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Physics
WAEC / WASSCE 2019
Current Electricity & Electric Circuits
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
Units, Measurements & Dimensional Analysis
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
Lenses & Optical Instruments
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
Current Electricity & Electric Circuits
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
Gas Laws & Heat Transfer
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
Current Electricity & Electric Circuits
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
Chemistry
UTME / JAMB 2019
Organic Chemistry II: Functional Groups, Polymers & Biomolecules
Solution
A secondary alkanol can be oxidized to give an?
A
alkanone
B
alkyne
C
alkanoic acid
D
alkanal
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
secondary alkanols (alcohols) can be oxidised to alkanones (ketones)
Source: Myschool Past Questions Archive (Record #55517)
Chemistry
UTME / JAMB 2019
Organic Chemistry I: Hydrocarbons & Crude Oil
Solution
Which of the following will give a precipitate with an aqueous solution of copper (I) chloride?
A
But-2-ene
B
But-1-ene
C
But-2-yne
D
But-1-yne
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
A hydrocarbon that will produce a red precipitate when reacted with an aqueous solution of copper(I) chloride is an alkyne. More specifically, an alkyne with a terminal triple bond, like ethyne (acetylene), will react to form a red precipitate of copper(I) acetylide. The most reactive alkynes are those where the triple bond is at the end of the carbon chain (terminal triple bond).
Source: Myschool Past Questions Archive (Record #55519)
Chemistry
UTME / JAMB 2019
Organic Chemistry II: Functional Groups, Polymers & Biomolecules
Solution
The IUPAC name for CH\(_3\)CH\(_2\)COOCH\(_2\)CH\(_3\) is
A
methylpropanoate
B
ethylethanoate
C
ethylpropanoate
D
methylethanoate
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
While naming the IUPAC nomenclature of hydrocarbon compounds, we usually start with the naming from the rare i.e the functional group from the right hand side.
Consider CH\(_3\)CH\(_2\)COOCH\(_2\)CH\(_3\), this is an ester obtained from two compounds propanoic acid and ethanol.
On splitting CH\(_3\)CH\(_2\)COOCH\(_2\)CH\(_3\), we have CH\(_3\)CH\(_2\)COO / CH\(_2\)CH\(_3\) dividing the compound into two; CH\(_3\)CH\(_2\)COO - Propanoate and CH\(_2\)CH\(_3\) - ethyl
So while naming as mentioned earlier on, we start the naming from the rare(RHS), and then the other(LHS). Thus, the name of the compound is; ethylpropanoate.
Source: Myschool Past Questions Archive (Record #55520)
Chemistry
UTME / JAMB 2019
Organic Chemistry I: Hydrocarbons & Crude Oil
Solution
Hydrocarbons which will react with Tollen's reagent conform to the general formula
A
C\(_n\)H\(_{2n}\)
B
C\(_n\)H\(_{2n + 2}\)
C
C\(_n\)H\(_{2n - 2}\)
D
C\(_n\)H\(_{2n + 1}\)OH
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
Only hydrocarbons containing an aldehyde group (RCHO) or a terminal alkyne (R-C≡C-H) will react with Tollens' reagent, leading to a silver mirror formation or a white precipitate, respectively. Aldehydes are oxidized by Tollens' reagent (ammoniacal silver nitrate solution) to form carboxylic acids, while the silver ions (Ag+) are reduced to metallic silver (Ag), which deposits as a silver mirror on the walls of the test tube.
Terminal alkynes react with Tollens' reagent, producing a white precipitate of silver acetylide.
Source: Myschool Past Questions Archive (Record #55523)
Chemistry
UTME / JAMB 2019
Organic Chemistry I: Hydrocarbons & Crude Oil
Solution
The IUPAC name of the compound CF\(_3\)CHBrCl is
A
2-chloro-2-bromo-1,1,1-trifluoroethane
B
2-bromo-2-chloro-1,1,1-trifluoroethane
C
1-bromo-1-chloro-2,2,2-trifluoroethane
D
1,1,1-trifluoro-2-bromo-2-chloroethane
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
The IUPAC name of the compound CF\(_3\)CHBrCl is 2-bromo-2-chloro-1,1,1-trifluoroethane.
This compound is famously known by its non-proprietary name, Halothane, and was widely used as an inhalation anesthetic.
Breakdown of the Name
To arrive at the correct IUPAC name, we follow standard nomenclature rules:
Identify the Parent Chain: The longest carbon chain contains two carbons, making the parent alkane ethane.
Numbering the Chain: We number the carbons to give the substituents the lowest possible locants.
If we start from the carbon with the three fluorine atoms, the locants are (1,1,1,2,2).
If we start from the carbon with bromine and chlorine, the locants are (1,1,2,2,2).
In this case, both directions yield the same set of locants. However, IUPAC rules state that if numbering is equivalent from both ends, we give the lower number to the substituent that comes first alphabetically.
Since Bromo and Chloro come before Fluoro, the carbon attached to Br and Cl is designated as Carbon-2 (wait, let me re-check that logic).
Actually, per the lowest locant set rule, we prioritize the side with more substituents to keep the numbers lower. The carbon with three fluorines gets "1" so the set is (1,1,1,2,2) rather than (1,2,2,2,2).
Correction: In the specific case of Halothane, the standard IUPAC assignment labels the carbon with the three fluorines as C1, resulting in 2-bromo-2-chloro-1,1,1-trifluoroethane
Source: Myschool Past Questions Archive (Record #55524)
Chemistry
UTME / JAMB 2019
Organic Chemistry II: Functional Groups, Polymers & Biomolecules
A synthetic rubber is not obtained from the polymerization of?
A
neoprene
B
isoprene
C
butandiene
D
hexane
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #55525)
Chemistry
UTME / JAMB 2019
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
Burning magnesium ribbon in air removes which of the following
(i) oxygen (ii) nitrogen (iii) argon and (iv) carbon(iv)oxide?
A
i and iv only
B
i, ii, iii and iv only
C
i only
D
i and ii only
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
The chemical reaction between magnesium ribbon and oxygen is a combustion reaction that produces magnesium oxide (MgO). Burning Magnesium ribbon in air removes oxygen, because as Magnesium burns in air, it reacts with oxygen in the air to form Magnesium oxide.
The chemical equation for this reaction is 2Mg(s)+O\(_2\)(g) → 2MgO(s)
Source: Myschool Past Questions Archive (Record #55526)
Chemistry
UTME / JAMB 2019
Non-Metals & Their Compounds
Solution
Which of the following is the best starting material for the preparation of oxygen? Heating of trioxonitrate (v) with
A
Pb
B
Zn
C
Ca
D
Na
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
The best starting material for the preparation of oxygen through heating trioxonitrate(V) (nitrates) would involve a metal that facilitates the decomposition of the nitrate to produce oxygen, coincidentally, Pb (Lead) falls into that category.
Lead(II) nitrate (Pb(NO\(_3\))\(_2\)) decomposes upon heating to produce oxygen, nitrogen dioxide, and lead(II) oxide. This reaction is often used in laboratory settings to generate oxygen.
2Pb(NO\(_3\))\(_2\) → 2PbO + 4NO\(_2\) + O\(_2\)
This makes lead (Pb) a suitable choice for generating oxygen from nitrates.
Source: Myschool Past Questions Archive (Record #55528)
Chemistry
UTME / JAMB 2019
Stoichiometry, Mole Concept & Volumetric Analysis
Which of the following metals is the most essential in the regulation of blood volume, blood pressure and osmotic equilibrium?
A
Zinc
B
Manganese
C
Sodium
D
Iron
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #55529)