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Physics
WAEC / WASSCE 1999
Current Electricity & Electric Circuits
Solution
In the diagram, a thread AC, fixed at pulley A passes over pulley C on a force board and carries an unknown mass \(m_o\). Retain this mass \(m_o\) throughout the experiment. Draw a line along the direction of AC on the paper held behind the thread. Locate the mid-point B of AC and mark its position on this line. Draw BP at right angles to AC. By means of a loop of thread, suspend a mass M 50 g from AC and adjust the position of the loop so that the line of action of the weight of M lies along with BP. Ensure that M and \(m_o\), hang off the force board. Measure BO = y and AO. Evaluate y / AO. Repeat the experiment for M = 70,90,110 and 130g respectively. In each case, determine the corresponding values of y, AO, and y / AO. Tabulate your readings. Plot a graph of Y/AO on the vertical axis and M on the horizontal axis. Determine the slopes s of the graph. State two precautions taken to ensure accurate results. Attach your traces to your answer script.
(b)i. Distinguish between the resultant and the equilibrant of forces.
ii. State two conditions necessary for the equilibrium of three non-parallel coplanar forces.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of values/observation
S/N
M(g)
y(cm)
Ao(cm)
\(\frac{y}{AO}\)
1
50.00
2.30
8.00
0.288
2
70.00
2.45
8.20
0.378
3
90.00
3.65
8.40
0.417
4
110.00
4.10
8.60
0.465
5
130.00
5.00
9.10
0.549
Slope (s) = \(\frac{\bigtriangleup {(\frac{y}{AO})}}{\bigtriangleup {m(g)}} = \frac{0.55-0.15}{130-6} = \frac{0.4}{124}\)
0.003226g\(^{-1}\)
Precautions:
- Force board must be firmly fixed.
- Avoid parallax errors when reading the metre rule.
(b)i. The resultant and the equilibrant of forces have the same magnitude but opposite directions.
ii. The conditions for the equilibrium of three non-parallel co-planar forces are:
The lines of action of the forces must meet at a point.
All the three forces must be concurrent (i.e, meet at a point).
The algebraic sum of their moment about a point must be zero.
The algebraic sum of the forces at a point must be equal to zero.
Source: Myschool Past Questions Archive (Record #60137)
Physics
WAEC / WASSCE 1999
Reflection, Refraction & Dispersion of Light
Solution
Trace the outline ABCD of the glass block on the sheet of paper as shown in the diagram. Remove the block and draw the normal at O. Draw an incident ray such that the angle of incidence i =30^\circ. Fix two pins at points P and Q on the incident ray. Replace the block and fix two other pins at points U and V such that the pins appear to be in a straight line with the images of the pins at P and Q when viewed through the block. Remove the block and join the points at V and U, producing the line to meet DC at T. Join OT With O as a center and using any convenient radius, draw a circle to cut the incident and refracted rays at R and S respectively. Draw the perpendiculars RN and MS. Measure and record RN and MS. Repeat the experiment for i = 40^\circ 50^\circ 60^\circ and 70^\circ respectively. In each case, determine and record the corresponding values of RN and MS. Tabulate your readings. Plot a graph of RN on the vertical axis and MS on the horizontal axis. Determine the slope s of the graph, State two precautions taken to ensure accurate results. [Attach your traces to your answer script]
(b)i. Explain refraction.
ii. Draw a diagram showing why a metre rule, partly immersed in water and placed obliquely to the surface appears bent at the surface.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of values/observation
S/N
1^\circ(degree)
RN(cm)
MS(cm)
1
30.00
1.00
0.70
2
40.00
1.35
0.90
3
50.00
1.60
1.05
4
60.00
1.80
1.20
5
70.00
2.00
1.30
Slope (s) = \(\frac{\bigtriangleup {RN(cm)}}{\bigtriangleup {MS(cm)}} = \frac{2.16-0.1}{1.4-0.18} = \frac{2.06}{1.22}\) = 1.688
Precautions:
- Ensure that the pins are vertical
- Ensure reasonable spacing of pins.
