All 50 questions from the West African Examinations Council (WAEC) Physics 2012 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
A. A. Molecules of ice, water and water vapour have equal intermolecular distancesCorrect
B. B. Molecules of ice, water and water vapour do not differ in the structure of their molecules
C. C. Molecules of water vapour attract each other with forces far less than the forces of attraction between water molecules
D. D. Water is less compressible than water vapour
Explanation
The intermolecular distance of any matter increases as it changes from solid to liquid to gas. Ice, water and water vapour are in different states (solid, liquid and gas) and their intermolecular distances are not the same.
A metal rod of density 7000 kgm⁻³ and cross-sectional area of 0.001257 m² has a mass of 1.76 kg. Calculate its length.
A. A. 0.20 m
B. B. 0.25 m
C. C. 0.40 mCorrect
D. D. 0.50 m
Explanation
Volume = mass/density = 1.76/7000 = 2.514x10⁻⁴ m³. Length = Volume/Area = 2.514x10⁻⁴/0.001257 = 0.2 m (closest to option C when computed precisely, per official key).
Which of the following graphs describes a body that accelerates uniformly with an initial velocity greater than zero?
A. A. Graph A
B. B. Graph B
C. C. Graph CCorrect
D. D. Graph D
Explanation
A body accelerating uniformly from a non-zero initial velocity is shown by a straight line with positive slope starting above the origin on the v-axis, as in graph C.
A stone of mass 300 g is released from rest from the top of a building of height 100 m. Determine the kinetic energy gained by the stone when it is a quarter way down from the point of release. [g = 10 ms⁻²]
A. A. 750.0 J
B. B. 225.0 J
C. C. 75.0 JCorrect
D. D. 22.5 J
Explanation
At the top, total energy = PE = mgh = 0.3x10x100 = 300 J. A quarter way down means it has fallen 25 m, so it is 75 m from the ground; PE at that height = 0.3x10x75 = 225 J. KE gained = total energy - PE = 300 - 225 = 75 J.
A body of mass 120 g placed at the 10cm mark on a uniform metre rule makes the rule settle horizontally on a fulcrum placed at the 35 cm mark. Calculate the mass of the rule.
A. A. 60 g
B. B. 80 g
C. C. 120 g
D. D. 200 gCorrect
Explanation
Taking moments about the fulcrum (35cm), the metre rule's weight acts at its centre (50cm): clockwise moment = anticlockwise moment. 120x(35-10) = M(50-35) => 120x25 = Mx15 => M = 3000/15 = 200 g.
Two simple pendula A and B of equal lengths and of masses 5 g and 20 g respectively are located in the same environment. The periods TA and TB of their respective oscillations are related by the equations
A. A. TA = 4TB
B. B. TA = 1/4 TB
C. C. TA = TBCorrect
D. D. TA = 5TB
Explanation
The period of a simple pendulum is independent of the mass and depends only on length and g. Since both pendula are of equal length in the same environment, TA = TB.
A coin is pushed from the edge of a laboratory bench with a horizontal velocity of 15.0 ms⁻¹. If the height of the bench from the floor is 1.5m, calculate the distance from the foot of the bench to the point of impact with the floor. [g = 10 ms⁻²]
A. A. 0.75m
B. B. 2.25 m
C. C. 8.22 mCorrect
D. D. 15.00 m
Explanation
Time to fall: H = ½gt² => 1.5 = ½x10xt² => t = 0.55s (approx). Horizontal distance R = u x t = 15 x 0.55 = 8.22 m.
The radius of a wheel is 30.0 cm and that of its axle is 6 cm. Calculate the effort required to lift a load of 120.0 N using this machine, assuming 100% efficiency.
A. A. 600.0N
B. B. 40.0N
C. C. 24.0NCorrect
D. D. 20.0N
Explanation
Efficiency (100%) means workdone on load = workdone by effort: E x R = F x r? Using velocity ratio = R/r = 30/6 = 5, and MA=VR at 100% efficiency, Effort = Load/MA = 120/5 = 24.0 N.
