10000Frs
Hand book
 Multiple Choice Questions.
 sLong Structure Questions.
Physics
A necessary tool for every Advanced Level Physics student
Tel : 675150172 / 675408009 /691203523/699221743
<<ADVANCED LEVEL PHYSICS
Page 1
PREFACE
The role of Physics in scientific research and technology cannot be overemphasized. For
this reason, those who have any dreams of contributing to the advancement of technology need to
be given a good base or foundation especially at the Advanced Level where they receive the basics
for future specialization in the various fields. Taking into consideration the fact that many students
consider Physics as a monster, this book becomes primordial in assisting Advanced Level Physics
students in establishing this foundation and preparing themselves to face the ever-changing
scientific world.
STRIKING FEATURES ABOUT THE BOOK
 G.C.E. questions from 2001 to 2018 together with their proposed solutions.
 Well calculated spaces to permit the student train himself before casting an eye on the
proposed solutions
Tel: 652787829//651586646
ACKNOWLEDGEMENT
We thank the Almighty God for the wisdom and the strength to have gone about the
development of this material. We believe it will go a long way to improve the
Performance of our students first in GBHS Dschang and all those who will put it into
productive use.
<<ADVANCED LEVEL PHYSICS
Page 2
Special thanks to Mr. Chi Emmanuel the brain behind this project and to the team of
Physics teachers from GBHS Dschang (Mr Lukong Lutu, Mr Tenonfo Zomessi Anaclet ,
Mrs Alanguo Mawamba, Mr Mbetem Simon
JUNE 2001
1. A 100 g calorimeter contains 300 g of water at room temperature. 50 g of ice is added to this calorimeter
and the equilibrium temperature recorded is 282.7 K. Calculate the room temperature. The specific heat
capacity of copper = 380 Jkg-1K-1, the specific latent heat of fusion of ice = 3.25 x 105 JK-1.
2. In figure 1, the meters labeled M1 and M2 each read 1.5 A when the switch K is closed.
M
R3
M
K
R2
100 Ω
S
Figure 1
The source S supplies 300 W to the resistor R1, R2 and 100 ohms respectively. Calculate
(i) The potential difference across R1 (ii) The value for the resistance R2
3. A car is normally threaded with wire loops. Donlop tyres have about 200 loops a tyre. A car running on
such tyres travels at speed of 16.7 ms-1 along a level road such that the magnetic field of the earth makes
an angle of 530 with the axis of the tyre at all the times. If the magnetic field is 1.2 x 10-5 T, calculate
i. The induced current through a loop if the tyre’s diameter is 0.65 m and the resistance per unit
length is 8.0 x 10-3 ohms per meter.
ii. How much power is generated on the tyre due to the motion in the earth’s magnetic field?
iii. What does it suggest to you about the usage of threaded tyres?
4. (a) In an experiment to determine the focal length of a convex lens, f, the magnification, m, was
calculated from different image distance, v. The results are displayed in figure 2.
<<ADVANCED LEVEL PHYSICS
Page 3
m
figure 2
3
2.5
2
1.5
1
20
22
24
26
28
30
32
34
36
38
v/m
40
Use the graph to determine the value for the focal length, f, of the convex lens.
b) State any two advantages that optical fibre has over copper cable used for the transmission of
information.
5. (a) Explain hysteresis curve for rubber.
b) Explain why in practice, car tyres are made with synthetic rubber which has a smaller area of
hysteresis as compared to natural rubber.
c) Sketch on the same set of axes stress – strain curves for the following materials
i) Iron
ii) Glass
6. The following expression gives the variation of electric charge Q through a capacitor C with time t.
−𝑡
𝑄 = 𝐴𝐶(1 − 𝑒 𝐵𝐶 )
A. and B are physical constants. Show that the units of A and B are the volt and the ohm respectively
7. Figure 3 shows a basic amplifier circuit with an n – p – n transistor.
9V
3 kΩ
50 kΩ
V0
Vin
ac
Figure 3
0V
If the voltage Vi is 2.0 V, ac and dc
gain for the transistor is 60. Calculate
i.
The base current
ii.
The collector current
iii.
The output voltage.
On the same axis, sketch graphs to show how
the input voltage and the output voltage vary
with time
8. (a) (i) state Coulomb’s law.
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Page 4
Figure 4 shows how the force, F, varies with the inverse of the square of separation between two equal
charges Q, placed in the medium.
7
F/N
6
5
4
3
2
1
r/10-5 m-2
0
0
1
2
3
4
5
(ii) Use the graph to obtain a value for the permittivity of the medium if Q has a value of 4.4 x 10-3 C.
(iii) Hence obtain a value for the dielectric constant
(b) Two particles carrying charges Q1 = 4.0 µC and Q2 = -3 µC are placed at a distance of 5.0 cm apart
(i) Sketch the electric field lines between the charges
(ii) Determine the point on the line passing through the two charges at which the resultant field is zero.
(c Explain why birds are not often electrocuted when they land on high voltage lines
(d) (i) State Newton’s law of gravitation
(ii) When a planet moves in a circular orbit of radius r, about the sun, the centripetal force is provided by
the gravitational attractive force. Show that the periodic time T of the planet is given by the expression
𝐺𝑀
𝑟3 = 2 𝑇2
4𝜋
𝐺 = Gravitational constant
𝑀 = mass of the sun
(iii) Calculate the period of rotation of the moon about the earth if the radius of the moon is 3.5 x 108 m
and the mass of the earth is 6.0 x 1024 kg
(e) Figure 5 shows how T2 varies with r3 for a planet of mass 7.0 x 1026 kg
<<ADVANCED LEVEL PHYSICS
Page 5
figure 5
T2 /1012 s2
2.5
2
1.5
1
0.5
0
0
1
2
r3 /1027 m3
3
(i) Use the graph to obtain a value for the universal gravitational constant
(f) The mass of the earth is 80 times that of the moon and the distance from the centre of the moon to that
of the earth is 3.5 x 105 m. Calculate the distance from the centre of the earth of the point on the line
joining the centre of the earth and that of the moon where the resultant gravitational field of the earth and
the moon is zero.
9. (a) A ball is projected with an initial speed, u, at an angle 𝜃 to the horizontal. Neglecting air resistance.
(i)
Describe qualitatively the motion of the ball
(ii)
How would air resistance affect the maximum horizontal displacement of the ball?
(b) A bullet is fired from the top of a tall building, 100 m above the ground at an angle of 300 to the
horizontal and at a speed of 300 ms-1. Calculate
(i) The distance of the bullet from the building when it reaches the ground
(ii) The magnitude and direction of the velocity of the bullet as it hits the ground.
