  • (A) 61,600 J
  • (B) 30,800 J
  • (C) 15,400 J
  • (D) 770 J
Show Answer

Answer: B – Q = mcΔT = 2 × 385 × 40 = 30,800 J.

Q5. A heat engine does 450 J of useful work in each cycle while absorbing 1500 J of heat from a hot reservoir. What is the efficiency of the engine?

  • (A) 70%
  • (B) 60%
  • (C) 15%
  • (D) 30%
Show Answer

Answer: D – Efficiency = W ÷ Qh = 450 ÷ 1500 = 0.3 = 30%.

Q6. An ideal Carnot engine operates between reservoirs at 800 K and 320 K. What is its maximum possible efficiency?

  • (A) 40%
  • (B) 30%
  • (C) None of the other options
  • (D) 60%
Show Answer

Answer: D – Carnot efficiency = 1 – Tc ÷ Th = 1 – 320 ÷ 800 = 1 – 0.4 = 0.6 = 60%.

Q7. The root-mean-square speed of the molecules in a gas sample is 300 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) 900 m/s
  • (B) 2700 m/s
  • (C) 450 m/s
  • (D) 300 m/s
Show Answer

Answer: A – vrms is proportional to √T, so the speed is multiplied by √9 = 3. New vrms = 300 × 3 = 900 m/s.

Q8. A gas is compressed slowly at constant temperature. What happens to the entropy of the gas?

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

Answer: A – Entropy measures disorder. Compression confines the molecules to a smaller volume, so entropy falls.

Q9. What is the internal energy of 4 mol of a monatomic ideal gas at 250 K? (Use R = 8.31 J/mol·K.)

  • (A) 24930 J
  • (B) 12465 J
  • (C) 6232.5 J
  • (D) 8310 J
Show Answer

Answer: B – For a monatomic ideal gas, U = (3 ÷ 2)nRT = 1.5 × 4 × 8.31 × 250 = 12465 J.

Unit 10: Electric Force, Field, and Potential

Q10. Two point charges of 4 μC and 2 μC are placed 0.4 m apart. What is the magnitude of the electrostatic force between them? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 0.45 N
  • (B) 0.9 N
  • (C) 0.22 N
  • (D) 1.8 N
Show Answer

Answer: A – F = kq1q2 ÷ r2 = 9.0 × 109 × (4 × 10-6) × (2 × 10-6) ÷ 0.42 = 0.45 N.

Q11. What is the magnitude of the electric field at a point 0.5 m from a point charge of 2 μC? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 72,000 N/C
  • (B) 144,000 N/C
  • (C) 36,000 N/C
  • (D) 288,000 N/C
Show Answer

Answer: A – E = kQ ÷ r2 = 9.0 × 109 × (2 × 10-6) ÷ 0.52 = 72,000 N/C.

Q12. What is the electric potential at a point 0.2 m from a point charge of 4 μC, taking potential at infinity to be zero? (Use k = 9.0 × 109 N·m2/C2.)

  • (A) 360,000 V
  • (B) 90,000 V
  • (C) 180,000 V
  • (D) 45,000 V
Show Answer

Answer: C – V = kQ ÷ r = 9.0 × 109 × (4 × 10-6) ÷ 0.2 = 180,000 V.

Q13. A charge of 5 μC is placed in a uniform electric field of 4000 N/C. What is the magnitude of the electric force on the charge?

  • (A) 0.02 N
  • (B) 0.2 N
  • (C) 0 N
  • (D) 0.04 N
Show Answer

Answer: A – F = qE = (5 × 10-6) × 4000 = 0.02 N.

Q14. Two parallel metal plates are 0.05 m apart and have a potential difference of 200 V between them. What is the magnitude of the uniform electric field between the plates?

  • (A) 8,000 V/m
  • (B) 2,000 V/m
  • (C) 10 V/m
  • (D) 4,000 V/m
Show Answer

Answer: D – E = V ÷ d = 200 ÷ 0.05 = 4,000 V/m.

Q15. How much work is done by the electric field when a charge of 4 μC moves through a potential drop of 300 V?

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

Answer: C – W = qΔV = (4 × 10-6) × 300 = 0 J.

Q16. An electron is placed in a uniform electric field pointing north. In which direction is the electric force on the electron?

