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Physics
WAEC / WASSCE 2006
Current Electricity & Electric Circuits
Solution
(a) Distinguish between the forces of adhesion and cohension
(b) Give one example each of the forces of adhesion and cohesion.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Forces of adhesion are forces between molecules of different substances while forcea of cohesion are forces between molecules of the same substance.
(b) Examples: Adhesion — force between water molecules and glass molecules.
Cohesion — force between mercury molecules.
Source: Myschool Past Questions Archive (Record #53724)
Physics
WAEC / WASSCE 2006
Elasticity, Density, Pressure & Fluids
Solution
Explain why water in a narrow glass tube has a concave meniscus while mecury in the same tube, has a convex meniscus
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
For water in a glass tube, the forces of adhesion between the water and glass molecules are greater than the force of cohesion between the water molecules hence water clings to glass forming a concave meniscus. On the other hand, the forces of cohesion between mercury molecules are greater than those between mercury molecules and glass molecules, so mercury molecules cling to each other forming convex meniscus.
Source: Myschool Past Questions Archive (Record #53725)
Physics
WAEC / WASSCE 2006
Current Electricity & Electric Circuits
Solution
(a) State two applications of electrolysis.
(b) Explain what is meant by the electrochemical equivalent of copper is 3.3 x 10\(^{-7}\) kgC\(^{-1}\)
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Applications of electrolysis:
— Electrotyping
— Electroplating
— Purification of metals
— Extraction of metals
— Callibration of ammeter
(b) It means that when a charge of one coulomb IC is passed through a copper voltameter 3.3 x 10\(^{-7}\) kg of copper is on the cathode.
Source: Myschool Past Questions Archive (Record #53726)
Physics
WAEC / WASSCE 2006
Atomic, Quantum, Nuclear Physics & Electronics
Solution
The uncertinty in determining the duration during which an electron remains in a particular energy level before returning to the ground state is 2.0 x 10\(^{-9}\)s. Calculate the uncertainty in determining its energy at that level [Take \(\frac{h}{2\pi} = h = 1.054 \times 10^{-34}\) Js]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
\(\Delta E \geq \frac{h}{2\pi} \times \frac{1}{2t} = 1.054 x 10^{-37} \times \frac{1}{2.0 \times 10^{-9}}\)
\(\Delta E \geq 5.27 \times 10^{-25}\)J
Source: Myschool Past Questions Archive (Record #53727)
Physics
WAEC / WASSCE 2006
Reflection, Refraction & Dispersion of Light
Solution
A paralIel beam of unpolarized light is incident on a plane glass of refractive index 1.60 at an angle to the normal. If the reflected beam is completely polarized, calculate the angle of incidence of the beam.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
tan\(\theta = \mu \text{(Brewsters law)}\)
\(\theta tan^{-1} = \mu = tan{-1} 1.60 = 58^o\)
Source: Myschool Past Questions Archive (Record #53729)
Physics
WAEC / WASSCE 2006
Waves, Sound & Vibrations
Solution
List three observations in support of the de-Broglie's assumptions that moving particles behave like waves.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
— Electrons show diffraction patterns
— They have frequencies or wavelengths
— They show fraction or reflection
— The molecular velocities can determined in gases
Source: Myschool Past Questions Archive (Record #53730)
Physics
WAEC / WASSCE 2006
Work, Energy, Power & Simple Machines
Solution
(a) State two;
(i) laws of solid friction;
(ii) advantages of friction;
(iii) methods of reducing friction.
(b) Draw and label a diagram of a pulley system with velocity ratio of 5.
(c)(i) Show that the efficiency L, force ratio M.A. and the velocity ratio V.R. of a machine are related by the equation \(E = \frac{M.A.}{V.R}\) x 100%.
