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Physics
WAEC / WASSCE 2009
Magnetism, Electromagnetism & AC Circuits
Solution
(a) State two essential differences between a moving coil galvanometer and a d.c. generator.
(b) Explain the term eddy currents and state two devices in which the currents are applied.
(c) State the principle on which the potentiometer is based when it is functioning.
(d) A source of e.m.f. 110 V and frequency 60Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 2A, the potential differences across the resistor is 80 V and that across the inductor is 40 V. Draw the vector diagram of the potential differences across the inductor, the capacitor and the resistor.
Calculate the:
(i) potential difference across the capacitor;
(ii) capacitance of the capacitor;
(iii) inductance of the inductor. [\pi = 3.14]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
a) Differences between a d. c. generator and a moving coil galvanometer.
D. C. generator
moving coil galvanometer
converts mechanical energy to electrical energy
converts electrical energy to mechanical energy.
uses split rings or commutator
uses hair springs.
rotation of coil is continuous
rotation of coil is incomplete
uses carbon brushes (as terminals)
uses jeweled bearings (as terminals)
(b) Eddy currents are currents induced in a conductor when subjected to varying magnetic field.
Any valid additional information e.g.
eddy current flows in a circular path or closed loops;
eddy current generates heat;
eddy current cannot flow through gaps or slots;
The currents move in such a direction as to oppose the change producing them.
Devices in which eddy currents are applied
pointers of sensitive electric meters
sensitive mass balances
brakes in large electric motors
speedometers in automobiles
detection of cracks in railway tracks
detection of metals
(c) Principle on which a potentiometer is based
When a steady current is allowed to pass through a uniform wire, equal lengths of the wire will have equal potential differences.
OR
The p.d across a length of a wire is (directly) proportional to the length provided the wire has a uniform cross section.
(d)
Vector diagram showing
V\(_L\) with arrow
V\(_C\) with arrow
V\(_R\) with arrow
(i) V\(^2\) = V\(_R^2\) + (V\(_L\) - V\(_C\))\(^2\)
110\(^2\) = 80\(^2\) + (40 – V\(_C\))\(^2\)
(40 – V\(_C\)) = \(\sqrt{(110 + 80) (110 – 80)}\)
40 – V\(_C\) = + 75.5
V\(_C\) = 40 + 75.5
V\(_C\) = 115.5V
(ii) X\(_c\) \(\frac{V_c}{I} = \frac{1}{2 \pi f c}\)
C = \(\frac{1}{2 \pi fV_c}\)
C \(\frac{2}{2 \times 3.14 \times 60 \times 115.5} = 45.9\muF\)
(iii) \(V_L = IX_L = I \times 2 \pi fL}\)
L = \(\frac{V_l}{I \times 1 \times 2 \pifVc}\)
C = \(\frac{2}{2 \times 3.142 \times 60 \times 2}\)
= 0.053H
Source: Myschool Past Questions Archive (Record #53505)
Physics
WAEC / WASSCE 2009
Atomic, Quantum, Nuclear Physics & Electronics
Solution
(a) Briefly explain the following terms:
(i) emission line spectra;
(ii) line absorption spectra.
(b) Draw a labeled diagram showing the structure of a simple type of photocell and explain its mode of operation.
(c) State two
(i) reasons to show that x-rays are waves;
(ii) uses of x-rays other than in medicine.
(d) An electron jumps from an energy level of -1.6eV to one of -1.4 eV in an atom. Calculate the energy and wavelength of the emitted radiation. [ h = 6.6 x 10\(^{-34}\) Js; c = 3.00 x 10\(^8\) ms\(^{-1}\); eV = 1.6 x 10\(^{-19}\) J ]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) Emission line spectra consist of distinct and separate bright lines of definite wavelengths on a dark background.
Any valid additional information e.g
- They are obtained when light from a luminous source undergoes dispersion and is observed directly.
- When an electron moves from one energy level to another energy level, line spectra are observed
- The colour of the spectra is a characteristic of the source.
(ii) Line absorption spectra are obtained when light passes through a cool gas and certain wavelengths of the light are absorbed. This gives series of dark lines, each corresponding to one of the wavelengths absorbed.
(b) Diagram
When radiation of appropriate frequency falls on the cathode (with photosensitive surface) electrons are emitted. Because the anode is positive with respect to the cathode, the electrons are attracted to the anode. The electrons flow completes the circuit and current flows.
(c)(i) Wave properties of X-rays
- diffraction
- interference
- reflection
- refraction
- polarization
(ii) Uses of X-rays other than in medicine
to study crystal structures
to detect presence of cracks in welded parts
in artistic works for determining the authenticity of such works
at security check points to scan
X-ray spectroscopy for identifying isotopes of elements.