  • (A) North
  • (B) East
  • (C) West
  • (D) South
Show Answer

Answer: D – The force on a positive charge is in the direction of the field, while the force on a negative charge is opposite to the field. The field itself points away from positive charges and toward negative charges.

Q17. The distance between two point charges is halved, with both charges unchanged. How does the electrostatic force between them change?

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

Answer: C – F is proportional to 1 ÷ r2, so halving r multiplies F by 4.

Q18. A electron starting from rest is accelerated through a potential difference of 250 V. What kinetic energy does it gain, in electronvolts?

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

Answer: B – Kinetic energy gained = qV. A proton or electron carries one elementary charge, so KE = 250 eV.

Unit 11: Electric Circuits

Q19. A potential difference of 36 V is applied across a 12 Ω resistor. What current flows through it?

  • (A) 0.33 A
  • (B) 432 A
  • (C) 48 A
  • (D) 3 A
Show Answer

Answer: D – I = V ÷ R = 36 ÷ 12 = 3 A.

Q20. Three resistors of 5 Ω, 15 Ω and 2 Ω are connected in series. What is their equivalent resistance?

  • (A) 11 Ω
  • (B) 22 Ω
  • (C) 6.82 Ω
  • (D) 32 Ω
Show Answer

Answer: B – For series resistors, Req = R1 + R2 + R3 = 5 + 15 + 2 = 22 Ω.

Q21. Two resistors of 12 Ω and 4 Ω are connected in parallel. What is their equivalent resistance?

  • (A) 16 Ω
  • (B) 6 Ω
  • (C) 3 Ω
  • (D) 4 Ω
Show Answer

Answer: C – For two parallel resistors, Req = (R1R2) ÷ (R1 + R2) = (12 × 4) ÷ (12 + 4) = 48 ÷ 16 = 3 Ω.

Q22. A current of 3 A flows through a 10 Ω heating element. What power is dissipated in the element?

  • (A) 30 W
  • (B) 3.33 W
  • (C) 900 W
  • (D) 90 W
Show Answer

Answer: D – P = I2R = 32 × 10 = 9 × 10 = 90 W.

Q23. A battery of 24 V is connected across two resistors of 4 Ω and 8 Ω in series. What is the potential difference across the 4 Ω resistor?

  • (A) 16 V
  • (B) 8 V
  • (C) 4 V
  • (D) None of the other options
Show Answer

Answer: B – The same current flows through both resistors, so V1 = V × R1 ÷ (R1 + R2) = 24 × 4 ÷ 12 = 8 V.

Q24. A 24 V battery is connected across two resistors in parallel. One branch has resistance 8 Ω. What current flows in that branch?

  • (A) 3 A
  • (B) 6 A
  • (C) 1.5 A
  • (D) 32 A
Show Answer

Answer: A – Each parallel branch has the full battery voltage, so I = V ÷ R = 24 ÷ 8 = 3 A.

Q25. A capacitor of capacitance 10 μF is charged to 20 V. How much energy is stored in it?

  • (A) 0.004 J
  • (B) 0.001 J
  • (C) 0.2 J
  • (D) 0.002 J
Show Answer

Answer: D – E = 0.5CV2 = 0.5 × (10 × 10-6) × 202 = 0.5 × 10-5 × 400 = 0.002 J.

Q26. Two capacitors of 6 μF and 3 μF are connected in series. What is their equivalent capacitance?

  • (A) 2 μF
  • (B) 9 μF
  • (C) 18 μF
  • (D) 4.5 μF
Show Answer

Answer: A – Series: 1 ÷ Ceq = 1 ÷ 6 + 1 ÷ 3 = 3 ÷ 6, so Ceq = 2 μF.

Q27. A cell of emf 9 V and internal resistance 2 Ω supplies a current of 1.5 A. What is the terminal potential difference of the cell?

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

Answer: B – Terminal voltage V = emf – Ir = 9 – (1.5 × 2) = 9 – 3 = 6 V.

Unit 12: Magnetism and Electromagnetism

Q28. A particle with charge 2 μC moves at 5,000 m/s perpendicular to a uniform magnetic field of 0.4 T. What is the magnitude of the magnetic force on the particle?

