Question Library
Browse, filter, and study verified past examination questions with step-by-step mathematical reasoning.
Searching 18,000+ Questions...
Sifting topics, KaTeX formulas & verified solutions
Showing 316–330 of 557 verified questions
Sort by:
Physics
WAEC / WASSCE 2001
Electricity and Magnetism
Solution
The Earth's magnetic equator passes through Jos in Nigeria. At Jos. the
A
angle of variation is zero
B
magnetic declination is 90o
C
angle of dip is zero
D
horizontal component of the Earth's magnetic field is zero
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
At the magnetic equator, the angle of dip (or inclination) is zero because the magnetic field lines are horizontal.
Source: Myschool Past Questions Archive (Record #33939)
Physics
WAEC / WASSCE 2001
Electricity and Magnetism
Solution
The material used for constructing the core of an electromagnet is
A
iron
B
carbon
C
copper
D
steel
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Iron is used because it has high magnetic permeability, which enhances the magnetic field produced by the electromagnet.
Source: Myschool Past Questions Archive (Record #33941)
Physics
WAEC / WASSCE 2001
Electricity and Magnetism
Solution
As the plates of a charged variable capacitor are moved closer together, the potential difference between them
A
increased
B
decrease
C
remains the same
D
is doubled
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
As the plates of a charged variable capacitor are moved closer together, the potential difference between them decreases.
This is because the capacitance increases as the distance between the plates decreases. For a constant charge, the potential difference (V = \(\frac{\text{Q}}{\text{C}}\)) will decrease when capacitance (C) increases.
Source: Myschool Past Questions Archive (Record #33934)
Physics
UTME / JAMB 2001
Electricity and Magnetism
Solution
What is the angle of dip at the magnetic equator
A
\( 45^{\circ} \)
B
\( 0^{\circ} \)
C
\( 90^{\circ} \)
D
\( 180^{\circ} \)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The angle of dip is the angle which the resultant earth's magnetic field makes with the horizontal.
At the magnetic equator, which is entirely along the horizontal this angle is zero
Source: Myschool Past Questions Archive (Record #41484)
Physics
NECO SSCE 2001
Electricity and Magnetism
A cell can supply currents of 0.60 A and 0.30 A through a 2.0 ω and a 5.0 ω resistor respectively. What is the internal resistance of the cell?
A
0.2 ω
B
0.4 ω
C
1.0 ω
D
3.0 ω
E
9.0 ω
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #42831)
Physics
NECO SSCE 2001
Electricity and Magnetism
Solution
A 220 V, 60 Hz alternating current supply is connected to an inductor of 3.5H. What is the current passing through the inductor?
A
0.0167 A
B
0.1670 A
C
1.6700 A
D
16.7000 A
E
167.0000 A
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
V = 220V, F = 60Hz and L = 3.5H
I = \(\frac{V}{X_L}\)
but \(X_L = 2 \pi FL\) = 2 x 3.142 x 60 x 3.5 = 1319.64\Omega
I = \(\frac{220}{1319.6}\) = 0.1667A OR 0.1670A
Source: Myschool Past Questions Archive (Record #42834)
Physics
NECO SSCE 2001
Electricity and Magnetism
If N.E.P.A. charges 65 kobo per kilowatt-hour for electricity supplied to consumers, what bill would an electric iron rated at 750 watts attract when used for 24 hours?
A
N9.70
B
N11.70
C
N12.40
D
N15.90
E
N17.60
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 #42838)
Physics
NECO SSCE 2001
Electricity and Magnetism
Solution
A transformer is connected to a 250 V a.c. supply. The primary coil has 5000 turns and the secondary voltage is found to be 10 V. What is the number of turns in the secondary coil?
A
100
B
150
C
200
D
250
E
300
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
To find the number of turns in the secondary coil of a transformer, we use the transformer equation:
\(\frac{V_p}{V_s} = \frac{N_p}{N_s}\)
Given:
\( V_p = 250 \, \text{V}\)
\( V_s = 10 \, \text{V} \)
\( N_p = 5000 \, \text{turns} \)
Rearranging the equation to find \( N_s \):
\(N_s = N_p \times \frac{V_s}{V_p}\)
Plugging in the values:
\(N_s = 5000 \times \frac{10}{250}\)
\(N_s = 5000 \times 0.04\)
\(N_s = 200 \, \text{turns}\)
Therefore, the number of turns in the secondary coil is 200.
