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Physics
WAEC / WASSCE 2007
Circular Motion, Gravitation & Simple Harmonic Motion
Solution
You are provided with a retort stand, clamp and boss, a pendulum bob, a piece of thread, and other necessary apparatus. Carry out the fo lowing experiment:
Set up the apparatus as shown in the diagram above:
measure and record the distance = 130cm from the centre of the bob to the point of suspension of the pendulum
displace the pendulum through a small angle and release. Allow the pendulum to oscillate freely;
determine the time t for 20 complete oscillations;
also, determine the period T of the oscillations
evaluate T\(^{2}\) and L = -30;
repeat the procedure for four other values of l = 110, 90, 70, and 50 cm
in each case, determine t and evaluate T,T\(^{2}\) and L. Tabulate your readings.
plot of a graph of T\(^{2}\) on the vertical axis against L on the horizontal axis, starting both axes from the origin (0,0)
determine the slope, s, of the graph. Also determine the intercept, C, of the graph on the T\(^{2}\) axis;
evaluate: i. \(k_1\) = \(\frac{4\pi^{2}}{s_{1}}\),
ii. \(k_2\) = \(\frac{c}{8}\) [Take \(\pi\) = \(\frac{22}{7}\)]
state two precautions taken to ensure accurate results.
(b)i. What is meant by the period of oscillation of an oscillating body?
i. Explain the acceleration of free fall due to gravity.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of value/ Observation
S/N
L(cm)
\(t_1\)(sec)
T
T\(^{2}\)
L
1
130
45.70
2.29
5.22
100
2
110
42.00
2.10
4.41
80
3
90
38.00
1.90
3.61
60
4
70
33.10
1.66
2.74
40
5
50
28.70
1.44
2.06
20
Slope (s) = \(\frac{\bigtriangleup {T}^{2}}{\bigtriangleup {L}} = \frac{5.2-2.0}{100-18}\)
S = \(\frac{3.2}{82}\)
= 0.039
Intercept (c) = 1.3
Evaluation
\(K_1\) = \(\frac{4 \pi ^{2}}{s}\) = \(\frac{4{(\frac{22}{7})^{2}}}{0.039}\)
= \(\frac{4 \times 9.878}{0.039} = (\frac{39.51}{0.039})\)g = 1013g
\(K_2\) = \(\frac{c}{s} = \frac{1.3}{0.039}\) = 33.33
Precautions:
l avoided conical oscillations.
l ensured firm suspension of pendulum (
l ensured reading is repeated (shown on the table).
I avoided draught.
I corrected for zero error of metre rule/stop clock
I avoided parallax errors in reading the metre rule/stop clock.
(b)i. The period of oscillation of a body is the time taken for the body to make one complete to and fro movement.
ii. When a body is falling under its weight (alone), the rate of increase of its velocity with time (as a result of the earth's gravitational pull) is the acceleration of free fall due to gravity.
Source: Myschool Past Questions Archive (Record #60039)
Physics
WAEC / WASSCE 2007
Reflection, Refraction & Dispersion of Light
Diagram
Solution
Using the above diagram as a guide, carry out the following experiment:
Place the equilateral triangular glass prism on the drawing paper. Trace the outline ABC of the prism.
remove the prism. Draw a line NO such that it makes an angle i = 25º with the normal at point O on side AB
fix two pins \(R_1\) and \(R_2\) vertically on line N0. Replace the prism on its outline;
place the reflecting surface of the plane mirror in contact with the face AC of the prism;
looking through the face BC of the prism, fix two other pins at \(R_3\) and \(R_4\) such that the pins appear to be in a straight line with the images of the pins at \(R_1\) and \(R_2\);
remove the prism, the mirror, and the pins. Draw a line to join points \(R_4\) and \(R_3\)
produce line \(R_4\)\(R_3\) to meet line NO produced at T;
measure and record the angle \(\theta\) at T and e at D;
repeat the procedure for four other values of i = 30^\circ, 35^\circ, 40^\circ, and 45^\circ. In each case, measure and record the corresponding values of \(\theta\) and e. Tabulate your readings;
plot a graph of e on the vertical axis and \(\theta\) on the horizontal axis;
evaluate k = s\(^{-1}\)
state two precautions taken to ensure accurate results. [Attach your traces to your answer booklet]
(b)i. State four characteristics of the image of an object formed by a plane mirror.