  • (A) 0.04 N
  • (B) 0.01 N
  • (C) 0 N
  • (D) None of the other options
Show Answer

Answer: C – F = qvB = (2 × 10-6) × 5,000 × 0.4 = 0 N.

Q29. A straight wire of length 0.8 m carries a current of 2 A perpendicular to a uniform magnetic field of 0.5 T. What is the magnitude of the magnetic force on the wire?

  • (A) 1.6 N
  • (B) 0.4 N
  • (C) 0.8 N
  • (D) 8 N
Show Answer

Answer: C – F = ILB = 2 × 0.8 × 0.5 = 0.8 N.

Q30. What is the magnitude of the magnetic field at a distance of 0.1 m from a long straight wire carrying a current of 8 A?

  • (A) 32 μT
  • (B) 16 μT
  • (C) 8 μT
  • (D) 160 μT
Show Answer

Answer: B – B = μ0I ÷ (2πr) = (2 × 10-7 × 8) ÷ 0.1 = 16 × 10-6 T = 16 μT.


Q31. A long solenoid has 1200 turns per metre and carries a current of 1.5 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) 4.52 mT
  • (B) 2.26 mT
  • (C) 1.13 mT
  • (D) 22.61 mT
Show Answer

Answer: B – B = μ0nI = 4π × 10-7 × 1200 × 1.5 = 2.26 × 10-3 T = 2.26 mT.

Q32. A metal rod of length 0.5 m moves at 2 m/s perpendicular to a uniform magnetic field of 0.8 T, with the rod perpendicular to both the field and its motion. What emf is induced between the ends of the rod?

  • (A) 1.6 V
  • (B) 0.4 V
  • (C) 8 V
  • (D) 0.8 V
Show Answer

Answer: D – Motional emf = BLv = 0.8 × 0.5 × 2 = 0.8 V.

Q33. The magnetic flux through a coil of 50 turns changes by 0.06 Wb in 0.3 s. What is the magnitude of the average induced emf in the coil?

  • (A) 20 V
  • (B) 5 V
  • (C) 10 V
  • (D) 1 V
Show Answer

Answer: C – Faraday’s law: emf = N × (Δflux ÷ time) = 50 × (0.06 ÷ 0.3) = 50 × 0.2 = 10 V.

Q34. A flat loop of area 0.4 m2 is placed in a uniform magnetic field of 0.6 T. The field makes an angle of 60° with the normal to the loop. What is the magnetic flux through the loop?

  • (A) 0.24 Wb
  • (B) 0.06 Wb
  • (C) 0.12 Wb
  • (D) None of the other options
Show Answer

Answer: C – Flux = BA cos θ = 0.6 × 0.4 × cos 60° = 0.24 × 0.5 = 0.12 Wb.

Q35. A bar magnet with its north pole facing a closed loop is pulled away from the loop. What magnetic pole does the side of the loop facing the magnet become?

  • (A) A north pole, so that it repels the receding magnet
  • (B) It does not become a pole at all
  • (C) A south pole, so that it attracts the receding magnet
  • (D) A south pole, so that it repels the receding magnet
Show Answer

Answer: C – 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 60 cm in front of a concave telescope mirror of focal length 20 cm. How far from the mirror is the image formed?

  • (A) 60 cm
  • (B) 30 cm
  • (C) 15 cm
  • (D) 40 cm
Show Answer

Answer: B – Mirror equation: 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ di = 1 ÷ 20 – 1 ÷ 60 = 0.03 cm-1, giving di = 30 cm.

Q37. An object is placed 30 cm in front of a converging lens of focal length 10 cm. How far from the lens is the image formed?

  • (A) 30 cm
  • (B) 7.5 cm
  • (C) 40 cm
  • (D) 15 cm
Show Answer

Answer: D – Lens equation: 1 ÷ f = 1 ÷ do + 1 ÷ di, so 1 ÷ di = 1 ÷ 10 – 1 ÷ 30, giving di = 15 cm.

Q38. An object is 30 cm from a mirror and its image is formed 10 cm from the mirror. What is the lateral magnification of the image?

  • (A) -0.33
  • (B) 0.33
  • (C) -0.67
  • (D) -0.17
Show Answer

Answer: A – Magnification m = -di ÷ do = -10 ÷ 30 = -0.33. 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 1.33 at an angle of incidence of 45°. What is the angle of refraction, to the nearest 0.1°?