(c) A wooden box of mass 0.80 kg is pushed along a horizontal floor by a force of 4.8 N. The motion of
the box is opposed by a frictional force of 1.5 N between the box and the floor and the air resistance Kv2
where K = 6.0 x 10-2 kg and v is the speed of the box.
(i) Sketch a diagram showing the forces acting on the wooden box
(ii) Calculate the maximum speed of the wooden box
(d) (i) state the zeroth law of thermodynamics
(ii) How does this law leads to the definition of temperature?
(e) What is meant by?
<<ADVANCED LEVEL PHYSICS
Page 6
(I) Primary energy sources (ii) Alternate energy sources
Discuss the use of primary and alternate energy sources in Cameroon.
(f) Sea water is trapped in a bay of area 4.0 x 107 m2. The difference in levels of the water in the bay
between
high and low tides is 10 m.
(I) Calculate the average power obtainable for a tidal period of 12 hours if the density of water is 1100
kgm-3.
(ii) Explain the factors which make coastlines not depend on tides as a source of energy.
10. (a) Describe an experiment to investigate how the intensity of α – rays vary with distance from the source
of emission
(b) 8.0 mg of radioisotope of half-life 30 minutes is used for twelve minutes.
i. Explain the terms in italics
ii. What is the amount of the radioisotope remaining?
(c) The fusion of tritium nucleus with a deuterium nucleus releases energy according to the following
equation.
3
2
4
1
1𝐻 + 1𝐻 → 2𝐻 + 0𝑛 + ∆𝐸
(i) Calculate the energy ∆𝐸 that is released.
(ii) Given that the mass of one mole of deuterium is 2.0 g, how much energy is released per kilogram of
deuterium fuel?
Mass of 𝟐𝟏𝑯 = 3.345 x 10-27 kg; Mass of 𝟑𝟏𝑯 = 5.008 x 10-27 kg; Mass of 𝟒𝟏𝑯 = 6.647 x 10-27 kg
Mass of 𝟏𝟎𝒏 = 1.675 x 10-27 kg; Avogadro’s constant = 6.02 x 1023 mol-1; Speed of light = 3.0 x 108 ms1
(d) Describe an experiment to demonstrate ohms law
(e) Explain the deference between
(i) Ohmic and non – ohmic materials
(ii) Electromotive force (emf) and potential difference
(f) Determine
(i) I1, I2, and I3 in figure 6
(ii) The potential difference between A and B.
A
I2
20Ω
6.0 V
24.0 V
I1
15Ω
5Ω
10Ω
I3
Figure 6
B
STUDENT’S PROPOSED ANSWERS TO JUNE 2001
<<ADVANCED LEVEL PHYSICS
Page 7
JUNE 2002
1. Optical fibres are being increasingly used in modern technology
i. What are optical fibres
ii. State the main physical property that makes optical so important today
iii. Describe the use of optical fibres in medicine.
2. The stability of a nucleus depends on the binding energy per nucleon for the particular nucleus
a) What is binding energy?
b) Sketch a graph to show the binding energy versus mass number for the natural existing
nuclei.
On the graph, indicate ranges for possible
i. Fission reactions
ii. Fusion reactions
3. A series of thermodynamic processes are shown in the PV diagram of figure 1. In process ab, 150 J of
heat are added to the system, and in process bd, 600 J of heat are added.
P/Pa
8.0 X 104
b
i.
Calculate the internal energy change in the
process ab.
d
ii.
3.0 X 104
c
c
a
2.0 X 10-3
4.
c
Figure 1
5.0 X 10-3
Calculate the internal energy change
in the process abd.
V/m
(a) Newton’s law of gravitation may be stated as
𝐺𝑚1 𝑚2
𝑟2
i. Describe the different symbols in the equation
ii. What are the units for the physical constant G?
(b) Explain why an equation which is
i.
Not homogeneous with respect to units must be wrong
ii. Homogeneous with respect to units may non – the less be wrong.
𝐹=
<<ADVANCED LEVEL PHYSICS
Page 8
5. Rayleigh scattering by molecules is one cause of signal attenuation. Attenuation due to Rayleigh
1
scattering depends on the wavelength (λ) of the signal and is proportional to 𝜆4 .
a) What do you understand by?
i. Scattering and
(ii) Signal attenuation
b) If a signal of wavelength 850 nm is attenuated by 2.0 dBkm-1 because of Rayleigh scattering
i) Calculate the attenuation of 1500 nm signal in the same medium.
ii) What physical quantity has as unit decibel?
6. (a) Describe in relation to molecular behavior
i. One way in which gases are similar to liquids but different from solids and
ii. One way in which liquids are similar to solids but different from gases
b) Describe one phenomenon in each case to demonstrates that
(i) Matter is made up of tiny particles (ii) These particles are in random motion
7. (a) Explain why a charge particle moving in a constant uniform magnetic field describes a circular path,
if its velocity is perpendicular to the field lines.
(b) An electron enters a uniform magnetic field of 0.5 T with a speed 3.0 x 105 ms-1.
i. Calculate the centripetal force experienced by the electron
ii. Suppose the electron starts losing speed, what are the consequences to the path and the
environment of the electron?
SECTION II
8. (a) (i) Define the specific latent heat of fusion of a substance.
(ii) Describe an experiment to show how the specific latent heat of fusion of ice can be determined
(b) (i) Ethyl alcohol has about one half, the specific heat capacity of water. If equal masses of ethyl
alcohol and water in separate beakers are supplied with the same amount of heat, compare the
temperature change for the two liquids.
(ii) 10 kg of molten lead at its melting point of 3270C and 1.0 kg of ice at 00C are placed inside an
insulated chamber where they reach a common final temperature. Calculate the final temperature and the
heat lost by the lead in the process.
Specific latent heat of fusion of ice = 3.34 x 105 Jkg-1; Specific heat capacity of water = 4.2 x 103 Jkg-10C1
Specific heat capacity of lead = 1.28 x 103 Jkg-10C-1; Specific latent heat of fusion of lead = 2.45 x 103
Jkg-1
(c) Why is it possible to hold a lit match, even when it is burning to within a few millimeters of your
fingertips?
(d) (i) Define Young’s modulus
(ii) Describe an experiment to show how the Young’s modulus for a metal wire can be determined.
(e) Figure 2 shows a chain hung from a support.
<<ADVANCED LEVEL PHYSICS
Page 9
Where is the maximum stress on the chain; A, B, or C? Explain.
f) A load of 102 kg is supported by a wire of length 2.0 m and the cross
section of the wire is 0.10 cm2. If the wire is stretched by 2.2 mm,
calculate
A
B
Figure 2
C
In the circuit in figure 3, the capacitor is fully charged by using a 6.0 V battery and the two way
switch K, it is then discharged. Figure 3 below shows how the charge, Q on the capacitor, C
6.0 with
V time
changes
discharge.
