Question Library
Browse, filter, and study verified past examination questions with step-by-step mathematical reasoning.
Searching 18,000+ Questions...
Sifting topics, KaTeX formulas & verified solutions
Showing 421–435 of 13612 verified questions
Sort by:
Physics
WAEC / WASSCE 2024
Reflection of Light
Diagram
Solution
(a) State the condition for light ray incident on a concave mirror to reflect through the principal focus of the mirror.
(b) Explain spherical aberration as a defect of curved mirrors
(c) The diagram above illustrates an image formed when an object is placed in front of a particular lens. Redraw the diagram and indicate the center of curvature, C, principal focus, F, and the principal axis. ( SEE THE DIAGRAM ABOVE)
(d) What is meant by accommodation in connection to the human?
(e) State quantitatively, the (i) values of the near and far point of a normal eye. (ii) Use the answer in 10c(i) to determine the change in optical power of the normal human eye when reading a book and viewing the sky.[lens-to-retina distance = 2.5 cm].
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) The incident light rays must be parallel and close to the principal axis
(b) Spherical aberration as a defect of curved mirrors: When a wide parallel beam is incident on a spherical mirror, it is not brought to the same focus. This makes the image distorted.
(c) SEE THE DIAGRAM ABOVE
(d) Accommodation is the ability of the eyes to alter its focal length to form a clear image
(e) Quantitatively, (i) far point = infinity, Near point = 25 cm
P\(_{near point}\) = \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)
(ii) for near point, u = 0.25 m
P\(_{near point}\) = \(\frac{1}{f}\) = \(\frac{1}{0.25}\) + \(\frac{1}{v}\)
similarly, for far point, u = infinity(\infty)
P\(_{far point}\) = \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)
P\(_{far point}\) = \(\frac{1}{\infty}\) + \(\frac{1}{v}\)
= 0 + \(\frac{1}{v}\)
So, change in optical power
\DeltaP = P\(_{near point}\) - P\(_{far point}\)
\DeltaP = 4 + \(\frac{1}{v}\) - \(\frac{1}{v}\)
\DeltaP = 4 dioptres.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72093)
Physics
WAEC / WASSCE 2024
Refraction of Light
Solution
A ray of light travels from a less dense medium to a denser medium. How are the speed and frequency of the light affected? The speed
A
Increases and the frequency increases
B
decreases and the frequency decreases
C
increases and the frequency remains the same
D
decreases and the frequency remains the same
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
When light travels from a less dense medium (like air) to a denser medium (like glass or water), its speed decreases due to the higher optical density of the new medium.
The frequency of light remains constant when it passes from one medium to another. Although the speed changes, the frequency is determined by the source of the light and does not change with the medium.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72026)
Physics
WAEC / WASSCE 2024
Refraction of Light
Solution
Chromatic aberration in lenses can be corrected by using a combination of
A
diverging and converging lenses.
B
an opaque material to cover the rims of lenses.
C
wide parallel beams on the lenses' surface.
D
a thin convex lens.
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Chromatic aberration occurs because different wavelengths of light are refracted by different amounts when passing through a lens.
Chromatic aberration can be corrected by using a combination of diverging (concave) and converging (convex) lenses, which help focus different colors of light to the same point.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72040)
Physics
WAEC / WASSCE 2024
Refraction of Light
Solution
For a convex lens to form a real image, the distance between the object and the screen will be:
A
equal to the focal length of the lens
B
less than the focal length of the lens
C
twice the focal length of the lens
D
four times the focal length of the lens
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Twice the focal length of the lens: this is because, for a real image to be formed, the object must be placed outside the focal length of the lens, typically at a distance greater than twice the focal length.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72052)
Physics
WAEC / WASSCE 2024
Refraction of Light
Solution
The inside portion of a hollow metal sphere of diameter 20 cm is polished. The portion will therefore form a
A
Concave mirror of focal length 5cm.
B
Concave mirror of focal length 10 cm.
C
Convex mirror of focal length 10 cm
D
Convex mirror of focal length 5cm
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Diameter = 20cm
The radius of curvature, R = 10cm
But, focal length, f = \(\frac{R}{2} = \frac{10}{2}\) = 5 cm.
Because the inside portion is polished, light reflects off the hollow, caved-in interior. A mirror that reflects light from its inner surface is concave.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72066)
Physics
WAEC / WASSCE 2024
Refraction of Light
Solution
The phenomenon used for explaining the formation of a rainbow in the sky is
A
interference.
B
refraction.
