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Physics
WAEC / WASSCE 2019
Mechanics and Properties of Matter
Solution
An electron of mass 9.1 x 10\(^{-31}\) kg moves with a speed of 2.0 x 10\(^6\) ms\(^{-1}\) round the nucleus of an atom in a Circular path of radius
6.1 x 10\(^{11}\) m. Calculate the centripetal force acting on the electron.
A
7.7 x 10\(^{47}\) N
B
6.0 x 10\(^{-8}\)N
C
3.0 x 10\(^{-14}\)N
D
1.3 x 10\(^{-26}\)N
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Centripetal force (Fr) = \(\frac{MV^2}{r}\)
i.e. Fr = \(\frac{Mv^2}{r}\)
= \(\frac{9.1 \times 10^{-31} \times 2.0 \times 10^6 \times 2.0 \times 10^6}{6.1 \times 10^{-11}}\) N
= (\(\frac{36.4 \times 10^{-19}}{6.1 \times 10^{-11}}\)) N
= 5.967 x 10\(^{-8}\) N
~ 6.0 x 10\(^{-8}\) N
Source: Myschool Past Questions Archive (Record #58977)
Physics
WAEC / WASSCE 2019
Heat Energy & Temperature
Solution
Which of the following liquids has the highest surface tension?
A
Soapy water
B
Cold water
C
Hot water
D
oily water
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Cold water has a higher surface tension than soapy water, hot water, and oily water. Soapy water has surfactants that reduce surface tension, while hot water has lower surface tension than cold water due to increased molecular motion. Oily water also has a lower surface tension than pure water.
Source: Myschool Past Questions Archive (Record #59026)
Physics
WAEC / WASSCE 2019
Heat Energy & Temperature
Solution
The nucleon and proton numbers of a neutral atom of an element are 238 and 92 respectively. Determine the number of neutrons in the atom.
A
119
B
330
C
165
D
146
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Neutron No (N) = 238 - 92
= 146
Source: Myschool Past Questions Archive (Record #59027)
Physics
WAEC / WASSCE 2019
Mechanics and Properties of Matter
The area under a velocity-time graph represents
A
speed
B
acceleration
C
moment
D
distance
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #59030)
Physics
WAEC / WASSCE 2019
Waves and Optics
Solution
When ultraviolet light is incident on certain metallic p articles are emitted. These particles are called
A
positrons
B
protons
C
photoelectrons
D
photons
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Photoelectrons are electrons that are ejected from a material (usually a metal) when it absorbs photons of light, such as ultraviolet light. This phenomenon is known as the photoelectric effect.
Source: Myschool Past Questions Archive (Record #59035)
Physics
WAEC / WASSCE 2019
Thermal Physics
Diagram
Solution
The graph above illustrates the variation of temperature \(\theta\) with time (t) for a solid that is being heated. Which processes take place at segments P and Q respectively?
Attached Figure / Diagram
A
Freezing and vaporization
B
Evaporation and solidification
C
Melting and boiling
D
Condensation and evaporation
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Since the substance is solid, it will need heat to melt to liquid and once the liquid starts boiling temperature will remain the same.
Source: Myschool Past Questions Archive (Record #59043)
Physics
WAEC / WASSCE 2019
Mechanics and Properties of Matter
Solution
(a) (i) Define atomic spectra.
(ii) Differentiate between emission spectra and absorption spectra.
(b)
The diagram above illustrates an electron transition from energy level n = 3 to n = 1. Calculate the:
(i) energy of the photon
(ii) frequency of the photon
(ii) wavelength of the photon [h = 6.6 x 10\(^{-34}\)J s, c = 3.0 x 10\(^8\) ms\(^{-1}\); 1 ev = 1.6 x 10\(^{-19}\) J]
c)(i)Differentiate between soft x-rays and hard x-rays
(ii) Draw the circuit symbol for a p-n junction diode.