(i) The
stress
(ii)during
Thethe
strain
(iii) Young’s modulus
9. (a)
C
K
R
Figure 3
Q/mC
figure 4
4
3
2
1
0
0
50
100
150
200
250
300
350
t/s
400
Use the graph to answer the following questions
i. What is the capacitance of the capacitor?
ii. Estimate the initial current through the resistor during the discharging process, hence calculate the
resistance of the resistor and the time constant
iii. On the same axes, draw graphs to show how the voltage Vc across the capacitor and VR across the
resistor varies with time during charging. Indicate values where appropriate
<<ADVANCED LEVEL PHYSICS
Page 10
12 V
9 pF
S1
2.81 mH
S2
(b)
Figure 5
Figure 5 is an LC with an inductance of 2.81 mH and a capacitance of 9.0pF. The capacitor is
initially charged with a 12 V battery switch S1 is opened and S2 closed. S1 is then closed and S2
opened. If the frequency of S1 and S2, opening and closing corresponds to the resonance
frequency for LC circuit.
(i) Calculate this frequency
(ii) What are the maximum values of the charge on the capacitor and current in the circuit?
(c) A car of mass 1000 kg is initially at rest, it then moves along a straight road for 20 minutes
and comes to rest again. The momentum – time graph for the motion is shown in figure 6
P/103 kgm-1
25
20
B
C
10
15
15
10
5
0
A
0
5
D
20
25
t/60s
Use the graph to answer the following questions
(i) What are the resultant forces acting on the car during the parts of the motion labeled AB, BC and
CD?
(ii) Calculate the total displacement of the car during the 20 minutes
(iii) Sketch a displacement – time graph for the car during the 20 minutes.
(d) If, when travelling at the maximum speed, the car had struck and remained attached to a stationary
vehicles of mass 1500 kg.
i. With what velocity would the interlocked vehicles have travelled immediately after the collision?
<<ADVANCED LEVEL PHYSICS
Page 11
ii.
Calculate the kinetic energy of the car just prior to the collision and the kinetic energy of the
interlocked vehicles just after the collision. Comment on the values obtained.
10. (a) Explain briefly the difference between an emission spectrum and an absorption spectrum. Describe an
observation to illustrate each of the spectra.
-0.54 eV
n=5
-0.85 eV
n=4
-1.51 eV
n=3
Figure
-3.40 7eV
-13.6e V
n=2
n=1
Figure 7 shows some of the possible energy levels of the hydrogen atom
(i) Why are the energy levels negative?
(ii) Explain briefly how figure 7 can be used to account for the fact that the ionization energy of the
hydrogen atom is 13.6 eV
(iii) Calculate the highest possible frequency in the line spectrum of hydrogen from the transition
between the five spectral lines shown in the diagram. In what region of the electromagnetic spectrum
does this transition lie?
(iv) Which transition in figure 7 corresponds to the maximum wavelength that would be visible to the
eye?
(c) in an α – scattering experiment, the fraction of incident alpha particles reflected back through more
than 900 is very small. How does this result lead to the idea that an atom has a nucleus?
i. Whose diameter is small compared with the atomic diameter and
ii. Which contains most of the atomic mass?
(d) Explain briefly the difference between electromagnetic waves and mechanical waves. Give one
example of each.
(e) Distinguish between stationary and progressive waves?
Describe briefly how best stationary waves can be produced from two progressive waves. Draw
diagrams to show the superposition of the waves involved to produce the resultant effect.
(f) In an experiment to investigate the properties of stationary waves, one end of a rubber cord is
attached to a vibrator, the frequency of which can be varied, and the other end to a rigid support.
Figure 8 is a diagram of the cord drawn to scale showing the cord vibrating at one of its harmonics.
<<ADVANCED LEVEL PHYSICS
Page 12
Figure 8
i.
ii.
iii.
What are harmonics
Determine the wavelength and amplitude of the wave portrayed
If the frequency of the vibrator is 600 Hz, calculate the wave speed, and the fundamental
frequency of the cord when supported in this manner.
STUDENT’S PROPOSED ANSWERS TO JUNE 2002
<<ADVANCED LEVEL PHYSICS
Page 13
JUNE 2003
1. Waves may be classified on the basis of their medium of transmission or mode of propagation.
(a) Which are these classes in terms of?
(i) Medium of transmission
(ii) Mode of propagation
(b) Give one example of each.
2.
A
Figure 1
R
Black box
E
S
Figure 1 shows a circuit which can be used to establish the I – V characteristic for three conductors,
copper, tungsten filament lamp and a junction diode
(i) Copy the diagram and insert the missing component
(ii) Sketch separate I – V characteristic for the three conductors.
3. A nuclide 220
86𝑋 decays to a nuclide Y by emission of two α – particles and two β – emissions
i. Write down the equation of this decay process
ii. The activity of radioactive carbon – 12 in living wood is 19 counts per minutes per gram.
Measuring the activity of the isotope in a piece of ancient wood gave an activity of 7 counts per
minute per gram. Given that the half – life of the isotope is about 6000 years. Estimate the age of
the piece of ancient wood.
4. A grating has 6000 lines per mm and is illuminated by light of wavelength 5.9 x 10-7 m. which is incident
normal to the grating.
i. Find the direction of the first order diffraction image
ii. Is it possible to obtain a third order image with this diffraction grating for this wavelength?
iii. What would be the effect on the number of orders, if the wavelength of the wave increases?
5. (a) Calculate the root mean square speed for the molecules of nitrogen at stp, if the density of nitrogen at
these conditions is 2.6 kgm-3.
(b) The speed of sound at stp is about 330 ms-1. Explain how this is related to the root mean square speed
of nitrogen.
6. Figure 6 shows a pendulum bob of mass 20 g attached to the roof of a train that is in motion. The
pendulum hangs when the train was stationary.
<<ADVANCED
LEVEL PHYSICS
Pendulum
bob
200
Page 14
Figure 2
(i) Draw a free body diagram, to show the forces acting on the pendulum bob when the train is in
motion.
(ii) Calculate the acceleration of the train
(iii) Describe what happens to the bob if the train attains constant velocity.
7. (a) Distinguish between crystalline , amorphous and polymeric solids
(b) Give examples of each type of solids.
8. (a) (i) Distinguish between specific latent heat of vaporization and latent heat of vaporization
(ii) Describe an experiment to determine the specific latent heat of vaporization of water.
(b) An electric heater rated at 2.0 Kw is used to heat 15 g of water in a kettle. The initial temperature of
the water is 200 C.
i. What time does it take to heat the water to its boiling point?
ii. Calculate the mass of the water that would have boiled away in five minutes.