(ii) The efficiency of a machine is 80%. Calcuate the work done by a person using the machine to raise a load of 300 kg through a height of 4 m.[ g = 10 ms\(^{-2}\) ]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) Laws of solid friction states that:
— friction opposes motion
— friction depends on the nature of e surfaces
— friction is independent of the area of the surfaces in contact
— frictional force F is proportional to the normal reaction R
— friction is dependent of relative velocity between the surfaces.
ii) Advantages of friction:
— it prevents man from ipping when walking
— it provides gripping effect of beIts in machines
— for grinding when using grinding ones
— used for the production of electric charges used in the brake systems of vehicles etc.
iii) Methods of reducing friction:
— by lubrication by streamlining
— by smoothing
— by using ball bearings or rollers.
b)
(c)(i) Efficiency = \(\frac{\text{Work output}}{\text{work output}}\) x 100%
= \(\frac{\text{load x distance by load}}{\text{effort x distance moved by effort}}\)
= M.A x \(\frac{1}{VR} \times \frac{100 %}{1}\)
(ii) \(e = \frac{\text{Work output}}{\text{Work input}}\) x 100
Work input = \(\frac{\text{Work output}}{E}\) x 100
= \(\frac{300 \times 10 \times 4 \times 100}{80}\)
= 15,000J or 15KJ
Source: Myschool Past Questions Archive (Record #53736)
Physics
WAEC / WASSCE 2006
Lenses & Optical Instruments
Solution
(a) With the aid of ray diagrams, explain total internal reflection.
(b) Describe, with the aid of a labelled diagram, the essential features of an astronomical telescope in normal adjustment.
(c) A converging lens forms a real image of a real object. If the magnification is 2 and the distance between the image and the object is 90.0 cm, determine the
(i) focal length of the lens;
(ii) object distance for which the image would be the same size as the object.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Consider a ray of light travelling from a denser medium (glass) to a less dense medium (air) at a small angle of incidence.
(i)
As angle of incidence increases i, the angle of refraction r also increases, at a point when r = 90^\circ, i = C (the critical angle)
If the angle of incidence i is made greater than the critical angle C no refraction in the less dense medium, the ray is totally internally reflected.
(b)
An astronomical telescope consists of two converging lenses i.e an objective lens of long focal length fo and an eye piece of short focal length fe. In normal adjustment the principal foci of the lenses coincide, the separation of the lenses d = fo + fe.
(c)(i) But M = \(\frac{v}{u} = 2 = \frac{v}{u}\)
v = 2u
u + v = 90
u + 2u = 90
u = 30cm
f = \(\frac{uv}{u + v} = \frac{u \times 2u}{u + 2u} = \frac{2u}{3}\)
= \(\frac{2 \times 30}{3}\) = 20cm
(iii) U = 2f = 2 x 20
= 40cm
Source: Myschool Past Questions Archive (Record #53737)
Physics
WAEC / WASSCE 2006
Current Electricity & Electric Circuits
Solution
(a)(i) Name and explain the common defects of a primary cell.
(ii) State two advantages of a secondary cell over a primary cell.
(b) Draw a labelled diagram to show the essential parts of a dry leclanche cell.
(c)(i) Explain why six accumulators each of e.m.f 2V connected in series can be used to start the engine of a car whereas eight dry cells each of e.m.f 1.5 V connected in series cannot be used.
(ii) Name the materials used for the positive terminal, the negative terminal and the electrolyte in a
I. leclanche cell;
II. charged lead acid accumulator.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) Primary cell defects:
— Polarization
— Local action Polarization
— This is due to hydrogen bubbles covering the positive electrode (copper plate). Local action
— This is due to the use of impure zinc or the gradual wearing away of the negative electrode zinc plate.
(ii) Advantages of secondary cell over a primary cell:
— a secondary cell can be recharged while primary cell cannot.
— secondary cell can supply current for a longer period while the primary cell has a short life-span.
(b)
(c)(i) The internal resistance of the dry cells is much larger than that of the accumulators, hence accumulators will produce larger currents needed to start the engine.