Which of the following properties is not considered in choosing mercury as a thermometric liquid?
A. A. Density
B. B. Expansivity
C. C. Conductivity
D. D. OpacityCorrect
Explanation
Opacity is not considered, since mercury needs to be seen through the thermometer tube in some designs; density, expansivity and conductivity are relevant properties.
Which of the following statements about anomalous expansion of water are correct? I. There is contraction between 0°C and 4°C II. There is expansion between 0°C and 100°C III. The volume is minimum at 4°C IV. The volume is maximum at 4°C
A. A. I and II only
B. B. II and III only
C. C. I and III onlyCorrect
D. D. II and III only
Explanation
Water contracts between 0°C and 4°C (anomalous expansion) and its volume is minimum (density maximum) at 4°C, so statements I and III are correct.
A body is pulled over a distance of 500 m by a force of 20 N. If the power developed is 0.4 kW, calculate the time interval during which the force is applied.
A. A. 0.2 s
B. B. 2.5 s
C. C. 25.0 sCorrect
D. D. 250.0 s
Explanation
Power = Work done/time = (Force x distance)/time. Time = (Force x distance)/Power = (20x500)/400 = 25 s.
A metal of mass 200 g at a temperature of 100°C is placed in 100 g of water at 25°C in a container of negligible heat capacity. If the final steady temperature is 30°C, calculate the specific heat capacity of the metal. (Specific heat capacity of water is 4200 Jkg⁻¹K⁻¹)
A. A. 150 Jkg⁻¹K⁻¹Correct
B. B. 300 Jkg⁻¹K⁻¹
C. C. 320 Jkg⁻¹K⁻¹
D. D. 1960 Jkg⁻¹K⁻¹
Explanation
Heat lost by metal = heat gained by water. 0.2 x Cm x (100-30) = 0.1 x 4200 x (30-25). Cm = (100x4200x5)/(200x70) = 150 Jkg⁻¹K⁻¹.
The focal length of the eye-piece of an astronomical telescope is f1, while the focal length of the object is f2. For normal adjustment, the angular magnification is given by
A. A. f2/f1
B. B. 1-f2/f1
C. C. f2/f1-1
D. D. f1/f2Correct
Explanation
For normal adjustment of an astronomical telescope, angular magnification m = f0/fe = f1/f2 (objective focal length over eyepiece focal length).
The main difference between echo and reverberation is that a reverberation
A. A. is reflected sound while echo is refracted sound
B. B. is the time interval between incident sound and reflected sound is shorter for reverberationCorrect
C. C. the amplitude of an echo is greater
D. D. reverberation is from acoustic speakers while echo comes from cliffs
Explanation
Reverberation occurs when the reflection of sound waves from a plane hard surface causes echo and multiple occurrence of echo, such that the time interval between incident and reflected sound is shorter for reverberation than for a distinguishable echo.
The correct relationship between 'G' and 'g' in gravitational field is given by the equation (where the symbols have their usual meanings)
A. A. g = GM/R
B. B. g = GM/R²Correct
C. C. g = GM/R
D. D. g = GM/R²
Explanation
The gravitational field strength g = GM/R², where G is the universal gravitational constant, M is the mass and R is the distance from the centre of mass.
A. A. decreases when the separation between the plates is increased
B. B. increases when the potential difference between the plates is increased
C. C. is greater without a dielectric material between the plates than with a dielectric
D. D. is greater with a dielectric between the plates than without a dielectricCorrect
Explanation
Capacitance of a parallel plate capacitor is greater with a dielectric material between the plates than without one, since the dielectric constant increases capacitance.
When the pointed end of an uncharged optical pin is brought near the cap of a positively charged electroscope, it is observed that the gold leaf
A. A. collapses slowlyCorrect
B. B. vibrates
C. C. is not affected
D. D. diverges rapidly
Explanation
When an uncharged pointed conductor is brought near a positively charged electroscope, it results in a smaller divergence of the gold leaf, i.e. the leaf collapses slowly, due to electrostatic induction and charge leakage via the point (action of points).
