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Physics
NECO SSCE 2004
Electricity and Magnetism
Solution
A resistor connected in parallel with a galvanometer so that a large amount of current can be measured is called a/an
A
ammeter
B
avometer
C
coil resistor
D
multiplier
E
shunt.
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
A galvanometer can be converted into ammeter by connecting a low resistance called shunt in parallel to the galvanometer.
Source: Myschool Past Questions Archive (Record #42995)
Physics
UTME / JAMB 2003
Electricity and Magnetism
A positive charged rod X is brought near as uncharged metal sphere Y and is then touched by a finger with X still in place. When the finger is removed, the result is that y has
A
no charge and a zero potential
B
a positive charge and zero potential
C
a negative charge and positive potential
D
a negative charge and a negative potential
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #12367)
Physics
UTME / JAMB 2003
Electricity and Magnetism
Diagram
I the diagram above, what would happen to the current I, if another resistor, R2, is connected in parallel to R1?
Attached Figure / Diagram
A
It will increase because the equivalent resistance will increase
B
It will decrease if R2 is greater than R1
C
It will increase because the effective resistance will decrease
D
It will decrease if R2 is less than R1
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 #12385)
Physics
UTME / JAMB 2003
Electricity and Magnetism
Solution
A wire of 5\Omega resistance is drawn out so that its new length is two times the original length. If the resistivity of the wire remains the same and the cross-sectional area is halved, the new resistance is
A
40Ω
B
20Ω
C
10Ω
D
5Ω
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Given:
- Original resistance \( R_1 = 5 \, \Omega \)
- The new length \( L_2 = 2L_1 \) (where \( L_1 \) is the original length)
- The new cross-sectional area \( A_2 = \frac{1}{2} A_1 \) (where \( A_1 \) is the original area)
Step 1: Recall the formula for resistance
The resistance \( R \) of a wire is given by:
\(R = \rho \frac{L}{A}\)
where:
- \( \rho \) is the resistivity,
- \( L \) is the length,
- \( A \) is the cross-sectional area.
Step 2: Calculate the new resistance
The new resistance \( R_2 \) can be calculated using the new length and new area:
\(R_2 = \rho \frac{L_2}{A_2}\)
Substituting for \( L_2 \) and \( A_2 \):
\(R_2 = \rho \frac{2L_1}{\frac{1}{2}A_1} = \rho \frac{2L_1 \cdot 2}{A_1} = \rho \frac{4L_1}{A_1}\)
Step 3: Relate \( R_1 \) and \( R_2 \)
From the original resistance \( R_1 \):
\(R_1 = \rho \frac{L_1}{A_1}\)
Now, we can express \( R_2 \) in terms of \( R_1 \):
\(R_2 = 4 \left(\rho \frac{L_1}{A_1}\right) = 4R_1\)
Step 4: Substitute the value of \( R_1 \)
Now substitute \( R_1 = 5 \, \Omega \):
\(R_2 = 4 \times 5 \, \Omega = 20 \, \Omega\)
Therefore, the new resistance is: \(\boxed{20 \, \Omega}\)
Source: Myschool Past Questions Archive (Record #12405)
Physics
UTME / JAMB 2003
Electricity and Magnetism
The most suitable cell used for short interval switches in electric bells is a
A
lead- acid accumulator
B
Daniel cell
C
Leclanche cell
D
nickel iron accumulator
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 #12416)
Physics
UTME / JAMB 2003
Electricity and Magnetism
Solution
In a semi conductor junction diode, as the depletion or barrier layer is forward-biased, the layer
A
widens
B
narrows
C
remains constant
D
widens then narrows
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
In a semiconductor junction diode, applying forward bias to the depletion or barrier layer causes the layer to B. narrow.
This occurs because the applied voltage lowers the potential barrier, facilitating the recombination of charge carriers and reducing the width of the depletion region.
Source: Myschool Past Questions Archive (Record #12761)
Physics
WAEC / WASSCE 2003
Electricity and Magnetism
Solution
A positively charged glass rod is placed by near the cap of a positively charge electroscope. The divergence of the leaf is observed to
A
decrease
B
increase
C
remain the same
D
increase and collapse immediately
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
One of the uses of the Gold leaf electroscope is to test charges whether the charge is positive or negative. The gold-leaf electroscope tests the sign of the charge of an object. If an unknown charge is brought near to a charged electroscope and the leaf diverges more, the unknown charge is similar to the charge on the electroscope and vice versa.
Source: Myschool Past Questions Archive (Record #34229)
Physics
WAEC / WASSCE 2003
Electricity and Magnetism
Solution
Two identical cells each of e.m.f 2v and internal resistance 1.0\(\Omega\) are connected in parallel. The combination is connected to an external load of 1.5\(\Omega\). Calculate the current in the circuit
A
0.57A
B
1.00A
C
1.14A
D
2.00A
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
I = \(\frac{E}{R + r} = \frac{2}{1.5 + 0.5}\)
= 1A
Source: Myschool Past Questions Archive (Record #34230)
Physics
WAEC / WASSCE 2003
Electricity and Magnetism
Which of the following factors does not affect the the electric resistance of a wire?
A
length
B
mass
C
temperature
D
cross sectional area
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 #34231)
Physics
WAEC / WASSCE 2003
Electricity and Magnetism
Solution
A current of 10A passes through a conductor for 10s, calculate the charge flowing through the conductor
A
100.0C
B
10.0C
C
1.0C
D
0.1C
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Q = it
= 10 x 10 = 100C
Source: Myschool Past Questions Archive (Record #34232)
Physics
WAEC / WASSCE 2003
Electricity and Magnetism
Solution
A lamp is rated 240V 60W, calculate the resistance of its filament
A
240\(\Omega\)
B
360\(\Omega\)
C
960\(\Omega\)
D
1440\(\Omega\)
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
R = \(\frac{V^2}{P} = \frac{240^2}{60} = 960\Omega\)
Source: Myschool Past Questions Archive (Record #34233)
Physics
WAEC / WASSCE 2003
Electricity and Magnetism
One difference between a reactance and a resistance in an alternating current circuit is
A
resistance is larger than reactance
B
energy is not diaaipated in a reactance
C
energy is stored in a resistance but not in a reactance
D
the current in a reactance is lower than in the resistance
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #34235)
Physics
WAEC / WASSCE 2003
Electricity and Magnetism
Solution
Which of the following statements about an electromagnet is not correct?
A
it is a temporary magnet
B
its strength depends on the current
C
its stength depends on the number of turns in its coil
D
it has permanent poles
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
properties of an electromagnet include, 1) its strength depends on current and the number of turns in the coil,2) it is temporary, 3)it does not have permanent poles, and 4) its direction of flow is determined by the direction of current flowing through the coil.
Source: Myschool Past Questions Archive (Record #34237)
Physics
WAEC / WASSCE 2003
Electricity and Magnetism
Diagram
The diagram above shows the resultant electric field pattern due to two electric point charges P and S. Which of the following statements is correct?
Attached Figure / Diagram
A
P is negatively charged while S is positively charged
B
P is positively charged while S is negatively charged
C
both P and S are positively charge
D
both P and Q are negatively charged
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #34265)
Physics
NECO SSCE 2003
Electricity and Magnetism
Calculate the amount of heat generated in an external load of resistance 8Ω if an alternating current of peak value 5 A is passed through it for 100s.
A
2.0 × 104J
B
1.0 × 104J
C
5.0 × 103J
D
4.0 × 103J
E
2.0 × 102J
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 #42908)