A ball, P, is released from the roof of a building at the same time as another ball, Q, is thrown vertically upward from the base of the building. Which of the following statements is correct?

C. The acceleration of P is equal to that of Q and both are in the same direction.

A. The acceleration of P is greater than that of Q and both are in the same direction.

B. The acceleration of P is less than that of Q and both are in the same direction.

C. The acceleration of P is equal to that of Q and both are in the same direction.

D. The acceleration of P is equal to that of Q and are in opposite directions.

When a light-weight body and a massive body are acted upon by the same force for the same period of time, the

D. momenta of both bodies remain the same.

A. massive body has a higher momentum.

B. massive body builds up a higher velocity.

C. lighter body has a higher momentum.

D. momenta of both bodies remain the same.

A solenoid is constructed by winding an insulated copper on a long test tube. A constant direct current is passed through the solenoid. Which of the following actions will not increase the strength of the magnetic field of the solenoid? Inserting a

D. Soft iron rod in the test tube

Which of the following equations is not a correct mirror formula? [The symbols have their usual meanings]

A. N = $\frac{360}{\theta}$ - 1

B. $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

The specific latent heat of vapourization of lead is 8.7 × $10^5 Jkg_1$. This means that 8.7 × $10^5$ J of thermal energy is required to change

B. 1 kg of molten lead to vapour at constant temperature and pressure.

A. a given mass of molten lead to vapour at constant temperature and pressure.

B. 1 kg of molten lead to vapour at constant temperature and pressure.

C. 1 kg of solid lead to vapour at constant temperature and pressure.

D. 1 kg of molten lead to vapour for every degree rise in temperature.

Which of the following statements about the properties of the electric field produced by charged parallel plates is not correct?

D. The electric field is uniform everywhere in the space between the parallel plates

A. The electric field in the region outside the parallel plate is zero, except for the edge effects

B. The magnitude of the electric field strength at any point between the plates, except near the edges, depends on the magnitude of the charge on each plate.

C. The magnitude of the electric field strength at any point between the plates is directly proportional to the magnitude of the charge on each plate.

D. The electric field is uniform everywhere in the space between the parallel plates

In a controlled thermal fission reactor, the use of control rods will not affect the

C. speed of the neutrons released on fission.

A. energy generated in the nuclear reactor

B. rate of production of the neutrons.

C. speed of the neutrons released on fission.

D. amount of radiation produced in the nuclear reactor.

The following statements are advantages of a solid dielectric material between the plates of a capacitor except that

B. it increases the electrical conduction of the capacitor.

A. the maximum potential difference the capacitor can withstand is increased.

B. it increases the electrical conduction of the capacitor.

C. the capacitance of the capacitor of given dimensions is much greater than that of the plates in a vacuum.

D. it keeps the plates at a fixed separation.

Which of the following statements about a ferromagnetic material is correct? It

B. attracts non-magnetic materials.

C. has low relative permeability.

D. repels other magnetic materials.

The purpose of commutators in a direct current motor is to

A. reverse the direction of the current after every half rotation

B. increase the number of rotations per second

C. enable a uniform couple to act on the coil.

D. increase the efficiency of the motor.

In the Rutherford scattering experiment, a beam of alpha particles was fired at a thin gold film and few of the particles were deflected considerably. This shows that the nucleus of an atom

A. is positively charged and is concentrated in a tiny volume

B. contains protons and electrons distributed in a tiny volume.

C. is concentrated in a tiny volume and contains alpha particles.

D. is distributed in a tiny volume and emits alpha particles.

A liquid is left in an open dish. After several days, there is less liquid in the dish. Which of the following statements explains this observation? The

A. most energetic molecules leave the surface and escape into the air.

B. least energetic molecules leave the surface and escape into air.

C. most energetic molecules leave the surface and return.

D. least energetic molecules leave the surface and return.

The conclusive test for magnetism in a steel bar suspended horizontally at its midpoint is when the bar

A. always settles in the magnetic N-S direction.

C. always settles in the magnetic E-W direction

An atom in an excited state is one whose

D. electrons have moved to a higher energy level.

A. electrons are in the conduction band.

B. potential energy is minimum.

C. potential energy is maximum.

D. electrons have moved to a higher energy level.

A direct current cannot pass through a parallel plate capacitor because

B. the dielectric between the plates is non-conducting.

A. the plates are non-conducting.

B. the dielectric between the plates is non-conducting.

C. there is a leakage of charge from the plates.

D. the plates have equal but opposite charges.

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WAEC (WASSCE) 2023

2023 WAEC (WASSCE) Physics Past Questions and Answers

The 2023 WAEC (WASSCE) Physics past questions with answers: 49 objective questions from the 2023 paper in the DeeepSolve question bank, each with its answer behind "Show answer". Try each question before you check, then practise the paper as a timed WAEC-style CBT.