ii. State two Conditions necessary for total internal reflection to occur in a medium.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of values (observation)
S/N
i^\circ
\(\theta\)^\circ
eº
1
25
62
28
2
30
54
32
3
35
45
35
4
40
40
37
5
45
30
41
Slope (s) = \(\frac{\bigtriangleup {e}}{\bigtriangleup {\theta}}\) = \(\frac{49 - 20}{83 - 10} = \frac{29}{73}\) = 0.397
Intercept (c) = 53
Evaluation: \(K_1\) = S\(^{-1}\) = 0.397\(^{-1}\) = 2.519
Precautions;
l ensured neat traces/sharp pencil (shown)
I ensured mirror is firmly attached to prism
I ensured pins are reasonably spaced (about 4 cm)
l avoided parallax error in reading protractor.
(b)i. Characteristics of images formed by a plane mirror
- Same size as object
- Same distance from mirror as object
- Virtual
- Erect
- Laterally inverted.
ii. For total internal reflection of light to occur:
(a)Light must be travelling from a more dense to a less dense medium
(b)Angle of incidence in the denser medium must be greater than the critical angle
Source: Myschool Past Questions Archive (Record #60041)
Physics
WAEC / WASSCE 2007
Current Electricity & Electric Circuits
Solution
You have been provided with an accumulator E, a standard resistor Rx, two resistance boxes R\(B_1\) and R\(B_2\), two keys \(K_1\) and \(K_2\) and other necessary apparatus.
Measure and record the e.m.f of the accumulator.
Connect a circuit as shown above.
Set the resistance R, in the resistance box such that R in R\(B_1\) = R in R\(B_2\) = 1\(\Omega\).
With \(K_1\) open and \(K_2\) closed, measure and record the potential difference V across the standard resistor Rx. (v) Close K, and K,. Read and record the potential difference \(V_o\) across Rx.
Evaluate V\(_{1}^{-1}\)
Repeat procedure (v) for four other values of R = 2, 3, 4 and 5\(\Omega\) respectively. In each case, ensure that the value of R in R\(B_1\) is equal to the value of R in R\(B_2\).
Evaluate V\(_{1}^{-1}\) in each case. Tabulate your readings.
Plot a graph of V\(_{1}^{-1}\) on the vertical axis against R on the horizontal axis starting both axes from the origin (0,0).
Determine the slope, s, of the graph and the intercept on the vertical axis.
Evaluate y = \(\frac{1}{s}\)
State two precautions taken to ensure accurate results
(b)i. Explain what is meant by the potential difference between two points in an electric circuit
ii. A cell has an e.m.f. of 3 V. When it is connected across a resistor of resistance 4\(\Omega\), a current 0.5A passes through the circuit. Calculate the internal resistance of the cell.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of values (Observation)
R (\(\Omega\)
\(V_1\)(v)
V\(_{1}^{-1}\)
1
1.50
0.667
2
1.30
0.769
3
1.15
0.870
4
1.10
0.909
5
0.90
1.111
Slope (s) = \(\frac{\bigtriangleup {V}_{1}^{-1}}{DR} = \frac{1.22 - 0.6}{6.5 - 0.3}\)
= \(\frac{0.62}{6.20}\) = 0.1
Intercept (c) = 0.57
Evaluation: y = \(\frac{1}{S} = \frac{1}{0.1}\) = 10
Precautions:
I removed the key when not taking readings
I ensured tight connections/clean terminals
I avoided parallax error in reading the voltmeter
I corrected for zero error on the voltmeter
(b)i. The potential difference (pd) (in volts) between two points in a circuit is the work done (in joules) when one unit of positive charge (coulomb) moves from one point to the other
ii. E = IR + Ir, 3 =0.5 x 4 + 0.5r, 3 = 2 +0.5r
r = \(\frac{3 - 2}{0.5}\) = 2\(\Omega\)
Source: Myschool Past Questions Archive (Record #60043)
Physics
UTME / JAMB 2007
Atomic, Quantum, Nuclear Physics & Electronics
Solution
A photon of wavelength 6.0 x 10\(^{-7}\) m behaves like a particle of a certain mass. The value of that mass is [h = 6.63 x 10\(^{-34}\) Js, c = 3.0 x 10\(^8\) ms\(^{-1}\)]
A
3.5 x 10\(^{-36}\)kg
B
2.2 x 10\(^{-35}\)kg
C
1.1 x 10\(^{-35}\)kg
D
2.2 x 10\(^{-7}\) kg
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
\(\lambda\) = \(\frac{h}{mv}\), then, h = mv\(\lambda\)
m = \(\frac{h}{v \lambda}\) = \(\frac{6.63 \times 10^{-34}}{6.0 \times 10^{-7} \times 3 \times 10^8}\) = 3.68 x 10\(^{-36}\) kg
OPtion A is the closest.