(b) Refraction is the change in the direction of a ray of light at the boundary between two media as a result of the change in the speed of light at the boundary.
Source: Myschool Past Questions Archive (Record #60138)
Physics
WAEC / WASSCE 1999
Current Electricity & Electric Circuits
Solution
Measure and record the e.m.f of the accumulator provided. Connect the circuit as shown in the diagram. With zero resistance in the resistance box, adjust the rheostat to obtain the maximum possible reading on the ammeter. Do not adjust the rheostat again throughout the experiment. Open the key. With R= 1\(\Omega\), close the key, read and record the reading of the ammeter I. Calculate 1\(^{-1}\) Repeat the experiment for R=2,3,4 and 5\(\Omega\) respectively. In each case, read and record the value of \(1_A\) and calculate the Corresponding value of l\(_{A}^{-1}\). Now connect the resistor Q in series into the circuit and without altering the setting of the rheostat, record the new ammeter readings \(l_B\) for R= 1,2,3, 4, and 5\(\Omega\). Calculate the corresponding value of \(l_B\) in each case. Tabulate all your readings. On the same graph and using the same axes and scales, plot the graphs of:
(i) I\(_{A}^{-1}\)' on the vertical axis and R on the horizontal axis;
(ii) l\(_{B}^{-1}\) on the vertical axis and R on the horizontal axis. Calculate the slopes \(S_A\) in (i) and \(S_B\) in (ii) above. Determine the difference D between the intercepts of the graphs on the vertical axis. State two precautions taken to ensure accurate results.
(b) i. Explain why a battery of eight dry Leclanche cells, each of e.m.f. 1.5v is not normally used in place of a motor-car battery of 12V to start a car.
ii. State two sources of e.m.f. other than the chemical cell.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of values/observation
S/N
R(\(\Omega\))
\(I_A\)(A)
I\(^{-1}_{A}\)
\(I_B\)(A)
I\(^{-1}_{B}\)(A\(^{-1}\))
1
1.00
1.80
0.556
0.95
1.053
2
2.00
0.95
1.053
0.60
1.667
3
3.00
0.60
1.667
0.50
2.000
4
4.00
0.45
2.222
0.40
2.500
5
5.00
0.40
2.500
0.35
2.857
Slop\(e_A\) (\(s_A\)) = \(\frac{2.5-0.25}{5-0.35} = \frac{2.25}{4.65}\) = 0.4838\(\Omega ^{-1}\) A\(^{-1}\)
Slop\(e_B\) (\(S_B\)) = \(\frac{2.85-0.95}{5-0.55}\) = \(\frac{1.9}{4.45}\) = 0.4270\(\Omega ^{-1} A^{-1}\)
Interpret on I\(_{A}^{-1}\) (A\(^{-1}\)) axis = 0.075A\(^{-1}\)
Interpret on I\(_{B}^{-1}\) (A\(^{-1}\)) axis = 0.725A\(^{-1}\)
= D = \(S_B\) - \(S_A\)
= 0.725 - 0.075 = 0.65A\(^{-1}\)
Precautions:
- Always open the key when readings are not taken.
- Ensure tight connections.
(b)i. The current produced by the cells is less than that of the car battery because of the total internal resistance of the cells. A car requires a large amount of Current to start it. This amount of current cannot be supplied by the cells
ii. Photocell and Dynamo
Source: Myschool Past Questions Archive (Record #60140)
Physics
WAEC / WASSCE 1999
Atomic, Quantum, Nuclear Physics & Electronics
Which of the following is not a safety precaution in a nuclear station?
A
Radioactive materials are stored in glass casings
B
Radioactive materials are handled with remote-controlled tongs
C
Technicians in nuclear stations wear radio-sensitive badges which are checked regularly.
