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Physics
WAEC / WASSCE 2003
Current Electricity & Electric Circuits
Solution
(a) Explain diffusion
(b) State one factor that can affect the rate of diffusion.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Diffusion is the tendency of a gas to mix with another and fill an empty space as a result of the constant random motion of the molecules
(b) Factors affecting rate of diffusiom are;
(i) mass
(ii) temperature
(iii) density
Source: Myschool Past Questions Archive (Record #54016)
Physics
WAEC / WASSCE 2003
Atomic, Quantum, Nuclear Physics & Electronics
Solution
(a) List two types of waves, apart from light, that can be plane polarized.
(b) State two applications of plane polarized light.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Plane polarized waves are:
(i) radio waves
(ii) x-rays.
(iii) infrared radiation etc
(b) Application of plane polarized light
– To reduce glare from reflecting surfaces.
– In sunglasses to reduce the intensity of incident sunlight.
– To determine sugar solution concentration etc.
Source: Myschool Past Questions Archive (Record #54017)
Physics
WAEC / WASSCE 2003
Scalars, Vectors & Relative Motion
Solution
Explain the rise of water in a glass capillary tube using the kinetic theory.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Glass and water are made up of molecules. The force of attraction between glass molecules and water molecules is stronger than that among water molecules, hence the resultant force acts upwards leading to the rise of water in the glass tube.
Source: Myschool Past Questions Archive (Record #54018)
Physics
WAEC / WASSCE 2003
Atomic, Quantum, Nuclear Physics & Electronics
Solution
(a) Explain the term electrolyte and give two examples.
(b) State the relationship between the mass of a substance liberated during electrolysis and the charge passed.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Electrolyte is a substance in liquid or molten state which conducts electricity and undergoes decomposition e.g. metallic salts, acids and alkalis.
(b) M \(\alpha\) Q
M = kQ. Mass m is proportional to charge Q where K is a constant of proportionality.
Source: Myschool Past Questions Archive (Record #54022)
Physics
WAEC / WASSCE 2003
Linear Motion, Projectiles & Kinematics
Solution
(a) What is the principle upon which the lighting in fluorescent tubes operate?
(b) State two factors which determine the colour of projection, g the acceleration of free fall due to gravity light from a fluorescent tube
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)The principles is electric discharge through low pressure gas.
(b) Colour of light is determined by
– Nature of gas in the tube
– Nature of coating inside the tube
Source: Myschool Past Questions Archive (Record #54023)
Physics
WAEC / WASSCE 2003
Elasticity, Density, Pressure & Fluids
Solution
Define;
(i) Elasticity
(ii) Young's modulus
(iii) Force constant
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(i) Elasticity is the ability of a material to regain its original shape and size after undergoing a stretching or compression.
(ii) Young's modulus is the ratio of tensile stress to tensile strain.
(iii) Force constant is the amount of force that causes unit extension.
Source: Myschool Past Questions Archive (Record #54025)
Physics
WAEC / WASSCE 2003
Elasticity, Density, Pressure & Fluids
Solution
A force of 40 N applied at the end of a wire of length 4m and diameter 2.00 mm produces an extension of 0.24 mm. Calculate the;
(a) stress on the wire;
(b) strain in the wire (\(\pi = 3.142\))
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Stress = \(\frac{F}{A} = \frac{40}{3.14(1 \times 10^{-3})^2} = 12.74 \times 10^{6} Nm^{-2}\)
(b) Strain = \(\frac{e}{I} = \frac{24 \times 10^{-5}}{4} = 6 \times 10^{-5}\)
Source: Myschool Past Questions Archive (Record #54029)
Physics
WAEC / WASSCE 2003
Waves, Sound & Vibrations
Solution
(a) What is meant by the wave-particles duality of matter?
(b) mention one physical phenomenon, in each case, that can be explained in terms of the wave nature and particle nature of light.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) It means that light behaves either as a wave or as a particle but not as both simultaneously.
(b) As wave nature e.g. diffraction, interference etc. As particle nature e.g. heated body radiation, photoelectric effect etc.
Source: Myschool Past Questions Archive (Record #54032)
Physics
WAEC / WASSCE 2003
Newton's Laws, Momentum & Collisions
Solution
(a) Define (i) Linear momentum; (ii) Impulse
(b) State the principle of conservation of linear momentum.
