WAEC Physics 2005 past questions
Ten questions from the 2005 WAEC Physics paper, with answers and explanations. We hold 45 questions from this paper, and the Skolra app has all of them.
WAEC Physics 2005 questions and answers
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1.Two capacitors if capacitance 0.4 and 0.5 are connected in parallel and charged to a p.d. of 50V, Determine the total charge acquired
- A22.5 μC
- B45.0 μCCorrect
- C90.0 μC
- D0.45 μC
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Answer: B
For capacitors in parallel, total capacitance is the sum of individual capacitances:
C_total = 0.4 + 0.5 = 0.9 μF
Total charge: Q = C_total × V = 0.9 × 10⁻⁶ × 50 = 45 × 10⁻⁶ C = 45 μC
- A
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2.A ray of light is incident normally on a plan mirror. If the incident ray is kept fixed while the mirror is rotated through an angle of 30o, determine the initial and final angles of reflection respectively
- A0° and 60°
- B30° and 60°
- C0° and 30°Correct
- D30° and 30°
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Answer: C
Initially: The ray is incident normally (along the normal), so the angle of incidence = 0° and the angle of reflection = 0°.
After rotating the mirror 30°: The normal to the mirror rotates by 30°. The fixed incident ray now makes an angle of 30° with the new normal, so the angle of incidence = 30° and the angle of reflection = 30°.
(Note: the reflected ray rotates by 2 × 30° = 60°, but the angle of reflection itself is 30°.)
- A
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3.A wheel and axle of radii 800mm and 200mm respectively is used to raise a body of weight 800N by the application of 250N. Calculate the efficiency of the machine
- A50%
- B60%
- C75%
- D80%Correct
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Answer: D
Velocity Ratio (VR) = Radius of wheel / Radius of axle = 800 / 200 = 4
Mechanical Advantage (MA) = Load / Effort = 800 / 250 = 3.2
Efficiency = (MA / VR) × 100% = (3.2 / 4) × 100% = 80%
- A
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4.An immersion heater rated 400W, 220V is used to heat a liquid of mass 0.5kg. If the temperature of the liquid increases uniformly at the rate of 2.5oC per second, calculate the specific heat capacity of the liquid. [Assume no heat is lost]
- A160 Jkg⁻¹K⁻¹
- B320 Jkg⁻¹K⁻¹Correct
- C640 Jkg⁻¹K⁻¹
- D1600 Jkg⁻¹K⁻¹
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Answer: B
The electrical power equals the rate of heat transferred to the liquid (no heat loss):
P = mc(ΔT/t)
where ΔT/t = 2.5°C s⁻¹ (rate of temperature rise)
400 = 0.5 × c × 2.5
400 = 1.25c
c = 400 / 1.25 = 320 Jkg⁻¹K⁻¹ - A
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5.The friction which exists between two layers of liquid in relative motion is called
- ACohesion
- BViscosityCorrect
- CCapillarity
- DSurface tension
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Answer: B
Viscosity is the internal friction between adjacent layers of a fluid that are in relative motion. It is a measure of a fluid's resistance to flow. Honey has high viscosity; water has low viscosity. Surface tension acts at the liquid surface; capillarity involves rise or fall of liquid in narrow tubes; cohesion is attraction between like molecules.
- A
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6.Which of the following statements is correct?
- ARadio waves are longitudinal waves
- BThe speed of electromagnetic waves in a vacuum is 3 × 10⁸ ms⁻¹Correct
- CSound waves can travel through a vacuum
- DLight waves require a material medium to propagate
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Answer: B
The speed of all electromagnetic waves in a vacuum is a universal constant: c ≈ 3 × 10⁸ ms⁻¹.
The other options are incorrect: sound cannot travel through vacuum (it is a mechanical wave); radio waves are transverse electromagnetic waves, not longitudinal; light propagates through vacuum (e.g., sunlight reaching Earth) so it does not require a material medium.
- A
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7.An electric lamp is rated 240V, 40W. What is the cost of running the lamp for 72 hours if the electricity authority charges N2.50 per kWh?
- AN14.40
- BN72.00
- CN0.72
- DN7.20Correct
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Answer: D
Power = 40 W = 0.04 kW
Energy consumed = Power × Time = 0.04 kW × 72 h = 2.88 kWh
Cost = Energy × Rate = 2.88 × N2.50 = N7.20
- A
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8.Water falls through a height of 50m. Determine the temperature rise at the bottom of the fall. [Neglect energy losses. Specific heat capacity of water = 4200Jkg-1 K-1, g = 10ms2]
- A0.24°C
- B0.10°C
- C0.05°C
- D0.12°CCorrect
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Answer: D
By conservation of energy, the potential energy converts entirely to heat energy:
mgh = mcΔθ
Dividing both sides by m:
gh = cΔθ
Δθ = gh/c = (10 × 50) / 4200 = 500 / 4200 ≈ 0.12°C - A
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9.Which of the following statements is not a reason for using soft iron in making the core of a transformer? It
- Ahas low hysteresis loss
- Bis easily magnetized and demagnetized
- Chas a high retentivityCorrect
- Dhas high magnetic permeability
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Answer: C
Soft iron is used in transformer cores because it has high permeability (easily magnetized), low hysteresis loss (energy lost per cycle is small), and it is easily magnetized and demagnetized with the alternating current.
High retentivity is NOT a desirable property; soft iron has low retentivity, which allows it to reverse magnetization rapidly with the alternating flux. High retentivity would mean the core stays magnetized and resists reversal, causing large energy losses.
- A
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10.In a nuclear reactor, chain reactions result from the release of
- Aelectrons
- Balpha particles
- Cprotons
- DneutronsCorrect
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Answer: D
In a nuclear reactor, a chain reaction is sustained by the release of neutrons during each fission event.
When a fissile nucleus (e.g., U-235) absorbs a neutron and splits, it releases 2–3 fast neutrons. These neutrons can go on to cause further fission reactions, sustaining the chain.
Control rods (e.g., boron or cadmium) absorb excess neutrons to regulate the reaction rate.
- A
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 45 questions we hold from the 2005 paper, on a timed CBT screen
Every question has the answer and an explanation, in the Skolra app.