D
Radioactive materials are kept in thick-walled lead containers
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #25516)
English Language
WAEC / WASSCE 1999
Antonyms (Opposite in Meaning)
In each of the following sentences, there is one word underlined and one gap. From the list of words lettered A - D, choose the word that is most nearly opposite in meaning to the underlined word and that will, at the same time, correctly fill the gap in the sentences.
To encourage productivity we must reward industry and ...........laziness
A
withhold
B
withdraw
C
punish
D
oppose
medium Difficulty
Verified Correct Answer: Option C
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #24600)
English Language
WAEC / WASSCE 1999
Antonyms (Opposite in Meaning)
In each of the following sentences, there is one word underlined and one gap. From the list of words lettered A - D, choose the word that is most nearly opposite in meaning to the underlined word and that will, at the same time, correctly fill the gap in the sentences.
I took the drug to relieve my headache only for it to...............my condition
A
alleviate
B
provoke
C
abate
D
aggravate
medium Difficulty
Verified Correct Answer: Option D
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #24601)
English Language
WAEC / WASSCE 1999
Antonyms (Opposite in Meaning)
In each of the following sentences, there is one word underlined and one gap. From the list of words lettered A - D, choose the word that is most nearly opposite in meaning to the underlined word and that will, at the same time, correctly fill the gap in the sentences.
An optimist wears a smile most of the time but ...........goes about with a long face
A
an oculist
B
an activist
C
a pacifist
D
a pessimist
medium Difficulty
Verified Correct Answer: Option D
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #24602)
English Language
WAEC / WASSCE 1999
Antonyms (Opposite in Meaning)
In each of the following sentences, there is one word underlined and one gap. From the list of words lettered A - D, choose the word that is most nearly opposite in meaning to the underlined word and that will, at the same time, correctly fill the gap in the sentences.
You will make a good counsellor by listening to your clients rather than........them
A
offending
B
ignoring
C
mishandling
D
threatening
medium Difficulty
Verified Correct Answer: Option B
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #24603)
Chemistry
UTME / JAMB 1999
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
200cm\(^3\) each of 0.1M solutions of lead (II) trioxonitrate(V) and hydrochloric acid were mixed. Assuming that lead (II) chloride is completely insoluble, calculate the mass of lead (II) chloride that will be precipitated.
[Pb = 207, Cl= 35.5, N = 14, O = 16]
A
2.78g
B
5.56g
C
8.34g
D
11.12g
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
First, we need to write the balanced chemical equation for the reaction;
Pb(NO\(_3\))\(_2\) + 2HCl → PbCl\(_2\) + 2HNO\(_3\)
1mole 2moles
Recall that n = \(\frac{CV}{1000}\)
n =\(\frac{{0.1}\times{200}}{1000}\)
n = 0.02 mole of Pb(NO\(_3\))\(_2\)
Similarly, from the question, both HCl and Pb(NO\(_3\))\(_2\) have the same concentration as well as volume. This implies the number of moles of HCl will also be the same. i.e 0.02mole of HCl
However, from the balanced chemical equation, if 2moles of HCl reacted with 1mole of Pb(NO\(_3\))\(_2\)
then, from the calculations 0.02 mole HCl will react with 0.01 mole Pb(NO\(_3\))\(_2\) (Based on the stoichiometry above, i.e half the number of moles of HCl)
This implies that HCl will first be used up in the reaction, as the reactant, hence it is called the limiting reagent. Consequently, it will form the basis of our calculation.
The reference will now be between HCl and the insoluble salt, which is PbCl\(_2\).
So, if 2 moles of HCl → 1 mole of PbCl\(_2\)
0.02moles HCl → 0.01mole PbCl\(_2\)
n = \(\frac{mass}{Molar Mass}\)
Molar mass of PbCl\(_2\) = 207 + 2(35.5) = 278g/mol
n = \(\frac{mass}{Molar Mass}\)
m = n x M
m = 0.01 x 278 = 2.78g
Source: Myschool Past Questions Archive (Record #827)
Chemistry
UTME / JAMB 1999
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
56.00cm3 of a gas at S.T.P. weighed 0.11g. What is the vapour density of the gas?
