2025 WAEC Physics Practical (Alternative B) Questions & Answers Now Available

Wednesday, 18th June 2025

Physics 3 (Practical) (Alternative B)
09:30am – 12:15pm (1st Set)

Physics 3 (Practical) (Alternative B)
12:40pm – 03:25pm (2nd Set)

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1(bi)
(i) When a loaded test tube floats vertically in water and if it is depressed slightly and released, the test tube oscillates because the restoring force (buoyant force) acts to return it to the equilibrium position.

(ii) The restoring force in the test tube is the buoyant force, which depends on the weight of the water displaced by the test tube.

(bii)
Potential energy stored in the stretched catapult = (1/2)kx², where k = 160 N/m and x = 0.1 m.
– Potential energy = (1/2) × 160 × (0.1)² = 0.8 J.
– This energy is converted into the kinetic energy of the stone, (1/2)mv², where m = 0.04 kg.
– Equating energies: 0.8 = (1/2) × 0.04 × v².
– Solving for v²: v² = (0.8 × 2) / 0.04 = 40.
– Thus, v = √40 ≈ 6.32 m/s.
The stone is propelled with a speed of approximately 6.32 m/s.

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(3bi)
A volt (V) is the unit of electromotive force (emf) and electric potential difference, defined as the potential difference that allows 1 ampere to dissipate 1 watt of power.

(3bii)
Given E = 12.0 V, r = 0.5 Ω, R = 2 Ω.
R_total = r + R = 2.5 Ω
I = E / R_total = 4.8 A
V = E – Ir = 12.0 V – 2.4 V = 9.6 V
The terminal voltage is 9.6 V.