Source: Myschool Past Questions Archive (Record #72219)
Chemistry
UTME / JAMB 2007
Oxidation, Reduction & Redox Reactions
Solution
Trioxonitrate (V) acid is not used for preparing hydrogen gas from metals because it
A
is strongly oxidizing
B
is volatile
C
renders the metal passive
D
forms a layer of oxide on the metal
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
Nitric acid is not used in the preparation of hydrogen because it is a strong oxidizing agent, meaning it readily oxidizes the hydrogen gas produced during the reaction with a metal, converting it to water and thus preventing the collection of hydrogen gas; instead, acids like dilute hydrochloric acid or sulphuric acid are preferred for hydrogen production as they do not oxidize hydrogen.
Source: Myschool Past Questions Archive (Record #3734)
Chemistry
UTME / JAMB 2007
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
An element used in the production of matches is
A
nitrogen
B
aluminium
C
copper
D
sulphur
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
Nitrogen prevents food from degrading by creating an unreactive atmosphere.
Aluminium is used in buildings because it's resistant to corrosion and thermally efficient.
Copper is a versatile metal used in many applications, including electrical wiring, plumbing, construction, and transportation.
Sulphur is commonly used as an ingredient in the head of a match, contributing to its ability to ignite easily when struck against a striking surface; most modern safety matches contain sulfur or charcoal alongside other chemicals like potassium chlorate and antimony sulphide. Sulphur acts as a fuel in the match head, readily combusting when combined with an oxidizer like potassium chlorate.
Source: Myschool Past Questions Archive (Record #3745)
Chemistry
UTME / JAMB 2007
Organic Chemistry I: Hydrocarbons & Crude Oil
Solution
When calcium oxide is heated with carbon, the oxide will
A
be converted to carbide
B
be converted to the trioxocarbonate (IV) salt
C
decompose to the metal
D
melt, producing carbon (IV) oxide
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
When calcium oxide (CaO) reacts with carbon (C), it produces calcium carbide (CaC\(_2\))and carbon monoxide (CO). This reaction occurs at very high temperatures, around 700^\circC.
CaO + 3C → CaC\(_2\) + CO
Source: Myschool Past Questions Archive (Record #3748)
Chemistry
UTME / JAMB 2007
Non-Metals & Their Compounds
Solution
A solution which on treatment with hydrochloric acid produces a colourless, odourless gas which turns lime water milky is most likely to be a
A
trioxonitrate (V)
B
chloride
C
trioxocarbonate (IV)
D
hydroxide
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
A solution containing carbonate when treated with hydrochloric acid, produces a colourless, odourless gas (carbon dioxide) that turns lime water milky.
For example, when hydrochloric acid (HCl) reacts with sodium carbonate (Na\(_2\)CO\(_3\)), it releases carbon dioxide (CO\(_2\)) gas, which is colourless and odourless.
Source: Myschool Past Questions Archive (Record #3751)
Chemistry
UTME / JAMB 2007
Metals, Transition Elements & Extraction
Solution
The soft solder used for welding and plumbing is alloy of
A
lead and tin
B
iron and tin
C
lead and zinc
D
iron and zinc
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
An alloy is a mixture of chemical elements of which in most cases at least one is a metallic element, although it is also sometimes used for mixtures of elements;
The soft solder used for welding and plumbing is alloy of Tin and Lead. The mixture of tin and lead creates a eutectic alloy, meaning it has a lower melting point than either metal alone, making it easy to work with and join metal parts together without excessively heating the base material, making it ideal for delicate applications like plumbing joints and electrical connections.