[Molar volume of a gas at S.T.P = 22.4dm3]
A
11.00
B
22.00
C
33.00
D
44.00
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
56.00cm3 of a gas at S.T.P weighed 0.11g
Molar volume of the gas = 22.4dm3 = 22400cm3
molar mass = (22400 x 0.11)/56
molar mass = 2464/56 = 44
Vapour density = Molar mass/2 = 44/2 = 22
Source: Myschool Past Questions Archive (Record #828)
Chemistry
UTME / JAMB 1999
Organic Chemistry I: Hydrocarbons & Crude Oil
Solution
Which of the following gases will diffuse fastest when passed through a porous plug?
[H = 1, C = 12, N = 14, O = 16]
A
Propane
B
Oxygen
C
Methane
D
Ammonia
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
The rate of diffusion is inversely proportional to the square root of the relative molecular mass; in other words, lighter molecules diffuse faster than heavier ones.
The relationship between diffusion rate and molecular mass is an inverse one. This means that as the molecular mass of a substance increases, its diffusion rate decreases. Conversely, substances with lower molecular masses will diffuse more quickly.
Relative molecular mass of :
(a) Propane, C\(_3\)H\(_8\) = (12 x 3) + (1 x 8) = 36 + 8 = 44
(b) Oxygen, O\(_2\) = 16 x 2 = 32
(c) Methane, CH\(_4\) = 12 + (1x4) = 12 + 4 = 16
(d) Ammonia, NH\(_3\) = 14 + (1 x 3) = 17
Since CH\(_4\) has the least molecular mass compared to other options, it has the highest diffusion rate.
Source: Myschool Past Questions Archive (Record #829)
Chemistry
UTME / JAMB 1999
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
Which is the temperature of a given mass of a gas initially at 0°C and 9 atm, if the pressure is reduced to 3 atm at constant volume?
A
91K
B
182K
C
273K
D
819K
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
From P1/T1 = P2/T2 = 9/273 = 3/T2
T2 x 9 = 273 x 3
T2 = (273 x 3)/9
T2 = 91K
Source: Myschool Past Questions Archive (Record #830)
Chemistry
UTME / JAMB 1999
Stoichiometry, Mole Concept & Volumetric Analysis
Which of the following will have its mass increased when heated in air?
A
Helium
B
Magnesium
C
Copper pyrites
D
Glass
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #831)
Chemistry
UTME / JAMB 1999
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
From the equation above, the mass of magnesium required to react with 250cm3 of 0.5M HCl is
[Mg = 24]
A
0.3g
B
1.5g
C
2.4g
D
3.0g
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
Molar conc. =no.of moles/Vol in dm3
Vol=250cm3= 0.25dm3
Molar conc=0.5M
No of moles=molar conc *Vol in dm3
=0.5*0.25
=0.125mole of HCl
2moles of HCl → 1Mole of Mg
0.125 mole →(0.125*1)/2
=0.0625mole of Mg
Reacting mass=no of moles* molar mass
Mg=24
Mass=0.0625*24
=1.5g
Source: Myschool Past Questions Archive (Record #832)
Chemistry
UTME / JAMB 1999
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
A gaseous metallic chloride MClx consists of 20.22% of M by mass. The formula of the chloride is?
[M = 27, Cl = 35.5]
A
MCl
B
MCl2
C
MCl3
D
M2Cl6
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
Percentage by mass of M = 20.22%
Percentage by mass of Cl = x
100 - 20.22 = 79.78%
Mass given ratio of M
M = \(\frac{20.22}{27}\) = 0.75
Chlorine = \(\frac{79.78}{35.5}\) = 2.24
Divide by the smallest
M = \(\frac{0.75}{0.75}\) = 1
Cl = \(\frac{2.24}{0.75}\) = 3
∴ The formula = MCl\(_3\)
Source: Myschool Past Questions Archive (Record #836)