(d)
\DeltaE = E\(_i\) - E\(_f\)
= - 1.6 – (- 10.4)
= 8.8 e V or 1.41 x 10\(^{-18}\) J
E = hf = hc/\lambda
or
\lambda = hc/E
= 6. 6 x 10\(^{-34}\) x 3 x10\(^{8}\)
8.8 x 1.6 x 10\(^{-19}\)
\lambda = 1.4 x 10\(^{-7}\) m
Source: Myschool Past Questions Archive (Record #53508)
Physics
WAEC / WASSCE 2009
Waves, Sound & Vibrations
Solution
Give three observations in support of de Broglie's assumption that moving particles behave like waves
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
- Moving electrons exhibits wavelength/frequency
- electrons show refraction
- electrons show reflection
- electrons show interference
- electrons show diffraction patterns
Source: Myschool Past Questions Archive (Record #53511)
Physics
WAEC / WASSCE 2009
Reflection, Refraction & Dispersion of Light
Solution
State three methods of polirizing an unpolarized light
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Method of polarizing an unpolarized light;
- by reflexion
- by double refraction
- selective absorption by crystals
- polarizing by pile of plates
- passing unpolarised light through a polariod
Source: Myschool Past Questions Archive (Record #53513)
Physics
WAEC / WASSCE 2009
Equilibrium of Forces, Moments & Flotation
Solution
You are provided with two meter rules and other necessary apparatus.
Place one of the rules on a knife edge and determine its centre of gravity C. Mark this pos with a piece of chalk.
Read and record the mass \(M_R\) of the metre rule written on the reverse side of it.
Attach the mass M= 100g firmly to the rule AB at C using sellotape.
Suspend the metre rule by two parallel threads of length h = 40 cm each at the 10 cm marks. Ensure that the graduated face to the metre rule is facing upwards.
Set the rule AB into a small angular oscillation about the vertical axis through its centre of gravity.
Determine the time, t for 20 complete oscillations. Evaluate the period T and T\(^{2}\)
Read and record the value of d in meters.
Keeping d constant throughout the experiment, repeat the procedure for other values of h = 50, 60, 70, and 80 cm. In each case determine the corresponding values of f T and T. Tabulate your reading. (x) Plot a graph of T on the vertical axis and h on the horizontal axis.
Determine the slope S, of the graph. Evaluate k =s, where Q =2 S, Q 250P
State two precautions taken to ensure accurate results.
(b)i. Define the term couple as it relates to rotational or oscillatory systems.
ii. Give two practical application of a couple in everyday life.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Center of gravity C = 49.5cm
Mass of metre rule (MR) = 135g, d = 80cm = 0.8m
h(cm)
d(m)
t(sec)
T=\(\frac{\text{t(sec)}}{20}\)
T\(^{2}\)(Sec\(^{2}\))
40
0.8
28.0
\(\frac{28.0}{20}\)=1.40
1.9600
50
0.8
29.0
\(\frac{29.0}{20}\)=1.45
2.1025
60
0.8
31.0
\(\frac{31.0}{20}\)=1.55
2.4025
70
0.8
34.0
\(\frac{34.0}{20}\)=1.70
2.8900
80
0.8
37.0
\(\frac{37.0}{20}\)=1.85
3.4225
Slope (s) = \(\frac{\bigtriangleup {T}^{2}}{\bigtriangleup {h(cm)}}\) = \(\frac{(3.42-0.40)sec^{2}}{(80-21)cm}\)
= \(\frac{3.02sec^{2}}{59cm}\)
= 0.0512sec\(^{2}\)cm\(^{-1}\)
K = \(\frac{S}{Q}\), Q = \(\frac{2}{25d^{2}}\)
K = \(\frac {\frac{S}{2}}{25d^{2}}\)
S = 0.0512sec\(^{2}\)cm\(^{-1}\), d = 50cm = 0.8cm
K = \(\frac{0.0512sec^{2}cm^{-1} \times 25 \times (0.80)^{2}}{2}\)
K = \(\frac{0.0512 \times 25 \times 0.6400sec^{2}cm^{-1}}{2}\) = 0.4096sec\(^{2}\)cm\(^{-1}\)m\(^{2}\)
Precautions: i
- ensured smooth and regular oscillations in a horizontal plane.
- ensured there was no error due to parallax when reading my stopwatch.
(b)i. Couple -This can be defined as two forces of the same magnitude acting in opposite directions but not on the same line.
ii. Practical application of couple include:
- Opening or closing taps.
- Turning a spanner.
- Setting a flywheel in motion about an axis through its center
- Motion of a spinning spindle.
- Riding/pedaling of bicycle
Source: Myschool Past Questions Archive (Record #60022)
Physics
WAEC / WASSCE 2009
Gas Laws & Heat Transfer
Solution
You are provided with a measuring cylinder, two different tins labeled C and D, a thermometer, and other necessary materials.
Use the measuring cylinder provided to measure 100 cm of water and pour it into the tin label.