C
reflection.
D
dispersion.
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The formation of a rainbow is primarily due to dispersion, as it involves the separation of light into its constituent colors when it passes through water droplets in the atmosphere, it is refracted and dispersed due to varying wavelengths.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72070)
Physics
WAEC / WASSCE 2024
Sound Wave
Solution
Which of the following factors does not affect the speed of sound?
A
Loudness.
B
Wind direction
C
Temperature
D
Material density
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Loudness does not affect the speed of sound.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72078)
Physics
WAEC / WASSCE 2024
Electricity and Magnetism
Solution
State three uses of ferromagnetic materials
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
The uses of ferromagnetic materials are:
(i) Electromagnets
(ii) Magnetic Storage
(iii) Magnetic Sensors
(iv) Magnetic Shielding
(v) Permanent Magnets
(vi) Inductive Components, others are transformers, electric bells, electric generators, etc.
Source: Myschool Past Questions Archive (Record #72088)
Physics
WAEC / WASSCE 2024
Mechanics and Properties of Matter
Solution
PART2
(a) State the effect of increasing temperature on the viscosity of a: (i) liquid, (ii) gas
(b) State two factors that determine the magnitude of a moment of a force
(c) A uniform stick AB of length, L, and mass, m, is balanced horizontally on a knife edge 10.0cm from A when an object of 400 g is suspended at A. When the knife edge is moved 5 cm further, the object has to be moved to a point 9.00 cm from A for the stick to balance.
(i) Represent the balance system with a suitable diagram
(ii) Determine the: I. mass, m of the stick; II. length, L.
(d) Explain in terms of air molecules why pressure at the top of a high mountain is less than at sea level.
(e) Mercury of density 13.6 x 10\(^3\)kgm\(^{-3}\) is poured in a container of uniform cross-sectional area 40cm \(^2\) to a height of 20 cm. The total pressure exerted on the base of the container is 1.42 x 10\(^5\)P. Calculate the: (i) mass of the mercury in the container and (ii) pressure exerted at the surface of the mercury. [g = 10ms\(^{-2}\)
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(ai) Increasing the temperature of a liquid decreases the viscosity of the liquid. (aii) Increasing the temperature of a gas increases the viscosity of the gas
(b) Factors that determine the magnitude of the moment of a force about a point are: 1. Angle between the distance from the turning point and the line of action of the force. 2. magnitude of the applied force. 3. perpendicular distance from the pivot to the line of action of the applied force.
(c) SEE THE DIAGRAM ABOVE.
(ii) Taking a moment about the support: clockwise moment = anti-clockwise moment
400 x 10 = m x (\(\frac{L}{2}\) - 10) -- ------- from figure 1 above
4000 = m x (\(\frac{L}{2}\) - 10) - - - -- - - (i)
400 x 6 = m x (\(\frac{L}{2}\) - 15) - - - - -- - - -(ii)
divide eqn i by ii
\(\frac{4000}{2400}\) = \(\frac{m \times \frac{L}{2} - 10)}{m \times \frac{L}{2} - 15)}\)
\(\frac{5}{3}\) = \(\frac{\frac{L}{2} - 10)}{\frac{L}{2} - 15)}\)
3 x (\(\frac{L}{2}\) - 10) = 5 x (\(\frac{L}{2}\) - 15)
\(\frac{3L}{2}\) - 30) = \(\frac{5L}{2}\) - 75)
\(\frac{5L}{2}\) - \(\frac{3L}{2}\) = 75 - 30 = 45
L = \(\frac{2L}{2}\) = 45 cm
From equation (ii) 400 x 6 = m x (\(\frac{L}{2}\) - 15)
2400 = m x (\(\frac{45}{2}\) - 15)
2400 = m x 7.5
m = \(\frac{2400}{7.5}\) = 320 g
Therefore, I. m = 320g and II. L = 45 cm
(d) Reasons for less pressure at the top of a mountain:
The lower pressure at the top of a mountain is due to the reduced number of air molecules in the atmosphere above that elevation, resulting in a lighter air column and fewer molecular collisions, which together create lower atmospheric pressure compared to sea level.
