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Physics
WAEC / WASSCE 2009
Electricity and Magnetism
When a conductor mounted on an insulating stand is charged and left for sometime, the conductor eventually loses all its charges. This is because the
A
electric charges evaporate from the surface of the conductor
B
charges are connected to earth
C
charges on the conductor are neutralized by opposite charges from the surrounding air
D
charges ionize the surrounding air
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 #35202)
Physics
WAEC / WASSCE 2009
Electricity and Magnetism
Solution
An isolated metal sphere of radius R, carrying an electric charge Q, is situated in the medium of relative permitivity, Er. A test charge is placed at a point p, distance r from the surface of the sphere. Let Eo represent the permitivity of free space. The electric potential at p is given by the expression
A
\(\frac{Q}{4 \pi E_o E_r}\)
B
\(\frac{Q}{4 \pi E_o E_r (R +r)}\)
C
\(\frac{Q}{4 \pi E_o E_r(R - r)}\)
D
\(\frac{Q}{4 \pi E_o E_rR}\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The electrical potential V at a point P located at a distance r from the surface of an isolated charged metal sphere is given by
V = \(\frac{\text{Q}}{4\pi\varepsilon}\).\(\frac{1}{\text{d}}\)
Where d = R + r is the distance from the center of the sphere to point P
\(\epsilon\) = \(\varepsilon_r\)\(\varepsilon_o\) is the permittivity of the medium, with \(\epsilon_r\) being the relative permittivity and \(\epsilon_o\) the permittivity of free space
\(\therefore\) V = \(\frac{\text{Q}}{4\pi \varepsilon_r \varepsilon_o}.\frac{1}{R + r}\)
Source: Myschool Past Questions Archive (Record #35205)
Physics
WAEC / WASSCE 2009
Electricity and Magnetism
An isolated metal sphere of radius R, carrying an electric charge Q, is situated in the medium of relative permitivity, Er. A test charge is placed at a point p, distance r from the surface of the sphere. Let Eo represent the permitivity of free space. The magnitude of the electric field intensity at P is given by the expression
A
\(\frac{Q}{4 \pi E_o E_r r^2}\)
B
\(\frac{Q}{4 \pi E_o E_r(R + r)^2}\)
C
\(\frac{Q}{4 \pi E_o E_r(R - r)^2}\)
D
\(\frac{Q}{4 \pi E_o E_rR^2}\)
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 #35206)
Physics
WAEC / WASSCE 2009
Electricity and Magnetism
A resistor of resistance R is connected to a battery of negligible internal resistance. If a siilar resistor is connected in series with it the
A
effective resistance of the circuit is halved
B
total power dissipated is doubled
C
total current in the circuit is halved
D
terminal voltage is halved
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 #35207)
Physics
WAEC / WASSCE 2009
Electricity and Magnetism
Solution
A cell of e.m.f. 1.5V is connected in series with a resistor of resistance 3.0\(\Omega\). A voltmeter connected across the cell registers 0.9V. Calculate the internal resistance of the cell
A
2.0\(\Omega\).
B
3.0\(\Omega\).
C
5.0\(\Omega\).
D
6.0\(\Omega\).
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
\(\frac{V}{R} = \frac{E}{R + r}\)
\(\frac{0.9}{3} = \frac{1.5}{3 + r}\)
r = 2\(\Omega\).
Source: Myschool Past Questions Archive (Record #35208)
Physics
WAEC / WASSCE 2009
Electricity and Magnetism
Solution
A battery of e.m.f. 12.0V and internal resistance0.5\(\Omega\) is connected to 1.5\(\Omega\) and 4.0\(\Omega\) series resistor. Calculate the terminal voltage of the battery.