(ii)
Positive terminal
Negative terminal
Electrolyte
Leclanche cell
Carbon rod
Zinc rod
Ammonium chloride (NH\(_4\)
charged lead acid accumulator
Lead (IV) oxide or Lead peroxide
Lead
Lead Tetra oxo-sulphate (VI) acid (H\(_2\)SO\(_4\))
Source: Myschool Past Questions Archive (Record #53738)
Physics
WAEC / WASSCE 2006
Atomic, Quantum, Nuclear Physics & Electronics
Solution
(a) Define
(i) proton number;
(ii) nucleon number;
(iii) isotopes.
(b) A nuclide \(^A_ZX\) emits \(\beta\)-particle to form a daughter nuclide Y. Write a nuclear equation to illustrate the charge conservation.
(c) The radioactive nuclei \(^{210}_{84}P_o\) emits an \(\alpha\) - particle to produce \(^{206}_{82}P_b\). Calculate the energy, in MeV, released in each disintegration.
Take the masses of \(^{210}_{84}P_o\) = 209.936730 u;
\(^{206}_{82}P_b\) = 205.929421 u;
\(^{4}_{2}He\) = 4.001504 u;
and that 1u = 931 MeV
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) Proton number
— the number of protons in the nucleus of an atom
(ii) Nucleon number — the total number of protons and neutrons in the nucleus of are atom.
(iii) Isotopes — are atoms or nuclides of the samc element having the same number of protons or atomic number but different neutrons or mass numbers
(b) \(^A_Z \to ^A_{Z+}Y + ^0_{-1}\)
\(^{210}_{84}Po \to ^{206}_{82}Pb + ^4_2He + Energy\)
Mass detect = 209.936730u - (205.929421 + 4.001504)u
209.9367330u - 209.930925u
= 5.805 x 10\(^{-3}\)U
Energy released = 5.805 x 10\(^{-3}\) x 931
= 5.40MeV
Source: Myschool Past Questions Archive (Record #53740)
Physics
WAEC / WASSCE 2006
Equilibrium of Forces, Moments & Flotation
Solution
(a) State the conditions for the equilibrium of a rigid body acted upon by parallel forces.
(b)(i) Describe an experiment to determine the mass of a metre rule using the principle of moments.
(ii) State two precautions necessary to obtain accurate results in the experiment described in (b)(i) above.
(c) A bullet of mass 120 g is fired horizontally into a fixed wooden block with a speed of 20 ms\(^{-1}\). If the bullet is brought to rest in the block in 0.1s by a constant resistance, calculate the (i) magnitude of the resistance; (ii) distance moved by the bullet in the wood.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) — The sum of clockwise moments about any point is equal to the sum of anticlockwise moments about the same point
— The sum of the forces in one recticn is equal to the sum of the forces in the opposite direction.
b)
Method: Suspend the metre rule on a thread loop from clamp. Adjust its position until the metre rule balances horizontally and obtain its centre of gravity G. Suspend known mass M at a point on one side of the rule and just the position of the loop until the rule balances horizontally. Read and record the distances y and x in the diagram above. sing the principle of moments Mgx = Mgy
m = \(\frac{My}{x}\)
Therefore the mass M of the metre rule is calculated.
ii) Precautions:
— Firmly fixed the loop
— Avoid draught
— Avoid parallax when reading metre rule.
.
c)(i) Declaration a = \(\frac{v - u}{t} = \frac{0 — 20}{0.1} = — 200m/s^2\)
Resistance F = Ma = 0.12 x 200 = 24N
ii) Distance S = \(\frac{v^2 - u^2}{2a} = \frac{0^2 - 20^0}{2 \times 200}\)
= \(\frac{400}{400} = 1.0m\)
Source: Myschool Past Questions Archive (Record #53741)
Physics
WAEC / WASSCE 2006
Current Electricity & Electric Circuits
Solution
You are provided with a constantan wire, a 2\(\Omega\) standard resistor, an accumulator E, an ammeter A, a key K, and other necessary apparatus.
Measure and record the e.m.f.of the accumulator provided.
Connect a circuit as shown in the diagram above.
Close the key, read and record the ammeter reading \(l_o\) when the crocodile clip is not in contact with the constantan wire.