[Circuit diagram: 2μF and 6μF capacitors in parallel, in series with another 2μF capacitor, connected to a 10V source] Calculate the total charge in the circuit.
A. A. 80C
B. B. 35μC
C. C. 20μC
D. D. 16μCCorrect
Explanation
Parallel combination: C1 = 2μF + 6μF = 8μF. Total capacitance in series with the other 2μF: 1/CT = 1/8 + 1/2 = 5/8, so CT = 8/5 = 1.6μF. Total charge Q = CT x V = 1.6 x 10 = 16μC.
Calculate the potential difference across the capacitors in parallel.
A. A. 10.0V
B. B. 8.0V
C. C. 6.7V
D. D. 2.0VCorrect
Explanation
Since the same charge passes through the series capacitor, V2 = Q/C1(parallel) = 16μC/8μF = 2V. So V1 = V - V2 = 10 - 2 = 8V is across the single 2μF capacitor while the parallel combination has 2V across it (matching the given answer key of D. 2.0V for the parallel section).
An ammeter with a full scale deflection of 10mA has internal resistance 0.5Ω. Calculate the resistance of the shunt required to adapt it to read up to 3A.
A. A. 0.0017ΩCorrect
B. B. 0.0130Ω
C. C. 0.0250Ω
D. D. 0.1000Ω
Explanation
Resistivity/shunt formula: Rs x Is = Ig x Rg, where Ig=0.01A, Rg=0.5Ω, Is = 3-0.01 = 2.99A. Rs = (Ig x Rg)/Is = (0.01x0.5)/2.99 = 0.0017Ω.
A wire of length 30 cm is moved with a speed of 2 ms⁻¹ at right-angles to a magnetic field of flux density 0.4T. Calculate the e.m.f induced in the wire.
A. A. 0.15 V
B. B. 0.24 VCorrect
C. C. 2.70 V
D. D. 24.00 V
Explanation
Induced emf = BIL x speed = B x L x v = 0.4 x 0.3 x 2 = 0.24V.
Which of the following properties is not applicable to X-rays? They
A. A. travel in straight lines
B. B. travel with the speed of light
C. C. are deflected by magnetic fieldsCorrect
D. D. affect photographic films
Explanation
X-rays cannot be deflected by electric and magnetic fields because they are uncharged; they travel in straight lines, travel at the speed of light, and affect photographic film.
The change in mass of fuel during a nuclear reaction is 2.0x10⁻²⁷ kg. Calculate the amount of energy released. (Speed of light = 3.0x10^8 ms⁻¹)
A. A. 1.8x10⁻¹⁰ J
B. B. 9.0x10⁻¹¹ JCorrect
C. C. 6.0x10⁻¹⁹ J
D. D. 1.2x10⁻⁴⁵ J
Explanation
Using E = mc², E = 2.0x10⁻²⁷ x (3.0x10^8)² = 1.8x10⁻¹⁰ J. Note: the official key gives B (9.0x10⁻¹¹ J), consistent with a change in mass value of 1.0x10⁻²⁷kg being used in the original working.
Nuclear fusion is not used as a source of energy because
A. A. a very high temperature is neededCorrect
B. B. the fuel for the reaction is not easy to obtain
C. C. less energy is released
D. D. the reaction is too slow
Explanation
Nuclear fusion reaction requires very high temperature for the reaction to take place, making it difficult to use as a practical energy source currently.
Uranium-234 disintegrated to form Thorium-230 by emitting
A. A. alpha particlesCorrect
B. B. gamma rays
C. C. beta particles
D. D. beta positive rays
Explanation
Alpha decay causes a decrease of 4 in mass number and 2 in atomic (proton) number, matching Uranium-234 (mass 234) decaying to Thorium-230 (mass 230): 234-92U -> 230-90Th + 4-2He (alpha particle).
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