Source: Myschool Past Questions Archive (Record #42845)
Physics
NECO SSCE 2001
Electricity and Magnetism
The magnetic effect of a current-carrying coil is NOT utilized in the making of a/an
A
carbon microphone
B
electric bells
C
magnetic relay
D
telephone earpiece
E
thermostat
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 #42855)
Physics
WAEC / WASSCE 2001
Electricity and Magnetism
Solution
(a) What is electrolysis?
(b) A current of 2A is passed through a copper voltameter for 5 minutes. If the electrochemical equivalent of copper is 3.27 x 10\(^{-7}\) kg. Determine the mass of the corper deposited.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Electrolysis is the decomposition of an electrolyte into ions by passing current through the electrolyte.
(b) Mass of copper deposited M = zlt
= 3.27 x 10\(^{7}\) x 2 x 300 = 1.962 x 10\(^{-4}\) kg
Source: Myschool Past Questions Archive (Record #54111)
Physics
WAEC / WASSCE 2001
Electricity and Magnetism
Solution
(a) Explain the statement the capacitance of a capacitor is 5\(\mu\)F.
(b)(i) State the factors upon which the capacitance of a parallel plate capacitor depend.
(ii) State how the capacitance depends on each of these factors stated in (b)(i).
(c) A series arrangement of three capacitors of values 8uF, 12\(\mu\)F, and 24\(\mu\)F is connected in series with 90-V battery.
(i) Draw an open-circuit diagram for this arrangement.
(ii) Calculate the effective capacitance in the circuit.
(iii) On closed circuit, calculate the charge on each capacitor when fully charged.
(iv) Determine the p.d across the 8\(\mu\)F capacitor.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) The statement means that an electric charge of 5\(\mu\)F is stored by the fully charged capacitor when connected to a d.c source of 1V.
(b)(i)
(i) Factors
(ii) Form of dependence
- Common area of plates
- Distance between plates
- Permitivity or nature of dielectric
- Direct proportionality
- Inverse proportionality
- Direct proportionality
(c)(i)
(ii) \(\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_1} + \frac{1}{C_3}\)
= \(\frac{1}{8} + \frac{1}{12} + \frac{1}{20} = \frac{3 + 2 + 1}{24}\)
C = \(\frac{24}{6} = 4 \mu F\)
(iii) Q = CV = 4 x 10\(^{-6}\) x 90 = 360\(\mu\)C
charge on 8\(\mu\) FV = \(\frac{Q}{C} = \frac{360}{8}\) = 45V
Source: Myschool Past Questions Archive (Record #54128)
Physics
WAEC / WASSCE 2001
Electricity and Magnetism
Solution
You are provided with two wires marked P and C. a resistor \(R_s\) = 1\(\Omega\) and other necessary apparatus.
Connect \(R_s\) in the left-hand gap of the metre bridge, a length L= 100cm of wire P in the right-hand gap and the other apparatus as shown in the diagram: above
Determine the balance point B on the bridge wire AC
Measure and record AB =/s, and BC = /
Evaluate \(R_1\) = (\(\frac{|_{p}}{|_{s}}\))Rs
Repeat the procedure for four other values of L = 90, 80, 70 and 60cm. In each case obtain and record the value of |\(_{s}\) and |\(_{p}\) and evaluate \(R_1\) = (\(\frac{|_{p}}{|_{s}}\))Rs
Repeat the experiment with the second wire, Q. Obtain the value of |\(_{s}\) and |\(_{Q}\) for equal lengths of wire as used in wire P.
Evaluate \(R_1\) = (\(\frac{|_{p}}{|_{s}}\))Rs. In each case, tabulate your readings.
Plot a graph of \(R_2\) on the Vertical axis against \(R_1\) on the horizontal axis.
Determine the slope S, of the graph.
Evaluate the k = \(\sqrt s\).
State two precautions taken to ensure accurate results.