Open the key. With the clip making contact with the wire, when I = 90cm, close the key. Read and record the ammeter reading /. Evaluate l\(^{-1}\).
Repeat the procedure for l= 80, 70, 60 and 50cm.
In each case, read and record the ammeter reading and evaluate l\(^{-1}\). Tabulate your readings.
Plot a graph of l on the vertical axis against l\(^{-1}\) on the horizontal axis.
Determine the slope, s, of the graph and its intercept, c on the vertical axis.
Evaluate k = \(\frac{c}{s}\).
Using your graph, determine the Current i when / = 55cm.
State two precautions taken to ensure accurate results.
(b)i. Explain what is meant by the potential difference between two points in an electric circuit.
ii. State two factors on which the resistance of a resistance wire depends.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of values (Observation)
\(l_o\) (amps)
l(cm)
l\(^{-1}\)
l(amps)
0.65
90
0.011
0.70
0.70
80
0.013
0.75
0.75
70
0.014
0.80
0.80
60
0.017
0.85
0.85
50
0.020
0.90
7.
(8). Slope (s) = \(\frac{\bigtriangleup {I}}{\bigtriangleup {/}^{-1}} = \frac{0.65}{0.024}\) = 27.1
(9). Intercept (c) on the vertical axis = 0.43 amps
(10). Evaluate, K = \(\frac{c}{s} = \frac{0.43}{27.1}\) = 0.016
(11) The value of the current l when l = 55cm, l\(^{-1}\) = 0.018cm,
Therefore; l on the graph = 0.875 amps
Precautions:
The key was removed when the reading were not being teken.
I ensured that terminals were clean.
I ensured that the connetions were tight.
Zero error was noted and corrected in the ammeter/ metre rule
Parallax error was avoided in anmmeter/metre rüle.
(b)i. The potential differenoe between two points in an electric ciroult the work done in Joules when one coulomb of electric charge moves from one point to the other
ii. Factors on which the resistance of a resistance wire depends are:
- resistivity of the wire/ nature of the wire
- length of the wire.
- cross sectional area of the wire /thickness of the wire.
- temperature
Source: Myschool Past Questions Archive (Record #60073)
Physics
WAEC / WASSCE 2006
Circular Motion, Gravitation & Simple Harmonic Motion
Solution
You have been provided with a metre rule, a clamp, and a set of masses.
Clamp the metre rule to the edge of the bench such that 90cm of the rule projects from the edge as shown in the diagram above. Ensure that the rule is capable of performing oscillatory motion.
Fix a mass M = 50g at the free end of the rule.
Deflect the rule slightly such that it performs vertical oscillation.
Determine the time t for 10 complete oscillations.
Calculate the period T of the oscillations and evaluate T\(^{2}\)
Repeat the procedure for four other values of M = 100, 150, 200, and 250g. In each case determine and record the corresponding values of t, T, and T\(^{2}\). Tabulate your readings.
Plot a graph of T\(^{2}\) on the vertical axis against M on the horizontal axis, starting both axes from the origin (0,0).
Determine the slopes, of the graph and its intercept C on the vertical axis.
Evaluate k = 4\(\pi\)/s. [Take \(\pi\) = \(\frac{22}{7}\)].
From your graph, determine the period T, when M= 180g.
State two precautions taken to ensure accurate results.
(b)i. Explain simple harmonic motion.
ii. Define period and frequency, with respect to a simple harmonic motion.
Precautions:
l ensured that the metre rule was firmly clamped
Readings were repeated
Parallax was avoided when readings on the stopwatch/clock were taken.
zero error was noted and corrected on the stopwatch/clock.
(b)i. Simple harmonic motion is a motion in which the acceleration is proportional to the displacement from a fixed point and is directed towards the point.
ii. Period is the time taken by an oscillatory body to make one complete oscillation.
Frequency: is the number of complete oscillations performed in one second.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of values (observation
Mass (M)g
Time (t) sec.
