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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
UTME / JAMB 2019
Heat Energy & Temperature
Solution
Which of the following is consistent with Charles' law?
A
I
B
II
C
III
D
IV
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
This is the correct graph. The graph is volume against temperature where temperature is in Celsius.
Source: Official Exam / Past Question Archive
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
Physics
UTME / JAMB 2018
Heat Energy & Temperature
Diagram
Solution
The efficiency of energy conversion on the energy flow through a hydroelectric power is?
Attached Figure / Diagram
A
\(\frac{E1}{E2}\)
B
\(\frac{E3}{E2}\)
C
\(\frac{E2}{E1}\)+ E2
D
\(\frac{E2}{E3}\) + E1
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Efficiency = \(\frac{\text{useful energy output from machine}}{\text{energy input into machine}}\)
= \(\frac{E3}{E2}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #47742)
Physics
WAEC / WASSCE 2018
Motion Force Circular Motion
Solution
A body of mass m moving around a circle of radius r with a uniform speed v experiences a centripetal force F. The work done by the centripetal force on it is
A
Fv
B
zero
C
Fmr
D
\(\frac{Fr}{m}\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Work is done whenever the given conditions are satisfied. (ii) there is a displacement of the body caused by the applied force along the direction of the applied force. No displacement takes place in this case. So, work done by a centripetal force is always zero.
The work done by centripetal force in the case of a uniform circular motion is 0 joules.
Source: WAEC / JAMB / NECO Classroom Archive (Record #49653)
Physics
UTME / JAMB 2018
Simple Machine
Diagram
Solution
The diagram shown represents a block-and-tackle pulley system on which an effort of W Newtons supports a load of 120.0N. If the efficiency of the machine is 40, then the value of W is?
Attached Figure / Diagram
A
28.0
B
48.0N
C
233.0N
D
50.0N
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Given Data:
Efficiency = 40% Load = 120N, V.R = 6 (Number of Pulleys)
Efficiency = \(\frac{M.A}{V.R} * 100\)
40 = \(\frac{M.A}{6} * 100\)
6 * 40 = 100 M.A
M.A = \(\frac{240}{100}\) → 2.4
Where M.A = \(\frac{Load}{Effort}\)
2.4 = \(\frac{120}{Effort}\)
: Effort = \(\frac{120}{2.4}\) → 50N
Source: WAEC / JAMB / NECO Classroom Archive (Record #47386)
Physics
WAEC / WASSCE 2018
Projectile Motion
Solution
A missile is projected so as to attain its maximum range. Calculate the maximum height attained if the initial velocity of projection is 200 ms\(^{-1}\). [g = 10ms\(^{-2}\)]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
H = \(\frac{U^2sin^2 \theta}{2g}\)
H = \(\frac{200^2(sin 45^o)^2}{20}\)
= 999.8m or 1000m
Source: WAEC / JAMB / NECO Classroom Archive (Record #52867)
Physics
WAEC / WASSCE 2018
Deformation of Solids
Solution
(a) Define strain.
(b) A rubber band is stretched to twice its original length. Calculate the strain on the rubber band.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Strain is the ratio of the extension to original length of an object.
(b) Strain = \(\frac{e}{l_o} = \frac{2l_o - l_o}{l_o}\) = 1
Source: WAEC / JAMB / NECO Classroom Archive (Record #52863)
Physics
UTME / JAMB 2018
Scalars and Vectors
Solution
A man walks 1km due east and then 1 km due north His displacement?
A
√2km N 45°E
B
1km N 30°E
C
1km N 15°E
D
√2km N 60°E
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
His movements 1km due east and then 1 km due north make a Right-Angled Triangle.
Using Pythagoras' theorem;
The Resultant becomes the Hypotenuse [h]
Hyp\(^2\) = Adj\(^2\) + \Opp(^2\)
Hyp\(^2\) = 1\(^2\) + 1\(^2\)
Hyp\(^2\) = 2 → Hyp = \sqrt2
: Resultant = \sqrt2
DIRECTION TAN Ø = \(\frac{OPP}{ADJ}\)
TAN Ø = \(\frac{1}{1}\)
Ø = TAN\(^{-1}\) [1]
Ø= 45^\circ
Displacement is \sqrt2km N 45^\circE
Source: WAEC / JAMB / NECO Classroom Archive (Record #47375)
Physics
UTME / JAMB 2018
Heat and Temperature
Solution
Which of the following best describes the energy changes which take place when a steam engine drives a generator which lights a lamp?
A
heat ----> Light----> Sound ----> Kinetic
B
Kinetic ----> Light ----> Heat ----> Electricity
C
heat ----> Kinetic ----> Electricity ----> Heat and Light
D
electricity ----> Kinetic ----> Heat ----> Light
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The steam engine uses the Heat supplied from steam to supply kinetic energy into the generator which converts it to electricity to light the lamp, the lamp gives out light and heat as it works
Source: WAEC / JAMB / NECO Classroom Archive (Record #47383)
Physics
UTME / JAMB 2018
Current Electricity
Solution
The electrochemical equivalent of silver is 0.0012g/C. If 36.0g of silver is to be deposited by electrolysis on a surface by passing a steady current for 5mins, the current must be?
A
6000A
B
100A
C
10A
D
1A
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Given Data: Mass of Ag = 36.0g,
Z which is electrochemical equivalent of Ag = 0.0012coulomb
t = 5mins → 5 * 60 = 300secs.
Using M =Z * I * t
make I the subject of the formula
I = \(\frac{M}{Z * t}\) → \(\frac{36}{0.0012 * 300}\)
Current (I) = 100A
Source: WAEC / JAMB / NECO Classroom Archive (Record #47389)
Physics
UTME / JAMB 2018
Current Electricity
Solution
The resistance of a 40W car head lamp, drawing current from a 12V battery is ____.
A
0.3Ω
B
3.3Ω
C
3.6Ω
D
4.8Ω
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Power delivered to head lamp = \(\frac{V^2}{R}\) = 40W
R = \(\frac{V^2}{40}\)
= \(\frac{(12 \times 12)}{40}\) = 3.6\Omega
Source: WAEC / JAMB / NECO Classroom Archive (Record #47699)
Physics
WAEC / WASSCE 2018
Current Electricity
Diagram
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.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
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: WAEC / JAMB / NECO Classroom Archive (Record #59891)