(iiii) Give the reason for doping a semiconductor material
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) The emission of light from a substance when its excited electrons move to a lower energy level.
OR
Distinct/separate bright lines of light of definite frequency/wavelength (emitted from vapourised atoms)
(ii) Emission spectra is the pattern of light displayed when electrons jump from a higher energy level to a lower energy level, while absorption spectra is the displayed pattern of wavelengths absorbed by the gas.
(b) \(\Delta\)E = E\(_i\) - E\(_f\)
= -1.51 + 13.6
= 12.09 eV
= 12.09 x 1.6 x 10\(^{-19}\) J
= 1.934 x 10\(^{-18}\) J
(ii) \(\Delta\)E- hf
f = \(\frac{\Delta E}{h}\) = \(\frac{1.934 \times 10^{-18}}{6.6 \times 10^{-34}}\)
f = 2.93 x 10\(^{15}\)Hz
(ii) c = f\(\lambda\)
\(\lambda\) = \(\frac{3 \times 10^8}{2.93 x 10^{13}}\)
= 1.023 x 10\(^{-7}\)m
(c) (I) Soft x-rays are those with longer wavelengths/lowNer penetrating/less energy while hard x-rays are those with shorter wavelengths/higher penetrating power more energy.
(iii) To increase the electrical conductivity/reduce electrical resistivity of the semiconductor.
Source: Myschool Past Questions Archive (Record #59307)
Physics
WAEC / WASSCE 2019
Mechanics and Properties of Matter
Solution
Two bodies of masses 3.0 kg and 2.0 kg are separated by a distance of 50 cm. Calculate the force of attraction between them. [G = 6.67 x 10\(^{-11}\) Nm\(^2\) kg\(^2\)]
A
1.6 x 10\(^{-9}\)N
B
1.3 x 10\(^{3}\)N
C
2.3 x 10\(^{3}\)N
D
5.0x 10\(^{3}\)N
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Using the formula
F = \(\frac{Gm_1m_2}{r^2}\)
= \(\frac{6.67 \times 10^{-11} \times 3.0 \times 2.0}{(50)^2}\)
= \(\frac{40.02 \times 10^{-11}}{2500}\)
= 1.6 x 10\(^{-9}\)N
Source: Myschool Past Questions Archive (Record #59019)
Physics
WAEC / WASSCE 2019
Mechanics and Properties of Matter
Diagram
Solution
The diagram above illustrates an object moving in a circular path at a constant speed. Which of the arrows indicates the direction of linear velocity?
Attached Figure / Diagram
A
z
B
X
C
Y
D
W
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
for an object moving along a circular path, it speed is constant but its linear velocity changes with time at every point on the circle.
Source: Myschool Past Questions Archive (Record #59022)
Physics
WAEC / WASSCE 2019
Waves and Optics
Diagram
Solution
In the diagram above, the time taken to trace a wave between P and Q is?
Attached Figure / Diagram
A
2 periods
B
1 period
C
1\(\frac{1}{2}\) periods
D
1\(\frac{1}{4}\) periods
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
frequency means no of oscillation per secs, so the time taken to trace a wave between p and q is 2secs.
Source: Myschool Past Questions Archive (Record #59024)
Physics
WAEC / WASSCE 2019
Electricity and Magnetism
Which of the following materials does not serve as a safety device in electrical circuits?
A
Connecting wires
B
Earth wire
C
Fuse
D
Switch
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #59025)
Physics
WAEC / WASSCE 2019
Mechanics and Properties of Matter
Diagram
Solution
You are provided with a stopwatch, a meter rule, a split cork, retort stand and clamp, a pendulum bob, a piece of thread, and other necessary apparatus.
i. Place the retort stand on a laboratory stool. Clamp the split cork.
ii. Suspend the pendulum bob from the split cork such that the point of support P of the bob is at height H = 100cm above the floor Q. The bob should not touch the floor and H should be kept constant throughout the experiment.