(e)(i) Density = \(\rho\) = \(\frac{m}{v}\)
m = v x \(\rho\), but v = A x h
m = A x h x \(\rho\) = 13.6 x 10\(^3\) x 40 x 20 x 10\(^{-6}\) = 10.88kg
(e)(ii) pressure exerted by the surface of the mercury
P\(_t\) = P\(_{atm}\) + pressure of mercury
P\(_{atm}\) = P\(_t\) - \(\rho\) x h x g = 1.42 x 10\(^5\) - 13.6 x 10\(^3\) x 0.2 x 10 = 1.148 x 10\(^5\)Pa
Source: Myschool Past Questions Archive (Record #72091)
Physics
WAEC / WASSCE 2024
Mechanics and Properties of Matter
Solution
(a)i State the condition for a charged particle to experience a force in magnetic field
(ii) State the expression for the magnetic force, F acting on a charged particle, Q in a magnetic field of flux density, \(\beta\) with speed, V
(iii) Prove quantitatively that there is no magnetic force moving along the direction of a magnetic field.
(b)i Mention the main parts of an electrical transformer.
(ii) Explain how an alternating potential difference applied to the primary coil gives rise to an induced e.m.f in the secondary coil of a transformer.
(c) Proton of mass 1.7 x 10(^{-27}\) kg enters a magnetic field of flux density 0.207 normally and followed a quarter circle path before existing with a constant speed of 4.5 x 10\(^6\)m/s
(i) Explain why the speed of the proton remains constant.
(ii) Calculate the
I. radius of the circular path
II. time taken by the proton to move through the magnetic field [e = 1.6 x 10\(^{-19}\) C, \(\pi\) = 3.14]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)i The charged particle must be in motion or have velocity and the particle's velocity must be inclined to the magnetic field at an angle \(\theta\)
(ii) F = Qv\(\beta\)sin\(\theta\)
(iii) for a charged particle moving parallel to the magnetic field,
\(\theta\) = 0 and F = Qv\(\beta\)sin\(\theta\)
F = Qv\(\beta\)sin 0 = 0
Thus, F = 0
(b)The main parts of an electrical transformer are the coil, Soft iron core
(ii) An alternating current in the primary coil produces a changing magnetic field in the primary coil. The changing magnetic field around the core gives rise to a changing magnetic flux linking the secondary coil thereby inducing an alternating e.m.f in the secondary coil as stated in Faraday's law.
c(i) Why the speed of the proton remains constant: The speed of the proton remains constant because the magnetic force acts perpendicular to its velocity, providing centripetal acceleration without doing work on it. This means the force changes the direction of the proton's motion but not its speed, resulting in constant kinetic energy and a uniform speed throughout its circular path.
(ii) I. radius of curved path
qv\(\beta\) = \(\frac{mv^2}{r}\)
r = \(\frac{mv}{q\beta}\)
r = \(\frac{1.7 \times 10^{-27}}{1.6 \times 10^{-19}}\) = 0.24 m.
II. time taken by the proton to move through the magnetic field,
t = \(\frac{\pi r}{2v}\) = \(\frac{3.14 \times 0.24}{2 \times 4.5 \times 10^6}\) = 8.37 x 10\(^{-8}\)s
Source: Myschool Past Questions Archive (Record #72094)
Physics
WAEC / WASSCE 2024
Mechanics and Properties of Matter
Diagram
Solution
(a) You are provided a retort stand, a spring balance, masses, a beaker containing water, another beaker containing a liquid labelled, L, and other necessary apparatus.
(i) Suspend the mass, m = 20 g on the spring balance and measure the weight in air, W\(_A\) in Newton.
(ii) Immerse the suspended mass completely in water and measure weight in water, W\(_W\) in Newton. Evaluate U\(_1\) = W\(_A\) - W\(_W\).
(iii) Immerse the suspended mass completely in the liquid L of the same volume with water and measure the weight in liquid W\(_L\). Evaluate U\(_2\) = W\(_A\) - W\(_L\).
(iv) Repeat the procedure for m = 40g, 60g, 80g, and 100g respectively. In each case, evaluate \(W_A\), W\(_W\), U\(_1\), W\(_L\), U\(_2\).
(v) Tabulate the readings.
(vi) Plot a graph of U\(_1\) on the vertical axis and U\(_2\) on the horizontal axis starting both axes from the origin (0, 0).
(vii) Determine the slope s of the graph
(viii) Evaluate K = \(\frac{1}{\text{s}}\)
(ix) State two precautions taken to ensure accurate results.
b(i) State the Archimedes' principle,
(ii) State two differences between density and relative density
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(v)
M(g)
W\(_A\)(N)
W\(_W\)(N)
U\(_1\) = W\(_A\) - W\(_W\).
