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Physics
UTME / JAMB 2024
Electromagnetic Field
Solution
As per Faraday's laws of electromagnetic induction, an e.m.f is induced in a conductor whenever
A
lies parallel to the magnetic flux
B
lies in a magnetic field
C
lies out side but close to a magnetic field
D
cuts magnetic flux
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Cuts magnetic flux: According to Faraday's law, when a conductor cuts through magnetic field lines (flux), an e.m.f is induced.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70692)
Physics
WAEC / WASSCE 2024
Electromagnetic Field
Solution
Which of the following modes is the most economical method of transmitting electrical power over a long distance?
A
Alternating current at high voltage and low current.
B
Alternating current at low voltage and high current.
C
Direct current at high voltage and low current.
D
Direct current at low voltage and high current.
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
When electricity is transmitted over long distances, power losses occur mainly due to the resistance of the transmission lines. These losses are proportional to the square of the current (I\(^2\)R losses). By using high voltage, the current can be kept low, significantly reducing these losses
Source: WAEC / JAMB / NECO Classroom Archive (Record #72071)
Physics
UTME / JAMB 2024
Electric Field
Solution
Find the value of a capacitor with voltage 5V and 30C.
A
7f
B
6f
C
15f
D
0.17f
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The capacitance of a capacitor is given by: C = \(\frac{Q}{V}\)
where Q = quantity of charges, and V = voltage
Q = 30c and V = 5v
C = \(\frac{Q}{V}\) = C = \(\frac{30}{5}\) = 6 farad
Source: WAEC / JAMB / NECO Classroom Archive (Record #70356)
Physics
UTME / JAMB 2024
Electric Field
Solution
The charge of magnitude 1.6 x 10 \(^{-19}\)C is placed in a uniform electric field of intensity 1200Vm\(^{-1}\). Calculate its acceleration, if the mass of the charge is 9.1 x 10\(^{-31}\)kg
A
2.11 x 10\(^{14}\)ms\(^{-2}\)
B
2.11 x 10\(^{13}\)ms\(^{-2}\)
C
1.20 x 10\(^{14}\)ms\(^{-2}\)
D
1.20 x 10\(^{13}\)ms\(^{-2}\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Electric field intensity (E) = \(\frac{\text{force}}{\text{charge}}\) = \(\frac{\text{F}}{\text{q}}\)
Given: E = 1200Vm\(^{-1}\), q = 1.6 x 10 \(^{-19}\)C, m = 9.1 x 10\(^{-31}\)kg
F = E x q
F = 1.6 x 10\(^{-19}\) x 1200 = 1.92 x 10\(^{-16}\)N
F = ma
a = \(\frac{F}{m}\) = \(\frac{1.92\times 10^{-16}}{9.1\times 10^{-31}}\) \(\approxeq\) 2.11 x 10\(^{14}\)ms\(^{-2}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #70510)
Physics
UTME / JAMB 2024
Electric Field
Diagram
Solution
The energy stored in the above capacitor is
Attached Figure / Diagram
A
2.25J
B
6.25J
C
18.25J
D
24.25J
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
The energy stored in the capacitor = \(\frac{1}{2}\)\(\frac{q^2}{C}\)
Where C = 2F, q = 3C
= \(\frac{1}{2}\)\(\frac{3^2}{2}\) = \(\frac{9}{4}\) = 2.25J
Source: WAEC / JAMB / NECO Classroom Archive (Record #70518)
Physics
UTME / JAMB 2024
Electric Field
Solution
Calculate the value of electric field intensity due to a charge of 4μC if the force due to the charge is 8N
A
2.0 x 10\(^6\)NC\(^{-1}\)
B
3.0 x 10\(^6\)NC\(^{-1}\)
C
4.0 x 10\(^6\)NC\(^{-1}\)
D
8.0 x 10\(^6\)NC\(^{-1}\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Electric field intensity(E) = \(\frac{\text{Force}}{\text{charge}}\) = \(\frac{F}{q}\)
q = 4\muC = 4.0 x 10\(^{-6}\)C
F = 8N
E = \(\frac{8}{4.0 \times 10^{-6}}\) = 2.0 x 10\(^6\)NC\(^{-1}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #70680)
Physics
UTME / JAMB 2024
Electric Field
Solution
Two capacitors of 0.0003μF and 0.0006μF are connected in series, find their combined capacitance.
