WAEC Physics 1995 past questions

Checked by the Skolra question team · Last updated 15 October 2026

Ten questions from the 1995 WAEC Physics paper, with answers and explanations. We hold 43 questions from this paper, and the Skolra app has all of them.

WAEC Physics 1995 questions and answers

1. 1.A charge of 1.6 x 10⁻¹⁰C is placed in a uniform electric field of intensity 2.0 x 10⁵ NC⁻¹. calculate the magnitude of the electric force exerted on the change
   1. A
      3.2 × 10⁻³ N
   2. B
      1.6 × 10⁻⁵ N
   3. C
      3.2 × 10⁻⁵ N
      Correct
   4. D
      6.4 × 10⁻⁵ N
   Show answerHide answer

   Answer: C

   The electric force on a charge is given by F = qE, where q is the charge and E is the electric field intensity.

   F = 1.6 × 10⁻¹⁰ C × 2.0 × 10⁵ N C⁻¹ = 3.2 × 10⁻⁵ N

2. 2.A wire is stretched between two points, 1m apart. If the speed of the wave generated on plucking the wire is 200ms⁻¹, what is the minimum frequency which will resonate with the wire?
   1. A
      50 Hz
   2. B
      400 Hz
   3. C
      100 Hz
      Correct
   4. D
      200 Hz
   Show answerHide answer

   Answer: C

   The minimum (fundamental) frequency corresponds to the longest standing wave that fits on a wire fixed at both ends: λ = 2L = 2 × 1 = 2 m.

   Therefore f = v/λ = 200 ÷ 2 = 100 Hz.

3. 3.A radioactive nuclide of mass 6.09 has a half-life of 8 days. Calculate the time during which 5.25g of the nuclide would have decayed
   1. A
      24 days
      Correct
   2. B
      16 days
   3. C
      8 days
   4. D
      32 days
   Show answerHide answer

   Answer: A

   The amount remaining after decay: N = 6.0 g − 5.25 g = 0.75 g (taking the intended initial mass as 6.0 g).

   Using N = N₀ × (½)ⁿ:
   0.75 = 6.0 × (½)ⁿ
   (½)ⁿ = 0.75/6.0 = 1/8 = (½)³
   So n = 3 half-lives.

   Time = 3 × 8 = 24 days.

4. 4.Which of the following cannot be explained by the molecular theory of matter?
   1. A
      Diffusion
   2. B
      Brownian motion
   3. C
      Radioactivity
      Correct
   4. D
      Surface tension
   Show answerHide answer

   Answer: C

   Radioactivity originates in the nucleus of atoms and is a nuclear phenomenon — it cannot be explained by the molecular (kinetic) theory of matter, which deals with intermolecular forces and the motion of molecules.

   Diffusion, Brownian motion, and surface tension are all successfully explained using molecular theory.

5. 5.Calculate the electric potential at a distance of 20.0cm from a point charge of 0.015C placed in air of permittivity (Take 1/4πε₀ as 9.0 x 10⁹ Nm²C⁻²)
   1. A
      6.75 × 10⁸ V
      Correct
   2. B
      1.35 × 10⁹ V
   3. C
      4.50 × 10⁸ V
   4. D
      3.375 × 10⁸ V
   Show answerHide answer

   Answer: A

   The electric potential at a distance r from a point charge Q in air is given by: V = Q/(4πε₀r) = kQ/r.

   Substituting: V = (9.0 × 10⁹ × 0.015) / 0.20 = 135 × 10⁶ / 0.20 = 6.75 × 10⁸ V.

6. 6.Which of the following is used for shielding radioactive fall-outs?
   1. A
      Aluminium
   2. B
      Concrete only
   3. C
      Copper
   4. D
      Lead
      Correct
   Show answerHide answer

   Answer: D

   Lead is the most effective material for shielding against radioactive fall-outs (particularly gamma radiation) because of its very high density and high atomic number (Z = 82), which provides a large cross-section for absorbing and attenuating ionizing radiation.

   Thick concrete is also used for structural shielding, but lead is the standard shielding material in portable and compact applications.

7. 7.A balloon containing 546cm³ of air is heated from 0°C to 10°C. If the pressure is kept constant, what will be its volume at 10°C?
   1. A
      556 cm³
   2. B
      526 cm³
   3. C
      576 cm³
   4. D
      566 cm³
      Correct
   Show answerHide answer

   Answer: D

   At constant pressure, Charles's Law applies: V₁/T₁ = V₂/T₂.

   Converting temperatures to Kelvin: T₁ = 0 + 273 = 273 K; T₂ = 10 + 273 = 283 K.

   V₂ = V₁ × T₂/T₁ = 546 × 283/273 = 546 × 1.0366 ≈ 566 cm³.

8. 8.At what angle to the horizontal must the nozzle of a machine gun be kept when firing to obtain a maximum horizontal range for the bullets?
   1. A
      30°
   2. B
      90°
   3. C
      60°
   4. D
      45°
      Correct
   Show answerHide answer

   Answer: D

   The horizontal range of a projectile is given by R = u² sin2θ / g.

   Range is maximum when sin2θ = 1, i.e., 2θ = 90°, so θ = 45° to the horizontal gives the maximum range.

9. 9.What is the escape velocity of a body on the surface of the earth of radius R if the gravitational constant is G and the mass of the earth is M? (Neglect energy losses to the surroundings)
   1. A
      √(GM/2R)
   2. B
      √(GM/R)
   3. C
      2GM/R
   4. D
      √(2GM/R)
      Correct
   Show answerHide answer

   Answer: D

   For a body to escape Earth's gravity, its kinetic energy must equal the gravitational potential energy: ½mv² = GMm/R.

   Solving for v: v_esc = √(2GM/R).

   Here G is the gravitational constant, M is Earth's mass, and R is Earth's radius.

10. 10.The angular speed of an object describing a circle of radius 4m with a linear constant speed of 10 ms⁻¹ is
    1. A
       1.5 rad s⁻¹
    2. B
       3.0 rad s⁻¹
    3. C
       2.0 rad s⁻¹
    4. D
       2.5 rad s⁻¹
       Correct
    Show answerHide answer

    Answer: D

    Angular speed is related to linear speed by ω = v / r.

    ω = 10 m s⁻¹ / 4 m = 2.5 rad s⁻¹

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 43 questions we hold from the 1995 paper, on a timed CBT screen

Every question has the answer and an explanation, in the Skolra app.

Download for WindowsDownload for Mac

About the app · Skolra for Windows · Skolra for Mac