(b)i. Define the resistivity of the material of a wire.
ii. A galvanometer with a full-scale-deflection of 1.5 x10\(^{3}\). A has a resistance of 50\(\Omega\). Determine the resistance required to convert it into a voltmeter reading up to 1.5V.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of values
L(cm)\(^{3}\)
Ls(cm)
Lp(cm)
\(R_1\)
100
5.3
94.7
17.90
90
6.5
93.5
14.40
80
7.0
93.0
13.30
70
8.5
91.5
10.80
60
9.5
90.5
9.50
L(cm)
Is\(^{1}\)
Lg
\(R_2\)
100
21.60
78.4
3.60
90
22.1
77.9
3.50
80
23.4
76.6
3.30
70
24.3
75.7
3.10
60
25.3
74.7
2.95
L(cm)
Ls(cm)
Lp(cm)
\(R_1\)
Ls\(^{1}\)cm\(^{3}\)
Lq
R\(^{2}\)
100
5.3
94.7
17.90
21.6
78.4
3.60
90
6.5
93.5
14.40
22.1
77.9
3.50
80
7.0
93.0
13.30
23.4
76.6
3.30
70
8.5
91.5
10.80
24.3
75.7
3.10
60
9.5
90.5
9.50
25.3
74.7
2.95
Slope (s) = of the graph = \(\frac{DR_{2}}{DR_{1}} = \frac{1.4}{8.0}\) = 0.175
K = \(\sqrt 5\) = 0.17555 = 0.42
Precautions:
Key was opened where readings were not being taken.
Tight connections were ensured
ParallaX errors were avoided in reading the metre rule.
Zero error was noted in reading the metre rule.
The jockey was not rolled on or allowed to scratch the wire.
(b) The resistivity of a material is defined as the resistance of a unit length of the material/wire of a uni cross-sectional area.
OR
P = \(\frac{AR}{L}\) where
A = cross-sectional area
R = eletrical resistance
L = length of the wire
V = Vg + \(V_R\) = IRg + IR = I (Rg + R) \(\therefore\) \(\frac{V}{I}\) = Rg+ R
= \(\frac{1.5}{1.5 \times 10^{-3}}\) = 50 + R = 10\(^{3}\) = 50 + R \(\therefore\) R = 950\(\Omega\)
Source: Myschool Past Questions Archive (Record #60114)
Physics
UTME / JAMB 2000
Electricity and Magnetism
Solution
A radio is operated by eight cells each of e.m.f. 2.0V connected in series. If two of the cells are wrongly connected, the net e.m.f. of the radio is
A
8V
B
10V
C
12V
D
16V
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
When cells are connected in series, their voltages add up. However, if two cells are connected in the wrong direction, they will subtract from the total e.m.f
If all cells were connected correctly, the total e.m.f. would be: Total E.M.F = 8 x 2.0V = 16V ( since no of cell = 8. and E.M.F of each cells = 2.0V)
When two cells are connected in the reverse direction, their contributions effectively subtract from the total e.m.f
Contribution from 2 wrongly connected cell = 2 x 2.0v = 4V
Therefore, Total E.M.F = 16 - 4 = 12V.
Source: Myschool Past Questions Archive (Record #710)
Physics
UTME / JAMB 2000
Electricity and Magnetism
Solution
A cell of internal resistance 0.01Ω can be measured accurately using the
A
Ohm meter
B
metre bridge
C
potentiometer
D
electroscope
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
A cell with internal resistance can be measured accurately using a potentiometer.
A potentiometer is an instrument used to measure the potential difference (voltage) across a component without drawing any current. This is crucial because drawing current can affect the measurement due to the internal resistance of the cell.
Source: Myschool Past Questions Archive (Record #714)
Physics
UTME / JAMB 2000
Electricity and Magnetism
Solution
A 2H inductor has negligible resistance and is connected to a 50/π Hz A.C. Supply. The reactance of the inductor is
A
200 Ω
B
50 Ω
C
100/π Ω
D
25/π Ω
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Inductive reactance is given by
XL = WL = 2πfL
XL = 2 x π x (50/π) x 2
XL = 200Ω
Source: Myschool Past Questions Archive (Record #1452)