iii. Adjust the length of the thread such that the center A of the bob is at a height y= AQ= 20cm from the floor.
iv. Displace the bob such that it oscillates in a horizontal plane.
v. Take the time t for 20 complete oscillations.
vi. Determine the period T of oscillation and evaluate T
vii. Repeat the procedure for four other values of y = 30cm, 40cm, 50cm, and 60cm. In each case, determine T and T.
viii. Tabulate the results.
ix. Plot a graph of T on the vertical axis and y on the horizontal axis, starting both axes from the origin (0,0).
x. Determine the slope, s, of the graph and the intercept c on the vertical axis.
xi. If in this experiment SR= c, calculate R.
x. State two precautions taken to ensure accurate results.
(b) i. The bob of a simple pendulum is displaced a small distance from the equilibrium position and then released to perform simple harmonic motion Identify where its:
(\(\propto\)) kinetic energy is maximum
(\(\beta\)) acceleration is maximum
ii. An object of weight 120N vibrates with a period of 4.0s when hung from a spring. Calculate the force per unit length of the spring. [g= 10ms\(^{-2}\), \(\pi\)=3.142]
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
H = 100cm
Table of value
SN
y(cm)
t(sec)
T
T\(^{2}\)
1
2
3
4
5
20
30
40
50
60
50.50
47.50
43.50
40.00
35.80
2.525
2.375
2.175
2.000
1.790
6.376
5.641
4.731
4.000
3.204
Slope (s) = \(\frac{\bigtriangleup T^{2}}{\bigtriangleup \text{y}}\) = \(\frac{604 - 2.0}{76 - 20}\)
= \(\frac{4.4}{56}\) = 0.079 = 0.08
Intercept on vertical axis = 8
Evaluate R = \(\frac{s}{c}\) = \(\frac{8.00}{0.08}\) = 100
Scale: Let 2cm represent 1 unit on the vertical axis and 1.5cm represent 10 units on the horizontal axis.
Precautions
- Avoided parallax error in reading stopwatch/clock/ meter rule.
- Noted/corrected/avoided zero error on stopwatch/ clock/meter rule.
- Avoided draught/switch off fans.
- Avoided conical oscillation Ensured that support was rigid/firm.
- Ensured bob of the pendulum was free from table/did not touch table
- Repeated reading shown on the table.
- Small angular displacement
(b) (\(\propto\)) The kinetic energy is maximum at the equilibrium position
(\(\beta\)) The acceleration is maximum at the point of maximum displacement.
ii. The period T is given by
T = 2\(\pi\)\(\sqrt\frac{M}{K}\)
T\(^{2}\) = 4\(\pi^2\)\(\frac{M}{K}\)
K = 4\(\pi^2\)\(\frac{M}{T^{2}}\)
= \(\frac{4x(3.14)^2(\frac{120}{10})}{4}^2\)
29.62NM\(^{-1}\)
OR
T = 2\(\pi\)\(\sqrt\frac{e}{g}\)
T\(^{2}\) = 4\(\pi^2\)\(\frac{e}{g}\)
e = \(\frac{T^{2}}g{4\pi^2}\) = 4.05
k = \(\frac{f}{e}\) = \(\frac{120}{4.05}\)
= 29.62NM\(^{-1}\)
Source: Myschool Past Questions Archive (Record #59765)
Physics
WAEC / WASSCE 2019
Waves and Optics
Diagram
Solution
You are provided with a converging lens and holder, a screen, a ray box containing an illuminated object pin, and a meter rule.(See illustration above)
i. Place the lens in its holder such that it is facing a distant object seen through a well-lit laboratory window. Move the screen to and fro until a sharp image of the distant object is formed on it. Measure the distance, \(f_0\), between the screen and the lens.
ii. Clamp the meter rule securely to the table. Place the illuminated object pin at the end R of the meter rule.
iii. Place the lens at a position P such that X = RP = 20cm.
