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Physics
WAEC / WASSCE 2018
Thermal Physics
Solution
a) What does the acronym LASER stand for?
b) What is a laser?
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) LASER - Light Amplification by Stimulated Emission of Radiation.
(b) A laser is a device that generates an intense beam of coherent monochromatic light.
Source: Myschool Past Questions Archive (Record #52869)
Physics
WAEC / WASSCE 2018
Mechanics and Properties of Matter
Diagram
Solution
(a) Define uniform acceleration.
(b) Forces act on a car in motion. List the
(i) horizontal forces and their directions;
(ii) vertical forces and their directions
(c) A car starts from rest and accelerates uniformly for 20s to attain a speed of 25 ms\(^{-1}\). It maintains this speed for 30s before decelerating uniformly to rest. The total time for the journey is 60 seconds.
(i) Sketch a velocity-time graph for the motion.
(ii) Use the graph to determine the (\(\alpha\)) total distance travelled by the car (\(\beta\)) deceleration of the car.
The figure above illustrates a force-extension graph for a stretched spiral spring. Determine the work done on the spring. SEE THE DIAGRAM ABOVE
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Uniform acceleration is the constant time rate of increase in velocity.
(b) (i) Horizontal Force;
Forward force - Thrust
backward force - Frictional force
(ii) Vertical Force
Upward force - Normal reaction.
Downward force - Weight of the car.
(c) SEE THE DIAGRAM ABOVE.
(ii) (\(\lambda\)) Total distance
= Area of trapezium OABC
= \(\frac{1}{2}\)[AB + OC] AD
= \(\frac{1}{2}\)[60+30] x 25
=1125m
(\(\beta\)) Deceleration = slope of BC
= \(\frac{BE}{CE} = \frac{25}{10}\)
= 25m/s\(^2\)
(d) Work done = \(\frac{1}{2}\)Fe
= \(\frac{1}{2}\) x 12 x \(\frac{0.5}{100}\)J
= 0.03J
Source: Myschool Past Questions Archive (Record #52870)
Physics
WAEC / WASSCE 2018
Mechanics and Properties of Matter
Solution
(a) List two factors each that affect heat loss by:
(i) radiation;
(ii) convection.
(b) State two factors that determine the quantity of heat in a body.
(c) Explain the statement: The vecilic latent heat of vaporization of mercury is 2.72 x 10\(^5\) Jkg\(^{-1}\).
(d)A jug of heat capacity 250 Jkg\(^{-1}\) contains water at 28^\circC. An electric heater of resistance 35\(\Omega\) connected to a 220 V source is used to raise the temperature of the water until it boils at 100^\circC in 4 minutes. After. another 5 minutes, 300 g of water has evaporated. Assuming no heat is lost to the surroundings, calculate the:
(i) mass of water in the jug before heating;
(ii) specific latent heat of vaporization of steam. [Specific heat capacity of water = 4200 kg\(^{-1}\)K\(^{-1}\)]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) (i) Radiation:
Surface area, Temperature.
(ii) Convection:
Nature, Density/Viscosity of the fluid, Thermal conductivity of fluid, Specific heat capacity of fluid and Exposed surface area.
(b) Factors that determine the quantity of heat in a body: Heat/thermal capacity and. Temperature
(c) It means that 2.72 x 105J of heat energy is required/ needed to change I kg of mercury at its boiling point to vapour (without temperature change)
(d) (i) heat supplied by heater = ivt = \(\frac{v^2}{R}\)t
heat gained by water = M\(_w\)C\(_w\)(\(\theta_2\) - \(\theta_1\))
where M\(_w\) = mass of water
C\(_w\) = s.h.c of water
\(\theta_2\) = final temp. of water
\(\theta_1\) = initial temp. of water
Heat gained by jug = C\(_j\) (\(\theta_2\) - \(\theta_1\))
\(\frac{v^2t}{R}\) = M\(_w\)C\(_w\) (\(\theta_2\) - \(\theta_1\))
= \(\frac{220^2 \times 4 \times 60}{35}\)
= M\(_w\) x 4200[100 - 28] + 250[100 - 28]
M\(_w\) = 1.038kg
Source: Myschool Past Questions Archive (Record #52871)
Physics
WAEC / WASSCE 2018
Waves and Optics
Solution
(a) Define dffraction.
(b)(i) Explain critical angle. The diagram here illustrates a ray of light passing through a rectangular transparent plastic block \(\alpha\) Determine the value of the critical angle. \(\beta\) Calculate the refractive index of the block.
