WAEC Physics 2002 past questions
Ten questions from the 2002 WAEC Physics paper, with answers and explanations. We hold 46 questions from this paper, and the Skolra app has all of them.
WAEC Physics 2002 questions and answers
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1.In a model of the hydrogen atom, the energy levels Wn are given by the formula Wn = -, where n is an integer and R is a constant. Determine the energy released in the transition from n = 3 to n = 2.
- AR/4
- B7R/36
- C5R/36Correct
- DR/9
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Answer: C
Energy at level n: Wn = −R/n²
W3 = −R/9 and W2 = −R/4
Energy released in transition from n = 3 to n = 2:
ΔE = W3 − W2 = −R/9 − (−R/4) = R(1/4 − 1/9) = R(9 − 4)/36 = 5R/36
- A
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2.In a compound microscope, the image formed by the objective lens is at a distance of 3.0cm from the eye lens. Calculate the focal length of the eye lens
- A8.0 cm
- B3.0 cmCorrect
- C5.0 cm
- D2.5 cm
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Answer: B
In a compound microscope at normal adjustment (final image formed at infinity for a relaxed eye), the intermediate image produced by the objective lens must fall exactly at the focal point of the eyepiece. Since this image is 3.0 cm from the eyepiece, the focal length of the eyepiece is fe = 3.0 cm.
- A
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3.The lead-acid accumulator consists of a
- AZinc and copper plates in dilute sulphuric acid
- BSilver and zinc plates in hydrochloric acid
- CLead plates and dilute sulphuric acidCorrect
- DCarbon electrodes in ammonium chloride paste
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Answer: C
The lead-acid accumulator (secondary cell) consists of a lead (Pb) anode, a lead dioxide (PbO₂) cathode, and dilute sulphuric acid (H₂SO₄) as the electrolyte. During discharge, both plates gradually convert to lead sulphate (PbSO₄).
- A
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4.Three capacitors each of capacitance 1.5 are connected in series. The total capacitance in the circuit is
- A3.0 μF
- B1.5 μF
- C0.5 μFCorrect
- D4.5 μF
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Answer: C
For capacitors in series: 1/Ctotal = 1/C₁ + 1/C₂ + 1/C₃
1/Ctotal = 1/1.5 + 1/1.5 + 1/1.5 = 3/1.5 = 2
Ctotal = 0.5 μF
- A
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5.Which of the following is usually used to cause fission in an atomic reactor?
- AAlpha particles
- BSlow (thermal) neutronsCorrect
- CFast neutrons
- DGamma rays
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Answer: B
In a nuclear reactor, slow (thermal) neutrons are used to cause fission of uranium-235. Fast neutrons from fission are slowed down by a moderator (e.g., graphite or heavy water) to thermal speeds so they can be efficiently absorbed by ²³⁵U nuclei and trigger further fission.
- A
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6.Which of the following statements correctly defines quantitatively the work done by a force? the product of the force and the distance moved
- Athe product of the force and the distance moved in the direction of the forceCorrect
- Bthe product of the force and the velocity of the object
- Cthe product of the force and the time taken
- Dthe product of the force and the total distance moved
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Answer: A
Work is defined quantitatively as W = Fd cosθ, where θ is the angle between the force and the displacement. In words, this is: work = force × distance moved in the direction of the force. Only the component of force parallel to the displacement does work; force × time is impulse, and force × velocity is power.
- A
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7.A sheet of paper is placed in the path of the following radiations: I. Alpha particles II. Beta particles III. Gamma rays. Which of the radiations will pass through the sheet of paper?
- AI and II only
- BI, II and III
- CIII only
- DII and III onlyCorrect
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Answer: D
Alpha particles (I) are stopped by a thin sheet of paper due to their large mass and double positive charge — they do not pass through.
Beta particles (II) are fast electrons; a single sheet of paper is insufficient to stop them — they pass through.
Gamma rays (III) are highly penetrating electromagnetic radiation — they also pass through paper.
Therefore only II and III pass through the sheet of paper.
- A
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8.A spiral spring of natural length 30.0cm and force constant of 20Nm-1 is compressed to 20.0cm. Calculate the energy stored in the spring
- A2.00 J
- B0.20 J
- C0.10 JCorrect
- D1.00 J
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Answer: C
Compression x = 30.0 − 20.0 = 10.0 cm = 0.10 m
Energy stored = ½kx² = ½ × 20 × (0.10)² = ½ × 20 × 0.01 = 0.10 J - A
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9.A solid weighs 45 N and 15 N respectively in air and water. Determine the relative density of the solid.
- A1.50Correct
- B2.50
- C3.00
- D2.00
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Answer: A
Upthrust (apparent loss in weight) = weight in air − weight in water = 45 − 15 = 30 N
Relative density = weight in air ÷ upthrust = 45 ÷ 30 = 1.50
This works because the upthrust equals the weight of an equal volume of water. - A
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10.When two notes of nearly equal frequencies are sounded together, a periodic rise and fall in sound intensity can be heard. This phenomenon can be called
- Aresonance
- Binterference
- CbeatsCorrect
- Ddiffraction
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Answer: C
Beats occur when two sound waves of slightly different frequencies f₁ and f₂ are superposed. The resultant amplitude — and hence intensity — varies periodically at the beat frequency = |f₁ − f₂|. The sound alternately reinforces (constructive interference) and cancels (destructive interference), producing the periodic rise and fall in loudness.
- 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 46 questions we hold from the 2002 paper, on a timed CBT screen
Every question has the answer and an explanation, in the Skolra app.