JAMB Physics 2021 past questions

Checked by the Skolra question team · Last updated 23 October 2026

Ten questions from the 2021 JAMB Physics paper, with answers and explanations. We hold 48 questions from this paper, and the Skolra app has all of them.

JAMB Physics 2021 questions and answers

1. 1.

   The velocities of light in air and glass are 3.0 x 10$^8$ms$^{-1}$ and 2.0 x 10$^8$ms$^{-1}$ respectively. If the angle of refraction is 30º, the sine of the angle of incidence is?

   1. A
      0.50
   2. B
      0.75
      Correct
   3. C
      1.50
   4. D
      0.33
   Show answerHide answer

   Answer: B

   Light is passing from air into glass, so the refractive index is found from the velocities: n = v(air) ÷ v(glass) = (3.0 × 10⁸) ÷ (2.0 × 10⁸) = 1.5.

   By Snell's law, n = sin(i) ÷ sin(r), so sin(i) = n × sin(r). With r = 30°, sin(30°) = 0.5, so sin(i) = 1.5 × 0.5 = 0.75. That makes 0.75 correct.

   A candidate who divides the wrong way round, using v(glass) ÷ v(air) × sin(r) = (2/3) × 0.5, lands on 0.33 — that comes from mixing up which velocity goes on top in the refractive index ratio.

   A candidate who forgets to bring in the refractive index at all and just copies sin(r) gets 0.50, which is only the sine of the angle of refraction, not the angle of incidence.

2. 2.

   A source of sound produces waves in air of wavelength1.65m. If the speed of soundin air is 330ms$^{-1}$, the period of vibration in air is?

   1. A
      2.5 × 10⁻³ s
   2. B
      2 × 10² s
   3. C
      5 × 10⁻³ s
      Correct
   4. D
      5.45 × 10² s
   Show answerHide answer

   Answer: C

   The wave equation gives speed as v = λ/T, so the period is T = λ/v.

   Substituting the given values: T = 1.65 m ÷ 330 m/s = 0.005 s = 5 × 10⁻³ s. This is the correct period of vibration.

   A candidate who divides the wrong way round (v ÷ λ instead of λ ÷ v) gets 330 ÷ 1.65 = 200, and may mistakenly attach seconds to it, landing on 2 × 10² s — but that value is actually the frequency in hertz, not the period.

   Multiplying instead of dividing (λ × v = 1.65 × 330 = 544.5) gives another trap answer, 5.45 × 10² s, which has no physical meaning here.

   Halving the correct period by mistake, perhaps confusing it with a half-cycle, gives 2.5 × 10⁻³ s, which is also wrong.

3. 3.

   Calculate the mass of ice that would melt when 2kg of copper is quickly transferred from boiling water to a block of ice without heat loss (specific heat capacity of copper = $400\text{ J kg}^{-1}\text{ K}^{-1}$, latent heat of fusion of ice = $3.3 \times 10^{5}\text{ J kg}^{-1}$)

   1. A
      0.36 kg
   2. B
      2.40 kg
   3. C
      0.24 kg
      Correct
   4. D
      0.12 kg
   Show answerHide answer

   Answer: C

   Step 1: The copper loses heat as it cools from the boiling water temperature (100°C) down to the ice temperature (0°C), a temperature drop of 100 K.

   Heat lost by copper = mcΔθ = 2 kg × 400 J kg⁻¹K⁻¹ × 100 K = 80,000 J

   Step 2: Since there is no heat loss, all of this heat goes into melting the ice. Using Q = mL:

   m = Q ÷ L = 80,000 ÷ (3.3 × 10⁵) = 0.24 kg

4. 4.

   Calculate the mass of ice that would melt when 2kg of copper is quickly transferred from boiling water to a block of ice without heat loss (specific heat capacity of copper = 400Jkg$^{-1}$k$^{-1}$, latent heat of fusion of ice = 3.3 x 10$^{5}$Jkg$^{-1}$)

   1. A
      0.484 kg
   2. B
      0.242 kg
      Correct
   3. C
      0.121 kg
   4. D
      2.420 kg
   Show answerHide answer

   Answer: B

   Since the copper loses heat and the ice gains it, with no heat lost to the surroundings, we equate the two:

   Heat lost by copper = mass × specific heat capacity × temperature change
   = 2 kg × 400 J/kg/K × (100 − 0) K = 80,000 J

   Boiling water is at 100°C and ice is at 0°C, so the temperature drop for the copper is 100 K.

   This is the correct answer. The value 0.121 kg comes from forgetting to multiply by the mass of the copper (using only c × ΔT ÷ L). The value 2.420 kg comes from a decimal-point slip, misplacing 3.3 × 10⁵ by a factor of ten. The value 0.484 kg results from mistakenly doubling the temperature change (using 200 K instead of 100 K).

5. 5.

   A man stands 4m in front of a plane mirror. If the mirror is moved 1m towards the man, the distance between him and his new image is?

   1. A
      5 m
   2. B
      8 m
   3. C
      6 m
      Correct
   4. D
      3 m
   Show answerHide answer

   Answer: C

   In a plane mirror, the image is formed as far behind the mirror as the object is in front of it. So the distance between an object and its image is always twice the object's distance from the mirror.

   Originally the man is 4 m from the mirror, so his image is 4 m behind it, giving 4 m + 4 m = 8 m between him and his image (this is a trap if you forget the mirror moved).

