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Physics
WAEC / WASSCE 2009
Waves and Optics
The angle of incidence of a ray of light on a plane mirror is 55o. Determine the angle between the reflected ray and the mirror
A
35o
B
45o
C
55o
D
110o
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 #35194)
Physics
WAEC / WASSCE 2009
Waves and Optics
A concave mirror forms a magnified and errect image only when the object is placed
A
at the centre of curvature
B
at the focus
C
between the focus and the pole of the mirror
D
beyound the radius of curvature
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 #35195)
Physics
WAEC / WASSCE 2009
Mechanics and Properties of Matter
Solution
A thin lens is placed 50cm from an illustrated object. The image produced has linear magnification of \(\frac{1}{4}\). Calculate the power of the lens in dioptres
A
2.5D
B
5.0D
C
10.0D
D
25.0D
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
\(\frac{1}{v} = \frac{1}{u}\)
m = \(\frac{v}{u} = \frac{1}{4} = \frac{v}{50}\)
v = 12.5
But \(\frac{1}{v} + \frac{i}{u} = \frac{1}{f}\)
= \(\frac{1}{50} + \frac{1}{12.5} = f\)
f = 10m
P = \(\frac{100}{10} = 10D\)
Source: Myschool Past Questions Archive (Record #35196)
Physics
WAEC / WASSCE 2009
Waves and Optics
Which of the following characteristics of light determines its colour?
A
velocity
B
wavelength
C
amplitude
D
intensity
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 #35197)
Physics
WAEC / WASSCE 2009
Electricity and Magnetism
Which of the following is not a part of the electro-magnetic spectrum?
A
x-rays
B
microwaves
C
infrared radiation
D
alpha rays
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 #35198)
Physics
WAEC / WASSCE 2009
Waves and Optics
Sound interference is necessary to produce the phenomenon of
A
beats
B
doppler effect
C
acoustic resonance
D
echo
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 #35199)
Physics
WAEC / WASSCE 2009
Mechanics and Properties of Matter
A satellite in circular motion around the earth does not have
A
a gravitational force acting on it
B
a uniform velocity
C
an acceleration
D
a centripetal force acting on it
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 #35200)
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
Heat Energy & Temperature
Solution
A wire of resistivity 4.40 x 10-5\(\Omega\) cm has a cross sectional area of 7.50 x 10-4 cm2. Calculate the length of the wire that will be required to make a 4.0\(\Omega\) resistor
A
82.50 cm
B
68.18cm
C
15.90cm
D
11.95cm
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
L = \(\frac{R.A}{e} = \frac{4 \times 7.50 \times 106{-4}}{4.40 \times 10^{-5}} = 68.18cm\)
Source: Myschool Past Questions Archive (Record #35209)
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)