(c) A tractor of mass 5.0 x 10\(^{3}\)kg is used a tow a car of mass 2.5 x 103 kg. The tractor moved with a speed of 3.0 ms\(^{-1}\) just before the towing rope becomes taut. Calculate the:
(i) Speed of the tractor immediately the rope becomes taut
(ii) Loss in kinetic energy of the system just after the car has started Moving;
(iii) Impulse in the rope when it jerks the car into motion.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) Linear momentum is the product of the mass of a body and its velocity.
(ii) Impulse is the product of a large force and short time for which the force acts.
(b) The principle of conservation of linear momentum states that in system of colliding bodies the total momentum of the system is constant provided no external forces act on the system.
(c) (i) \(m_1u_1 + m_2U_2 = v(m_1 + m_2)\) 5 x 10\(^{3}\) x 3 + 0
= v(5 x 10\(^{3}\) + 2.5 x 10\(^{3}\)) v = \(\frac{15 \times 10^3}{7.5 \times 10^3}\) = 2m/s
(ii) Initial K.E = \(\frac{1}{2}mv^2 = \frac{1}{2} MV^2 = \frac{1}{2} times 5 \times 10^{3} \times 3^2\)
= 2.25 x 10\(^4\)
Final K.E = \(\frac{1}{2}mv^2 = \frac{1}{2} (5 + 2.5) \times 10^3 \times 2^2\)
= 1.5 x 10\(^{4}\)J
Loss in K.E of the couple = (2.25 - 1.5) x 10\(^4\)
= 75 x 10\(^3\)J
(iii) Impulse = (\(mu_1U_1 + m_2U_2\)) - v
(\(mu_1 +m_2) (0 + 5 \times 10^3) - 2 (2.25 + 5) 10^3\) = 0Ns
Source: Myschool Past Questions Archive (Record #54033)
Physics
WAEC / WASSCE 2003
Work, Energy, Power & Simple Machines
Solution
(a) Using the kinetic theory of matter, explain why;
(i) Evaporation causes cooling
(ii) Boiling water changes to steam without any change in temperature, although heat is being supplied to the water.
(b) (i) State Boyle's law.
(ii) With the aid of a labelled diagram, describe an experiment to illustrate the relationship between the volume and pressure of a given mass of gas at constant temperature. (iii) State two precautions necessary to obtain accurate results.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) Liquids consist of molecules in random motion having different velocities, hence near the surface molecules with higher velocities and kinetic energy escape from the surface. Therefore the average kinetic energy of the remaining liquid is reduced. Since the average kinetic energy is a measure of temperature, therefore temperature is lowered. Hence evaporation causes cooling.
(ii) The heat supplied to boiling water increases the kinetic energy of the molecules and also weaken the bonds between the molecules. Hence at boiling the heat supplied is used in breaking the bonds, therefore molecules escaped as steam without change in temperature.
(b) (i) Boyle's law states that the volume of a fixed mass of gas at constant temperature varies inversely as the pressure upon it.
(ii)
Introduce dry air into burette (B) and close the tap. Read and record the barometer reading H i.e. a.p., keeping B steady, raise the reservoir R, read the mercury head h from the scale, at the same time the corresponding volume V of air from the Repeat burette. Repeat the experiment with various values of H, and determine the corresponding values of h and v. Evaluate P = H \pm h and V. Graph of P against V\(^{-1}\)
The graph P against V\(^{-1}\) is a straight line graph passing through the origin, showing that pressure is inversely proportional to volume.
(iii) Precautions;
– I ensured that dry air is used.
– I avoided parallax error when reading metre rule.
Source: Myschool Past Questions Archive (Record #54038)
Physics
WAEC / WASSCE 2003
Waves, Sound & Vibrations
Solution
a) (i) Illustrate, using a ray diagram, how an image can be formed by a convex mirror.
(ii) State one advantage rid one disadvantage of using a convex mirror as a iving mirror.
(iii) Explain the action of a compound microscope.
(b) Illustrate using labelled diagrams only, sonometer wire of length I, vibrating at its fundamental (ii) first overtone (iii) second overtone
(c) A tuning fork vibrating at a frequency of 512 Hz is held over the top of a jar filled with water and fitted with a tap at the buttom. If the jar is 60 cm tall and the speed of sound is 350 ms\(^{-1}\), determine the possible resonance position(s).
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i)
(ii) Advantages of using a convex mirror as driving mirror.
– It has wide field of view
– Image is always erect.
Disadvantages;
– Image is always smaller than the object
– It gives a false impression of the distance as image seems further away.
(iii) The objective lens forms the image of an object placed beyond its focal length within the focal length of eye piece. This image is real inverted and magnified. This image serves as object to the eye piece which functions as a simple microscope to form a final virtual and magnified image.