  • (A) 42.1°
  • (B) 22.1°
  • (C) 57.9°
  • (D) 32.1°
Show Answer

Answer: D – Snell law: n1 sin θ1 = n2 sin θ2, so sin θ2 = (1 × sin 45°) ÷ 1.33 = 0.53. Hence θ2 = 32.1°.

Q40. What is the critical angle for light travelling from water of refractive index 1.33 into air, to the nearest 0.1°?

  • (A) 48.8°
  • (B) 58.8°
  • (C) 38.8°
  • (D) 41.2°
Show Answer

Answer: A – At the critical angle, sin θc = n2 ÷ n1 = 1 ÷ 1.33 = 0.75, so θc = 48.8°.

Q41. Light travels at 2.25 × 108 m/s in a transparent material. What is the refractive index of the material? (Take c = 3.00 × 108 m/s.)

  • (A) 2.66
  • (B) 1.33
  • (C) 0.67
  • (D) 0.75
Show Answer

Answer: B – n = c ÷ v = (3.00 × 108) ÷ (2.25 × 108) = 1.33.

Q42. Light of wavelength 660 nm in air enters a medium of refractive index 1.5. What is its wavelength in the medium?

  • (A) 660 nm
  • (B) 880 nm
  • (C) 440 nm
  • (D) 220 nm
Show Answer

Answer: C – Wavelength in a medium: λ = λ0 ÷ n = 660 ÷ 1.5 = 440 nm. The frequency stays the same, so the wavelength shortens.

Q43. An object is placed between the focal point and a concave mirror. Which description best fits the image formed?

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

Answer: D – 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 10 Hz and wavelength 8 m. What is its speed?

  • (A) 160 m/s
  • (B) 80 m/s
  • (C) 40 m/s
  • (D) 18 m/s
Show Answer

Answer: B – v = fλ = 10 × 8 = 80 m/s.

Q45. A string of length 1.5 m is fixed at both ends. Waves travel along it at 300 m/s. What is the frequency of its second harmonic?

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

Answer: D – For a string fixed at both ends, fn = nv ÷ (2L) = 2 × 300 ÷ (2 × 1.5) = 200 Hz.

Q46. An air column of length 0.85 m is closed at one end. Taking the speed of sound as 340 m/s, what is its fundamental frequency?

  • (A) 100 Hz
  • (B) 200 Hz
  • (C) 50 Hz
  • (D) 400 Hz
Show Answer

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

Q47. Two tuning forks of frequencies 440 Hz and 436 Hz are sounded together. What is the beat frequency heard?

  • (A) 4 Hz
  • (B) 8 Hz
  • (C) 6 Hz
  • (D) 876 Hz
Show Answer

Answer: A – Beat frequency = |f1 – f2| = |440 – 436| = 4 Hz.

Q48. In a double-slit experiment, light of wavelength 500 nm passes through slits 0.2 mm apart and forms fringes on a screen 1.5 m away. How far from the central bright fringe is the first bright fringe, in millimetres?

  • (A) 3.75 mm
  • (B) 7.5 mm
  • (C) 1.88 mm
  • (D) 37.5 mm
Show Answer

Answer: A – Fringe position y = mλL ÷ d = 1 × (500 × 10-9) × 1.5 ÷ (0.2 × 10-3) = 3.75 × 10-3 m = 3.75 mm.

Q49. A sound source emitting 600 Hz moves at 34 m/s away from a stationary observer. Taking the speed of sound as 340 m/s, what frequency does the observer hear, to the nearest hertz?

  • (A) 585 Hz
  • (B) 505 Hz
  • (C) 545 Hz
  • (D) 600 Hz
Show Answer

Answer: C – Doppler shift for a moving source: f’ = f × v ÷ (v + vs) = 600 × 340 ÷ 374 = 545 Hz to the nearest hertz.

Q50. A listener moves to three times the original distance from a point source of sound. How does the sound intensity change?

  • (A) It becomes 1/3 of the original intensity
  • (B) It becomes 1/9 of the original intensity
  • (C) It becomes 3 times the original intensity
  • (D) It becomes 1/4 of the original intensity
Show Answer

Answer: B – I proportional to 1 ÷ r2, so tripling r divides I by 9.