(c) Estimate how long it will take all the water to evaporate. State any assumption that you have made
in your calculations.
Specific heat capacity of water is 4200 JK-1K-1; Heat capacity of copper is 400 JK-1
Specific latent heat of water is 2.0 x 106 Jkg-1
(d) (i) State the law of conservation of linear momentum
(ii) Describe an experiment to verify the law of conservation of linear momentum.
(e) Two bodies A and B with masses 2m and m, respectively, make a head on collision. The bodies
move in the same direction with a velocity of 5.0 ms-1 and 2.0 ms-1 respectively. If the velocity of A
after collision with B is 3.0 ms-1.
(i) Calculate the velocity of B after the collision
(ii) IS the collision elastic or inelastic? Explain.
(f) A particle of mass m hits a rigid wall and bounces back with the same speed. Explain whether the
law of conservation of linear momentum is satisfied or not?
9. (a) (i) What is the photoelectric effect?
(ii) What are the experimental observations of the photoelectric effect?
(iii) How do these observations compare the classical theory?
(b) Figure 3 shows how the stopping Vs varies with the frequency, f, of the incident radiation in a
photoelectric investigation of metals.
<<ADVANCED LEVEL PHYSICS
Page 15
the
Use the graph in figure 3 to calculate values for
i. The plank’s constant
(ii) The work function of this metal
(c) (i) Explain the meaning of thermionic emission
(ii) Make a sketch of an electron gun
(d) Using either force and kinetic energy or separation and motion of particles, differentiate between
different states of matter.
<<ADVANCED LEVEL PHYSICS
Page 16
(e) (i) use the graph in figure 4 to describe the forces that exist between the two molecules
(ii) How can this graph be used to explain Hookian behavior of matter
(iii) Sketch and energy – separation curve from this graph
(iv) Estimate the amount of work done in separating the molecules from the equilibrium position to
infinity.
(v) What is the physical significance of this energy?
10. (a) (i) Distinguish between the conduction mechanism for copper and silicon.
(ii) Describe how an n – type semiconductor may be produced. Hence explain the formation of a p – n
junction.
(b)
6V
R
c
b
e
0V
Figure 5
From the circuit in figure, calculate
i. The base emitter voltage of the transistor at saturation
ii. If the base emitter resistance is 100 Ω. Calculate a value for R at saturation.
(c) (i) Draw a circuit diagram of a bridge rectifier to convert a.c to d.c.
(ii) Sketch the output characteristic of a transistor in the C.E mode.
Radio frequency/106 Hz
Aerial
Tuning circuit
Sound
Demodulator
Amplifier
Speaker
103 Hz
Figure 6
<<ADVANCED LEVEL PHYSICS
Page 17
(d) (i) Figure 6 shows a simple radio receiver. The tuning circuit selects one station. Explain why this
happens.
(ii) Explain the functions of the demodulation and the amplifier
(e) (i) Briefly explain the difference between FM and AM transmissions
(ii) If the capacitance C of the capacitor is 2 µF and the circuit is tuned at a frequency of 106 Hz.
Calculate the inductance of the inductor.
(iii) How could the tuning circuit be altered so that it could select other frequencies.
(f) A satellite of mass (m) is launched from the earth’s surface to cycle the plane of the equator?
(i) Explain the conditions under which such a situation is possible.
(ii) Explain the height of the satellite orbit above the earth’s surface, if the radius of the earth is 6400 km.
STUDENT’S PROPOSED ANSWERS TO JUNE 2003
JUNE 2004
1. Figure 1 shows a point charge of value -25 nC at the centre of an earth hollow metal shell.
25nC
Figure 1
(i) Copy the diagram and show the charge distribution on the shell. Explain your diagram.
(ii) Sketch a diagram to show how the electric field intensity varies with distance from the point charge.
2. Figure 2 shows a body of mass 4.0 kg being raised by two unequal forces of 25 N and 40 N. The 25 N
force acts at an angle of 300 to the vertical and the 40 N acts at an angle of 600 to the vertical.
300
600
25 N
4.0 kg
40 N
Figure 2
<<ADVANCED LEVEL PHYSICS
Page 18
Calculate the acceleration of the mass
3. Figure 3 shows how two circuits which could be used to determine the resistance of a conductor.
State and explain which circuit could be appropriate to determine the resistance R, for
(i) Large R
(ii) Small R
A
A
R
R
V
V
y
Figure 3
x
(b)Describe how you would use the CRO to measure time.
4. A closed pipe contains air at 240 C. If the fundamental note emitted from the pipe is 512 Hz, determine
the length of the pipe.
(Speed of sound in are at 00 C is 340 ms-1)
5. Figure 4 shows a simple pendulum bob of mass 50 g displaced to a height of 1.5 cm above the
equilibrium. When the bob is released, it makes an elastic head on collision with a body of mass = 50 g
placed at the equilibrium position.
1.5 cm
m
Figure 4
i.
ii.
Calculate the velocity of the body after collision
If the body moves a distance of 50 cm before coming to rest, calculate the frictional force acting
on the body.
6. (a) Draw a block diagram of the major components of a hydroelectric plant. Explain the role of each
component.
(b) Outline two environmental hazards associated with a hydroelectric plant.
7. (a) Define the term half life as used in radioactivity.
<<ADVANCED LEVEL PHYSICS
Page 19
(b) The half-life of carbon 14 is 5730 years. If the count rate of carbon – 14 in 2 kg of living bone is 3 x
104 per minute, determine the count rate in 0.3 kg of an ancient bone that is 25000 years old.
8. (a) Forces could be classified as contact forces or action at a distance forces. Explain the meaning of the
phrases in italics.
(b) Describe an experiment to determine the acceleration due to gravity of a body undergoing free fall.
(c) A car driver stops at a traffic light when the light shows red. A truck driver arrives at the traffic light
just when it shows green. He continues driving at a constant speed of 36 kmh-1. Suppose the road beyond
the traffic light is straight and flat. How long will the car driver take to catch up with the truck, if he is
driving at a constant acceleration of 2 ms-2 in the direction of the truck? Assume the average reaction
time for a driver is 0.7 s.
(d) Waves may be classified as transverse and longitudinal or mechanical and electromagnetic. Explain
giving one example in each case, of the terms in italics.
(e) Describe an experiment to measure the velocity of sound in free air.
(f) Figure 5 shows a graph of the variation of height of tides with time at a particular harbor. The
variation in height could be regarded as a result of the superposition of two waves. Estimate the
amplitude and frequency of the waves.
9. (a) (i) Distinguish between self induction and mutual induction.