iv. Move the screen to a position Q to receive a sharp image of the object. Measure the distance Y = PQ.
v. Evaluate Z = (X+Y)
vi. Repeat the procedure for five other values of x = 25cm. 3Ocm, 35cm, 40cm and 45cm. In each case, record X,Y and evaluate Z.
vii. Tabulate the results.
viii. Plot a graph with Z on the vertical axis and X on the horizontal axis. Draw a smooth curve through the points.
ix. Determine from your graph the minimum value of Z=\(Z_0\) and its corresponding distance
x. Evaluate W = ½ (\(\frac{Z_0}{4} + \frac{X_0}{2}\))
xi. State two precautions taken to ensure accurate results.
(b) i. Draw a ray diagram to show how a Convex lens forms an image of magnification less than one.
ii. Name two pairs of features in the human eye and a lens camera that performs similar functions.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
F = 13.0cm
Table of value
S/N
Xcm
Ycm
Z(x+y)cm
1
2
3
4
5
6
20
25
30
35
40
45
60
38
30
26
24
21
80
63
60
61
64
61
See graph above
Let 1cm represent5 units on the vertical axis and 1cm represent 5 units on the horizontal axis.
Minimum value of Z=\(Z_0\) from the graph = 60cm
Corresponding distance of \(X_0\) = 30cm
Evaluate W = ½ (\(\frac{Z_0}{4} + \frac{X_0}{2}\))
= ½(\(\frac{60}{40} + \frac{30}{2}\)
= ½(\(\frac{60 + 60}{4}\) = ½\(\frac{120}{4}\)
= ½(30) = 15
Precautions
- Coaxial arrangement of optional instruments/ray box, len and screen on a straight line.
- Avoided parallax error in reading metre rule.
- Lens kept uptight
- Repeated reading shown on table
- Surface of lens cleaned
- Noted/corrected/avoided zero error on metre rule.
(b)i.
See the diagram above(bi)
ii. Features performing the same function
Human eye
Lens camera
Retina
Film/screen
Eye lens
Camera lens
Iris
Diaphragm
Pupil
Aperture
Source: Myschool Past Questions Archive (Record #59766)
Physics
WAEC / WASSCE 2019
Electricity and Magnetism
Solution
You are provided with a battery of e.m.f. E, a key K, a voltmeter, a standard resistor \(R_0\) = 2\(\Omega\), a resistance box R, and some connecting wires.
i. Measure and record the e.m.f. E of the battery.
i. Set up a circuit as shown in the diagram above with the key open.
iii. Set the resistance on the resistance box to R 22.
iv. Close the key, read and record the potential difference V on the voltmeter.
v. Evaluate V\(^{-1}\)
vi. Repeat the procedures for five other values of R = 5\(\Omega\) 10\(\Omega\),12\(\Omega\),15\(\Omega\) and 20\(\Omega\). In each case, record V and evaluate V\(^{-1}\)
vii. Tabulate the results.
viii. Plot a graph with R on the vertical axis and V\(^{-1}\) l on the horizontal axis, starting both axes from the origin (0,0).
ix. Determine the slope, s, of the graph and the intercept c on the vertical axis.
x. Calculate \(\propto\) and \(\beta\) from the equations s = \(R_0\) \(\propto\) and c= - (\(R_0\)+B).
xi. State two precautions taken to obtain accurate results.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
\(R_0\) = 2\(\Omega\), emf of the battery = 2.0V
Table of value
S/N
R\(\Omega\)
V
V
1
2
3
4
5
6
2
5
10
12
15
20
4.70
2.40
1.75
1.40
1.10
0.76
0.214
0.417
0.571
0.714
0.909
1.320
Slope = \(\frac{\bigtriangleup R}{\bigtriangleup V^{-1}}\) = \(\frac{30.00-0.00}{1.50-0.15}\)
= \(\frac{30.00}{1.35}\) = 22
Intercept on the horizontal axis = 0.15
Calculation of \(\propto\)
\(\propto\) = \(\frac{S}{R_0} = \frac{22}{2}\) = 11
Calculation of \(\beta\)
\(\beta\) = -(\(R_0\) + C) = - (2 + 0.15) = -(2.15) = -2.15
Precuations
- Key opened in between reading/key opened when readings were not taken.