A
0.0009μF
B
0.0002μF
C
0.0005μF
D
0.0001μF
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Combined capacitance in series = \(\frac{1}{C_T}\) = \(\frac{1}{C_1}\) + \(\frac{1}{C_2}\)
The two capacitors have capacitance of 0.0003\muF and 0.0006\muF respectively
\(\frac{1}{C_T}\) = \(\frac{1}{0.0003}\) + \(\frac{1}{0.0006}\) = \(\frac{0.0002 + 0.0001}{0.0006}\)
\(\frac{1}{C_T}\) = \(\frac{0.0003}{0.0006}\)
C\(_T\) = 0.0002\muF.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70682)
Physics
WAEC / WASSCE 2024
Electric Field
Solution
Two isolated charged spheres each of magnitude + 1.0 x 10\(^{-6}\) C are seperated by a distance of 2.0 x 10\(^{-1}\) m. Calculate the magnitude of the electrostatic force between them [K = \(\frac{1}{4\pi \epsilon_o}\) = 9 x 10\(^9\)Nm\(^2\)C\(^{-2}\)]
A
2.25 x 10\(^{-1}\)N
B
4.50 x 10\(^{-1}\)N
C
2.25 x 10\(^1\)N
D
4.50 x 10\(^1\)N
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
F = \(\frac{Kq_1q_2}{r^2}\)
F = \(\frac{9 \times 10^9 \times (1.0 \times 10^{-6})^2}{(0.2)^2}\)
= 2.25 x 10\(^{-1}\)N
Source: WAEC / JAMB / NECO Classroom Archive (Record #72047)
Physics
WAEC / WASSCE 2024
Electric Field
Diagram
Solution
In the circuit diagram above, which lamp causes the other to go off?
Attached Figure / Diagram
A
P
B
Q
C
R
D
S
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
If lamp S goes off, all other lamps will go off. This is because lamp S is required to complete the whole circuit.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72049)
Physics
WAEC / WASSCE 2024
Electric Field
Solution
A capacitor described as a mica capacitor has
A
both plates are made of mica.
B
a mica dielectric.
C
a dielectric made of mica
D
the same capacitance as that of mica
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
A mica capacitor has mica as its dielectric material. Known for excellent stability, low loss, and high insulation resistance, they are used in precision applications, particularly in RF circuits and high-frequency settings.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72064)
Physics
UTME / JAMB 2024
Current Electricity
Solution
Electrolysis can be investigated using
A
Voltmeter
B
Ammeter
C
Galvanometer
D
Voltameter
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
A voltameter measures the amount of substance that is deposited during electrolysis, allowing for the study of the electrochemical process. It provides information about the quantity of electric charge passed through the electrolyte and the resulting chemical changes. Voltameter is also called an Electrolytic cell.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70367)
Physics
UTME / JAMB 2024
Current Electricity
Solution
If a sheet of an insulating material is inserted between the plates of an air capacitor, the capacitance will
A
become zero
B
decrease
C
increase
D
remain unchanged
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
If a sheet of an insulating material (dielectric) is inserted between the plates of an air capacitor, the capacitance will increase.
When a dielectric material is inserted: The permittivity \(\epsilon\) increases (since the permittivity of the dielectric is greater than that of air),
This results in an increase in capacitance.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70373)
Physics
UTME / JAMB 2024
Current Electricity
Solution
A cell of internal resistance of 2\(\Omega\) supplies current through a resistor, X if the efficiency of the cell is 75%, find the value of X.
A
2\(\Omega\)
B
4\(\Omega\)
C
6\(\Omega\)
D
8\(\Omega\)
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The efficiency of a cell = \(\frac{\text{Ext. R}}{\text{Ext.R + int. r}}\) x 100
Ext. R = X
int. r = 2\(\Omega\)
Efficiency = 75% = 0.75
0.75 = \(\frac{\text{X}}{\text{X} + 2}\)
0.75( X + 2 ) = X
0.75x + 1.5 = X
0.25X = 1.5
X = \(\frac{1.5}{0.25}\) = 6\(\Omega\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #70426)
Physics
UTME / JAMB 2024
Current Electricity
Solution
In electrolysis, when same quantity of electricity is passed through different electrolytes, mass of substances deposited is proportional to
A
Molecular weight
B
Chemical equivalent
C
Atomic number
D
Chemical equivalent per valency
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Faraday's second law states that the masses of different substances deposited or liberated by the same quantity of electricity are directly proportional to the chemical equivalents of the substances.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70436)
Physics
UTME / JAMB 2024
Current Electricity
Solution
The capacitance of a capacitor, C, is inversely proportional to
A
area of the plate
B
dielectric substance
C
distance between plate
D
direction of the plate
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The capacitance of a capacitor, C, is inversely proportional to:
Distance between plates: Capacitance decreases as the distance increases.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70492)