An inductor with an iron core is connected in series with a milliameter and a 1.5 V supply. Figure 6
shows the way the current varies with time when the switch is closed.
<<ADVANCED LEVEL PHYSICS
Page 20
Figure 6
(ii) Sketch a circuit diagram from which such results could have been obtained.
(iii) How long does it take for the current to reach maximum value?
(iv) How would the graph in figure 6 be modified if a resistor replaced the inductor?
(b) In an experiment to investigate the behavior of a charge on a small insulated charged ball of mass 50
g, the ball is suspended by an insulated thread between two vertical and parallel plates which are 5 cm
apart. When a potential difference of 600 V is applied between the plates, the ball is pulled such that the
thread makes an angle of 600 to the horizontal.
(i) Sketch a diagram showing all the forces acting on the ball. State the origin of these forces with respect
to the fundamental forces.
(ii) Calculate the electric charge on the ball.
(c) (i) Distinguish between potential difference and electromotive force.
In an experiment to determine the internal resistance of a cell, current through the cell was made to vary
with the voltage across the cell as shown in figure 7
<<ADVANCED LEVEL PHYSICS
Page 21
(ii) Sketch a circuit diagram from which such results could have been obtained.
(iii) Determine using the graph of figure 7, values for the
(a) Emf of the cell
(b) Internal resistance of the cell
(d) A heating coil of power 10 W is required when the p.d across it is 20.0 V. Estimate the length of the
copper wire that would be needed to make the coil, if its cross sectional area is 1.0 x 10-7 m2 and its
resistivity is 1.0 x 10-6 Ωm.
(e) How much would it cost to use the coil in (d) for thirty days if AES – SONEL charges 60 frs per
KWh
10. (a) Figure 8 shows a circuit diagram that can be used to measure the charge on a conducting sphere by
transferring the charge to the capacitor
+6 V
A
R
V
3.2 µC
4.7 µF
0V
Figure 8
i.
ii.
What is the reading of the ammeter, if the base emitter voltage is 0.6 V
What is the reading of the voltmeter, if the voltage gain is 20?
(b) An LED is mounted on the dashboard of a car and is used as an indicator for a car alarm. The car
battery supply is 12 V and the LED requires 10 mA to run correctly. What is the value of the protective
resistance needed for the correct functioning of the LED?
(c) Figure 9 shows a basic circuit that can be used to provide a back – up energy source for a solar
powered wrist watch.
<<ADVANCED LEVEL PHYSICS
Page 22
Solar
cell
C = 0.33
µF
Black
Figure 9
In direct sunlight the solar cell supplies power to the watch mechanism and charges the capacitor C to a
voltage of V = 2.4 V. Calculate
i. The charge stored on the capacitor
ii. The energy stored in the capacitor
(d) In conditions of poor light, the voltage produced by the solar cell drops to zero and the watch
mechanism will cease to function if the voltage across it falls to a value below 1.0 V. The capacitor
acts as a backup power supply discharging through the watch mechanism.
i.What is the purpose of the diode in the circuit
ii.Calculate the charge which would have flowed through the watch mechanism when the voltage
across the capacitor falls to 1.0 V
iii.The watch mechanism is designed to draw current of 1.0 µA, as long as the voltage across it is
greater than 1.0 V. use this fact and your answer in d (ii) to estimate for how many hours the
capacitor can back up the watch mechanism.
(e) Figure 10 shows a section through a possible crystalline structure for a metal. Each circle
represents an atom of the metal
d
Figure 10
i.
ii.
(f) Calculate the spacing d, between the centers of adjacent atoms, if the molar mass of the
metal is 6.4 x 10-2 kg, density of the metal is 8.9 x 103 kgm-3, Avogadro’s constant is 6.0 x 1023
mol-1
Figure 11 shows how the force, F, between a pair of atoms in a solid varies with their separation
<<ADVANCED LEVEL PHYSICS
Page 23
F
0
x
Figure 11
Copy the diagram and mark on it the distance d, calculated in a (i) above. Explain why you have chosen
the indicated point
(f) Explain briefly with the aid of a diagram what you would expect to happen to a nearly spherical
droplet of water resting on a horizontal surface, if a tiny droplet were added to it. How would you
account for the change that might occur?
(i) Define surface tension
(ii) Give a brief explanation, in terms of intermolecular forces, of the origin of surface tension
(iii) In terms of intermolecular forces explain how the surface of a liquid differs from the bulk of the
liquid.
(h) The two vertical arms of a manometer containing water have different internal radii of 10-3 m and 2.0
x 10-3 m respectively. What is the difference in height of the two liquids when the arms are open to the
atmosphere? (Density of water = 103 kg and surface tension = 7.0 x 10-2 Nm-1
STUDENT’S PROPOSED ANSWERS TO JUNE 2004
JUNE 2005
1. (a) Explain what is meant by the homogeneity of a physical equation.
(b) Show that the expression 𝑐 2 𝜇0 𝜀0 = 1 is homogeneous where 𝜇0 is the permeability of free space, 𝜀0 is
the permittivity of free space and c is the speed of light.
(c) Given that𝜇0 = 4𝜋 × 10−7 𝐻𝑚−1, calculate the value of 𝜀0 .
2. Figure 1 shows a graph of the square of the frequency against the inverse of the length for a simple
pendulum.
<<ADVANCED LEVEL PHYSICS
Page 24
(i) Use the graph to determine the acceleration due to gravity
(ii) Calculate the length for which the pendulum would have a frequency of 20.0 Hz
3.
600
B
A
𝜃
Figure 2
Figure 2 shows snooker ball A moving with velocity of 5.5 ms-1, which hits a stationary snooker ball B.
after collision A moves with a velocity of 2.5 ms-1 at 600 to its original path. Calculate the velocity of B
after collision
4. Figure 3 shows two light beams X and Y of wavelength 450 nm travelling in air and incident on a
composite crystal of thickness 20 µm. The refractive index of P is 1.40 and that of Q is 1.45
X
Y
P
Q
a) Determine which will first emerge from the crystal
b) If X and Y are in phase as they enter the crystal,
calculate their phase difference as they leave the
crystal.
Figure 3
5. A converging lens of focal length 20.0 cm is placed 25.0 cm away from a screen on which an image is
formed. A biconcave lens of focal length 30. 0cm is now placed between the converging lens and the
screen so that it is 10.0 cm from the converging lens. Calculate how far the screen has to be moved to
focus the new image.
<<ADVANCED LEVEL PHYSICS
Page 25
6. Figure 4 shows an electrical circuit.
12 V
10 Ω
I1
Determine the
(i) Currents I1 , I2 and I3
(ii) Pd between X and Y
5Ω
6V
5Ω
I3
Figure 4
Y
7.