A
13.0V
B
11.0V
C
3.0V
D
1.0V
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(\frac{V}{R} = \frac{E}{R + r} = \frac{V}{5.5} = \frac{12}{6}\)
v = 11.0V
Source: Myschool Past Questions Archive (Record #35210)
Physics
WAEC / WASSCE 2009
Electricity and Magnetism
If the direction of the curent in a straight wire is reversed, the magnetic field
A
remains the same
B
becomes parallel to the wire
C
ceases to exist
D
is oppositely directed
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 #35211)
Physics
WAEC / WASSCE 2009
Electricity and Magnetism
Solution
An ammeter connected to an a.c circuit records 5.5. The peak current in the circuit is
A
7.8A
B
7.1A
C
3.8A
D
3.5A
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
\(I_o = I_{rms} \times \sqrt{2} = 5.5 \times 1.41 = 7.8A\)
Source: Myschool Past Questions Archive (Record #35213)
Physics
NECO SSCE 2009
Electricity and Magnetism
In the electrical method of making a magnet, the polar the magnet formed depends on the
A
amount of current passed
B
direction of the current
C
nature of the material used
D
orientation of the magnetic material in space
E
size of the magnetic material
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 #43061)
Physics
NECO SSCE 2009
Electricity and Magnetism
A transformer has an efficiency of 92.5%. the ratio number of turns in the primary coil to that in the secondary coil is 128 : 45. If the current passing through the secondary coil is 9.0A, calculate the current passing through the primary-coil.
A
2.93A
B
3.16A
C
3.42A
D
25.60A
E
27.68A
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 #43066)
Physics
NECO SSCE 2009
Electricity and Magnetism
Solution
When a resistor of resistance R is connected across a.c. the terminal p.d. of the cell is reduced to three-quarter its e.m.f. The cell’s internal resistance in terms of R is
A
\(\frac{R}{4}\)
B
\(\frac{R}{3}\)
C
\(\frac{R}{2}\)
D
2\(\frac{R}{3}\)
E
R
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Given that the terminal potential difference (p.d.) of the cell is reduced to three-quarters of its e.m.f. when a resistor \( R \) is connected, we can express this as:
\(V = \frac{3}{4} E\)
The relationship between terminal voltage \( V \), e.m.f. \( E \), internal resistance \( r \), and current \( I \) is given by:
\(V = E - Ir\)
The current \( I \) can be expressed as: \(I = \frac{V}{R}\)
Substituting for \( I \) in the voltage equation:
\(V = E - \left(\frac{V}{R}\right) r\)
Substituting \( V = \frac{3}{4} E \):
\(\frac{3}{4} E = E - \left(\frac{\frac{3}{4} E}{R}\right) r\)
Rearranging gives: \(\frac{3}{4} E = E - \frac{3Er}{4R}\)
Simplifying: \(\frac{3}{4} E - E = -\frac{3Er}{4R}\)
This leads to: \(-\frac{1}{4} E = -\frac{3Er}{4R}\)
Dividing both sides by \( -E \): \(\frac{1}{4} = \frac{3r}{4R}\)
Cross-multiplying results in: \(R = 3r\)
Finally, solving for \( r \): \(r = \frac{R}{3}\).
Source: Myschool Past Questions Archive (Record #43070)
Physics
NECO SSCE 2009
Electricity and Magnetism
Solution
Which of the following cells does not require a dipolar for effective and efficient delivery of current?
A
Daniel cell
B
Dry Leclanche cell
C
Nickel-iron accumulator
D
Voltaic cell
E
Wet Leclanche cell
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The dry Leclanché cell's use of a paste electrolyte allows it to function effectively without the need for a dipolar structure, making it advantageous for portability and stability in various applications
Source: Myschool Past Questions Archive (Record #43071)
Physics
NECO SSCE 2009
Electricity and Magnetism
An effort P applied at one end of a crowbar just overcomes the resistance W at the lid of a tin. The mechanical advantage of the crowbar is expressed as
A
W + P
B
P - W
C
\(\frac{w}{p}\)
D
\(\frac{p}{w}\)
E
wp
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 #43082)
Physics
NECO SSCE 2009
Electricity and Magnetism
The phenomenon of photoelectric effect is used in the following EXCEPT
A
automatic doors
B
bugler alarm
C
electric fire alarm
D
sound production from film tracks
E
television cameras
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 #43083)
Physics
NECO SSCE 2009
Electricity and Magnetism
Solution
Electric charges on the surface of a conductor would be most evenly distributed if the shape of the conductor is
A
conical
B
cubical
C
cylindrical
D
parabolic
E
spherical
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
Excess charges placed on a spherical conductor repel and move until they are evenly distributed.
Source: Myschool Past Questions Archive (Record #43088)