W\(_L\)(N)
U\(_2\) = W\(_A\) - W\(_L\)
20.0
0.20
0.16
0.04
0.18
0.02
40.0
0.40
0.33
0.07
0.36
0.04
60.0
0.60
0.49
0.11
0.54
0.06
80.0
0.80
0.66
0.14
0.72
0.08
100.0
1.00
0.82
0.18
0.90
0.10
(vi) SEE THE GRAPH ABOVE
(vii) s = \(\frac{\Delta U_1}{\Delta U_1}\) = \(\frac{0.162 - 0.2}{0.09 - 0.067}\) = \(\frac{0.042}{0.023}\) = 1.826
(viii) K = \(\frac{1}{\text{s}}\) = \(\frac{1}{1.826}\) = 0.5476
(ix)(I) I avoided errors due to parallax when reading the spring balance. (II) I cleared the mass thoroughly before immersing it in liquid L
b(i) Archimedes's principle states that when a body is completely or partially immersed in a fluid, it experiences an upthrust which is equal to the weight of fluid displaced.
(ii) Density has a unit whereas relative density has none. (II) The value of density is different in different systems of measurement, whereas relative density is the same in all systems of measurement.
Source: Myschool Past Questions Archive (Record #72107)
Physics
WAEC / WASSCE 2024
Electricity and Magnetism
Diagram
Solution
(a) You are provided with two resistance wires labelled: A and B, standard resistor, R\(_x\) = 1 \(\Omega\), metre bridge, cell of emf,E, Rheostat R\(_h\), galvanometer and apparatus as shown.
Use the circuit diagram above as a guide to perform the experiment.
(i) Connect R\(_x\) in the left-hand gap of the metre bridge, a length L = 100 cm of the wire in the right-hand gap and the other apparatus shown.
(ii) Determine and record the balance point P on the metre bridge wire NQ
(iii) Measure and record L\(_x\) = NP and L\(_y\) = PQ
(iv) Evaluate R\(_1\) = \(\frac{L_y}{L_x}R_x\)
(v) Repeat the procedure for four other values of L = 90cm, 80cm, 70cm, and 60cm. In each case, determine and record the balance point, P and evaluate R\(_1\)
(vi) Repeat the experiment with the second wire B. Obtain the balance point P and evaluate R\(_2\) in each case
(vii) Tabulate the readings
(viii) Plot a graph of R\(_2\) on the vertical axis and R\(_1\) on the horizontal axis
(ix) Determine the slope, s, of the graph
(x) Evaluate K = \(\sqrt{s}\)
(xi) State two precautions taken to ensure accurate results.
b(i) State two advantages of potentiometer over voltmeter for measuring potential difference
(ii) Define internal resistance of a cell.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(vii)
L(cm)
L\(_x\)(cm)
L\(_y\)(cm)
R\(_1\) = (\(\frac{L_y}{L_x})R_x\)\(\Omega\)
L\(_{x2}\)
L\(_{y2}\)
R\(_2\) = \(\frac{L_{y2}}{L_{x2}}R_x\)\(\Omega\)
100.0
5.2
94.8
18.2
21.5
78.5
3.6
90.0
6.0
94.0
15.7
22.5
77.5
3.4
80.0
6.8
93.2
13.7
23.4
76.6
3.3
70.0
7.6
92.4
12.2
24.5
75.5
3.1
60.0
8.4
91.6
10.9
25.6
74.4
2.9
(viii) SEE THE GRAPH ABOVE
(ix) Slope = \(\frac{\Delta R_2}{\Delta R_1}\) = \(\frac{3.5 - 3.3}{17.0 - 14.6}\) = \(\frac{0.2}{2.4}\) = 0.083
(x) K = \(\sqrt{0.083}\) = 0.2881
(xi) I Tight connection/clean terminal was ensured II. Key removed when readings were not being taken.
b(i) I. A potentiometer can be used for measuring the internal resistance of a cell and also to compare the e.m.f. of two cells. II. A potentiometer gives the exact value of potential difference across any two points in a circuit.
(ii) It refers to the opposition to the flow of current offered by the cells and batteries themselves resulting in the generation of heat.
Source: Myschool Past Questions Archive (Record #72108)
Physics
WAEC / WASSCE 2024
Heat Energy & Temperature
Solution
(a) You are provided with two metre rules, two retort stands, a mass, m = 100g, thread, and other necessary apparatus
(i) Place one of the metre rules on a knife edge and determine the centre of gravity, C
(ii) Measure and record the mass, M\(_R\) of the metre rule
(iii) Attach the mass = 100g firmly to the metre rule AB at C using paper tape
(iv) Suspend the metre rule by two parallel threads of length, h = 40cm each at 10 cm and 90cm mark. Ensure that the graduated face of the metre rule is facing upwards.