4 x 108
C
-10 x 10-8C
B
A
2 x 108
C
P
Figure 5
Figure 5 shows three charges A, B and P placed in a straight line. The charge at A is 4 x 10-8C,
that at B is -10 x 10-8C and that at P is 2 x 10-8C
a) Calculate the force on the charge at P due to the charges A and B.
b) Show that the resultant force on P cannot be zero, if P is placed between A and B.
8. (a) (i) State Newton’s laws of motion
(ii) Show how the principle of conservation of momentum could be derived from the second and third
laws of motion.
(b) Describe an experiment to verify the principle of conservation of linear momentum.
(c) Distinguish between conservative force and non – conservative forces, giving one example of each.
(d) (i) State Kirchhoff’s laws
(ii) Explain how each of the laws is essentially a statement of either the conservation of energy or the
conservation of electric charge.
(e) Describe an experiment to investigate how the pd across a wire filament varies with current through it
at constant temperature.
(f) Given the circuit in figure 6.
12 V
8Ω
Calculate
6V
6Ω
12 Ω<<ADVANCED
Figure 6 LEVEL PHYSICS
a) The current in the 6 Ω resistor
b) The pd across the 6 Ω resistor
Page 26
9.
(a) State the assumptions used in deriving the kinetic theory equation. From these assumptions, derive
the kinetic theory equation
1
𝑃 = 𝜌𝑐̅̅̅2
3
Where P = pressure of the ideal gas
𝜌 = density of the gas molecules.
(b) Figure 7 shows how the pressure of oxygen at temperature T and 300 K varies with density.
Use the graph to
i. Calculate the value for the r.m.s speed of oxygen molecules at 300 K
ii. Explain whether T is higher or lower than 300 K
(c) (i) On the same axis sketch labeled graphs to show how the speed of the molecules in an ideal gas
are distributed at temperatures of 300 K and 600 K.
ii) On one of the graphs, show the position of average velocity, r.m.s speed and most probable speed.
(d) Materials could be classified as crystalline, amorphous or polymeric. Define the terms in italics. Give
one example for each of the terms.
(e) An aluminum wire and a glass thread are subjected to linear stress until they break. On the same axis,
sketch graph of stress – strain to show the behavior of each material.
(f) Figure 8 shows a graph of extension, e, against force, F, for a certain nylon climbing robe.
<<ADVANCED LEVEL PHYSICS
Page 27
e/m
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
1
2
3
4
figure 8
5
6
7
8
9
10
11 12 13
F/KN
A climber of mass 80 kg attached to a 10 m length of robe can withstand a force from the rope of not
more than 7.5 KN without the risk of serious injury. Use the graph to
(i) Estimate the maximum energy stored in the rope when climber is not at risk.
(ii) Explain how you would determine a value for Young’s modulus for a given extension.
10. (a) Differentiate between interference and diffraction of light
(b) A parallel beam of light of wavelength 5.5 x 10-2 m in air is incident on the slits in Young’s double
slit experiment. A thin film of transparent plastic of refractive index 1.48 and thickness 5.2 x 10-6 m is
placed over one of the slits.
i. Determine the increase in the optical path of the light passing through the thin film. Hence
determine the number of fringes by which the central bright fringe is displaced.
ii. Explain how the fringe spacing would change if the slit separation were increased, the slit –
screen distance were increased.
(c) Light of wavelength 5.5 x 10-7 m falls on a single slit of width 0.15 mm. A screen is placed 1.2 m
beyond the slit.
i. Sketch a graph showing the light pattern observed on the screen
ii. Calculate the width of the central fringe
(d) State the following laws
(i) Newton’s law of gravitation (ii) Coulomb’s law.
(e) Explain in what way the Coulomb force between two charge particles is
i. Different
ii. Similar to the gravitational force between two masses
(f) The gravitational potential energy U of a mass, m, a distance h above the surface of the earth is
<<ADVANCED LEVEL PHYSICS
Page 28
𝐺𝑀𝑚
(𝑅 + ℎ)
Where M is the mass of the earth, R the radius of the earth and G, the gravitational constant (R = 6.4
x 106)
i. Show that this expression is equivalent to U = mhg usually quoted in elementary physics
courses where g is the gravitational force per unit mass near the surface of the earth.
ii. How much kinetic energy must a 100 kg space craft have at the surface of the earth to be able
to leave the earth completely?
iii. Explain what would happen if the spacecraft had
- Less energy, More energy
𝑈= −
STUDENT’S PROPOSED ANSWERS TO JUNE 2005
<<ADVANCED LEVEL PHYSICS
Page 29
JUNE 2006
1. (a) The force F acting on a metal conductor of length, l, placed in a magnetic field of flux density, B, and
carrying a current I is given by
𝐹 = 𝐵𝐼𝑙𝑠𝑖𝑛𝜃
Where 𝜃 is the angle between B and I. Show that the equation is physically correct?
(b) The force between two equal charges, Q, separated by a distance, r, is given by
𝑄2
𝐹=
4𝜋𝜀𝑜 𝑟 2
What are the base units of 𝜀𝑜
2. Sketch on the same axes graphs to illustrate the temperature distribution along a metal bar heated at one
end when the bar is
a) Lagged and the other end dipped in melting ice
b) Unlagged and the other end dipped in melting ice.
In each case assume that steady state has been attained.
3. The maximum energy of photoelectrons ejected from a tungsten metal surface by monochromatic light of
wavelength 248 nm was found to be 8.6 x 10-20 J. calculate
(i) The work function of the tungsten in eV
(ii) The threshold frequency of the tungsten
4. Differentiate between progressive waves and stationary waves in terms of
i. Amplitude of vibration of the particles of the medium
ii. Phase of vibration of the particles of the medium
5. A ‘supper man’ is sitting on a tree 98 m high with a baby has rescued from the claws of a tiger.
Unfortunately, the child slips and falls with an initial velocity of zero. ‘Supper man’ realized what was
happening 2 seconds later and flew to catch the child at a constant speed of 39.2 ms-1. Calculate
i. The time ‘supper man’ will fly before catching the child
ii. The distance ‘supper man’ will fly to catch the child.
6.
12 V
S
1
3 µF
4 µF
1 µF
R
5Ω
<<ADVANCED LEVEL PHYSICS
S2
Page 30
Figure 1 show how capacitors, switches, a resistor and batteries are connected with S1 and S2
open.
i. Calculate the quantity of charge on the 4µF capacitor with S1 closed and S2 opened.
ii. If S1 is now opened and S2 closed, calculate the current through R.
7. A rubber tyre of mass 15 kg is suspended with a rope 7.5 m long at a fixed support. A girl of mass 30 kg
sits on the tyre and is made to swing. The speed of the girl at the lowest point of the swing is 3 ms-1.
