WAEC Physics 2025 Theory Past Questions

Model answer

R = u²sin2θ/g, where u = initial velocity (LT⁻¹) and g = acceleration due to gravity (LT⁻²). sin2θ is dimensionless. R = (LT⁻¹)² / (LT⁻²) = L²T⁻² / (LT⁻²) = L. Therefore the dimension of Range, R, is [L].

Model answer

(i) Gas laser: used for surgical procedures/skin treatment, missile guidance, reading barcodes, target designation, and making holograms. (ii) Chemical laser: used for cutting metals, drilling materials, military weapon manufacturing, spectroscopy and military defence. (iii) Dye laser: used for isotope separation, detecting pollutants, studying absorption/emission spectra, and atomic vapour laser isotope separation.

Model answer

The path of the object is a parabola (projectile motion). On a graph of vertical distance (y) against horizontal distance (x/m): the object leaves the origin with initial velocity u at angle θ to the horizontal, rises smoothly to a peak at maximum height H (at the midpoint of the horizontal range), then descends symmetrically, landing back on the horizontal (x) axis at a horizontal distance R (the range) from the point of projection. The curve is symmetric about the vertical line through the point of maximum height.

Model answer

For circular orbital motion, v = 2π(R+h)/T. 5.8 = [2×3.14×(6400+h)] / (20.2×3600) 421776 / (2×3.142) = 6400+h 67161.78 = 6400 + h h = 67161.78 − 6400 ≈ 60,761.78 km ≈ 60,762 km above the earth's surface.

Model answer

By Hooke's law, f = ke, so f/e is constant (same spring/cord). e₂ = (f₂/f₁)×e₁ = (8/6)×0.04 = 0.053 m. New length, l₂ = l₀ + e₂ = 5 + 0.053 = 5.053 m.

Model answer

Laser light: consists of a single wavelength/colour (monochromatic); the light waves are in phase with each other (coherent); it is directional, travelling in a narrow beam; it is very intense and has a high concentration of energy. White light: consists of a broad spectrum of wavelengths (polychromatic); the waves are not in phase (incoherent); it is non-directional and scatters in various directions; it is less intense than laser light.

Model answer

(i) Retentivity is the ability of a magnetic material to retain its magnetism even after the external magnetic field that magnetized it has been removed. (ii) Materials with high magnetic susceptibility include: Iron, Nickel, Cobalt, Magnetite and Alnico (any one).

Model answer

(i) A semiconductor is said to be doped when impurities are deliberately added to it. (ii) I. Intrinsic semi-conductor: a pure semiconductor material, i.e. a semiconductor with an equal number of holes and electrons. II. Extrinsic semi-conductor: a semiconductor material that has been doped with impurities, so that the numbers of holes and electrons differ.

Model answer

In forward bias, current, I, rises slowly at first then increases sharply (almost exponentially) once the applied voltage, V, exceeds the diode's threshold/turn-on voltage. In reverse bias, only a very small (near-zero) leakage/reverse-saturation current flows as V is increased in the negative direction, until the breakdown voltage is reached, beyond which current increases sharply again.

Model answer

(i) K.E. = ½mv² → v = √(2×K.E./m) = √(2×1.5×10⁻¹⁹/9.1×10⁻³¹) = √(3/9.1×10¹²) = 5.74×10⁵ m/s. (ii) Using Einstein's photoelectric equation, hf = W₀ + K.E. → W₀ = hf − K.E. = (hc/λ) − K.E. = [(6.6×10⁻³⁴×3×10⁸)/(3.0×10⁻⁷)] − 1.5×10⁻¹⁹ = 6.6×10⁻¹⁹ − 1.5×10⁻¹⁹ = 5.1×10⁻¹⁹ J.

Model answer

Stopping potential is the negative potential of the anode in a photocell which is just sufficient to stop the most energetic photoelectron from reaching it; i.e. the negative voltage applied at the anode (with respect to the cathode) that just prevents the most energetic electron in a photocell from reaching the anode.

Model answer

Deformation is the change in the dimensional shape and size of an object due to an applied external force. Elasticity is the property/ability of a material under an applied force to regain its original shape or size once the force is removed.