Practise this paper as a timed WAEC-style CBT All WAEC (WASSCE) Physics years

Questions

1.
A ball, P, is released from the roof of a building at the same time as another ball, Q, is thrown vertically upward from the base of the building. Which of the following statements is correct?
1. A.The acceleration of P is greater than that of Q and both are in the same direction.
2. B.The acceleration of P is less than that of Q and both are in the same direction.
3. C.The acceleration of P is equal to that of Q and both are in the same direction.
4. D.The acceleration of P is equal to that of Q and are in opposite directions.
Show answer
Answer: C. The acceleration of P is equal to that of Q and both are in the same direction.
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2.
When a light-weight body and a massive body are acted upon by the same force for the same period of time, the
1. A.massive body has a higher momentum.
2. B.massive body builds up a higher velocity.
3. C.lighter body has a higher momentum.
4. D.momenta of both bodies remain the same.
Show answer
Answer: D. momenta of both bodies remain the same.
AI explanation
Analyze the question: The same force acts on two bodies of different masses for the same time. We are asked to compare momentum (and velocity).
Concept: Newton's second law in terms of momentum says force equals the rate of change of momentum:
F = Δp / Δt, so Impulse = F × t = Δp
Solution:
Both bodies feel the same force F for the same time t.
Impulse on each body = F × t, which is the same for both.
Both bodies start from rest, so the change in momentum equals the final momentum.
Therefore p₁ = p₂ = Ft. Their momenta are equal.
Checking the options:
A. Wrong. The massive body does not gain more momentum.
B. Wrong. Since p = mv, v = Ft/m, so the massive body gets the lower velocity.
C. Wrong. The lighter body has higher velocity, but not higher momentum.
D. Correct. The momenta gained are equal, since impulse is equal.
Tip: When you see "same force, same time," think impulse = Ft = change in momentum. Momentum gained is equal, while velocity gained depends on mass (v = Ft/m). Lighter bodies speed up more, but momentum stays equal.
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3.
A lead bullet of mass 0.05 kg is fired with a speed of 200 m/s into a lead block of mass 0.95 kg at rest. Given that the lead block moves after the impact, determine its kinetic energy.
1. A.150J
2. B.200J
3. C.50J
4. D.100J
Show answer
Answer: C. 50J
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4.
A 100 kg device is pulled up a plane inclined at 30° to the horizontal with a force of 1000 N. If the coefficient of friction between the device and the surface is 0.25, determine the total force opposing the motion. [ g = 10 $ms^{-2}$ ]
1. A.283.5N
2. B.975.0N
3. C.216.5N
4. D.716.5N
Show answer
Answer: D. 716.5N
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5.
A solenoid is constructed by winding an insulated copper on a long test tube. A constant direct current is passed through the solenoid. Which of the following actions will not increase the strength of the magnetic field of the solenoid? Inserting a
1. A.steel rod in the test tube
2. B.Nickel rod in the test tube
3. C.Brass rod in the test tube
4. D.Soft iron rod in the test tube
Show answer
Answer: C. Brass rod in the test tube
AI explanation
Analyze the question: We want the material that will not strengthen the magnetic field of a current-carrying solenoid when inserted into it.
Concept: The magnetic field inside a solenoid is B = μnI, where μ = μ₀μᵣ.
n = turns per unit length
I = current
μᵣ = relative permeability of the core material
A core with a high relative permeability (μᵣ ≫ 1) greatly increases B. These are ferromagnetic materials: iron, steel, nickel and cobalt. The domains in these materials line up with the solenoid's field and add to it.
A non-magnetic material such as brass (an alloy of copper and zinc) has μᵣ ≈ 1. It behaves almost like air, so it adds essentially nothing to the field.
Checking each option:
A. Steel rod: ferromagnetic, so the field increases (steel keeps its magnetism well).
B. Nickel rod: ferromagnetic, so the field increases.
C. Brass rod: non-magnetic, μᵣ ≈ 1, so there is no significant increase.
D. Soft iron rod: very high permeability, so it gives the biggest increase. This is why it is used in electromagnets.
Answer: The brass rod will not increase the strength of the field, so the answer is C.
Exam tip: Remember the ferromagnetic materials as "Iron, Cobalt, Nickel, Steel". Brass, copper, aluminium, wood and plastic are not good core materials. When a question asks for what will not increase the field, look for the non-magnetic material.
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6.
A cell of emf 2 V has an internal resistance of 0.5 Ω. A current of 0.4 A passes through an external resistor, P. Determine the resistance of P.
1. A.4.5Ω
2. B.5Ω
3. C.1Ω
4. D.4Ω
Show answer
Answer: A. 4.5Ω
AI explanation
Analyze the question: We have a cell with emf E = 2 V and internal resistance r = 0.5 Ω. The current is I = 0.4 A through external resistor P. We need R.
Concept: For a cell in a circuit,
E = I(R + r)
so R = E/I − r
Solution:
Total resistance of the circuit = E/I = 2 / 0.4 = 5 Ω
This total is R + r, so R = 5 − 0.5 = 4.5 Ω
Check: I = 2 / (4.5 + 0.5) = 2/5 = 0.4 A ✔
Answer: The resistance of P is 4.5 Ω, which is option A.
Common mistake: Option B (5 Ω) is the trap. It is the total resistance, and students forget to subtract the internal resistance.
Tip: Remember "E = V + Ir", or "emf = I × (total resistance)". Always ask yourself: is the resistance I found the total or just the external part? If the question asks for the external resistor, subtract r.
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7.
The instrument used to measure relative humidity is a
1. A.Hygrometer
2. B.Barometer
3. C.Hydrometer
4. D.Manometer
Show answer
Answer: A. Hygrometer
AI explanation
Step 1: Analyze the question The question asks which instrument measures relative humidity, which is the amount of water vapour in the air compared with the maximum amount the air can hold at that temperature (usually given as a percentage).
Step 2: Explain the concept Each instrument in the options measures something different:
Hygrometer: measures humidity (moisture in the air). Examples are the hair hygrometer and the wet-and-dry bulb hygrometer.
Barometer: measures atmospheric pressure.
Hydrometer: measures the density or relative density of liquids, such as acid in a car battery or milk.
Manometer: measures the pressure of gases or liquids, usually in a U-tube.
Step 3: Solution Relative humidity is a measure of moisture in the air, so the instrument must be a hygrometer. The prefix "hygro-" comes from the Greek word for moisture or wetness.
In the wet-and-dry bulb hygrometer, two thermometers are used. The wet bulb is cooled by evaporation, so the difference between the two readings is used with a table to find the relative humidity.
Step 4: Answer The correct option is A. Hygrometer.
Step 5: Exam tip Use the prefixes to remember the instruments:
Hygro = moisture, so hygrometer measures humidity
Baro = pressure (weight), so barometer measures atmospheric pressure
Hydro = liquid, so hydrometer measures density of liquids
Mano = gas or liquid pressure, so manometer measures pressure
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8.
Which of the following equations is not a correct mirror formula? [The symbols have their usual meanings]
1. A.N = $\frac{360}{\theta}$ - 1
2. B. $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
3. C.m = $\frac{v}{u}$
4. D.f = $\frac{1}{D}$
Show answer
Answer: D. f = $\frac{1}{D}$
AI explanation
Analyzing the question: We need to find the equation that is NOT a correct formula connected with mirrors. Let's check each one.
Checking each option:
A. N = 360/θ − 1: This is the formula for the number of images formed by two plane mirrors inclined at an angle θ. It is correct.
B. 1/f = 1/u + 1/v: This is the standard mirror formula (real-is-positive convention). It is correct.
C. m = v/u: Magnification is the ratio of image distance to object distance (the sign depends on convention, with m = -v/u in the Cartesian convention). In WASSCE, m = v/u is accepted as the magnification formula for mirrors, so it is treated as correct.
D. f = 1/D: This looks like the power of a lens, P = 1/f, where the power D is in dioptres. As written, f = 1/D is actually mathematically equivalent to D = 1/f. So it is not strictly wrong in form, but it is a lens-power relation, not a mirror formula.
Note on option D: Strictly, f = 1/D is algebraically valid when D is the power. The question asks for the equation that is "not a correct mirror formula," and D is the only one that doesn't describe mirror behavior (it is the power of a lens). A, B, and C all relate to mirrors (plane mirror images, the mirror equation, and magnification).
Answer: D
Tip: In "which is NOT correct" questions, test each option against the topic. Here, the lens power formula (P = 1/f) is the odd one out among mirror formulas. Also remember: number of images from inclined plane mirrors is N = (360/θ) − 1.
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9.
Which of the following devices is associated with alternating currents only?
1. A.Step-up transformer
2. B.Induction coil
3. C.Moving coil galvanometer
4. D.Lead-acid accumulator
Show answer
Answer: A. Step-up transformer
AI explanation
Analyze the question: We need the device that works with alternating current (AC) only, meaning it cannot work with steady direct current (DC).
Concept:
A transformer works by mutual induction. The changing current in the primary coil produces a changing magnetic flux, which induces an e.m.f. in the secondary coil. Steady DC gives a constant flux, so no e.m.f. is induced. A transformer therefore works only with AC.
An induction coil changes low DC voltage into a very high voltage. It uses a make-and-break interrupter to keep interrupting the DC supply, so it is powered by DC, not AC only.
A moving coil galvanometer detects small DC currents. The deflection depends on the direction of the current, so it is for DC.
A lead-acid accumulator is a secondary cell. It supplies and is charged with DC.
Step by step:
Lead-acid accumulator: DC only, so eliminate D.
Moving coil galvanometer: DC, so eliminate C.
Induction coil: runs on a DC battery with an interrupter, so eliminate B.
Step-up transformer: needs a changing current and so works only on AC. This matches the question.
Answer: A. Step-up transformer
Tip: Remember "transformer = AC only." If a device needs a changing flux without any mechanical interrupter, it needs AC. The induction coil is a common trap, because it gives high voltage like a transformer but is fed by DC through its interrupter.
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10.
A planet has mass, M, and radius R, If the universal gravitational constant is G, what is the expression for the escape velocity of an object on the planet?
1. A. $\frac{\sqrt{GR}}{2}$
2. B. $\frac{\sqrt{2GM}}{R}$
3. C. $\frac{\sqrt{2G}}{R}$
4. D. $\frac{\sqrt{GM}}{R}$
Show answer
Answer: B. $\frac{\sqrt{2GM}}{R}$
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11.
In a mass spectrometer, an ion of charge, q, and mass, m, moving in a path of radius, r, in a field of flux density, B, has a speed of
1. A. $\frac{qBr}{m}$
2. B. $\frac{mr}{qB}$
3. C. $\frac{qm}{Br}$
4. D. $\frac{mq}{r}$
Show answer
Answer: A. $\frac{qBr}{m}$
AI explanation
Analyze the question: An ion moves in a circle inside a magnetic field. We need its speed v in terms of q, m, B and r.
Concept: A charged particle moving perpendicular to a magnetic field feels a magnetic force that points toward the centre of the circle. This force supplies the centripetal force.
Magnetic force: F = qvB
Centripetal force: F = mv²/r
Solution:
Equate the two forces:
qvB = mv²/r
Divide both sides by v:
qB = mv/r
Make v the subject:
v = qBr/m
Check the options:
A. qBr/m, which matches our result.
B. mr/(qB) has the wrong arrangement (it is not dimensionally a speed).
C and D do not contain the right combination of terms.
Unit check: q·B·r/m = C · T · m / kg. Since T = kg/(C·s), this gives (C · kg/(C·s) · m)/kg = m/s ✓.
Tip: For any "charge in a magnetic field" circular-motion question, write qvB = mv²/r immediately. From it you get v = qBr/m and r = mv/(qB). Remember that r is directly proportional to the speed and mass, and inversely proportional to the charge and field strength.
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12.
A body starts with an initial velocity of 20 m/s and moves with a uniform acceleration, a, for 15 seconds. Which of the following equations represents the distance, S, traveled?
1. A.300 + 56 $\frac{a}{2}$
2. B.300 + $\frac{225a}{2}$
3. C.300 + $\frac{225}{2a}$
4. D.300a + $\frac{225}{2}$
Show answer
Answer: B. 300 + $\frac{225a}{2}$
AI explanation
Analyze the question: We have an initial velocity u = 20 m/s, uniform acceleration a, and time t = 15 s. We need the distance S.
Concept: For uniform acceleration, the second equation of motion gives the distance:
S = ut + ½at²
Solution:
Substitute u = 20 and t = 15: S = (20)(15) + ½ · a · (15)²
Calculate each term: • ut = 20 × 15 = 300 • ½at² = ½ · a · 225 = 225a/2
Combine: S = 300 + 225a/2
Check the options:
A: 56(a/2) is wrong because 56 is not 15² (225).
B: 300 + 225a/2 matches our result.
C: 225/(2a) puts a in the denominator, which is wrong.
D: 300a + 225/2 puts a on the wrong term.
Tip: When you see "initial velocity, acceleration, time, find distance," go straight to S = ut + ½at². Substitute carefully, and remember to square the time (15² = 225) before multiplying by ½ and a.
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13.
The specific latent heat of vapourization of lead is 8.7 × $10^5 Jkg_1$ . This means that 8.7 × $10^5$ J of thermal energy is required to change
1. A.a given mass of molten lead to vapour at constant temperature and pressure.
2. B.1 kg of molten lead to vapour at constant temperature and pressure.
3. C.1 kg of solid lead to vapour at constant temperature and pressure.