- Tight connections ensured
- Avoided parallax error when taking reading on voltmetre
- Repeated readings shown on table
- Noted/corrected zero error on voltmetre
- Clean terminals ensured.
(b) i. R\(_{AB}\) = \(\frac{4 \times 5}{4+5} = \frac{20}{9}\)
P\(_{AB}\) =\(\frac{V_{AB^2}}{R_{AB}}\)
= \(\frac{4^2 \times9}{20}\) = 7.2 W
ii. I = \(\frac{p}{v}\)=\(\frac{3.6\times10^3}{240}\)
= 15A
The breaker will not open since the current drawn (15A) is less than (20A).
Source: Myschool Past Questions Archive (Record #59786)
Physics
WAEC / WASSCE 2019
Heat Energy & Temperature
Solution
You are provided with a battery of e.m.f. E, a key K, a voltmeter, a standard resistor \(R_0\) = 2\(\Omega\), a resistance box R, and some connecting wires.
i. Measure and record the e.m.f. E of the battery.
i. Set up a circuit as shown in the diagram above with the key open.
iii. Set the resistance on the resistance box to R 22.
iv. Close the key, read and record the potential difference V on the voltmeter.
v. Evaluate V\(^{-1}\)
vi. Repeat the procedures for five other values of R = 5\(\Omega\) 10\(\Omega\),12\(\Omega\),15\(\Omega\) and 20\(\Omega\). In each case, record V and evaluate V\(^{-1}\)
vii. Tabulate the results.
viii. Plot a graph with R on the vertical axis and V\(^{-1}\) l on the horizontal axis, starting both axes from the origin (0,0).
ix. Determine the slope, s, of the graph and the intercept c on the vertical axis.
x. Calculate \(\propto\) and \(\beta\) from the equations s = \(R_0\) \(\propto\) and c= - (\(R_0\)+B).
xi. State two precautions taken to obtain accurate results.
A
Connecting wires
B
Earth wire
C
Fuse
D
Switch
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
\(R_0\) = 2\(\Omega\), emf of the battery = 2.0V
Table of value
S/N
R\(\Omega\)
V
V
1
2
3
4
5
6
2
5
10
12
15
20
4.70
2.40
1.75
1.40
1.10
0.76
0.214
0.417
0.571
0.714
0.909
1.320
Slope = \(\frac{\bigtriangleup R}{\bigtriangleup V^{-1}}\) = \(\frac{30.00-0.00}{1.50-0.15}\)
= \(\frac{30.00}{1.35}\) = 22
Intercept on the horizontal axis = 0.15
Calculation of \(\propto\)
\(\propto\) = \(\frac{S}{R_0} = \frac{22}{2}\) = 11
Calculation of \(\beta\)
\(\beta\) = -(\(R_0\) + C) = - (2 + 0.15) = -(2.15) = -2.15
Precuations
- Key opened in between reading/key opened when readings were not taken.
- Tight connections ensured
- Avoided parallax error when taking reading on voltmetre
- Repeated readings shown on table
- Noted/corrected zero error on voltmetre
- Clean terminals ensured.
(b) i. R\(_{AB}\) = \(\frac{4 \times 5}{4+5} = \frac{20}{9}\)
P\(_{AB}\) =\(\frac{V_{AB^2}}{R_{AB}}\)
= \(\frac{4^2 \times9}{20}\) = 7.2 W
ii. I = \(\frac{p}{v}\)=\(\frac{3.6\times10^3}{240}\)
= 15A
The breaker will not open since the current drawn (15A) is less than (20A).
Source: Official Exam / Past Question Archive