Model answer

(i) The plot of F(N) against extension(cm) is a straight line passing through the origin, showing that extension is directly proportional to the applied force (Hooke's law). (ii) Slope, s = ΔF/Δe = (5−1)/(25−5) = 4/20 = 0.2 N/cm. (iii) The slope represents the force constant (spring constant) of the spiral spring.

Model answer

Elastic potential energy of the rubber = Kinetic energy of the stone: ½ke² = ½mv² → v = √(ke²/m) = √(350×(0.04)²/0.04) = √14 = 3.74 ms⁻¹.

Model answer

(i) Bulk modulus is the ratio of the very small increase in pressure applied to a material to the resulting relative decrease in its volume: k = −ΔP/(ΔV/V) (Pa), where ΔP is the change in pressure, ΔV is the change in volume, and V is the original volume. The negative sign shows that an increase in pressure leads to a decrease in volume. (ii) F = mg = 2.5×10⁻³×10 = 0.025 N; e = 0.02 m. Energy gained, W = ½Fe = ½×0.025×0.02 = 2.5×10⁻⁴ J.

Model answer

(i) First overtone is the next frequency of vibration immediately above the fundamental frequency in a system of a vibrating string or air column. (ii) End correction is the difference between the effective length of a vibrating air column in a pipe and the actual physical length of the pipe.

Model answer

For a closed pipe at fundamental frequency: l + e = λ/4 → λ = 4(l+e) = 4(40+3.2) = 4×43.2 = 172.8 cm = 1.728 m. f = v/λ = 330/1.728 ≈ 191 Hz.

Model answer

Number of images, N = 360/β − 1. 8 = 360/β − 1 → 9 = 360/β → β = 40°.

Model answer

Mirage: involves gradual refraction at boundaries as light passes through air layers of varying density; occurs on hot road surfaces or in deserts due to a temperature difference in the air layers; often observed on a hot road or in a desert. Total internal reflection: involves sudden reflection of light at a single boundary; occurs at an angle of incidence greater than the critical angle; observed in fibre optics and prisms.

Model answer

For the final image to form at infinity, the image formed by the objective lens must lie at the focal point of the eyepiece lens. This means the image distance for the objective lens, v₀ = 20 − 6 = 14 cm. Using the lens formula, 1/f₀ = 1/u₀ + 1/v₀ → 1/3 = 1/u₀ + 1/14 → 1/u₀ = 1/3 − 1/14 = 11/42 → u₀ = 42/11 ≈ 3.82 cm.

Model answer

Both processes: are surface phenomena; involve the escape of particles (electrons or molecules) from a material; require energy input for the particles to escape.

Model answer

Photon: an electromagnetic radiation (a quantum of light); carries no electric charge; has no rest mass; its energy depends on its frequency. Photoelectron: a particle of matter (an electron); is negatively charged; has a rest mass (9.1×10⁻³¹ kg); its (kinetic) energy depends on the work function of the metal and the frequency/energy of the incident radiation.

Model answer

(i) Xc = 1/(2πfC) = 1/(2×(1500/π)×π×4×10⁻⁶) = 1/(2×1500×4×10⁻⁶) = 1/0.012 ≈ 83.3 Ω. Impedance, Z = √(R²+Xc²) = √(400²+83.3²) = √166,938.9 ≈ 408.6 Ω. (ii) Current, I = Vrms/Z = 60/408.6 ≈ 0.1469 A. Potential difference across capacitor, Vc = I×Xc = 0.1469×83.3 ≈ 12.24 V.

Model answer

(i) Magnetic lines of force point away from the North pole and towards the South pole outside the magnet; they are closer together where the field is stronger; they never intersect one another; they form closed loops. (ii) A solenoid is a long wire wound in the shape of a helix (spiral), usually around a cylindrical core, used to produce a uniform magnetic field when current flows through it.

