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Physics
WAEC / WASSCE 2021
Electricity and Magnetism
Diagram
Solution
(a) You are provided with a battery, an ammeter, a voltmeter, a resistance box, a key and connection wires.
(i) Set up circuit as shown in the diagram above.
(ii) With the key opened, measure and record the e.m.f. E\(_o\) of the battery
(iii) With the key closed, select the resistance R=1 on the resistance box. Read and record the current, 1\Omega.
(iv) Evaluate I\(^{-1}\).
(v) Repeat the procedure for five other values of R=2\Omega,3\Omega, 4\Omega, 5\Omega, and 6\Omega.
In each case, record I and evaluate I\(^{-1}\) results.
(vii) Plot a graph with R on the vertical axis and I\(^{-1}\) on the horizontal axis.
(Viii) Determine the slope, s, of the graph.
(ix) Determine the intercept C, On the vertical axis.
(x) State two precautions taken to ensure accurate results.
(b)(i) Define potential difference in an electric field.
(ii) A piece of resistance wire of diameter 0.2 mm and length 25 cm has a resistance of 7\Omega. Calculate the resistivity of the wire of the battery. [ \pi = 22\7]
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
SEE GRAPH ABOVE
(vii) Slope = \(\frac{\Deltad}{\DeltaI^{-1}}\) = \(\frac{d_2 - d_1}{I_2 - I^{-1}\)
→ \(\frac{80 - 30}{3.333 - 15}\} = \(\frac{50}{1.833}\)
Slope = 27.2cm / A
Precaution:
We should avoid parallax error by placing both eyes at the measured level.
We should always switch the fan off to ensure no external environment interferes with the result.
Diameter d = 0.6cm = 0.006m
: r = \(\frac{0.006}{2}\) = 0.003m
Resistivity = 1.0 x 10\(^{-6}\)
R = 4\Omega
R = \(\frac{ρL}{A}|) → L = \(\frac{RA}{ρ}\)
L = \(\frac{ 4 x 22 x 0.0000009}{7 x 1.0 x 10^{-6}}\)
L = 6.289 x 2 x 9 x 10\(^{-6}\) x 10\(^{-6}\)
L = 113.14 x 10\(^{-6 + 6}\)
L = 113.14
Source: Myschool Past Questions Archive (Record #62351)
Physics
UTME / JAMB 2021
Heat Energy & Temperature
Solution
A 2H inductor has negligible resistance and is connected to a \(\frac{50}{\pi}\) Hz A.C supply. The reactance of the inductor is?
A
200\(\Omega\)
B
50\(\Omega\)
C
\(\frac{100}{\pi}\)
D
\(\frac{25}{\pi \Omega}\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Given;
L = 2H
F = \(\frac{50}{\pi}\) \(H_2\)
\(X_L\) = ?
\(X_L\) = 2\(\pi\)fL = 2\(\pi\) x \(\frac{50}{\lambda}\) x 2
= 200\(\Omega\)
Source: Official Exam / Past Question Archive
Physics
WAEC / WASSCE 2021
Heat Energy & Temperature
Solution
(a) You are provided with a battery, an ammeter, a voltmeter, a resistance box, a key and connection wires.
(i) Set up circuit as shown in the diagram above.
(ii) With the key opened, measure and record the e.m.f. E\(_o\) of the battery
(iii) With the key closed, select the resistance R=1 on the resistance box. Read and record the current, 1Ω.
(iv) Evaluate I\(^{-1}\).
(v) Repeat the procedure for five other values of R=2Ω,3Ω, 4Ω, 5Ω, and 6Ω.
In each case, record I and evaluate I\(^{-1}\) results.
(vii) Plot a graph with R on the vertical axis and I\(^{-1}\) on the horizontal axis.
(Viii) Determine the slope, s, of the graph.
(ix) Determine the intercept C, On the vertical axis.
(x) State two precautions taken to ensure accurate results.
(b)(i) Define potential difference in an electric field.
(ii) A piece of resistance wire of diameter 0.2 mm and length 25 cm has a resistance of 7Ω. Calculate the resistivity of the wire of the battery. [ \pi = 22\7]
A
Isaac Newton
B
Ernest Rutherford
C
Louis de Broglie
D
J.J. Thompson
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
SEE GRAPH ABOVE
(vii) Slope = \(\frac{\Deltad}{\DeltaI^{-1}}\) = \(\frac{d_2 - d_1}{I_2 - I^{-1}\)
→ \(\frac{80 - 30}{3.333 - 15}\} = \(\frac{50}{1.833}\)
Slope = 27.2cm / A
Precaution:
We should avoid parallax error by placing both eyes at the measured level.