Calculate the tension in the rope as it goes through the lowest point.
8. (a) Describe one method you can use to produce a uniform magnetic field in the laboratory.
(b) A beam of electrons are accelerated through a pd V, and enters a uniform magnetic field, B, with
velocities at right angles to the field. The electrons move in a circular path. Use the above information to
2𝑉
show that the expression for the specific charge is given by: specific charge =𝐵2 𝑟 2, where r is the radius
of the circular path.
(c) Describe an experiment you would carry out in the laboratory using a beam of electrons following
circular path in a magnetic field to determine the specific charge of an electron.
(d) Neon ions each of mass 3.3 x 10-26 kg are accelerated through a Pd of 1400 V. the ions then enters a
region of space where there are uniform magnetic and electric fields acting at right angles to each other
and to the original direction of motion of the ions.
i. Calculate the speed of the accelerated ions just before they enter the B and E fields
ii. Calculate the magnitude of the electric field for the ions to go through the B and E fields un –
deflected. The flux density is 0.4T
(e) Define surface tension
Describe an experiment to measure the surface tension of water at room temperature.
(f) Two drops of mercury – one small and the other large are placed on a smooth polished surface. Sketch
and explain the shapes.
(g) A circular ring of thin wire of mean radius 1 cm is suspended horizontally by a thread passing through
the 5 cm mark on a ruler pivoted at its centre and the ring is balanced by a 5 g mass suspended at 60 cm
mark. A beaker of liquid is then placed so that the ring just parts the surface. Find the surface tension of
the liquid.
9. (a) (i) Explain what is meant by the resistivity of a material?
(ii) Sketch on the same set of axes graphs to show how the resistivity of a conductor, semiconductor, and
insulator vary with temperature.
(b) The graph in figure 2 shows the results of an experiment to determine the resistivity of a wire of
length 80.0 cm.
<<ADVANCED LEVEL PHYSICS
Page 31
V/V
8
7
6
5
4
3
2
1
0
0
0.2
0.4
0.6
figure 2
0.8
1
1.2
I/A
1.4
i.
ii.
Draw an appropriate circuit that could have been used to obtain such results?
Use the graph to calculate a value for the resistivity of the wire if its diameter is 5.0 mm
(c) A car battery with a capacity of 60 ampere – hour is used to deliver current when a pd across its
terminals is 12 V. How much electrical energy is available from such a battery?
(d) (i) Explain what is meant by a material is elastic.
(ii) When a piece of rubber is extended and allowed to contract, energy is dissipated in the process.
Draw a force extension graph for the extension and contraction of rubber. Explain how the enrgy
dissipated can be obtained from your graph.
(e) The graph in figure 3 shows the results of an experiment to measure the elasticity for a piece of
wire 80.0 cm long.
<<ADVANCED LEVEL PHYSICS
Page 32
F/KN
7
6
5
4
3
2
1
0
1
1.5
2
2.5
3
e/m3.5
figure 3
i.
Sketch an appropriate set – up from which such results could have been obtained
ii. Use the graph to calculate Young’s modulus for the wire if its diameter is 15 mm.
iii. Calculate the energy stored in the wire for the extension.
(f) A lift of mass 450 kg is designed to contain a maximum of 10 people each of mass 75 kg. the
distance
from the top of the floor to the ground the floor is 30 m. calculate the minimum radius , the
cable should have so as to just support these people (tensile stress is 4.0 x 108 Nm-2
10. (a) Describe the formation of
i. Line emission spectrum (ii) Line absorption spectrum
(b) By using either line emission spectrum or line absorption spectra
i. Describe how the presence of any particles in space could be detected
ii. Describe how the presence of different types of elements in a sample of matter could be
identified.
iii. The spectrum from from a sodium flame showed two prominent yellow lines of wavelengths
589.0 nm and 589.6 nm. Using energy level diagram, explain how this is possible. Calculate the
energy the energy difference between the sodium lines
(c) In a nuclear reactor, the following processes occur – nuclear fission and controlled chain reaction
i. Explain these terms
ii. Draw an energy flow diagram for energy conversion in a nuclear reaction to produce electricity
from nuclear fission.
(d) Describe the formation of
i. N-type semiconductor
(ii) p-n junction.
(e) (i) What is the effect of temperature change on the conductivity of an intrinsic semiconductor.
(ii) What are the important characteristics that distinguish the depletion layer in p-n junction from the
n- and p – regions?
<<ADVANCED LEVEL PHYSICS
Page 33
+6 V
Loud speaker
10 kΩ
R
npn transistor
Y
0V
Figure 4
Figure 4 shows a simple alarm circuit.
The device Y could be a
(i) Thermistor (ii) LDR
Explain how each of these could be used to make
figure 4 functions as an alarm
The resistance of the LDR in position Y for a
given light intensity is 200 Ω. Explain whether the
alarm in figure 4 will be on or not.
STUDENT’S PROPOSED ANSWERS TO JUNE 2006
<<ADVANCED LEVEL PHYSICS
Page 34
JUNE 2007
1. The speed of light, c is related to the permeability μo and the permittivity, εo by the expression
1
c=
√μo εo
i.
Show that this equation is homogeneous
ii. Calculate the magnitude of εo
2. In figure 1 below, the current in the 3 Ω resistor and R are 1.5 A and 0.5 A respectively.
R
3Ω
E
Calculate
i.
ii.
4Ω
The emf of the battery
The resistance of R
Figure 1
(i) Explain why it is preferable to describe elastic behavior of materials in terms of stress – strain rather
than force extension.
(ii) Figure 2 is a graph of the extension and contraction of a rubber band. Calculate the work done in this
process.
F/N
3.
12
B
10
8
C
6
4
A
2
0
0
0.5
1
figure 2
1.5
2
2.5
x/m
3
4. A drill using a current of 1.5 A when connected to a mains supply of 240 V makes a round hole in a piece
of iron of mass M. in one minute 75 % of the electrical energy is converted to internal energy of the iron
which cause a rise in temperature of 200C. if the specific heat capacity of iron is 460 Jkg-1K-1
<<ADVANCED LEVEL PHYSICS
Page 36
i. Calculate the mass M of the piece of iron
(ii)State any assumption
5. (a) sketch
i. The transfer (ii) The input (iii) The output
Characteristic for an npn transistor.
