Students attempting this WAEC Physics 2023 paper will encounter problems testing thermal physics, geometrical and wave optics, electrodynamics, and nuclear reactions. A close review of the questions highlights recurring concepts in energy conservation and latent heat, magnetic fields generated by currents, and RLC circuit behavior at resonance. Make the most of this practice session by analyzing the underlying formulas and theoretical explanations for each problem rather than relying on rote memorization.
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The instrument used to measure relative humidity is a
Instruments like barometers, hydrometers, and manometers measure atmospheric pressure, liquid density, and pressure differences respectively, which can easily confuse students. To find water vapor content, a specific meteorological tool is required. A hygrometer measures the amount of water vapor in the air relative to the maximum possible at that temperature, indicating relative humidity. Common mistake: Confusing hygrometers with hydrometers due to similar spelling.
Which part of the human eye performs the same function as the diaphragm in a lens camera?
Structures like the choroid, cornea, or retina serve nutritive, protective, or light-detecting roles rather than controlling exposure. Regulation of light entering the optical system is managed by a pigmented muscular tissue. The iris adjusts the pupil size to control the amount of light entering the eye, similar to how a camera diaphragm regulates light exposure for clear imaging. Common mistake: Confusing the light-regulating iris with the light-sensing retina.
An electron moves from an energy level of -8.6 eV to another level of 3.5 eV. Determine the frequency of the absorbed photon [1 eV = 1.6 × 10⁻¹⁹ J, h = 6.6 × 10⁻³⁴ Js]
Students might incorrectly subtract the energy levels or misapply the electron volt conversion factor to joules. The correct approach finds the difference between the final and initial states: 3.5 minus negative 8.6 yields 12.1 eV. Converting this energy to joules by multiplying by 1.6 times 10 to the minus 19 gives 1.936 times 10 to the minus 18 J. Dividing this result by Planck's constant results in approximately 2.93 times 10 to the 15 Hz. Common mistake: Sign errors when calculating the energy level difference across negative values.
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⁻¹. Determine H [g = 10 m s⁻²]
One might use the wrong kinematic equation or miscalculate gravity, leading to incorrect height values. Using the formula v squared equals u squared plus 2gh with an initial velocity of zero, final velocity of five, and acceleration due to gravity of ten, we get twenty-five equals twenty H. Solving this yields a height of 1.25 m. Common mistake: Forgetting to square the velocity term in the kinematic equation.
Which of the following statements about latent heat are correct? I. When ice melts, latent heat is absorbed II. When water freezes, latent heat is given off III. When water evaporates, latent heat is absorbed IV. When water vapour condenses, latent heat is given off
Students often remember that phase changes involve latent heat but forget whether the energy is absorbed or released during specific transitions. Melting and evaporation require energy input, meaning latent heat is absorbed, while freezing and condensation release energy to the surroundings. Latent heat is absorbed during melting and evaporation, but given off during freezing and condensation. Common mistake: Assuming latent heat is always absorbed during any phase change regardless of direction.
Increasing the frequency of a sound wave produces a sound with
Some might associate frequency changes with loudness or wavelength in the wrong direction due to wave equation confusion. Sound quality and perceived highness are directly tied to how fast the wave oscillates. Pitch is determined by frequency; higher frequency results in higher perceived pitch, while wavelength decreases. Common mistake: Confusing wave frequency changes with changes in sound loudness.
A body of mass 2 kg is moving with a velocity of 3 m s⁻¹. What is its momentum?
Students might mistakenly choose option B by confusing momentum with velocity, or option D by incorrectly multiplying numbers differently. To find the correct momentum, multiply the mass of 2 kg by the velocity of 3 m s⁻¹, resulting in 6 kg m s⁻¹. Common mistake: Multiplying or adding mass and velocity incorrectly instead of using the formula p = mv.
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
Students might mistakenly select option A or D by misplacing the radius or magnetic flux density during algebraic rearrangement. Equating the magnetic force to the centripetal force gives Bqv = mv² / r, which simplifies directly to v = Bqr / m upon solving for speed. Common mistake: Rearranging the balance of forces with inverted variables.
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]
Students might mistakenly choose option D by confusing wavelength with frequency or misinterpreting the distance given in the problem. The distance of 100 m does not affect the wavelength calculation. Dividing the speed of sound 331 m s⁻¹ by the frequency 512 Hz yields approximately 0.646 m. Common mistake: Attempting to incorporate the 100 m listener distance into the wave speed formula.
