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Physics
WAEC / WASSCE 1999
Electricity and Magnetism
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
UTME / JAMB 1998
Electricity and Magnetism
The iron core of an induction coil is made from bundles of wire so as to
A
minimize eddy-currents
B
generate eddy-current
C
prevent sparking at the contact breaker
D
get the greatest posible secondary voltage
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 #27665)
Physics
UTME / JAMB 1998
Electricity and Magnetism
Solution
A transformer which can produce 8V from a 240v a.c. supply, has an efficiency of 80%. If the current in the secondary coil is 15A, calculate the current in the primary coil.
A
0.625A
B
1.600A
C
2.500A
D
6.250A
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Efficiency = \(\frac{I_sV_s}{I_pV_p}\) x 100
Source: Myschool Past Questions Archive (Record #27667)
Physics
UTME / JAMB 1998
Electricity and Magnetism
Solution
The electrochemical equivalent of a metal is 1.3 x 10-7kgC-1. The mass of the metal which 2.0 x 104C of electricity will deposit from a suitable electrolyte is
A
6.5 x 10-2kg
B
2.6 x 10-2kg
C
6.5 x 10-3kg
D
2.6 x 10-3kg
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
m = QZ
= 2 x 104 x 1.3 x 10-7
= 2.6 x 10-3kg
Source: Myschool Past Questions Archive (Record #27672)
Physics
WAEC / WASSCE 1998
Electricity and Magnetism
In the electrical method of magnetisation, the polarity of the magnet depends on the
A
amount of current passed
B
direction of current
C
magnetic material used
D
size of the magnetic material
E
orientation of the magnetic material in space
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 #33669)
Physics
WAEC / WASSCE 1998
Electricity and Magnetism
Solution
A cell of e.m.f. 1.5v is connected in series with a resistor of 3\(\Omega\). A high resistance voltmeter connected across the cell registers only 0.9v. Calculate the internal resistance of the cell
A
5.0\(\Omega\)
B
4.5\(\Omega\)
C
2.4\(\Omega\)
D
2.0\(\Omega\)
E
1.8\(\Omega\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
E.M.F (E) = 1.5v, Internal Resistance (r) = ?, Potential Difference (V) = 0.9v, Load resistance (R) = 3\Omega,
Current (I) = V / R = 0.9 / 3 = 0.3A
USING : E = I ( R + r )
r = (E / I) - R = (1.5 / 0.3) - 3
r = 5 - 3
r = 2\Omega,
Source: Myschool Past Questions Archive (Record #33672)
Physics
WAEC / WASSCE 1998
Electricity and Magnetism
Solution
Calculate the resistance of the filament of lamp rated at 240V 60W
A
240\(\Omega\)
B
360\(\Omega\)
C
960\(\Omega\)
D
144\(\Omega\)
E
2880\(\Omega\)
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
P = \(\frac{V^2}{R} = \frac{240^2}{60} = 960\)
Source: Myschool Past Questions Archive (Record #33674)
Physics
WAEC / WASSCE 1998
Electricity and Magnetism
Solution
An electron is accelerated from rest through a potential difference of 70kV in a vacuum. Calculate the maximum speed acquired by the electron (electronic charge = -1.6 x 10-19; mass of an electron = 9.1 x 10-31kg)
A
3.00 x 108ms-1
B
2.46 x 108ms-1
C
1.57 x 108ms-1
D
1.32 x 108ms-1
E
1.11 x 108ms-1
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
ev = \(\frac{1}{2}\)mv2
v2 = \(\frac{ev}{\frac{1}{2}m}\)
V = \(\sqrt{\frac{ev}{\frac{1}{2m}}} = \sqrt{\frac{1.6 \times 10^{-19} \times 70 \times 10^3}{\frac{1}{2} \times 9.1 \times 10^{-31}}}\)
= \(\sqrt{246.2 \times 10^{14}}\) = 1.57 x 108
Source: Myschool Past Questions Archive (Record #33694)
Physics
WAEC / WASSCE 1998
Electricity and Magnetism
The direction of the magnetic field at a point in the vicinity of a bar magnetic is
A
along the line joining the point to the neutral point
B
always away from the south pole of the magnet
C
opposite the direction of theresultant field at that point
D
always towards the north pole of the magnet
E
the direction towards which the north pole of a compass needle would point
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option E.
Source: Myschool Past Questions Archive (Record #33682)
Physics
UTME / JAMB 1998
Electricity and Magnetism
Diagram
Solution
The total resistance measured at PQ in the diagram above is
Attached Figure / Diagram
A
18.0Ω
B
11.0Ω
C
4.0Ω
D
2.0Ω
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The three 2 ohms are in series, therefore the total resistance is
\(R=R_1+R_2+R_3\)
\(R_ {total} =2+2+2= 6\) ohms.
resketching circuit;
6 ohms is used to replace the three 2 ohms which is therefore placed by the 3 ohms resistor.