We should always switch the fan off to ensure no external environment interferes with the result.
Diameter d = 0.6cm = 0.006m
: r = \(\frac{0.006}{2}\) = 0.003m
Resistivity = 1.0 x 10\(^{-6}\)
R = 4Ω
R = \(\frac{ρL}{A}|) → L = \(\frac{RA}{ρ}\)
L = \(\frac{ 4 x 22 x 0.0000009}{7 x 1.0 x 10^{-6}}\)
L = 6.289 x 2 x 9 x 10\(^{-6}\) x 10\(^{-6}\)
L = 113.14 x 10\(^{-6 + 6}\)
L = 113.14
Source: Official Exam / Past Question Archive
Physics
WAEC / WASSCE 2020
Heat Energy & Temperature
Solution
The length of an iron bar is 100cm at 20°C. At what temperature will its length increase by 0.01%? [Linear expansivity of iron = 1.2 x 10\(^{-5}\)°C]
A
48.0°C
B
38.0°C
C
28.3°C
D
23.0°C
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
GIVEN DATA:
where \DeltaL is the increase in length = 0.01
L1 is the initial length = 100
T1 = 20
T2 = ?
\alpha is the linear expansivity of iron = 1.2 x 10\(^{-5}\)^\circC
Formula
\DeltaL = L1\alpha\DeltaT
\DeltaT = \(\frac{\DeltaL}{L1\alpha}\)
\DeltaT is the change in temperature = T2 - T1
T2 = \(\frac{\DeltaL}{L1\alpha}\) + T1
T2 = \(\frac{0.01}{100 \times 1.2 x 10^{-5}}\) + 20
T2 = 8.3 + 20
T2 = 28.3
Source: WAEC / JAMB / NECO Classroom Archive (Record #59417)
Physics
WAEC / WASSCE 2020
Simple Machine
Solution
Which of the following machines does not apply the lever principle?
A
Claw hammer
B
Wheelbarrow
C
Single pulley
D
sugar tong
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
A single pulley is a device composed of a wheel on an axle over which a rope, cord, or similar implement is looped.
Source: WAEC / JAMB / NECO Classroom Archive (Record #59367)
Physics
UTME / JAMB 2020
Heat and Temperature
Solution
Two divers G and H are at depths 20m and 40m, respectively below the water surface in a lake. The pressure on G is p\(_1\) while the pressure on H is P\(_2\), If the atmospheric pressure is equivalent to 10m of water, then the value of P\(_2\)/P\(_1\) is?
A
0.50
B
0.60
C
1.67
D
2.00
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Atmospheric pressure equivalence of 10m of water
: \(\frac{p_2+10}{p_1+10}\)
= \(\frac{40+10}{20+10\)
= \(\frac{50}{30}\) = 1.67
Source: WAEC / JAMB / NECO Classroom Archive (Record #56921)
Physics
WAEC / WASSCE 2020
Measurement of Heat Energy
Solution
Two identical cups containing the same volume of water at 45°C and 5°C, are left in a room at 25°C. Which of the following graphs correctly illustrates the variation of temperature with time?
A
A
B
B
C
C
D
D
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The cup containing water at 45∘C loses heat to the room since temperature of the room is lower than its temperature, hence its temperature reduces gradually, the cup containing water at 5∘C gains heat from the room since its temperature is lower than that of the room. hence, its temperature increases gradually, the lines are curved because as the temperature of the water changes the temperature difference between the water and surroundings also changes
Source: WAEC / JAMB / NECO Classroom Archive (Record #59402)
Physics
WAEC / WASSCE 2020
Electromagnetic Field
Solution
An electrical device has 50 turns in its primary coil and 20 turns in the secondary coil. The device can be a\an?
A
step-up transformer
B
step-down transformer
C
d.c generator
D
a.c generator
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
In a step down transformer, there are more turns in the primary coil than in the secondary coil and more current flows in the secondary circuit than in the primary.
Source: WAEC / JAMB / NECO Classroom Archive (Record #59388)
Physics
WAEC / WASSCE 2020
Current Electricity
Diagram
Solution
Which of the current-voltage characteristics shown above is that exhibited an ohmic conductor?
Attached Figure / Diagram
A
A
B
B
C
C
D
D
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The V-I graph of an ohmic material is a straight line passing through the origin.
Source: WAEC / JAMB / NECO Classroom Archive (Record #59439)
Physics
WAEC / WASSCE 2020
Simple Ac Circuit
Solution
In an electric circuit, an inductor of inductance 0.5 H and resistance 50\(\Omega\) is connected to an alternating current source of frequency 60 Hz. Calculate the impedance of the circuit.