4. D.1 kg of molten lead to vapour for every degree rise in temperature.
Show answer
Answer: B. 1 kg of molten lead to vapour at constant temperature and pressure.
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14.
In an RLC series circuit at resonance, the voltages across the resistor and the inductor are 30 V and 40 V respectively. Determine the voltage across the capacitor.
1. A.50V
2. B.30V
3. C.70V
4. D.40V
Show answer
Answer: D. 40V
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15.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.
16.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.
17.
When an electron jumps directly from an orbit of n = 4 to n = 2,
1. A.a photon is absorbed
2. B.a photon is emitted
3. C.two photons are emitted.
4. D.two photons are absorbed
Show answer
Answer: B. a photon is emitted
AI explanation
Analyze the question: An electron moves from a higher energy level (n = 4) to a lower one (n = 2) in a single, direct jump. We need to say what happens to the energy.
Concept: In the Bohr model, the energy of an electron in the nth orbit is
Eₙ = −13.6 / n² eV (for hydrogen)
Moving to a higher level (e.g., n = 2 → n = 4) requires energy, so a photon is absorbed.
Moving to a lower level (e.g., n = 4 → n = 2) releases energy, so a photon is emitted.
Solution:
E₄ = −13.6 / 16 = −0.85 eV
E₂ = −13.6 / 4 = −3.40 eV
Energy change: ΔE = E₂ − E₄ = −3.40 − (−0.85) = −2.55 eV
The negative sign shows the electron loses 2.55 eV. This energy leaves as one photon, with energy hf = 2.55 eV. This is a line in the Balmer series (visible light, blue-green, about 486 nm).
Because the jump is direct (4 → 2 in one step), only one photon is emitted. Two photons would be emitted only if the electron went in two steps (4 → 3, then 3 → 2).
Answer: Option B, a photon is emitted.
Tip: Remember "Down = Disappears energy (emit), Up = Used energy (absorb)". Also, count the number of jumps: one direct jump means one photon.
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18.
A sound of frequency 512 Hz reaches a listener 100 m away. If the sound travels in air at s.t.p., what is its wavelength? [speed of sound in air = 331 m/s]
1. A.1.550m
2. B.0,646m
3. C.5.120m
4. D.0.195m
Show answer
Answer: B. 0,646m
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19.
A ball of mass 0.8 kg is dropped from a height, H. Just before hitting the ground, its velocity is 5.0 m/s. Determine H. [g = 10 $ms^{-2}$ ]
1. A.1.25m
2. B.4.00m
3. C.6.25m
4. D.1.60m
Show answer
Answer: A. 1.25m
AI explanation
Analyze the question: A ball is dropped (so it starts from rest, u = 0). It hits the ground at v = 5.0 m/s. We need the height H it fell through. Concept: For free fall, we can use either the equation of motion or conservation of energy. The mass (0.8 kg) is extra information and does not affect the answer, because it cancels out. Solution: Using the equation of motion: v² = u² + 2gH 5² = 0 + 2(10)H 25 = 20H H = 25 / 20 = 1.25 m Check using energy conservation: Potential energy lost = Kinetic energy gained mgH = ½mv² gH = ½v² 10H = ½(25) = 12.5 H = 1.25 m ✔ Both methods give H = 1. Answer: Option A, 1.25 m Tip: In free-fall questions, the mass cancels out, so ignore it when it is given as a distractor. Remember the shortcut H = v²/(2g). Here that is 25/20 = 1.25 m.
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20.
The type of mirror used by dentists to inspect the teeth of patients is a
1. A.Plane mirror
2. B.cylindrical mirror
3. C.concave mirror
4. D.convex mirror
Show answer
Answer: C. concave mirror
AI explanation
Step 1: Analyze the question We need to identify which mirror a dentist uses to look at a patient's teeth.
Step 2: Explain the concept
Plane mirror: gives an image the same size as the object, so there is no magnification.
Convex mirror: always gives a diminished, virtual image. It is used as a driving mirror and for security, because it widens the field of view.
Concave mirror: when an object is placed inside the focal length (closer than F), it gives an erect, virtual, magnified image. It is used as a shaving mirror, a makeup mirror, in torches and headlamps as a reflector, and by dentists.
Cylindrical mirror: not a standard mirror for this purpose.
Step 3: Apply to the dentist's case A dentist holds the small mirror close to the tooth, so the tooth is within the focal length of the mirror. A concave mirror then produces an enlarged, upright, virtual image. This lets the dentist see small details such as cavities and cracks clearly.
Step 4: Answer The mirror used is a concave mirror, which is option C.
Note on a common confusion: Many dental mirrors are sold as flat mirrors, which is why option A can tempt you. But for WASSCE, the expected answer is the concave mirror, because of its magnifying property.
Exam tip: Remember the uses this way:
Concave = Close-up, magnifying (dentist, shaving, makeup)
Convex = wide view (driving mirror, shop security)
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21.
Which of the following statements about the properties of the electric field produced by charged parallel plates is not correct?
1. A.The electric field in the region outside the parallel plate is zero, except for the edge effects
2. B.The magnitude of the electric field strength at any point between the plates, except near the edges, depends on the magnitude of the charge on each plate.
3. C.The magnitude of the electric field strength at any point between the plates is directly proportional to the magnitude of the charge on each plate.
4. D.The electric field is uniform everywhere in the space between the parallel plates
Show answer
Answer: D. The electric field is uniform everywhere in the space between the parallel plates
AI explanation
Analyze the question: We need the statement that is NOT correct about the field between charged parallel plates.
Concept: For two large parallel plates carrying equal and opposite charges:
Field between the plates: E = σ/ε₀ = Q/(ε₀A), which is uniform in the central region.
Field outside the plates: approximately zero, because the fields of the two plates cancel.
Near the edges, the field lines bulge outwards (fringing or edge effect), so the field is not uniform there.
Checking each option:
A. Correct. Outside the plates the fields cancel, so E ≈ 0 apart from edge effects.
B. Correct. Since E = Q/(ε₀A), the field depends on the charge on each plate (away from the edges).
C. Correct. E = Q/(ε₀A), so E ∝ Q when the plate area is fixed.
D. Not correct. It says the field is uniform everywhere between the plates. Near the edges the field lines curve outwards (fringing), so the field is not uniform there. Options A and B both carry the "except near the edges" qualification, but D does not, which makes D false as written.
Answer: D
Exam tip: In "which is NOT correct" questions, look for absolute words like "everywhere," "always," or "never." The examiner often hides the false statement behind one of them. For parallel plates, remember that the field is uniform only in the central region, not at the edges.
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22.