Source: Myschool Past Questions Archive (Record #3752)
Chemistry
UTME / JAMB 2007
Metals, Transition Elements & Extraction
Solution
The reddish-brown rust on iron roofing sheets consists of
A
Fe3+(H2O)6
B
FeO.H2O
C
Fe2O3.3H2O
D
Fe3O4.2H2O
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
The chemical formula for rust is Fe\(_2\)O\(_3\).xH\(_2\)O, which represents hydrated iron(III) oxide, where "x" indicates the variable amount of water molecules associated with the iron oxide.
Source: Myschool Past Questions Archive (Record #3755)
Chemistry
UTME / JAMB 2007
Chemical Energetics & Thermochemistry
Solution
Na\(_2\)S\(_2\)O\(_3\)(aq) + 2HCl(aq) → 2NaCl(aq) + H\(_2\)O(l)
Which of the following would introduce the greatest increase in the rate of the chemical reaction above?
A
An increase in temperature and decrease in the concentration of the reactant
B
A decrease in the volume and increase in the pressure of the reactant
C
A decrease in the temperature and an incease in the concentration of the reactant
D
An increase in temperature and an increase in the concentration of the reactant
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
Generally speaking, all neutralization reactions are considered exothermic, meaning they release heat during the reaction; this is because the formation of water molecules from the hydrogen ions of an acid and hydroxide ions of a base is a highly stable bond formation, releasing energy in the process.
In every exothermic reaction, a decrease in temperature will favour the forward reaction,likewise increasing the concentration of the reactant. Option C
Source: Myschool Past Questions Archive (Record #3768)
Chemistry
UTME / JAMB 2007
Organic Chemistry II: Functional Groups, Polymers & Biomolecules
Solution
The tincture of iodine means iodine dissolved in
A
ethanol and water
B
bromine chloride
C
chlorine water
D
water
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
Tincture of iodine is iodine dissolved in ethanol and water. It's also known as iodine tincture or weak iodine solution. Tincture of iodine is a topical antiseptic solution that contains iodine, alcohol, and water. It's used to treat wounds, prevent infections, and disinfect surfaces.
Source: Myschool Past Questions Archive (Record #3770)
Chemistry
UTME / JAMB 2007
Reaction Rates & Chemical Equilibrium
Solution
Cu\(^{2+}\)(aq) + 4NH\(_3\)(g) ↔ [Cu(NH\(_3\))\(_4\)]\(^{2+}\)(aq)
In the reaction above, what is the effect of precipitating Cu\(^{2+}\)(aq) as CuS(s)?
A
NH3(g) concentration will decrease
B
more NH3(g)
C
The equilibrium will shift to the right
D
There will be no effect
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
The precipitation of Cu\(^{2+}\) as CuS decreases the concentration of Cu\(^{2+}\) in the solution. According to Le Chatelier's principle, the equilibrium will shift to the left to counteract this decrease.
Precipitating Cu\(^{2+}\) as CuS shifts the equilibrium to the left, decreasing the concentration of [Cu(NH\(_3\))\(_4\)]\(^{2+}\). Hence more of the reactant will be formed.i.e NH\(_3\)
Source: Myschool Past Questions Archive (Record #3772)
Chemistry
UTME / JAMB 2007
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
Copper and silver metals are commonly used as coin because they are
A
easily reduced
B
not easily reduced
C
easily oxidized
D
not easily oxidized
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
Copper and Silver metals are commonly used as coins because they are not easily oxidized. They are resistant to oxidation, meaning they don't easily react with oxygen in the air to form oxides. This property helps preserve the coin's appearance and structure over time.
Source: Myschool Past Questions Archive (Record #3774)
Chemistry
UTME / JAMB 2007
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
Which of the following sets of metals dissolves in dilute HCl?
A
Fe, Sn and Cu
B
Mg, Fe and Cu
C
Mg, Sn and Fe
D
Cu, Mg and Sn
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
Mg, Sn and Fe all dissolve in dilute HCl because they are more reactive and have a greater tendency to lose electrons to form ions. In contrast, Cu does not readily dissolve in dilute HCl.
Source: Myschool Past Questions Archive (Record #3784)