(c) \(\lambda = \frac{v}{f} = \frac{350}{512} = 0.68m\)
1st resonance position = \(\frac{\lambda}{f} = \frac{0.68}{4} = 0.17m\)
2nd resonance position = \frac{3\lambda}{4} \frac{3 \times 0.68}{4} = 0.51m\)
3rd resonance position = \(\frac{5\lambda}{4} = \frac{5 \times 0.85}{4} = 0.85m\)
Since 0.85m is greater than the length of the jar (0.6m), therefore only the first and second resonance positions are possible.
Source: Myschool Past Questions Archive (Record #54040)
Physics
WAEC / WASSCE 2003
Magnetism, Electromagnetism & AC Circuits
Solution
(a)(i) What is meant by neutral point in a magnetic field?
(ii) Draw and label a diagram to show the pattern and direction of the magnetic field produced around a straight current-carrying wire.
(b) When is an ammeter said to be (i) Sensitive (ii) accurate?
(c)(i) Explain, using a labelled diagram, how a delicate magnetic material could be protected, from the Earth's magnetic field.
(ii) A charge of 1.6 x 10\(^{-19}\) C enters a magnetic field of flux density 2.0 T with a velocity of 2.5 x 10\(^7\) ms\(^{-1}\) at an angle of 30^\circ with the field. Calculate the magnitude of the force exerted on the charge by the field.
(d) State the laws of electro-magnetic induction.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) (i) Neutral point is a point at which the resultant magnetic flux density is zero.
(ii)
(b)(i) (ii) An ammeter is accurate if the current it mea-sures is close to the true value of the current flowing through the instrument.
(c)
The material is surrounded by a thick wall of ferro-magnetic material which does not allow magnetism to pass through them. This shields or protects the material from the earth's magnetic field.
(ii) F = qvBsin\(\theta\) = 1.6 x 10\(^{-19}\) x 2.5 x 10\(^{7}\) x 2 x 0.5 = 4 x 10\(^{2}\) N
(d) Electromagnetic induction laws are
- Whenever there is a change in the magnetic flux linked with a circuit, an e.m.f is induced, the magnitude of which is proportional to the rate of change of the flux linked with the circuit.
- The direction of the induced current is always such as to oppose the change producing it.
Source: Myschool Past Questions Archive (Record #54042)
Physics
WAEC / WASSCE 2003
Atomic, Quantum, Nuclear Physics & Electronics
Solution
(a)(i) Explain photoelectric emission.
(ii) State four applications of photoelectric emission.
(b) Draw and label a diagram showing the structure of a simple type of a photocell and explain its mode of operation.
(c) In a photocell, no electrons are emitted until the threshold frequency of light is reached.
(i) Explain what happens to the energy of the light before emission of electrons begin.
(ii) State one factor that may affect the number of emitted electrons.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) Photoelectric emission is the ejection of electrons from the surface of a metal when light of appropriate wavelength and frequency is incident on the metal.
(ii) Applications of photoelectric emission i.e. photo cells:
- Burglar alarms
- Photo meters
- Automatic doors
- Sound production from film tracks
- T.V. camera etc
(b)
When light of a set wavelength and frequency falls on the cathode, electrons are emitted. The electrons flow toward the anode. The flow of these electrons constitute current.
(c) (i) For energy of light with frequency less than threshold frequency, complete absorption takes place. The energy absorbed is used to raise the energy of electrons or overcome the potential barrier in the photoelectric material.
(ii) The number of emitted electrons are affected by:
- intensity
- Number of photons.
Source: Myschool Past Questions Archive (Record #54047)
Physics
WAEC / WASSCE 2003
Equilibrium of Forces, Moments & Flotation
Solution
You are provided with a metre rule, a knife edge, two pieces of thread and two masses \(m_1\) and \(m_2\)
Record the values of \(m_1\) and \(m_2\).
Balance the metre rule horizontally on the knife edge and record the balance point G.
With the knife edge at the 60 cm mark of the metre rule, suspend \(m_1\) at the 20 cm mark and \(m_2\) at a suitable mark such that the rule balances horizontally as illustrated in the diagram above.
Record the positions Y of \(m_1\) and Q of \(m_2\).
Evaluate 1= P - Y and d = Q - P
Repeat the procedure for four other positions of m, at 18, 16, 14 and 12 cm marks.
In each case, evaluate and record l and d.
Tabulate your readings.
Plot a graph of l on the vertical axis against d on the horizontal axis.
Determine the slope of the graph.
State two precautions taken to ensure accurate results.
