WAEC Physics 2022 past questions

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

Ten questions from the 2022 WAEC Physics paper, with answers and explanations. We hold 53 questions from this paper, and the Skolra app has all of them.

WAEC Physics 2022 questions and answers

1.
An example of a mechanical wave is---------
1. A
  Radio wave
2. B
  X-ray
3. C
  Sound wave
  Correct
4. D
  Light wave
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Answer: C
Mechanical waves are waves that require a material medium (solid, liquid or gas) to travel; they cannot pass through a vacuum because they move by making the particles of the medium vibrate.
A sound wave is a mechanical wave because it travels by vibrating air (or other) particles, which is why sound cannot travel through empty space.
Light, radio waves and X-rays are all electromagnetic waves. Electromagnetic waves do not need a medium — they can travel through a vacuum, which is exactly why they do not count as mechanical waves. A candidate who mixes up "does not need a medium" with "mechanical" will wrongly pick one of these.
2.
The diameter of a brass ring at 30 °C is 50.0 cm. To what temperature must this ring be heated to increase its diameter to 50.29 cm? [ linear expansivity of brass = 1.9 x 10 $^{-5}$ K $^{-1}$ ]
1. A
  335 °C
  Correct
2. B
  183 °C
3. C
  33 °C
4. D
  305 °C
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Answer: A
Given: initial diameter L₀ = 50.0 cm, final diameter L = 50.29 cm, initial temperature = 30 °C, linear expansivity α = 1.9 × 10⁻⁵ K⁻¹.
Change in length: ΔL = 50.29 − 50.0 = 0.29 cm.
Using ΔL = L₀αΔT:
ΔT = ΔL / (L₀α) = 0.29 / (50.0 × 1.9 × 10⁻⁵) = 0.29 / 9.5 × 10⁻⁴ = 305.26 K.
Since the ring started at 30 °C, the final temperature is 30 + 305.26 ≈ 335 °C, which is the correct answer.
3.
One major reason why electrical appliances in homes are normally earthed is that the-----------
1. A
  electrical circuit of the appliance is completed so that it can function
2. B
  appliance is prevented from short-circuiting under normal working conditions
3. C
  current consumption of the appliance is reduced, thereby saving electricity bills
4. D
  metal casing does not become live and cause electric shock in case of a fault
  Correct
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Answer: D
Earthing (grounding) a metal-cased appliance connects its outer casing to the earth through a low-resistance wire.
Why this is done: If a fault occurs and a live wire inside touches the metal casing, the earth wire provides a very low-resistance path for the fault current to flow to the ground instead of through a person who touches the casing. This large fault current also blows the fuse quickly, cutting off the supply. So the metal casing not becoming live and causing electric shock is the correct reason.
Reducing current consumption is wrong — earthing has nothing to do with how much current the appliance normally draws; that is determined by its power rating.
4.
The effective potential energy, E, of a lunar satellite of mass, m, moving in an. elliptical orbit around the moon of mass, m, is given by
E = $\frac{K^{2}}{2m_{1}r^{2}} - \frac{Gm_{1}m_{2}}{r}$ where r is the distance of the satellite from the mooń and G is the universal gravitational constant of dimensions, M $^{-1}$ L $^{3}$ T $^{2}$ .
Ďetermine the dimensions of the angular momentum, K, of the satellite using dimensional analysis.
1. A
  ML²T⁻²
2. B
  MLT⁻¹
3. C
  M²L⁴T⁻²
4. D
  ML²T⁻¹
  Correct
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Answer: D
The equation E = K²/(2m₁r²) − Gm₁m₂/r is dimensionally consistent, so every term must carry the dimension of energy, ML²T⁻².
Using the K² term: K²/(2m₁r²) = E, so K² = E·m₁·r² = (ML²T⁻²)(M)(L²) = M²L⁴T⁻².
Taking the square root gives K = ML²T⁻¹, which is correct — this is exactly the known dimension of angular momentum (mass × velocity × radius = M·LT⁻¹·L).
5.
Arrange the following radiations in order of increasing ionization of air.
I. Alpha
Il. Gamma
IIl. Beta
1. A
  I, II, III
2. B
  I, III, II
3. C
  II, III, I
  Correct
4. D
  III, I, II
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Answer: C
Increasing order of ionization: Gamma, Beta, Alpha (II, III, I)
Alpha particles carry a double positive charge and are relatively heavy (helium nuclei), so as they move through air they collide with and strip electrons from many gas molecules — this makes them the strongest ionizers, though they only travel a short range.
Beta particles are fast, light electrons carrying a single charge. They interact with air molecules less strongly than alpha particles do, so their ionizing power is moderate — less than alpha but more than gamma.