(c) A pipe closed at one end has fundamental frequency of 200Hz. The frequency of the first overtone of the closed pipe is equal to the frequency of the first overtone of an open pipe. Calculate the:
(i) fundamental frequency of the open pipe;
(ii) length of the closed pipe;
(iii) length of the open pipe. [Speed of sound in air = 330 ms\(^{-1}\)]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Diffraction is the bending or spreading of a wave round an obstacle or as it passes through an opening.
(b) (i) Critical angle is the angle of incidence in the denser medium when an angle of refraction in the lens dense medium is 90^\circ.
(ii) (\(\alpha\))Critical angle = 90^\circ -.44^\circ = 46^\circ
(\(\beta\))n = \(\frac{1}{sin c} = \frac{1}{\sin 46^o}\) = 1.39
(c) (i) Closed Pipe
1st Overtone = 3f\(_o\) = 3 x 200 = 600Hz
Open Pipe
All harmonics are possible overtone = 2f\(_o\)
2f\(_o\) = 600
f\(_o\) = 300Hz
(ii) Closed Pipe
f\(_o\) = \(\frac{v}{4l}\)
l = \(\frac{v}{4f_o}\)
\(\frac{330}{4 \times 200}\)
= 0.412m
Open Pipe
\(f_1\) = \(\frac{v}{l}\)
600 = \(\frac{300}{l}\)
l = 0.55m
or
\(f_1\) = \(\frac{v}{2l}\)
300 = \(\frac{300}{2 \times l}\)
l = \(\frac{330}{600}\)
= 0.55m
Source: Myschool Past Questions Archive (Record #52872)
Physics
WAEC / WASSCE 2018
Mechanics and Properties of Matter
Solution
(a) Define:
(i) reactance;
(ii) impedance in an a.c.
The diagram here illustrates an a.c. generator. When the coil is rotated, an e.m.f is induced in the coil.
(i) Explain why an e.m.f. is induced.
(ii) State the purpose of the slip-rings.
(iii) Name and state the law used to determine the direction of the induced current.
(iv) State two ways to increase the induced e.m.f.
(c) A lamp is rated 12 V. 6 W. Calculate the amount of energy transformed by the lamp in 5 minutes.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) Reactance is the opposition to the flow of a.c offered by a capacitor or an inductor or both.
(ii) Impedance is the overall opposition of a mixed circuit containing a resistor, an inductor and/or a capacitor.
(b) (i) Emf is induced because the coil rotates at a steady speed about a fixed axis in the magnetic field of the elctromagnet, it cuts through the magnetic lines of force.
(ii) The direction of current is reversed in each half of the rotation of the coil.
(iii) The law used to determine the direction of the induced current is Lenz's law. It states that the direction of the induced current is such as to oppose the change producing it.
(iv) The induced e.m.f can be increased by any two of the following: increased speed of rotation, efficient core design, increased number of turns in the coil and increased area of coil.
(c) E = Power x Time
= 6 x 5 x 60
=1800J
Source: Myschool Past Questions Archive (Record #52873)
Physics
WAEC / WASSCE 2018
Mechanics and Properties of Matter
Solution
(a) Define binding energy in an atom.
(b) List three evidence to support the claim that X-rays are electromagnetic waves.
(c) List three peaceful uses of nuclear energy.
(d) Light of wavelength 4.5 x 10\(^{-7}\) in is incident on a metal resulting in the emission of photo electrons. If the work function of the metal is 3.0 x 10\(^{-9}\) J, calculate the:
(i) frequency of the incident light;
(ii) energy of the incident light;
(iii) energy of the photoelectrons. [Speed of light = 3.0 x 10\(^8\) ms\(^{-1}\), h = 6.6 x 10\(^{-34}\) Js]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Binding energy is the minimum energy required to seperate the nucleus of an atom.
(b) Three evidences in support of the claim that x -rays are electromagnetic waves are:
(i) They can be polarized.
(ii) They can travel through a vacuum.
(iii) They travel with the speed of light.
(iv) They are not affected by electric field.
(v) They are not affected by magnetic field.
(c) Three peaceful uses of nuclear energy are:
(i) It is used in the treatment of tumour.
(ii) It is used to power submarines.
(iii) It is used to generate electricity.