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
Students might mistakenly pick option A by assuming electrons share a tiny nuclear volume with protons. The rare and considerable deflections of alpha particles occur because a concentrated positive charge of protons within an extremely small nuclear space electrostatically repels those positive alpha particles. Common mistake: Forgetting that electrons reside outside the tiny central nucleus.
A charge of 10 C is transferred across a potential difference of 22 V. Determine the work done
Students might mistakenly choose option A by dividing the charge by the potential difference, or option B through arbitrary arithmetic errors. The work done, representing electrical energy transferred, is calculated by multiplying the transferred charge of 10 C by the 22 V potential difference to get 220 J. Common mistake: Dividing the charge by potential difference instead of multiplying them.
A solenoid is constructed by winding an insulated copper wire 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
Students might mistakenly select nickel, soft iron, or steel because they are ferromagnetic materials that strongly enhance a magnetic field. Brass is a non-magnetic material with low permeability, meaning inserting it into the test tube will fail to increase the solenoid's field strength. Common mistake: Assuming all metal rods placed inside a solenoid will increase its magnetic field.
The following statements are advantages of a solid dielectric material between the plates of a capacitor except that
Students might mistakenly choose option A or B because keeping plates separated and increasing maximum withstand voltage are indeed valid advantages of a solid dielectric. Dielectrics are insulators that increase capacitance without conducting; increased electrical conduction would lead to leakage and compromise the capacitor. Common mistake: Confusing insulation with electrical conduction in dielectrics.
Which of the following statements about a ferromagnetic material is correct? It
Students might mistakenly choose option B by assuming materials like iron possess low permeability instead of high. Ferromagnetic materials are characterized by a high magnetic susceptibility, which enables them to become strongly magnetized even in weak magnetic fields. Common mistake: Confusing high susceptibility with low permeability or non-magnetic attraction.
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?
Students might mistakenly select option C by thinking that an upward thrown object experiences an upward acceleration. Both falling and rising objects near the Earth's surface experience a uniform downward acceleration of approximately 9.8 m/s² due to gravity, regardless of initial velocity directions. Common mistake: Assuming the acceleration changes direction when an object is thrown upward.
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?
Students might mistakenly pick option C by confusing gravitational potential energy with escape velocity formulas. Equating kinetic energy to the gravitational potential energy loss to infinity yields the expression for escape velocity as the square root of 2GM divided by R. Common mistake: Omitting the factor of 2 inside the square root.
The specific latent heat of vaporization of lead is 8.7 × 10⁵ J kg⁻¹. This means that 8.7 × 10⁵ J of thermal energy is required to change
Students might mistakenly choose option B by stating solid lead instead of molten lead, or option D by adding a temperature dependency. Specific latent heat of vaporization refers specifically to the thermal energy needed to change 1 kg of liquid or molten lead into vapour at a constant temperature and pressure. Common mistake: Confusing the phase change from solid to liquid with liquid to vapour.
A 100 kg device is pulled up a plane inclined at 30° to the horizontal with a force of 100 N. If the coefficient of friction between the device and the surface is 0.25, determine the total force opposing the motion [g = 10 m s⁻²]
Students might mistakenly choose option A by calculating only the friction component or only the gravitational component. The total opposing force combines the gravitational pull down the plane (mg sinθ = 500 N) and the frictional force (μ mg cosθ ≈ 216.5 N), summing to 716.5 N. Common mistake: Forgetting to add both the frictional force and the component of gravity acting down the incline.
Two objects, A and B, accelerate from rest at uniform rates. B accelerates twice as much as A. Compared to A, B travels
Students might mistakenly pick option A or C by failing to account for the halved time it takes for B to reach the same velocity given double the acceleration. Since time is halved when acceleration is doubled, evaluating the distance equation results in B traveling the exact same distance as A. Common mistake: Assuming distance always scales directly with acceleration without factoring in time.
A 500 W electric heater operates for 2 hours. How much energy does it consume?
Students might mistakenly choose option A or C by forgetting to convert hours into seconds when computing electrical energy. Multiplying the power of 500 W by the total time in seconds (2 hours × 3600 seconds) gives an energy consumption of 3.6 × 10⁶ J. Common mistake: Multiplying power by hours directly without converting time to seconds.
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