6.
When the direction of vibration of the particles of a medium is perpendicular to the direction of propagation of a wave, the wave is said to be--------
1. A
  transverse
  Correct
2. B
  longitudinal
3. C
  progressive
4. D
  stationary
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Answer: A
A wave is classified by comparing the direction in which the particles of the medium vibrate to the direction in which the wave itself travels.
When the particles vibrate perpendicular to the direction of propagation, the wave is a transverse wave — think of a rope shaken up and down while the wave moves along its length, or light and water waves.
The longitudinal option is the classic mix-up: in a longitudinal wave (like sound), particles vibrate parallel to the direction of travel (compressions and rarefactions), which is the opposite condition to the one described here.
7.
Gamma rays are produced when?
1. A
  a radioactive nucleus de-excites from a higher to a lower energy state
  Correct
2. B
  a fast-moving electron is suddenly decelerated at a metal target
3. C
  a substance is heated to a very high temperature
4. D
  an electron jumps from a higher to a lower energy level in an atom
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Answer: A
Gamma rays are electromagnetic radiation that come from the nucleus, not the electron shells. When a radioactive nucleus emits an alpha or beta particle, it is often left in an excited (higher energy) state. It then settles down to a lower, more stable energy state, and in doing so it releases the extra energy as a gamma-ray photon. This is why the correct description is that a radioactive nucleus de-excites from a higher to a lower energy state.
The distractor about an electron jumping between energy levels in an atom describes how visible light or X-ray line spectra are produced, not gamma rays — that process involves the electron cloud, not the nucleus.
8.
If the kinetic energy of an electron is 100 eV, what is the wavelength of the de-Broglie wave associated with it? [h = 6.6 x 10 $^{-34}$ J s, e = 1.6 x 10 $^{19}$ C, m $_{e}$ , = 9.1 x 10 $^{-31}$ kg]
1. A
  1.23 × 10⁻¹⁰ m
  Correct
2. B
  2.46 × 10⁻¹⁰ m
3. C
  6.15 × 10⁻¹¹ m
4. D
  1.73 × 10⁻¹⁰ m
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Answer: A
Step 1: Convert the kinetic energy to joules.
1 eV = 1.6 × 10⁻¹⁹ J, so KE = 100 × 1.6 × 10⁻¹⁹ = 1.6 × 10⁻¹⁷ J.
Step 2: Find the momentum from the kinetic energy.
Since KE = p²/2m, the momentum is p = √(2mKE) = √(2 × 9.1 × 10⁻³¹ × 1.6 × 10⁻¹⁷) = √(2.912 × 10⁻⁴⁷) = 5.4 × 10⁻²⁴ kg m/s.
Step 3: Apply the de-Broglie relation.
λ = h/p = (6.6 × 10⁻³⁴) / (5.4 × 10⁻²⁴) ≈ 1.23 × 10⁻¹⁰ m, which is the correct wavelength.
9.
Three cells each of emf, 1.0 V, and internal resistance, 2 $\Omega$ , are connected in parallel across a 3 $\Omega$ resistor. Determine the current in the resistor.
1. A
  0.20 A
2. B
  0.27 A
  Correct
3. C
  1.50 A
4. D
  0.82 A
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Answer: B
Combining the cells: When identical cells of the same emf are connected in parallel, the combination behaves as a single cell with the same emf (1.0 V) but a smaller internal resistance, given by r/n = 2 Ω ÷ 3 = 2/3 Ω.
Total resistance in the circuit: R + r(effective) = 3 Ω + 2/3 Ω = 11/3 Ω.
Current in the 3 Ω resistor:
I = emf ÷ total resistance = 1.0 ÷ (11/3) = 3/11 ≈ 0.27 A.
10.
The efficiency of a wheel and axle is 100% and the ratio of their radii is 5:1. calculate the effort required to lift a load of mass 20kg using this machine, [g = 10 m s $^{-2}$ ]
1. A
  40 N
  Correct
2. B
  1000 N
3. C
  4 N
4. D
  200 N
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Answer: A
For a wheel and axle, the velocity ratio is the ratio of the radii: VR = R/r = 5/1 = 5.
Since the efficiency is given as 100%, the mechanical advantage equals the velocity ratio, so MA = 5.
First find the load (weight): Load = mg = 20 kg × 10 m s^-2 = 200 N.
Effort = Load ÷ MA = 200 N ÷ 5 = 40 N, and that is the correct answer.
200 N is simply the load itself, mistaken for the effort by forgetting to divide by the mechanical advantage. 1000 N comes from multiplying the load by the radii ratio instead of dividing. 4 N results from mixing up the wheel and axle radii, using r/R = 1/5 as the mechanical advantage instead of R/r.

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WAEC Physics 2022 past questions

WAEC Physics 2022 questions and answers

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