(d) (1) c = f\(\lambda\) or f = \(\frac{c}{\lambda}\)
f = \(\frac{3.0 \times 10^8}{4.5 \times 10^{-7}}\)
f = 6.67 x10\(^{14}\)Hz
(ii) E = hf
= 6.67 x10\(^{-34}\) x 6.67 x10\(^{14}\)J
= 4.4 x10\(^{-19}\)J
(iii) K.E = E- W
= 4.4 x 10\(^{-19}\)
=1.4 x 10\(^{-19}\) - 3.0 x 10\(^{-19}\)
=1.4 x 10\(^{-19}\)J
Source: Myschool Past Questions Archive (Record #52874)
Physics
UTME / JAMB 2018
Heat Energy & Temperature
Solution
The critical angle for light traveling from a transparent medium to air s measured as 34º. The refractive index of the medium is
A
0.56
B
1.50
C
1.79
D
2.02
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
N = \(\frac{1}{sin\theta}\)
= \(\frac{1}{sin34}\)
= 1.79
Source: Official Exam / Past Question Archive
Physics
WAEC / WASSCE 2018
Heat Energy & Temperature
Solution
You are provided with an ammeter, a resistor, key, metre bridge, and other necessary apparatus.
i. Connect a circuit as shown in the diagram above.
ii. Close the key and use the jockey to make contact with AB at N such that AN = d = 25cm
iii. Read and record the ammeter reading.
iv. Evaluate 1\(^{-1}\).
v. Repeat the procedures for values of d = 35cmm, 50cm, 65cm and 80cm. In each case, record I and determine 1\(^{-1}\)
vi. Tabulate your results.
vii. Plot a graph with log on the vertical axis and d on the horizontal axis.
viii. Determine the slope, s of the graph.
ix. State two precautions taken to obtain accurate results.
(b)i. Use your graph to determine the value of d = l = 1.5A.
ii. State two factors that affect the resistance of a wire.
A
0.5
B
0.8
C
1.2
D
1.5
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
S/N
d(cm)
I(A)
I\(^{1}\)(A\(^{-1}\))
Log I\(^{-1}\)
1
25
0.65
1.54
0.19
2
35
0.55
1.82
0.26
3
50
0.44
2.50
0.40
4
65
0.30
3.33
0.52
5
80
0.23
4.35
0.64
vii. Scale: 2cm to 10 units on x-axis
2cm to 0.1 unit on y-axis
viii. Slopes(s) = \(\frac{x_{2} - x_{1}}{y_{2} - y_{1}}\)
= \(\frac{80 -25}{0.64 - 0.20}\)
= \(\frac{55}{0.44}\)
= 125
(ix) Precautions
(a) Key opened in between reading or keys opened when readings were not taken
(b) Tight connections ensured
(c) Avoided parallax error when taking reading on ammeter or metre rule
(d) Repeated readings shown
(e) Avoided zero error on ammeter or metre rule
(f) Clean terminals ensured.
(b)i. 1\(^{-1}\) = \(\frac{1}{1.5}\)
= 0.67
Log(1\(^{-1}\)) =-0.174
ii. Factors affecting the resistance of a wire
- Length of wire
- Cross-sectional area/ thickness/ diameter/ radius of wire
- Resistivity or nature of wire
- Temperature.
Source: Official Exam / Past Question Archive
Physics
WAEC / WASSCE 2018
Mechanics and Properties of Matter
Diagram
Solution
You are provided with a pendulum bob, a metre rule, a stopwatch, a retort stand with clamp, and other necessary apparatus.
i. Suspend the pendulum bob from the clamp as illustrated in the diagram.
ii. Adjust the pendulum such that AC=L= 90 cm
iii. Displace the pendulum bob slightly such that it oscillates in a vertical plane.
iv. Measure and record the time t for 20 complete oscillations.
v. Evaluate T and \(\sqrt L\)
vi.Repeat the procedure for four others values of L= 80 cm,70 cm, 60 cm, and 50cm.
vii. Tabulate your readings
viii. Plot a graph with T on the vertical axis and \(\sqrt L\) on the horizontal axis.
ix. Determine the slope, s, of the graph.
x. Evaluate g= \(\frac{4\pi^{2}}{5^{2}}\)
xi. State two precautions taken to ensure accurate results.
(b) i. Determine from your graph, the period of the pendulum for L= 75 cm.
ii. A simple pendulum bob is set into simple harmonic motion. Sketch a diagram of the setup and indicate on it; the positions of:
(a) maximum velocity.