   When the mirror is moved 1 m towards the man, his new distance from the mirror becomes 4 m − 1 m = 3 m. His image is now 3 m behind the mirror, so the new distance between the man and his image is 3 m + 3 m = 6 m.

6. 6.

   The temperature gradient across a copper rod of thickness 0.02m, maintained at two temperature junctions of 20º and 80ºC respectively is?

   1. A
      4000°C/m
   2. B
      1500°C/m
   3. C
      3000°C/m
      Correct
   4. D
      6000°C/m
   Show answerHide answer

   Answer: C

   Temperature gradient is the rate of change of temperature with distance, given by:

   Temperature gradient = (T₂ − T₁) / thickness

   Here, T₂ = 80°C, T₁ = 20°C, and thickness = 0.02 m.

   Temperature gradient = (80 − 20) / 0.02 = 60 / 0.02 = 3000°C/m

   The option giving 3000°C/m is correct. A candidate who mistakenly uses only the hotter junction's temperature (80) instead of the difference gets 4000°C/m. Someone who uses half the actual temperature difference (30 instead of 60) lands on 1500°C/m. Doubling the thickness error the other way (using 0.01 m instead of 0.02 m) gives 6000°C/m. Always subtract the two junction temperatures first, then divide by the actual thickness of the rod.

7. 7.

   If the fraction of the atoms of a radioactive material left after 120 years is 1/64, what is the half-life of the material?

   1. A
      20 years
      Correct
   2. B
      40 years
   3. C
      15 years
   4. D
      24 years
   Show answerHide answer

   Answer: A

   The amount of a radioactive material left after each half-life is halved. So after n half-lives, the fraction remaining is (1/2)ⁿ.

   Here the fraction remaining is 1/64. Since 64 = 2⁶, we have 1/64 = (1/2)⁶, meaning the material has undergone 6 half-lives in the 120 years given.

   Half-life = total time ÷ number of half-lives = 120 ÷ 6 = 20 years.

8. 8.

   0.5kg of water at 10ºC is completely converted to ice at 0ºC by extracting (88000) of heat from it. If the specific heat capacity of water is 4200Jkg$^{-1}$ Cº. Calculate the specific latent heat of fusion of ice.

   1. A
      9.2 × 10⁴ Jkg⁻¹
   2. B
      2.18 × 10⁵ Jkg⁻¹
   3. C
      1.76 × 10⁵ Jkg⁻¹
   4. D
      1.34 × 10⁵ Jkg⁻¹
      Correct
   Show answerHide answer

   Answer: D

   Extracting heat from the water happens in two stages: first the water is cooled from 10°C down to 0°C (this removes sensible heat), then at 0°C it freezes into ice (this removes latent heat). The total 88000 J covers both stages.

   Step 1 — sensible heat to cool the water to 0°C:
   Q = mcΔT = 0.5 × 4200 × 10 = 21000 J

   Step 2 — heat left over to freeze the water at 0°C:
   Q_latent = 88000 − 21000 = 67000 J

   Step 3 — specific latent heat of fusion:
   L = Q_latent ÷ m = 67000 ÷ 0.5 = 134000 Jkg⁻¹ = 1.34 × 10⁵ Jkg⁻¹, and this is correct.

9. 9.

   An astronomical telescope is said to be in normal adjustment when the?

   1. A
      image formed by the objective lies beyond the focal point of the eyepiece
   2. B
      focal length of the eyepiece equals the focal length of the objective
   3. C
      distance between the objective and the eyepiece equals the sum of their focal lengths
      Correct
   4. D
      final image is formed at the focal point of the objective
   Show answerHide answer

   Answer: C

   An astronomical telescope is in normal adjustment when it is set up so that a person with a relaxed (unstrained) eye can view the final image, which means the final image is formed at infinity.

   For this to happen, the real image formed by the objective lens (which forms at its own focal point, f_o, from the objective) must fall exactly on the focal point of the eyepiece, f_e, so that the eyepiece then produces parallel rays. Since the objective's image point and the eyepiece's focal point coincide, the distance separating the two lenses must equal the sum of their focal lengths: L = f_o + f_e.

10. 10.

    Calculate the heat energy required to vaporize 50g of water initially at 80ºCif the specific heat capacity of water is 4.23jg$^{-1}$k$^{-1}$ (specific latent heat of vaporization of water is 2260jg$^{-1}$

    1. A
       117,230J
       Correct
    2. B
       4,230J
    3. C
       113,000J
    4. D
       134,150J
    Show answerHide answer

    Answer: A

    This is a two-stage heat problem. Water starts at 80°C, so it must first be heated to its boiling point (100°C) before it can vaporize.

    Stage 1 – heating the water (sensible heat):
    Q₁ = mcΔT = 50g × 4.23Jg⁻¹K⁻¹ × (100 − 80)K = 50 × 4.23 × 20 = 4230J

    Stage 2 – vaporizing the water at 100°C (latent heat):
    Q₂ = mL = 50g × 2260Jg⁻¹ = 113000J

    Total heat energy:
    Q = Q₁ + Q₂ = 4230J + 113000J = 117,230J

The questions come from past JAMB, WAEC and NECO papers. We use software to help write the answer options and explanations, and every answer is checked twice before it goes on the site. How we check

Spotted a wrong answer? Tell us from the app or on the contact page and we check it.

Practise all 48 questions we hold from the 2021 paper, on a timed CBT screen

Every question has the answer and an explanation, in the Skolra app.

Download for WindowsDownload for Mac

About the app · Skolra for Windows · Skolra for Mac