In a controlled thermal fission reactor, the use of control rods will not affect the
1. A.energy generated in the nuclear reactor
2. B.rate of production of the neutrons.
3. C.speed of the neutrons released on fission.
4. D.amount of radiation produced in the nuclear reactor.
Show answer
Answer: C. speed of the neutrons released on fission.
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23.
The following statements are advantages of a solid dielectric material between the plates of a capacitor except that
1. A.the maximum potential difference the capacitor can withstand is increased.
2. B.it increases the electrical conduction of the capacitor.
3. C.the capacitance of the capacitor of given dimensions is much greater than that of the plates in a vacuum.
4. D.it keeps the plates at a fixed separation.
Show answer
Answer: B. it increases the electrical conduction of the capacitor.
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24.
The ratio of maximum range to maximum height of a projectile is 4. Determine the angle of projection.
1. A. $30^0$
2. B. $60^0$
3. C. $45^0$
4. D. $0^0$
Show answer
Answer: C. $45^0$
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25.
The process of producing electric current by a changing magnetic field is referred to as
1. A.magnet induction
2. B.electromagnetic induction
3. C.flux linkage
4. D.electrostatic induction.
Show answer
Answer: B. electromagnetic induction
AI explanation
Step 1: Analyze the question The question asks for the name of the process in which a changing magnetic field produces an electric current.
Step 2: Explain the concept When the magnetic flux through a conductor or coil changes, an e.m.f. is induced in it. If the circuit is closed, a current flows. This is called electromagnetic induction, discovered by Michael Faraday.
Faraday's law: the induced e.m.f. is proportional to the rate of change of magnetic flux linkage: E = −N(ΔΦ/Δt)
Lenz's law: the induced current flows in a direction that opposes the change producing it (this is the meaning of the negative sign).
This is the principle behind generators, transformers and induction coils.
Step 3: Check each option
A. Magnet induction: not a standard physics term.
B. Electromagnetic induction: correct. A changing magnetic field produces an e.m.f. and current.
C. Flux linkage: this is a quantity (N × Φ), not a process.
D. Electrostatic induction: this is the redistribution of charge on a conductor by a nearby charged body. It involves electric charges, not a changing magnetic field.
Step 4: Answer The correct option is B. electromagnetic induction.
Exam tip: Match the keywords. "Changing magnetic field → current" means electromagnetic induction. "Charged body → charge separation" means electrostatic induction. Also remember that flux linkage is something you calculate, not a process.
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26.
Which of the following statements about a ferromagnetic material is correct? It
1. A.has a high susceptibility.
2. B.attracts non-magnetic materials.
3. C.has low relative permeability.
4. D.repels other magnetic materials.
Show answer
Answer: A. has a high susceptibility.
AI explanation
Step 1: Analyze the question We need to pick the true statement about ferromagnetic materials such as iron, cobalt and nickel.
Step 2: The concept Ferromagnetic materials are strongly attracted by magnets. Inside them, small regions called domains line up easily with an applied magnetic field. This gives them these properties:
High magnetic susceptibility (χm): the material becomes strongly magnetised in an applied field. Susceptibility measures how easily this happens.
High relative permeability (μr): μr = 1 + χm, so a large χm means a large μr, often in the hundreds or thousands.
They are attracted by magnets. They do not repel.
Step 3: Check each option
A. Has a high susceptibility. Correct. Ferromagnets are very easily magnetised, so χm is large and positive.
B. Attracts non-magnetic materials. Wrong. A non-magnetic material, such as wood or plastic, is not attracted by a ferromagnet.
C. Has low relative permeability. Wrong. Ferromagnets have high relative permeability. This is why iron is used as a core in electromagnets and transformers.
D. Repels other magnetic materials. Wrong. Ferromagnets are attracted to magnetic materials. Repulsion happens only between like poles of two magnets, and it is not a general property of ferromagnetic materials. (Diamagnetic materials are the ones weakly repelled.)
Step 4: Answer The correct option is A: a ferromagnetic material has a high susceptibility.
Exam tip: Remember "Ferro = Fantastically magnetic." Ferromagnets have high χm, high μr, and strong attraction. Diamagnetic materials have small negative χm, and paramagnetic materials have small positive χm.
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27.
What is the name given to the force of attraction between the nucleons in the nucleus of an atom?
1. A.Gravitational force.
2. B.Electrostatic force.
3. C.Magnetic force.
4. D.Nuclear force.
Show answer
Answer: D. Nuclear force.
AI explanation
1. Analyze the question The question asks for the name of the force that holds nucleons (protons and neutrons) together inside the nucleus.
2. Explain the concept The nucleus contains protons, which are positively charged, packed into a tiny space. Since like charges repel, the protons should push each other apart through the electrostatic (Coulomb) force. Yet the nucleus stays together, so there must be a much stronger attractive force acting between nucleons.
This force is called the nuclear force, also known as the strong nuclear force. Its features are:
It is very strong, much stronger than the electrostatic repulsion between protons at nuclear distances.
It has a very short range, about 10⁻¹⁵ m (1 femtometre).
It acts between all nucleons, whether proton-proton, neutron-neutron, or proton-neutron.
It does not depend on electric charge.
3. Work through the options
A. Gravitational force: This acts between masses but is extremely weak at the nuclear scale, so it cannot hold the nucleus together.
B. Electrostatic force: This is repulsive between protons, so it tends to break the nucleus apart rather than bind it.
C. Magnetic force: This acts on moving charges and is not the binding force in the nucleus.
D. Nuclear force: This is the strong, short-range attractive force between nucleons, so it is the correct answer.
4. Answer The correct option is **D. Nuclear force.
5. Exam tip Remember this phrase: "Nucleons are bound by the nuclear force." The name of the force comes from where it acts. If a question mentions a force inside the nucleus that overcomes proton-proton repulsion, the answer is the strong nuclear force.
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28.
The purpose of commutators in a direct current motor is to
1. A.reverse the direction of the current after every half rotation
2. B.increase the number of rotations per second
3. C.enable a uniform couple to act on the coil.
4. D.increase the efficiency of the motor.
Show answer
Answer: A. reverse the direction of the current after every half rotation