(b) maximum acceleration of the bob
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
S/N
L(cm)
t(s)
T = \(\overline{20}\)(S)
\(\sqrt L\)
1
90
38
1.90
9.49
2
80
36
1.80
8.94
3
70
34
1.70
8.37
4
60
32
1.50
7.73
5
50
30
1.50
7.07
viii. Scale: 2cm rep 0.05 unit on y-axis
2cm rep 1.0 unit on x-axis
ix. Slopes(s) = \(\frac{X_2 - X_1}{Y_2 - Y_1}\)
= (\(\frac{9.9 - 5.7}{2.0 - 1.25}\))
= \(\frac{4.2}{0.75}\)
= 5.6cm
x. Evaluate g = \(\frac{4\pi{^2}}{5^{2}}\)
= \(\frac{4(3.143)^{2}}{(5.6)^{2}}\)
= \(\frac{9.878}{31.36}\)
= 0.315
xi. Precuations:
i. Avoided zero error on stop clock/ metre rule.
ii. Avoided draught.
iii. Avoided conical oscillation
iv. Ensured that support was rigid or firm
v. Ensured bob/pendulum was free from table.
(b)i. \(\sqrt{75cm}\) = 8.66
ii.
Source: Myschool Past Questions Archive (Record #59880)
Physics
WAEC / WASSCE 2017
Electromagnetic Field
Solution
A transformer connected to a 240\(V_{r.m.s}\) has 3000 turns in its primary coil and 500 turns in its secondary coil. Calculate the output voltage.
A
4000V
B
400V
C
40V
D
4V
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
\(N_{p} = 3000 turns\)
\(N_{s} = 500 turns\)
\(E_{p} = 240V_{r.m.s}\)
\(E_{s} = ?\)
\(\frac{N_{s}}{N_{p}} = \frac{E_{s}}{E_{p}}\)
\(\frac{500}{3000} = \frac{E_{s}}{240}\)
\(\implies E_{s} = \frac{240\times500}{3000} = 40V\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #50080)
Physics
WAEC / WASSCE 2017
Electric Field
Diagram
Solution
The diagram above illustrates two charged bodies, X and Y separated by a distance r. Each body has charge q and mass m. Which of the following statements about the electrostatic force \(F_{e}\) and gravitational force \(F_{g}\) acting between the bodies is correct?
Attached Figure / Diagram
A
\(F_{e}\) is attraction while \(F_{g}\) is repulsion
B
Both \(F_{e}\) and \(F_{g}\) are that of repulsion
C
Both \(F_{e}\) and \(F_{g}\) is that of attraction
D
\(F_{e}\) is repulsion while \(F_{g}\) is attraction
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
electrostatic force can be repulsive or attractive and gravitational force is only attractive.
Source: WAEC / JAMB / NECO Classroom Archive (Record #50092)
Physics
WAEC / WASSCE 2017
Measurement
Diagram
Solution
The diagram above is a micrometer screw gauge. Determine its reading.
Attached Figure / Diagram
A
2.44mm
B
2.46mm
C
2.86mm
D
2.96mm
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
- Main scale: 2 mm
- Thimble scale: 46 divisions \times 0.01 mm = 0.46 mm
Total reading = 2 + 0.46 = 2.46 mm
Source: WAEC / JAMB / NECO Classroom Archive (Record #49910)
Physics
UTME / JAMB 2017
Heat Energy & Temperature
Solution
Which is the incorrect formula for a body accelerating uniformly?
A
a = v2 − \(\frac{u^2}{2}\)
B
v2 = u2 + 2as
C
s = ut + \(\frac{1}{2}\) at2
D
v2 − u2= 2as
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Look through the equations of motion
Source: Myschool Past Questions Archive (Record #46372)
Physics
UTME / JAMB 2017
Thermal Physics
Solution
Which of the following is not an electromagnetic radiation?
A
x - ray
B
Radio waves
C
sunlight
D
sound waves
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Electromagnetic waves do not require any material medium, but sound waves is a mechanical wave that require a material medium
Source: Myschool Past Questions Archive (Record #46373)
Physics
UTME / JAMB 2017
Electricity and Magnetism
Solution
Calculate the electric field intensity between two plates of potential difference 6.5V when separated by a distance of 35cm.
A
18.57NC\(^{-1}\)
B
53.06N C\(^{-1}\)
C
2.28NC\(^{-1}\)
D
0.80NC\(^{-1}\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Electric Field Intensity (E) = \(\frac{v}{d}\)
= \(\frac{\text{potential difference}}{\text{distance}}\)
= \(\frac{6.5v}{35cm}\)
= \(\frac{6.5}{35 \times 10^{- 2}}\)
= 18.57NC\(^{-1}\)
Source: Myschool Past Questions Archive (Record #46376)