Heat the water in the tin almost to boiling (90^\circC).
Remove the tin and place it on a cork or wooden stand.
Insert the thermometer into the tin and record the temperature of the water every minute starting from 85^\circC until the temperature falls to 60^\circC.
Repeat the experiment with the tin labeled D using exactly the same volume of water and temperature range. Tabulate your readings.
On the same graph sheet and using the same axis and scales, plot two graphs of temperature on the vertical axis and time on the horizontal axis from the readings obtained using tins C and D.
Label the graphs appropriately as C and D to correspond with the tins used.
From each graph, read off the time taken to cool from 85^\circC to 65^\circC.
State two precautions taken to ensure accurate results.
(b)i. Explain how heat losses by radiation and convection are minimized in a vacuum flask.
ii. State four factors that affect the rate of evaporation of a liquid in an open container.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of value (observation)
S/N
t(min)
Tc(^\circC)
To(^\circC)
1
0.00
85.00
85.00
2
1.00
83.00
84.00
3
2.00
81.00
83.00
4
3.00
79.00
81.00
5
4.00
78.00
79.00
6
5.00
75.00
78.00
7
6.00
73.00
76.00
8
7.00
71.00
75.00
9
8.00
70.00
73.00
10
9.00
68.00
71.00
11
10.00
66.00
69.50
12
11.00
65.00
68.00
13
12.00
63.00
66.00
14
13.00
62.00
64.00
15
14.00
60.00
62.00
16
15.00
-
61.00
17
16.00
-
60.00
Time taken to cool from 85^\circC to 65^\circC
Tin C = 11 minutes
Tin D = 12.40 minutes
Precautions: i
- ensured there was no error due to parallax when reading my thermometer.
- ensured that the liquid in the tins was gently stirred.
(b)i. Silvered wall minimized radiation.
Insulating cork minimized convection.
(i) Factors affecting the rate of evaporation.
- Temperature
- Nature of liquid.
- Humidity
- Area of liquid surface exposed
- Draught
Source: Myschool Past Questions Archive (Record #60026)
Physics
WAEC / WASSCE 2009
Current Electricity & Electric Circuits
Diagram
Solution
1. Connect the circuit as shown in the diagram above. PQ is a potentiometer wire 100 cm long and R is a standard resistor of 5\(\Omega\).
2. With the jockey J not making contact with PQ, close the switch. Read and record the ammeter reading I. Open the switch.
3. Use the jockey to make contact with PQ at the 20cm mark such that PJ = I = 20 cm. Close the switch, read and record the value \(I_i\) of the ammeter. Evaluate I\(^{-1}\).
4. Repeat the procedure for other values of I = 35, 50, 65, and 80 cm. In each case, determine the corresponding values of \(I_i\), and I\(^{-1}\). Tabulate your readings.
Plot a graph of I\(^{i}\) on the vertical axis and \(I_i\), on the horizontal axis, starting both axes from the origin (0, 0).
From your graph deduce the value, of \(I_o\) of \(I_i\), when I\(^{-1}\)= 0.
Evaluate \(I_o\)e
State two precautions taken to ensure accurate results.
(b)) Define the e. m.f. of a battery
ii. A cell X e.m.f. 1.00 V is balanced by a length of 40.0 cm on a potentiometer wire. Another cell Y is balanced by a length of 60.0 cm on the same wire. Calculate the e.m.f. of Y.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
S/N
L(cm)
Ii(A)
L\(^{-1}\)(cm\(^{-1}\))
1
20
1.00
0.050
2
35
0.90
0.029
3
50
0.85
0.020
4
65
0.80
0.015
5
80
0.75
0.013
When L\(^{-1}\) = 0, Ii = 0.75A
\(I_o\) = 75A
\(\frac{I_{o}}{I}= \frac{0.75A}{0.20A}\) = 37.5
SEE THE GRAPH ABOVE...
Precautions: i
- ensured there was no error due to parallax when reading my ammeter.
- ensured that I removed the key after each reading.
(b)i. E.m.f is the total work done in pushing one coulomb of electricity round an external resistor and the internal resistor of the source, or the total energy obtained per coulomb from a cell or battery
ii. \(\frac{Ey}{Ex} = \frac{Iy}{Ix}\)
Iy = 60cm
Ex = 1.00v
Ix = 40.cm
Ey = ?
\(\frac{Ey}{1.00}=\frac{60}{40}\)
Ey = \(\frac{60 \times 1.00}{40}\) = 1.50.v
Source: Myschool Past Questions Archive (Record #60030)
Physics
UTME / JAMB 2008
Calorimetry, Latent Heat & Change of State
Solution
In a good thermos flask, heat loss is reduced through the following features EXCEPT?