(v) Set the metre rule into small angular oscillation about the vertical axis through the centre of gravity.
(vi) Determine the time, t for 20 complete oscillations. Evaluate the period T and T\(^2\).
(vii) Read and record the value of d in metres.
(viii) Keeping d, constant, repeat the procedure for four other values of h = 50, 60, 70, and 80 cm. In each case, determine t and evaluate T and T\(^2\).
(ix) Tabulate the readings
(x) Plot a graph of T\(^2\) on the vertical axis and h on the horizontal.
(xi) Determine the slope, s of the graph
(xii) Evaluate K = \(\frac{\text{S}}{\text{Q}}\) where Q = \(\frac{2}{25d^2}\)
(xiii) State two precautions taken to ensure accurate results
b(i) Define couple as it relates to oscillatory motion
(ii) Give two practical applications of a couple in everyday life.
A
the variation of the volume of a gas with increasing temperature
B
expansion curve of mercury
C
expansion curve of liquid oxygen
D
anomalous expansion of water
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
(a)i c = 50.0cm
(ii) M\(_R\) = 90.0 g
(vii) d = 0.8m
(ix)
S/N
h(cm)
t(s)
T= \(\frac{t}{20}\)s
T\(^2\)s\(^2\)
1
40.0
26.12
1.306
1.7056
2
50.0
27.80
1.390
1.9321
3
60.0
30.68
1.534
2.3532
4
70.0
34.60
1.730
2.9929
5
80.0
35.80
1.790
3.2041
(x) SEE THE GRAPH ABOVE
(xi) Slope, s = \(\frac{\DeltaT^2}{\Delta h}\) = \(\frac{3.0 - 1.75}{75 - 46}\) = \(\frac{1.25}{29}\) = 0.431s\(^2\)cm\(^{-1}\)
(xii) K = \(\frac{\text{S}}{\text{Q}}\)
where, Q = \(\frac{2}{25d^2}\) = \(\frac{2}{25 \times (0.8)^2}\) = 0.125
K = \(\frac{\text{S}}{\text{Q}}\) = \(\frac{0.431}{0.125}\) = 0.3448
b(i) A couple, as related to oscillatory motion, refers to a pair of forces that are equal in magnitude but opposite in direction, and act on opposite sides of a pivot or fulcrum, causing rotation or torsion.
b(ii) I. Wrenches: When using a wrench to tighten or loosen a bolt, the force applied creates a couple that produces rotational motion. II. Steering Wheels: Turning the steering wheel of a car applies a couple that rotates the front wheels, allowing for directional change. III. Door Handles: When pushing or pulling on one side of a door handle, a couple is created that rotates the door about its hinges. IV. Bicycle Pedals: When pedaling a bicycle, the force applied to the pedals generates a couple that turns the crankshaft, propelling the bike forward.
Source: Official Exam / Past Question Archive
Chemistry
WAEC / WASSCE 2024
Acids, Bases, and Salts
Solution
Which of the following gases could be collected by downward displacement of air?
A
Chlorine
B
Hydrogen
C
hydrogen chloride
D
Sulphur(IV) oxide
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
Gases that are less dense than air are collected by downward displacement of air. Examples include ammonia and hydrogen.
Sulphur dioxide (SO\(_2\)) is not collected by downward displacement of air. It is collected by upward displacement of air because it is heavier than air.Also, chlorine is collected by upward displacement of air, not downward displacement. This is because although, it is heavier and denser than air, it is soluble in water. Instead, chlorine is collected by upward displacement of air.
Source: Myschool Past Questions Archive (Record #72167)
Chemistry
WAEC / WASSCE 2024
Atomic Structure and Periodic Table
Solution
Which of the following substances has the lowest boiling point?
A
Aqueous sodium chloride
B
Ethanol
C
Tetrachloromethane
D
Water
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
High boiling point is primarily caused by strong intermolecular forces, particularly hydrogen bonding which is the main reason for water's high boiling point, as water molecules can form multiple hydrogen bonds with each other, requiring more energy to separate them and vapourize. Consequently, water, ethanol and aquous sodium chloride would have relatively high boiling points, owing to the presence of hydrogen bonds while tetrachloromethane has no hydrogen bonding, hence, it is has the lowest boiling point.
Source: Myschool Past Questions Archive (Record #72168)