The 3 ohms and 6 ohms resistors are in parallel, so the combined resistance is
\(R=\frac{R_1xR_2}{R_1+R_2}\)
\(R= \frac{3\times6}{3+6}\)
\(R=\frac{18}{9}=2\) ohms.
Resketching circuit:
The 2 ohms is used to replace the 3 ohms and 6 ohms laying side-by-side.
Therefore, the 3 resistors (1 ohm, 1 ohm, and 2 ohms) are in series.
Thus, the total resistance is
\(R= (R_1+R_2+R_3)\)
\(R_{total} =1+1+2=4\) ohms.
Therefore, the total resistance measured at PQ is 4 ohms.
Source: Myschool Past Questions Archive (Record #41744)
Physics
WAEC / WASSCE 1998
Electricity and Magnetism
Solution
(a)(i) Describe, with the aid of a circuit diagram, an experiment to measure the resistance of a wire given an ammeter of low resistance, a battery, a key, a rheostat, va high-resistance voltmeter and some connecting wires.
(ii) State two precautions necessary to obtain an accurate result.
(b) Using the experimental result and any necessary measurements, explain how the resistivity of the wire may be determined.
(c) Two cells, each of e.m.f. 2V and internal resistance 0.552, are connected in series. They are made to supply current to a combination of three resistors, one of resistance 20 connected in series to a parallel combination of two other resistors each of resistance 3Q. Draw the circuit diagram and calculate the:
(i) current in the circuit
(ii) potential difference across the parallel combination of the resistors
(iii) lost volts of the battery.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)
Description: The circuit is set up as shown above. The key (k) is closed. The rheostat (R) is adjusted so that lowest current is flowing. Ammeter and voltmeter readings (I and V) are recorded. Adjust rheostat to increase current. Ammeter and voltmeter readings are recorded for other positions of the rheostat. Values of (are plotted against values of I. Slope, which is the value of the resistance is determine.
Precautions;
i) Kcy should be removed when readings are not taken
ii) Connections should be tight
(iii) Parallax' error when reading ammeter and voltmeter should be avoided
(iv) Only low current should be used
v) Zero error of the meters should be avoided. Note: Any correct two) Resistance, R = P'/A where = length of wire, A = cross sectional area, p = resistivity )f wire. But slope, S of graph = R. Hence S = P'/A
Measure I with a metre rule. Measure diameter with a micrometer screw gauge and determine A. Substitute p = \(\frac{SA}{I}\)
c) Diagram
i) Total e.m.f. = 4V
Total resistance = 1 + 2 + \(\frac{3 \times 3}{ 2 \times 3}\)
= 3 + 1\(\frac{1}{2}\)
= \(\frac{1}{2} \Omega\)
E = IR\(_T\),
= E/R
:.Current, I = 4/9 x 2/1
= 8/9;
A = 0.9A
(ii) P.d = IR = 8/9 x 3/2 = 4/3 = 1.3V
(iii) Lost volts = Ir = \(\frac{8}{9}\) x 1 = \(\frac{8}{9} = 0.9V\)
Source: Myschool Past Questions Archive (Record #54711)
Physics
WAEC / WASSCE 1997
Electricity and Magnetism
Lenz’s law of electromagnetic induction is essentially a statement of the
A
Inverse square law of gravitation
B
Inverse square law of magnetism
C
Inverse square law of electrostatics
D
Law of conservation of momentum
E
Law of conservation of energy
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option E.
Source: Myschool Past Questions Archive (Record #26314)
Physics
WAEC / WASSCE 1997
Electricity and Magnetism
Solution
The relationship between the directions of magnetic field, the current-carrying wire in the field is easily remembered using
A
Octet rule
B
Right hand grip rule
C
Maxwell’s screw rule
D
Ampere’s swimming rule
E
Fleming’s left hand rule
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The Right Hand Grip Rule (or Right Hand Rule) is used to determine the direction of the magnetic field around a current-carrying conductor. If you grip the wire with your right hand such that your thumb points in the direction of the current, your fingers will curl in the direction of the magnetic field lines.
Source: Myschool Past Questions Archive (Record #26315)
Physics
WAEC / WASSCE 1997
Electricity and Magnetism
When a resistor of resistance R is connected across a cell, the terminal p.d of the cell is reduced to three quarters of its e.m.f. The internal resistance of the cell is expressed as
A
R/4
B
R/3
C
R/2
D
3R/4
E
3R
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 #26319)
Physics
WAEC / WASSCE 1997
Electricity and Magnetism
The current in a series R-L-C circuit attains its maximum value when the
A
Impedance is greater than the capacitive reactance
B
Inductive reactance is equal to the capacitive reactance
C
Inductive reactance is greater than the capacitive reactance
D
capacitive reactance is less than resistance
E
Inductive reactance is greater than the capacitive reactance
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 #26328)