A
the lens surface
B
the insulating cork
C
silvered walls
D
the cork lid
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
The silver coating on the inner bottle of a thermos flask is used to prevent heat transfer by radiation
Source: WAEC / JAMB / NECO Classroom Archive (Record #14044)
Physics
UTME / JAMB 2008
Circular Motion, Gravitation & Simple Harmonic Motion
Solution
If a wheel 1.2 m in diameter rotates at one revolution per second, calculate the velocity of the wheel.
A
3.6 ms-1
B
3.77 ms-1
C
4.0 ms-1
D
7.5 ms-1
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
V = wr = 2\pif r = 2 x (22)/7 x 1 x 0.6
= 3.77ms-1
Source: WAEC / JAMB / NECO Classroom Archive (Record #14017)
Physics
UTME / JAMB 2008
Circular Motion, Gravitation & Simple Harmonic Motion
Solution
A short chain is sometimes attached to the back of a petrol tanker to?
A
generate more friction
B
ensure the balancing of the tanker
C
caution the driver when overspeeding
D
conduct excess charges to the earth
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The friction between the tanker and air creates charges which the chain conducts to the ground.
Source: WAEC / JAMB / NECO Classroom Archive (Record #14064)
Physics
UTME / JAMB 2008
Circular Motion, Gravitation & Simple Harmonic Motion
Solution
A force F required to keep a 5 kg mass moving round a cycle of radius 3.5 m at a speed of 7 ms-1. What is the speed, if the force is tripled?
A
4.0 ms-1
B
6.6 ms-1
C
12.1 ms-1
D
21.0 ms-1
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The force which is centripetal
F = \(\frac{MV^2}{R}\)
= \(\frac{5 \times 7^2}{3.5}\)
= 70
Since force is 70 N when speed is 7m/s
Speed = ? when force 70N is tripled → 70 * 3 = 210N
210 = \(\frac{5\timesv^2}{3.5}\)
210 \times 3.5 = 5 * v\(^2\)
735 \div 5 = v\(^2\)
v\(^2\) = 147
v =12.12m/s
Source: WAEC / JAMB / NECO Classroom Archive (Record #14484)
Physics
UTME / JAMB 2008
Circular Motion, Gravitation & Simple Harmonic Motion
Solution
A body of mass 4 kg resting on a smooth horizontal plane is simultaneously acted upon by two perpendicular forces 6N and 8N. Calculate the acceleration of the motion
A
2.5 ms-1
B
3.0 ms-1
C
4.0 ms-1
D
4.5 ms-1
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Smooth plane has negligible friction
∴ only the resultant of the perpendicular forces is acting on the body, which is given as
F = \(\sqrt{ 6^2 + 8^2}\)
= \(\sqrt{64 + 36}\)
= \(\sqrt{100}\)
= 10N
But F = Ma
∴ a = F/M = 10/4 = 2.5 ms-1
Source: WAEC / JAMB / NECO Classroom Archive (Record #14485)
Physics
WAEC / WASSCE 2008
Work, Energy, Power & Simple Machines
Solution
A ball is dropped and it hits the floor at a point A. It rebounds upward to a point B. While moving from A to B its?
A
kinetic energy is increasing
B
potential energy is increasing
C
potential energy is decreasing
D
kinetic energy remains constant
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The Kinetic Energy of a body is energy possessed by a body as a result of its motion.
Hence, while in motion the kinetic energy of a body changes into potential energy as it rebounds from the floor
Source: WAEC / JAMB / NECO Classroom Archive (Record #35026)
Physics
UTME / JAMB 2008
Work, Energy, Power & Simple Machines
Diagram
Solution
The efficiency of the pulley system shown above is 80%. find the effort E required to lift the load of 1200N
Attached Figure / Diagram
A
275N
B
325N
C
375N
D
573N
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
VR = No. of pulleys = 4
Efficiency = \(\frac{M.A}{V.R}\)
cross multiply
\(\frac{80}{100} * V.R\) = M.A
\(\frac{80}{100} * 4\) = M.A
M.A = 3.25
where M.A = \(\frac{load}{effort}\)
∴ Effort = \(\frac{1200}{3.2} * V.R\)
= 375N
Source: WAEC / JAMB / NECO Classroom Archive (Record #14487)
Physics
UTME / JAMB 2008
Equilibrium of Forces, Moments & Flotation
Solution
I. All the three forces must be concurrent.
II. The upward force is equal to the downward force.
III. The algebraic sum of the moment about any point must be zero.
which of the above conditions must hold for a body acted upon by a system of three coplanar forces equilibrium?
A
I and II only
B
I and III only
C
II and III only
D
I, II and III
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The condition (ii) is a condition for an equilibrium of parallel forces only while conditions (i) and (iii) are correct for both parallel and non- parallel coplanar forces.
No specifications are made, so all conditions apply.
Source: WAEC / JAMB / NECO Classroom Archive (Record #14018)