A
50.0\(\Omega\)
B
450.5 \(\Omega\)
C
195.0 \(\Omega\)
D
1950.1 \(\Omega\)
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The impedance of a R-L circuit is given by:
Z = \sqrt(R\(^2\) + XL\(^2\)) ----------- eqn(1)
where Z is the impedance,
R is the resistance, and
XL is the inductive reactance
XL = 2\pifL ------------ eqn(2)
where f is the frequency
From eqn(2),
XL = 2 * \pi * 60 * 0.5
= 60\pi
Also, from eqn(1),
Z = \sqrt(50\(^2\) + (60\pi)\(^2\))
Z = 195.0\Omega
Source: WAEC / JAMB / NECO Classroom Archive (Record #59467)
Physics
WAEC / WASSCE 2020
Refraction of Light
Solution
An image formed on a screen is always?
A
inverted
B
magnified
C
upright
D
virtual
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
A real image is defined as An image that is formed when the light rays from an object converge and meet each other and can be projected onto a screen. Real images are always inverted, Real images can be projected onto pieces of paper or screens. An example of a real image is the image formed on a cinema screen.
A virtual image is defined as An image that is formed when the light rays from an object do not meet but appear to meet behind the lens and cannot be projected onto a screen. Virtual images are always upright, Virtual images cannot be projected onto a piece of paper or a screen. An example of a virtual image is a person's reflection in a mirror.
Source: WAEC / JAMB / NECO Classroom Archive (Record #59365)
Physics
WAEC / WASSCE 2020
Refraction of Light
Solution
A ray of light traveling from glass into ethyl alcohol is incident at the boundary at an angle of incidence 30\(^o\). Calculate the angle of refraction.
[Refractive index of glass =1.5; refractive index of ethyl alcohol = 1.36]
A
27\(^o\)
B
33.5\(^o\)
C
51.7\(^o\)
D
72.8\(^o\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The angle of incidence l = 30º
The angle of refraction , r =?
The refractive index of glass ng = 1.5
The refractive index of ethyl alcohol, ne = 1.36
from snell's law :
\(\frac{n_e}{n_g} = \frac{sini}{sinr}\)
\(\frac{1.36}{1.5} = \frac{sin30º}{sinr}\)
1.36sinr = 1.5 x sin30º
sinr = \(\frac{1.5 \times sin30º}{1.36}\)
sinr = 0.5515
r = \(sin^{-1}\)(0.5515)
r = 33.5º
Source: WAEC / JAMB / NECO Classroom Archive (Record #59411)
Physics
UTME / JAMB 2020
Sound Wave
Solution
If a source of solid is moving, a stationary listener will hear a sound of different frequency. This is called?
A
doppler effect
B
resonance
C
ultra sound
D
rarefaction
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
The Doppler effect or Doppler shift is the change in frequency of a wave in relation to an observer who is moving relative to the wave source.
The drop in pitch of ambulance sirens as they pass by and the shift in red light are common examples of the Doppler Effect.
Source: WAEC / JAMB / NECO Classroom Archive (Record #56901)
Physics
WAEC / WASSCE 2020
Sound Wave
Solution
What factors determine the frequency of a note emitted by a vibrating string?
A
Amplitude of vibration, force constant of string and length of string
B
Amplitude of vibration, force constant of string, and tension in string
C
Mass per unit length of string, tension in string and length of string
D
Force constant of string, tension in string, and length of string.
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The fundamental frequency of vibrations of a string is inversely proportional to the square root of mass per unit length (linear density),
if the tension and vibrating length of the string are constant.
Source: WAEC / JAMB / NECO Classroom Archive (Record #59456)
Physics
WAEC / WASSCE 2020
Application of Lens
Diagram
Solution
The diagram above illustrates a beam of parallel rays from a distant object O, incident on one side of the total reflecting prism. Which diagram does not represent the correct path of the beam when it emerges from the prism?
Attached Figure / Diagram
A
W
B
X
C
Y
D
Z
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
When drawing a ray diagram for instance drawing two rays from the head of the object, the point of intersection of the two rays represents the position of the head of the image, but in the case of parallel rays of light drawn from the object such that the image is likewise formed at infinity, the ray from the head of the object forms the head of the image, similarly, the ray from the leg of the image. from the diagrams drawn, it is only in W that the ray from the head forms the leg of the image, others are correctly drawn
Source: WAEC / JAMB / NECO Classroom Archive (Record #59379)