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29.
In the Rutherford scattering experiment, a beam of alpha particles was fired at a thin gold film and few of the particles were deflected considerably. This shows that the nucleus of an atom
1. A.is positively charged and is concentrated in a tiny volume
2. B.contains protons and electrons distributed in a tiny volume.
3. C.is concentrated in a tiny volume and contains alpha particles.
4. D.is distributed in a tiny volume and emits alpha particles.
Show answer
Answer: A. is positively charged and is concentrated in a tiny volume
AI explanation
Step 1: Analyze the question Alpha particles were fired at a very thin gold foil. Most passed straight through, but a few were deflected through large angles (some even bounced back). We need to find what this tells us about the nucleus.
Step 2: The concept Rutherford's observations and what each one means:
Most particles passed straight through. Most of the atom is empty space.
A few were deflected considerably. Alpha particles are positively charged, so only a strong positive charge can repel them strongly. The charge must be concentrated in a very small region. If the positive charge were spread out through the atom, as in Thomson's "plum pudding" model, no large deflections would occur.
Very few bounced straight back. The nucleus is extremely small and dense (a tiny volume holding most of the atom's mass).
Step 3: Evaluate the options
A. Positively charged and concentrated in a tiny volume. Correct. Large deflections come from repulsion between the positive alpha particles and a small, dense, positive nucleus.
B. Contains protons and electrons. Wrong. The experiment says nothing about electrons being in the nucleus, and they are not.
C. Contains alpha particles. Wrong. The nucleus contains protons and neutrons. The alpha particles came from the source.
D. Distributed in a tiny volume and emits alpha particles. Wrong. "Distributed" contradicts the idea of concentration, and the gold nucleus did not emit the alpha particles.
Step 4: Answer The correct option is A.
Tip: Remember the Rutherford summary as "Most through, few deflected, very few back". This gives "empty space, small positive nucleus, dense nucleus". Whenever you see "deflected considerably," think "positive and concentrated."
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30.
An electron moves from an energy level of -8.6 eV to another level of -3.5 eV. Determine the frequency of the absorbed radiation. [1 eV = 1.6 × $10^{19}$ J, h = 6.6 × $10^{34}$ Js]
1. A. $2.93 \times 10^{15}$ Hz
2. B. $3.59 \times 10^{16}$ Hz
3. C. $5.39 \times 10^{15}$ Hz
4. D. $1.24 \times 10^{15}$ Hz
Show answer
Answer: D. $1.24 \times 10^{15}$ Hz
AI explanation
Analyze the question: An electron absorbs radiation and moves up from -8.6 eV to -3.5 eV. We need the frequency of that radiation.
Concept: The energy absorbed equals the difference between the two levels, and E = hf, so f = ΔE / h.
(The question prints the constants as 1.6 × 10^19 and 6.6 × 10^34. The signs are missing. They should be 1.6 × 10⁻¹⁹ J and 6.6 × 10⁻³⁴ Js.)
Solution:
Energy difference: ΔE = -3.5 - (-8.6) = 5.1 eV
Convert to joules: ΔE = 5.1 × 1.6 × 10⁻¹⁹ = 8.16 × 10⁻¹⁹ J
Find the frequency: f = ΔE / h = (8.16 × 10⁻¹⁹) / (6.6 × 10⁻³⁴) f ≈ 1.236 × 10¹⁵ Hz ≈ 1.24 × 10¹⁵ Hz
Tip: In energy-level questions, always subtract the lower level from the higher one (final − initial for absorption) to get a positive energy. Also convert eV to joules before dividing by h. A quick check is that 5.1 eV is roughly 1.2 × 10¹⁵ Hz, since 1 eV corresponds to about 2.4 × 10¹⁴ Hz.
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31.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.
32.
The most appropriate value of fuse for a kettle rated 220 V, 1.06 kW is
1. A.3A
2. B.6A
3. C.5A
4. D.4A
Show answer
Answer: C. 5A
AI explanation
Analyze the question: We need a fuse rating for a kettle rated 220 V and 1.06 kW. The fuse must be slightly higher than the normal operating current, so the kettle works normally but the fuse blows if the current becomes too large.
Concept: Power P = IV, so the working current is I = P / V.
Solution:
Convert the power to watts: 1.06 kW = 1060 W
Calculate the current: I = 1060 / 220 ≈ 4.82 A
Choose a fuse rating just above 4.82 A.
Checking the options:
A (3 A) is below the working current, so it would blow immediately.
D (4 A) is also below 4.82 A, so it would blow.
C (5 A) is just above 4.82 A, so it is suitable.
B (6 A) is safe, but it is not the closest rating above the working current, so it is less appropriate.
The most appropriate fuse is therefore 5 A, which is option C.
Exam tip: For fuse questions, always calculate I = P/V first, then pick the smallest fuse rating that is greater than that current. Never choose a fuse below the working current, and avoid ones far above it, because they give poor protection.
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33.
A car, P, with a velocity of 30 m/s is traveling in the same direction as a car, Q, with a velocity of 20 m/s. The velocity of P relative to Q is
1. A. $50ms^1$
2. B. $45ms^1$
3. C. $20ms^1$
4. D. $10ms^1$
Show answer
Answer: D. $10ms^1$
AI explanation
Analyze the question Two cars move in the same direction. We want the velocity of P as seen by someone in car Q (relative velocity).
Concept The velocity of A relative to B is:
v(A relative to B) = v_A − v_B
Both velocities are measured in the same direction. If the cars move the same way, you subtract. If they move in opposite directions, you add the magnitudes.
Solution
v_P = 30 m/s
v_Q = 20 m/s
v(P relative to Q) = v_P − v_Q = 30 − 20 = 10 m/s
This means that from Q's point of view, P pulls ahead at 10 m/s.
Checking the options
A (50 m/s) is what you get by adding, which is only correct if the cars moved in opposite directions.
B (45 m/s) and C (20 m/s) do not come from any valid operation on these numbers.
Answer: D, 10 m/s
Exam tip: For "relative to" questions, first check the directions. Same direction means subtract. Opposite directions means add. Remember "A relative to B = A minus B."
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34.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.
35.
The sagging of overhead electrical wires is caused by
1. A.radiation of heat.
2. B.rough handling.
3. C.gravitational attraction
4. D.Linear expansion
Show answer
Answer: D. Linear expansion
AI explanation
Analyze the question: Overhead electrical wires (like the ones on NEPA/PHCN poles) are often seen hanging loosely between poles. We need to identify the physical cause of this sagging.
Explain the concept: Most solids, including metals such as copper and aluminium, undergo linear expansion when heated. The increase in length is given by:
ΔL = L₀ α ΔT
where
L₀ = original length
α = coefficient of linear expansion
ΔT = change in temperature
Step-by-step reasoning:
During the day, especially in hot weather, the sun heats the wires. Current flowing through them also produces heat.
The wires get hotter, so their length increases (linear expansion).