Model answer

(i) Similarities: they can cause biological damage to body cells; they can penetrate materials (to varying extents); they are all emitted by radioactive materials. (ii) Natural radioactivity is the spontaneous disintegration/decay of an unstable atomic nucleus, with the emission of alpha, beta and/or gamma radiation (singly or in combination), accompanied by a release of energy. (iii) Alpha decay: ᴬₓX → ⁴₂He + ᴬ⁻⁴ₓ₋₂Y. (iv) Example: Uranium–238 decaying into Thorium–234 (²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He); or Radium–226 decaying into Radon–222.

Model answer

(i) From Einstein's photoelectric equation, hf = W₀ + K.E. → f = (W₀+K.E.)/h = (1.2×10⁻¹⁶+1.8×10⁻¹⁶)/6.6×10⁻³⁴ = 3.0×10⁻¹⁶/6.6×10⁻³⁴ ≈ 4.55×10¹⁷ Hz. (ii) A cathode ray tube: operates in a high-vacuum environment to minimize electron scattering; uses magnetic/electrostatic fields to deflect the electron beam; is coated with phosphorus on the inside of the screen; uses the electron beam to create images on the screen.

Model answer

(i) I. ¹³₇G → ⁴₂He + ⁹₅Y. II. Daughter nuclide has mass number 9 and atomic number 5 (as given in the WAEC marking scheme). (ii) By conservation of mass number: 238 = 206 + 4n → 4n = 32 → n = 8. Therefore, eight (8) alpha particles are emitted in the decay series.

Model answer

Sample results (n = 10 oscillations): M(g): 150, 200, 250, 300, 350; t(s): 6.000, 6.500, 7.200, 7.800, 8.400; T = t/n (s): 0.600, 0.650, 0.720, 0.780, 0.840; log T: −0.222, −0.187, −0.143, −0.108, −0.076; log M: 2.176, 2.301, 2.398, 2.477, 2.544. A graph of log T (y-axis) against log M (x-axis) gives a straight line. Slope, s = ΔlogT/ΔlogM = (0.08−(−0.19))/(2.54−2.30) ≈ 0.27/0.24 ≈ 0.458 ≈ 0.46 (taking the magnitude). For M₀ = 280g, log M₀ = log280 ≈ 2.45. Reading off the graph, log T₀ ≈ −0.12, so T₀ ≈ 0.759 s, giving t₀ = 10×0.759 ≈ 7.59 s. Precautions: avoided draughts throughout the experiment; avoided parallax error when taking readings on the stopwatch and metre rule scale; ensured smooth and regular oscillations of the rule in the vertical plane.

Model answer

(i) The volt is the SI unit of electric potential difference; it is (one volt) defined as the potential difference between two points in an electric circuit when one Joule of work is done to move one coulomb of charge from one point to the other. (ii) I = E/(R+r) = 12/(2+0.5) = 12/2.5 = 4.8 A. Terminal voltage, V = IR = 4.8×2 = 9.6 V.

Model answer

(i) Magnetic lines of force point away from a North pole and towards a South pole; they are closer together where the magnetic field is stronger; they do not intersect one another; they form closed loops. (ii) A solenoid is a long wire wound in the shape of a helix (spiral), usually around a cylindrical core.

Model answer

log RT = 0.17 → RT = 10^0.17 ≈ 1.48 Ω. (xi) Precautions taken: ensured tight connections of the wire, key and other component parts; avoided error due to parallax when taking readings on ammeter and voltmeter respectively.

Model answer

Sample results: I(A): 0.20, 0.30, 0.40, 0.50, 0.60; V(V): 0.13, 0.20, 0.27, 0.34, 0.40; log I: −0.70, −0.52, −0.40, −0.30, −0.22; log V: −0.90, −0.70, −0.60, −0.50, −0.40. Slope, s = ΔlogV/ΔlogI = (−0.28−(−0.70))/(−0.1−(−0.5)) = 0.42/0.4 ≈ 1.04. Intercept on the log I axis ≈ 0.17. Since log RT = c, RT = 10^0.17 ≈ 1.48 Ω. Precautions: ensured tight connections of the wires, key and other component parts of the circuit; avoided errors due to parallax when taking readings on the ammeter and voltmeter.