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Physics
WAEC / WASSCE 1998
Lenses & Optical Instruments
What is the image distance of an object placed at a distance of 2f from a converging lens of focal length f?
A
f/4
B
f/2
C
f
D
2f
E
4f
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 #33602)
Physics
WAEC / WASSCE 1998
Current Electricity & Electric Circuits
The images formed by diverging lenses are always
A
diminished, virtual and inverted
B
diminished, virtual and errect
C
diminished inverted and real
D
magnidied, virtual and errect
E
magnified, real and inverted
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 #33603)
Physics
WAEC / WASSCE 1998
Lenses & Optical Instruments
Solution
A converging lens of focal length 5cm is used as a magnifying glass by a man whose nearpoint is 35cm. Calculate the magnification given by the lens
A
8
B
7
C
6
D
5
E
4
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
\(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\)
= \(\frac{1}{5} - \frac{1}{35} = \frac{30}{175}\)
U = \(\frac{175}{30}\) = 5.8m
\(\frac{v}{u} = \frac{35}{5.8}\)
= 6
Source: Myschool Past Questions Archive (Record #33633)
Physics
WAEC / WASSCE 1998
Lenses & Optical Instruments
An astronomical telescope having an objective of focal length 100cm and an eyepiece of focal length 10cm is used in normal adjustment, calculate the separation of the lenses
A
0.10m
B
0.90cm
C
1.10m
D
1.80m
E
2.20m
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 #33636)
Physics
WAEC / WASSCE 1998
Lenses & Optical Instruments
Solution
A lantern gives an image of 3m square of a slide 7.62cm square on a screen. If the screen is 10m from the projection lens of the lantern, calculate the focal length of the lens
A
40.3cm
B
26.1cm
C
24.8cm
D
0.7cm
E
0.3cm
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
M = \(\frac{V_H}{U_H} = \frac{300}{7.62}\)
= 39.3
U = \(\frac{V}{M} = \frac{1000}{39.3}\) = 25.44
\(\frac{1}{f} = \frac{I}{U} + \frac{I}{V} = \frac{I}{25.44} = \frac{I}{1000} = \frac{1025.44}{25440}\)
f = \(\frac{25440}{1025.44} = 24.8\)
Source: Myschool Past Questions Archive (Record #33638)
Physics
WAEC / WASSCE 1998
Current Electricity & Electric Circuits
Solution
The intensity of light falling on the film in a camera depend on the i. brightness of the object. ii. diameter of the aperture stop. iii. speed of the shutter. Which of the statements is/are correct/
A
i, ii and iii
B
i and iii only
C
i and ii only
D
ii and iii only
E
iii only
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
iii. Speed of the shutter: This directly controls how long the film is exposed to the incoming light. A faster shutter speed reduces the time light falls on the film, thereby reducing the intensity (light per unit time) effectively recorded.
Source: Myschool Past Questions Archive (Record #33640)
Physics
WAEC / WASSCE 1998
Waves, Sound & Vibrations
Solution
the shortest length of the air column in a resonance tube at resonance is 0.12m and the next resonance length is 0.37m. calculate the frequency of vibration given that the speed of sound in air is 340m-1
A
1360Hz
B
694Hz
C
680Hz
D
34Hz
E
340Hz
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
f = \(\frac{v}{2(l_2 - l_1)} = \frac{340}{2(0.37 - 0.12)} = \frac{34}{0.5}\) = 680
Source: Myschool Past Questions Archive (Record #33645)
Physics
WAEC / WASSCE 1998
Waves, Sound & Vibrations
Solution
A sonometer wire under a tension of N10N, produces a frequency of 250Hz when plucked. Keeping the length of the wire constant, the tension is adjusted to produce a new frequency of 350Hz. Calculate the new tension
A
39.2N
B
19.6N
C
14.2N
D
7.4N
E
5.1N
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
f \(\alpha\) \(\sqrt{T}\)
\(\frac{f_1}{f_2} = \sqrt{\frac{T_1}{T_2}}\)
\(\frac{250}{350} = \sqrt{\frac{10}{T_2}}\)
\(\frac{5}{7} = \sqrt{\frac{10}{T_2}}\)
\(\frac{25}{49} = \frac{10}{T_2}\)
T2 = \(\frac{49 \times 10}{25} = 19.6\)
Source: Myschool Past Questions Archive (Record #33651)
Physics
WAEC / WASSCE 1998
Waves, Sound & Vibrations
Solution
Calculate he wavelength of a note which is one octave lower than a note of 256Hz in a medium in which the speed of sound is 352ms-1
A
0.69m
B
1.38m
C
2.75m
D
5.50m
E
9.30m
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(\lambda\) = \(\frac{v}{f} = \frac{352}{256}\)
= 1.38
Source: Myschool Past Questions Archive (Record #33657)
Physics
WAEC / WASSCE 1998
Atomic, Quantum, Nuclear Physics & Electronics
Which of the following are both mechanical and transverse?
A
infra-red rays
B
gamma rays
C
sound waves
D
water waves
E
micro waves
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 #33659)
Physics
WAEC / WASSCE 1998
Waves, Sound & Vibrations
Solution
An electromagnetic wave of frequency 5.0 x 1014Hz, is incident on the surface of water of refractive index \(\frac{4}{3}\). Taking the speed of the wave in air as 3.0 x 108ms-1, calculate the wavelength of the wave in water.
A
2.2 x 10-6m
B
1.7 x 10-7m
C
8.0 x 10-7m
D
6.0 x 10-7m
E
4.5 x 10-7m
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
\(_an_w = \frac{\lambda_a}{\lambda_w}\)
\(\frac{4}{3} = \frac{\lambda_a}{\lambda}\)
\(\lambda_w = \frac{v}{f}\)
= \(\frac{3.0 \times 10^8}{5.0 \times 10^{14}} = 0.6 \times 10^{-6}\)
\(\lambda_w = \frac{3}{4} \lambda_a = \frac{3}{4} \times 0.6 \times 10^{-6}\)
= 4.5 x 10-7m
Source: Myschool Past Questions Archive (Record #33666)
Physics
WAEC / WASSCE 1998
Circular Motion, Gravitation & Simple Harmonic Motion
Solution
The gravitational force of the moon is one-sixth that of the earth. If a body weighs 6.0N on the moon. Calculate its weight on the earth
A
36.0N
B
12.0N
C
6.0N
D
2.0N
E
1.0N
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
\(\frac{1}{6}F_E = F_{MOON}\)
FE = R = 6Fm
= 6 x 6 = 36N
Source: Myschool Past Questions Archive (Record #33668)
Physics
WAEC / WASSCE 1998
Current Electricity & Electric Circuits
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
Current Electricity & Electric Circuits
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
Current Electricity & Electric Circuits
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)