The wires are fixed between two poles, so the distance between the supports does not change. The extra length has nowhere to go but to hang down, which produces the sag.
This is why engineers deliberately leave wires slightly slack when installing them. It allows room for expansion in hot weather and prevents the wires from contracting and snapping in cold weather (harmattan nights).
Checking the other options:
A. Radiation of heat: Radiation is just a way heat is transferred. It is not what makes the wire sag.
B. Rough handling: This is not a scientific cause, and the sag is a general, expected effect.
C. Gravitational attraction: Gravity does pull the wire downward, and a wire has weight. However, the question asks what causes the sagging as a thermal effect. Gravity acts all the time, even on a tight wire. The change in length due to temperature is what allows the sag to develop or increase. The standard textbook answer for this question is linear expansion.
D. Linear expansion: Correct, as explained above.
Answer: D. Linear expansion
Tip: When you see "overhead wires," "railway tracks," "bridges," or "gaps in concrete roads" in a question, think thermal expansion. Engineers leave gaps or slack in all of these for the same reason.
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36.
Which of the following devices is used to compare the relative magnitudes of charges on two given bodies?
1. A.Ebonite rod
2. B.Proof Plane
3. C.Gold Leaf Electroscope
4. D.Electrophorus
Show answer
Answer: C. Gold Leaf Electroscope
AI explanation
Step 1: Analyze the question We need the device that compares the relative magnitudes (sizes) of charges on two bodies. It is a detector and comparer of charge, not a device for producing or transferring charge.
Step 2: Review each option
A. Ebonite rod: This is a material that gains negative charge when rubbed with fur. It is used to produce charge, not to compare it.
B. Proof plane: This is a small metal disc on an insulating handle. It transfers a sample of charge from a body to an electroscope. It only carries charge and does not measure or compare it.
C. Gold leaf electroscope: This detects charge. When a charged body is brought near or touched to the cap, the leaf diverges from the stem. A larger charge produces a larger divergence, so two charges can be compared by how far the leaf rises. It can also test the sign of a charge and whether a body is a conductor or insulator.
D. Electrophorus: This generates charge repeatedly by induction. It does not compare charges.
Step 3: Solution To compare two charges, you would use a proof plane to take a sample of charge from each body to the electroscope in turn. The electroscope's leaf divergence shows which charge is larger. The comparing is done by the gold leaf electroscope, since the proof plane only helps carry the charge.
Answer: C. Gold Leaf Electroscope
Exam tip: Remember the roles:
Ebonite rod and electrophorus produce charge.
Proof plane carries charge.
Gold leaf electroscope detects and compares charge.
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37.
A liquid is left in an open dish. After several days, there is less liquid in the dish. Which of the following statements explains this observation? The
1. A.most energetic molecules leave the surface and escape into the air.
2. B.least energetic molecules leave the surface and escape into air.
3. C.most energetic molecules leave the surface and return.
4. D.least energetic molecules leave the surface and return.
Show answer
Answer: A. most energetic molecules leave the surface and escape into the air.
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38.
The conclusive test for magnetism in a steel bar suspended horizontally at its midpoint is when the bar
1. A.always settles in the magnetic N-S direction.
2. B.is repelled by a magnet.
3. C.always settles in the magnetic E-W direction
4. D.is attracted by a magnet.
Show answer
Answer: B. is repelled by a magnet.
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39.
The property of a body that determines the direction of heat flow when the body is in contact with another body is the
1. A.Heat capacity
2. B.mass
3. C.Power
4. D.Temperature
Show answer
Answer: D. Temperature
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40.
An atom in an excited state is one whose
1. A.electrons are in the conduction band.
2. B.potential energy is minimum.
3. C.potential energy is maximum.
4. D.electrons have moved to a higher energy level.
Show answer
Answer: D. electrons have moved to a higher energy level.
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41.
The colours of the visible light region of the electromagnetic spectrum that have the shortest and longest wavelengths respectively are
1. A.red and blue.
2. B.indigo and violet
3. C.violet and red.
4. D.yellow and blue.
Show answer
Answer: C. violet and red.
AI explanation
Step 1: Analyze the question You are asked which colours of visible light have the shortest and the longest wavelengths, in that order.
Step 2: The concept White light splits into seven colours (ROYGBIV). Each colour has a different wavelength:
| Colour | Approx. wavelength | |---|---| | Red | ~700 nm (longest) | | Orange | ~620 nm | | Yellow | ~580 nm | | Green | ~530 nm | | Blue | ~470 nm | | Indigo | ~445 nm | | Violet | ~400 nm (shortest) |
Wavelength and frequency are inversely related (c = fλ). Violet has the highest frequency and so the shortest wavelength. Red has the lowest frequency and so the longest wavelength.
Step 3: Check each option
A. Red and blue: this gives the order backwards, and blue is not the shortest. Wrong.
B. Indigo and violet: both are at the short-wavelength end, and neither is the longest. Wrong.
C. Violet and red: violet is shortest and red is longest, in the correct order. Correct.
D. Yellow and blue: these are middle colours. Wrong.
Answer: C. violet and red.
Exam tip: Remember ROYGBIV. Going from Red to Violet, wavelength decreases while frequency and energy increase. The ends of the rainbow are the extremes: Red is longest wavelength and Violet is shortest. Also watch the order in the question ("shortest and longest respectively"), because option A uses the right idea in the wrong order.
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42.
The range of wavelengths of the infrared waves in an electromagnetic spectrum is $10^{-3}$ m to $10^{-6}$ m. The wavelength of the radiation is
1. A.longer than $10^{-6} m$ .
2. B.shorter than $10^{-6} m$ .
3. C.shorter than $10^{3} m$ .
4. D.longer than $10^{3} m$ .
Show answer
Answer: A. longer than $10^{-6} m$ .
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43.
Which of the following diagrams correctly illustrates the mode of oscillation of air in a tube for the first overtone? ( The diagrams above are meant to help you in answering this question)
1. A.A
2. B.B
3. C.C
4. D.D
Show answer
Answer: C. C
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44.
Two objects, A and B, accelerate from rest at uniform rates. B accelerates twice as much as Compared to A, B travels
1. A.at the same distance as A