(b}i. With the aid of a diagram, indicate the forces acting on the metre rule in the experimental set-up above.
ii. Define moment of a force about a point and state its S.1. unit.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
\(M_1\) = 20.0g
\(M_2\) = 50.0g
C.G = 50CM
S/N
L=P - Y (cm)
d = =Q - P (cm)
Y (cm)
1
40.00
37.00
20.00
2
42.00
37.50
18.00
3
44.00
38.30
16.00
4
46.00
38.70
14.00
5
48.00
39.30
12.00
Slope (s) = \(\frac{\bigtriangleup {L}}{\bigtriangleup {d}} = \frac{48-4}{39.2-28.2} = \frac{44}{10.4}\) = 4.23
Precautions:
Ensure that there is no air interference when determining the position of the centre of gravity (G) of the metre-rule.
Ensure that the suspended mass did not touch the table
Avoid error due to parallax when reading the metre rule.
Zero error of the metre rule must be noted.
(b)i.
ii. The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force. It is also defined as the turning effect of the force about a point. It is measured in Nm - SI unit.
Source: Myschool Past Questions Archive (Record #60082)
Physics
WAEC / WASSCE 2003
Reflection, Refraction & Dispersion of Light
Solution
Using the diagram above as a guide, carry out the following instructions:
Fix. the drawing paper provided on the drawing board
Place the mirror vertically with its longer side resting on the drawing paper. Trace the outline AB of the mirror. Remove the mirror.
Draw a normal PQ to meet the outline at the middle Q.
Draw a straight line through A to meet the outline of the mirror at the right angle.
Trace the incident ray, Q with pins \(P_1\) and \(P_2\) so that it meets the perpendicular line through A at C such that CA=X= 1.0m.
Replace the mirror on its outline. Locate the images of \(P_1\)\(P_2\) through the mirror using two other pins \(P_3\) and \(P_4\) so that \(P_3\) and \(P_4\) and the images of \(P_1\) and \(P_2\) are in a straight line.
Remove the mirror and pins \(P_3\) and \(P_4\).
Draw a straight line through the pin points to meet AB at Q and CA produced at D.
Measure and record angle ACQ as è\(_{1}\), and angle ADCQ as è\(_{2}\). Also, record the value of x. Evaluate è = (è\(_{1}\) + è\(_{2}\)), x\(^{-1}\) and tan è.
Repeat the procedure for four other values of x = 2.0, 3.0, 4.0 and 5.0cm. Tabulate your readings.
Plot a graph of tan è on the vertical axis against x on the horizontal axis.
Determine the slope, s, of the graph. Evaluate k = 2s
State two precautions taken to ensure accurate results. Attach your traces to your answer booklet.
(b)i. Distinguish between regular and diffused reflections.
ii. An object is situated 25cm in front of a plane mirror. Determine the distance of the image from the object. What is the size of the image relative to the object?
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of values / observations
S/N
\(\theta_{1}^{o}\)
\(\theta_{2}^{o}\)
x(cm)
x\(^{-1}\)(cm\(^{-1}\))
\(\theta\)=½(\(\theta_{1}\) + \(\theta_{2}\))
tan\(\theta\)
1
83.0
81.0
1.0
1.00
82.0
7.115
2
76.0
74.0
2.0
0.50
75.0
3.732
3
69.0
70.0
3.0
0.33
69.5
2.675
4
62.0
63.0
4.0
0.25
62.5
1.921
5
57.0
58.0
5.0
0.20
57.5
1.570
Slope (s) = \(\frac{\bigtriangleup {tan \theta}}{\bigtriangleup {x}^{-1}} = \frac{7.2-1.2}{1-0.14} = \frac{6}{0.86}\) = 6.977 \(\approx\) 6.98
K = 25 = 2x 6.977 = 13.954, \(\approx\) 14
Precautions: i
The optical pins must be straight and not bent. i.e pins are vertical
Avoid error due to parallax
Ensure reasonable spacing of pins
Ensure that the object pins and the image pins are in a straight line.
Ensure the Use of a sharp-pointed pencil so as to obtain neat traces and accurate angular measurement.
(b)i. Regular reflection occurs on a smooth surface giving rise to uniform (same) turning back of light rays when they fall on a surface. Diffused reflection occurs on a rough surface giving rise to a non-uniform (different) turning back to light rays when they fall on a surface.
ii.
Distance of the image from the object = 25-(-25)
= 25 + 25 =50cm
The size of the image is the same as the object.
Source: Myschool Past Questions Archive (Record #60083)