Q51. A diffraction grating has 400 lines per millimetre. Light of wavelength 500 nm is incident normally on it. What is the angle of the first-order maximum, to the nearest 0.1°?

  • (A) 21.5°
  • (B) 1.5°
  • (C) 78.5°
  • (D) 11.5°
Show Answer

Answer: D – Grating spacing d = 1 ÷ 400 mm = 2500 nm. For the first order, d sin θ = λ, so sin θ = 500 ÷ 2500 = 0.2, giving θ = 11.5°.

Unit 15: Modern Physics

Q52. What is the energy, in electronvolts, of a photon of wavelength 500 nm? (Use hc = 1240 eV·nm.)

  • (A) 2.48 eV
  • (B) 4.96 eV
  • (C) 1.24 eV
  • (D) 3.72 eV
Show Answer

Answer: A – E = hc ÷ λ = 1240 ÷ 500 = 2.48 eV.

Q53. Light of photon energy 5 eV falls on a metal surface whose work function is 2 eV. What is the maximum kinetic energy of the emitted photoelectrons?

  • (A) 7 eV
  • (B) 5 eV
  • (C) 3 eV
  • (D) 6 eV
Show Answer

Answer: C – Einstein photoelectric equation: Kmax = hf – work function = 5 – 2 = 3 eV.

Q54. A metal has a work function of 2.5 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) 12.07 × 1014 Hz
  • (B) 6.03 × 1014 Hz
  • (C) 3.02 × 1014 Hz
  • (D) 9.05 × 1014 Hz
Show Answer

Answer: B – Threshold: hf0 = work function = 2.5 × 1.60 × 10-19 J = 4 × 10-19 J. So f0 = (4 × 10-19) ÷ (6.63 × 10-34) = 6.03 × 1014 Hz.

Q55. An electron moves at 2 × 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) 0.36 nm
  • (B) 0.73 nm
  • (C) 0.18 nm
  • (D) 3.64 nm
Show Answer

Answer: A – λ = h ÷ p = h ÷ (mv) = (6.63 × 10-34) ÷ (9.11 × 10-31 × 2 × 106) = 0.36 × 10-9 m = 0.36 nm.

Q56. A hydrogen atom makes a transition from energy level n = 4 to n = 2. How much energy is released as a photon, in electronvolts? (Use En = -13.6 ÷ n2 eV.)

  • (A) 5.1 eV
  • (B) 1.27 eV
  • (C) 11.05 eV
  • (D) 2.55 eV
Show Answer

Answer: D – Energy released = E4 – E2 in magnitude = 13.6 × (1 ÷ 22 – 1 ÷ 42) = 13.6 × 0.19 = 2.55 eV.

Q57. A radioactive sample initially contains 120 g of an isotope. After 2 half-lives, how much of the isotope remains?

  • (A) 60 g
  • (B) 30 g
  • (C) 15 g
  • (D) 90 g
Show Answer

Answer: B – After each half-life half remains, so after 2 half-lives the fraction left is (1 ÷ 2)2 = 1 ÷ 4. Remaining mass = 120 ÷ 4 = 30 g.

Q58. In a nuclear reaction the total mass decreases by 0.05 u. How much energy is released? (Use 1 u = 931.5 MeV/c2.)

  • (A) 93.15 MeV
  • (B) 23.29 MeV
  • (C) 465.75 MeV
  • (D) 46.58 MeV
Show Answer

Answer: D – E = Δmc2 = 0.05 × 931.5 = 46.58 MeV.

Q59. A clock on a spacecraft moving at 0.8c ticks 9 s as measured in the spacecraft frame. How much time passes on Earth while the spacecraft clock ticks 9 s?

  • (A) 9 s
  • (B) 15 s
  • (C) 30 s
  • (D) 18 s
Show Answer

Answer: B – For v = 0.8c, the Lorentz factor is 1.67. Dilated time = 1.67 × 9 = 15 s.

Q60. A spacecraft of proper length 150 m travels past Earth at 0.8c. What length does an observer on Earth measure for the spacecraft?

  • (A) 150 m
  • (B) 180 m
  • (C) 90 m
  • (D) 75 m
Show Answer

Answer: C – Length contraction: L = L0 ÷ Lorentz factor = 150 ÷ 1.67 = 90 m.

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