2. B.six times as far.
3. C.two times as far
4. D.three times as far
Show answer
Answer: C. two times as far
AI explanation
The question is incomplete as written. It doesn't say how long the objects accelerate, so I need to fill in the missing part. The usual version of this question reads: "Two objects, A and B, accelerate from rest at uniform rates. B accelerates twice as much as A. Compared to A, B travels ___ in the same time." I'll solve it with that assumption.
Step 1: Concept For an object starting from rest with uniform acceleration, the equation of motion is:
s = ut + ½at²
Since u = 0, this becomes:
s = ½at²
So for the same time t, the distance travelled is directly proportional to the acceleration.
Step 2: Apply to A and B
A: s_A = ½ a t²
B: acceleration = 2a, so s_B = ½ (2a) t² = a t²
Step 3: Compare s_B / s_A = (a t²) / (½ a t²) = 2
So B travels two times as far as A in the same time.
Answer: C. two times as far
Note on the assumption: If the question meant the same time, the answer is C, as shown above. If it had meant something else, such as the same distance, the result would differ. The "same time" reading is the standard one and is the only one that fits the options given.
Exam tip: From rest, s = ½at², so for equal times, distance ∝ acceleration. Double the acceleration means double the distance. If the question compared equal distances instead, time would be ∝ 1/√a. Always check which quantity is held constant.
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45.
A charge of 10 C is transferred across a potential difference of 220 V. Determine the work done.
1. A.22J
2. B.2200J
3. C.210J
4. D.230J
Show answer
Answer: B. 2200J
AI explanation
Analyze the question: We are given a charge (Q = 10 C) moved across a potential difference (V = 220 V), and we need the work done (W).
Concept: Potential difference is defined as the work done per unit charge:
V = W / Q
so
W = Q × V
Solution:
W = Q × V W = 10 C × 220 V W = 2200 J
Checking the options:
A. 22 J: this would come from dividing 220 by 10 (V ÷ Q), which is the wrong operation.
B. 2200 J: matches our working.
C. 210 J: this would come from adding 10 and 200, which has no basis in the formula.
D. 230 J: this would come from adding 10 and 220, which is also wrong.
Tip: Remember the units: 1 volt = 1 joule per coulomb (1 V = 1 J/C). So multiply volts by coulombs to get joules. A good memory aid is "W = QV", similar to "Work = Quantity × Voltage". Avoid dividing here, since that is the trap behind option A.
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46.
A direct current cannot pass through a parallel plate capacitor because
1. A.the plates are non-conducting.
2. B.the dielectric between the plates is non-conducting.
3. C.there is a leakage of charge from the plates.
4. D.the plates have equal but opposite charges.
Show answer
Answer: B. the dielectric between the plates is non-conducting.
AI explanation
Analyze the question: We want to know why a steady (DC) current cannot flow through a parallel plate capacitor.
Concept: A parallel plate capacitor has two metal plates separated by an insulating material called a dielectric (air, paper, mica, etc.). The plates themselves are conductors. Charge flows onto the plates during charging, but no charge can cross the gap between them.
Solution:
A. The plates are non-conducting. Wrong. Capacitor plates are made of metal, so they are good conductors.
B. The dielectric between the plates is non-conducting. Correct. The insulating dielectric blocks any flow of charge across the gap. When a DC source is connected, current flows only briefly while the plates charge up. Once the capacitor's p.d. equals the supply voltage, the flow stops. No steady current passes through the capacitor.
C. There is a leakage of charge from the plates. Wrong. Leakage is a small imperfection in real capacitors. It does not explain why DC is blocked.
D. The plates have equal but opposite charges. Wrong. This describes the state of a charged capacitor. It is a result of charging, not the reason charge cannot cross the gap.
Exam tip: Remember "a capacitor blocks DC, passes AC." With AC, the plates repeatedly charge and discharge, so current appears to flow in the circuit, but no charge actually crosses the dielectric.
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47.
Increasing the frequency of a sound wave produces a sound with
1. A.higher pitch
2. B.longer wavelength
3. C.more overtones
4. D.reduced loudness
Show answer
Answer: A. higher pitch
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48.
Which part of the human eye performs the same function as the diaphragm in a lens camera?
1. A.Choroid
2. B.Retina
3. C.Iris
4. D.Cornea
Show answer
Answer: C. Iris
AI explanation
Step 1: Analyze the question The question asks which part of the eye works like the camera's diaphragm. The diaphragm in a camera controls how much light enters the lens by changing the size of the aperture.
Step 2: Explain the concept The eye and the camera work in similar ways:
| Camera part | Function | Eye part | |---|---|---| | Lens | Focuses light | Eye lens (and cornea) | | Film / sensor | Receives the image | Retina | | Diaphragm (aperture) | Controls the amount of light entering | Iris (with the pupil) | | Light-tight box | Prevents stray light | Choroid (dark pigmented layer) |
Step 3: Go through the options
A. Choroid: This is a dark layer that absorbs stray light inside the eye. It is like the black inner lining of a camera, not the diaphragm.
B. Retina: This is the light-sensitive screen where the image forms. It is like the film or sensor.
C. Iris: This is the coloured muscular part that changes the size of the pupil. In bright light the pupil gets smaller, and in dim light it gets bigger. This is exactly what the diaphragm does by adjusting the aperture.
D. Cornea: This is the transparent front layer that does most of the initial bending (refraction) of light. It is not a light-regulating part.
Step 4: Answer The iris controls the amount of light entering the eye, just as the diaphragm does in a camera.
Exam tip: Match the parts by function:
Iris = diaphragm
Retina = film
Lens = lens
Choroid = black inner walls
Remember "Iris = Intensity control" (how much light gets in).
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49.
The following statements are observations made about a towing vehicle and a bus that is being towed I. The force exerted on the bus by the towing vehicle is of the same magnitude as the force that the bus exerts on the road. II. The force exerted on the bus by the towing vehicle is of the same magnitude as the force exerted on the towing vehicle by the bus. III. The towing vehicle exerts a force on the bus greater than the force exerted by the bus on the towing vehicle. IV. The towing vehicle exerts a force on the bus that is the same as the force the road exerts on the towing vehicle. Which of the following statements is/are correct?
1. A.II, III and IV
2. B.II only
3. C.I, II and III only
4. D.I only
Show answer
Answer: B. II only
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