T
T\(^{2}\)
1
3.5
0.35
0.123
2
4.2
0.42
0.176
3
4.9
0.49
0.240
4
5.5
0.55
0.303
5
6.0
0.60
0.360
Slope (s) = \(\frac{y}{x} = \frac{\bigtriangleup {T}^{2}}{\bigtriangleup{M}} = \frac{0.26}{220}\) = 0.001
Intercept (c) on the vertical axis = 0.65
Evaluate, K = \(\frac{4 \pi}{S} = \frac{4 \times {\frac{22}{7}}}{0.001}\)
\(\frac{12.57}{0.0001}\) = 12571
From the graph the value of period T, where M = 180g is;
= \(\sqrt 0.275\) = 0.524, i.e T = 0.524
Source: Myschool Past Questions Archive (Record #60062)
Physics
WAEC / WASSCE 2006
Reflection, Refraction & Dispersion of Light
Solution
You have been provided with a rectangular glass prism, optical pins, and other necessary apparatus. Using the above diagram as a guide, carry out the following instructions:
Fix the drawing paper provided to the drawing board
Place the glass prism on the drawing paper and trace the outline, ABCD of the prism
Remove the prism, mark a point O on AB such that AO is about one-quarter of AB
Draw a normal through point O. Also draw an incident ray to make an angle i = 25 with the normal at O. Fix two pins at \(P_1\) and \(P_2\) On the incident ray.
Replace the prism. Fix two other pins at \(P_3\) and \(P_4\) such that the pins appear to be in a straight line with the images of the pins at \(P_1\) and \(P_2\) when viewed through the block along DC
remove the prism. Join points Pa and P4 and produce it to meet DC at 1. Also, draw a line to join Ol (
With O as center and using any Concinient radius, draw a circle to Cut the incident ray and the refracted ray at E and H respectively. Maintain this radius throughout the experiment
Draw the perpendiculars EF and GH. Measure and record d= EF and I= GH.
Repeat the procedure for four other values of i = 35^\circ, 45, 55^\circ, and 65^\circ respectively. In each case measure and record d and I
Plot a graph of d on the vertical axis against I on the horizontal axis
Determine the slope of the graph
State two precautions taken to ensure accurate results. [Attach your traces to your answer booklet)
(b)i. State Snell's law.
ii. Calculate the critical angle for a water-air interface. [refractive index of water = \(\frac{4}{3}\)]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of values (Observation)
S/N
iº
d(cm)
i (cm)
1
25^\circ
0.85
0.55
2
35^\circ
1.15
0.75
3
45^\circ
1.50
0.95
4
55^\circ
1.65
1.10
5
65^\circ
1.85
1.20
Slope of the graph = \(\frac{y}{x} = \frac{\bigtriangleup {d}}{\bigtriangleup {L}} = \frac{1.4}{0.9}\) = 1.56
Precautions:
I ensured that pins were vertical.
l ensured that pins were reasonably spaced (at least 1 cm apart) and pinpoints were seen on traces.
I ensured neat traces (sharp pencils)
I noted and corrected zero errors on the metre rule.
l avoided parallax using metre rule/protractor
(b) Snell's law states that for a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is always a constant. Or The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for the media concerned. Or
\(\frac{\text {sin i}}{\text {sin r}}\) = a constant for a given pair of media where
i = angle of incidence and
r = angle of refractions.
ii. \(\frac{1}{\text { sin C}}\) = \(_{a}\)\(n_w\)
Where C = critical angle, and
\(_{a}\)\(n_w\) = refractive index of water.
i.e. sin C = \(\frac{3}{4}\) = 0.75
C = sin\(^{-1}\), 0.75 = 48.6^\circ
Source: Myschool Past Questions Archive (Record #60068)
Physics
UTME / JAMB 2006
Atomic, Quantum, Nuclear Physics & Electronics
Diagram
Solution
Use the figures above to answer the question below
Which of the following diagrams represents correctly an n-p-n transistor?
Attached Figure / Diagram
A
A
B
B
C
C
D
D
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
.
Source: Myschool Past Questions Archive (Record #72264)