6. Figure 3 shows the path a ray of light would follow in an optical fibre whose core has a refractive index
n1 and the cladding has refractive index n2.
n2
na
𝜃i
𝜃c
n1
core
Cladding
Figure 3
The angle of incidence and the critical angles are respectively θi and θc
i. What is meant by critical angle?
ii. State and explain whether n1 is less than or greater than n2
iii. The refractive index for glass is 1.5, calculate θc
7. The mercury in glass thermometer and the constant volume gas thermometer can be used to measure
temperature.
i. Explain why the constant volume gas thermometer could give readings in degree Celsius and
mercury in glass thermometer in degree Celsius too.
ii. The two thermometers may give different readings when immersed in a volume of liquid. Explain
why?
8. (a) Describe an experiment to show that for a constant force, the mass of a body is inversely proportional
to its acceleration. State clearly how you would minimize errors in measurements and how you would
arrive at the required results from your measurements
(b) A ball X of mass 400 g travelling at 2.5 ms-1 makes elastic and head on collision with a second
identical, stationary ball Y. they remain in physical contact for 60 µs.
(i) What does elastic collision mean?
(ii) Calculate the velocities of X and Y after the collision
(iii) Find the average force exerted by X during the collision.
(c) Figure 4 shows a ball propelled from a point A. the ball moves with constant velocity of, hits a wall a
B and moves back to A with the same velocity. The ball is in physical contact with the wall for a time
interval∆𝑡. Sketch a graph of the momentum of the ball against time for the movement of the ball.
A
Figure 4
v
wall
<<ADVANCED LEVEL PHYSICS
Page 37
(d) Describe how you would measure the specific heat capacity of a liquid. Describe the procedure you
would use to make allowance for heat losses, and how you would derive the specific heat capacity from
you measurements.
(e) The kinetic theory of ideal gases leads to the equation
1
𝑃 = 𝜌𝑐̅̅̅2
3
2
̅̅̅
Where P is the pressure, 𝜌 is the density and 𝑐 is the mean square speed of the molecules.
(i) State the assumptions used to derive this result.
(ii) Hence derive the equation
9. (a) A radioactive source emits both alpha and beta radiations.
(i) What does it mean for a substance to be radioactive?
(ii) State and explain how you would distinguish between the two types of radiations
(b) Figure 5 shows a graph of the natural logarithm of the activity of a radioactive element plotted against
time in minutes. Sketch the set – up from which such results would have been obtained.
8
ln(A)
7
6
5
t/min
4
0
1
2
figure3 5
4
5
6
7
(c) (i) Use the graph to obtain a value for the half – life of the sample
(ii) Use the graph to calculate the initial activity of the sample
(d)(i) What is a capacitor?
(ii) In what ways is a capacitor?
(a) Similar to a diode.
<<ADVANCED LEVEL PHYSICS
Page 38
(b) Different from a diode.
(e) A capacitor, charged fully with a battery of 10 V is discharged through a resistor. Figure 6 shoows
how the current varies with time.
I/A
100
90
80
70
60
50
40
30
20
10
0
0
50
100
150
200
250
300
350
400
450
500t/s
figure 6
(i) Sketch an electric circuit from which such results would have been obtained
(ii) Use the graph to estimate the initial charge on the capacitor and hence, or otherwise, estimate its
capacitance.
(iii) Calculate the time constant for the capacitor.
(f) How will the graph be affected if the resistance R in the circuit is doubled? Explain your answer.
10. (a) Explain what is meant by the terms:
(i) Displacement,
(ii) Wave speed for a mechanical wave
(b) Distinguish clearly between stationary wave and progressive wave with reference to the following
characteristics of the wave
(i) Amplitude (ii) Frequency (iii) Wavelength ( iv) phase (v) wave form (vi) Energy transmitted
Diffraction and interference are phenomena exhibited by wave. State clearly the difference between these
phenomena
(c) A laser is used to produce young fringes with slits separated by 0.05 mm. The screen is 1.0 m from
the slits and 10 fringe separations occupy 12.5 mm. What is the wavelength of the laser light?
(d) Electrons can be emitted from the surface of zinc by ultraviolet light.
(i) Explain why visible light cannot cause electrons to be emitted from the surface of zinc whereas
ultraviolet light does?
<<ADVANCED LEVEL PHYSICS
Page 39
(ii) If both metals were illuminated with ultraviolet light of the same frequency, how will the energies of
electrons emitted from the zinc and potassium surfaces differ?
(e) Explain each of the following
(i) If the intensity of the ultraviolet light directed at a piece of zinc is doubled, the number of electrons
leaving the surface per second also doubles but the maximum kinetic energy is unchanged.
(ii) The maximum kinetic energy of photoelectrons is directly proportional to the difference between the
frequency of light falling on the surface and the threshold frequency of that metal.
(iii) Gamma photons are more harmful to people than infrared photons.
(f) Calculate the wavelength of photons emitted when an electron makes a quantum jump from n = 3 state
to the ground state of the hydrogen atom. The energy at the state n = 3 is -1.5 ev and the ground state
energy is – 13.6 eV.
STUDENT’S PROPOSED ANSWERS TO JUNE 2007
<<ADVANCED LEVEL PHYSICS
Page 40
JUNE 2008
1. One condition for an equation to be physically correct is that the equation should be homogeneous
i. What is the meaning homogeneous as used in the above statement?
ii.
4
Show that the equation 𝜂𝜋𝑎𝑣 = 3 𝜋𝑎3 (𝜌 − 𝜌′ )𝑔 is homogeneous, where 𝜌 is the density of a
sphere of radius a falling steadily through a liquid of density 𝜌′ with speed v. 𝜂 is the coefficient
of viscosity of the liquid with units kgm-1s-1
2. (a) Figure 1 shows a graph of image distance, v, versus magnification, m, for a convex lens.
(i) Use the graph to find a value for the focal length of the lens.
v/cm
30
29
28
27
26
25
24
23
22
21
m
20
1
1.125
1.25
1.375
1.5
1.625
1.75
1.875
figure 1
(b) List two advantages which optical fibres have over copper cables when used in telecommunication
3. A charged capacitor of capacitance, C, can be discharged through a resistor R. At a time t after the start
t
of discharge, the charge, Q, remaining on the capacitor can be given by the expressionQ = Q0 e−RC,
where Qo is the initial charge on the capacitor.
i. Use the above equation to obtain an expression for the half – life, 𝑇1 of the discharge process.
2
<<ADVANCED LEVEL PHYSICS
Page 42
ii.
iii.
Use the equation to define the time constant, 𝜏, of the discharge.
Compare the values of 𝑇1 and 𝜏.
2
4. Define simple harmonic motion (SHM).
For a body executing SHM, the displacement y, is given by the equation𝑦 = 𝑟𝑠𝑖𝑛𝜔𝑡. What do r, 𝜔 and t
represent in the above equation?
5. Light is incident normally on a diffraction grating of 500 lines per centimeter and a second order image is
obtained at an angle of 360.
(a) Calculate the wavelength of the light used
(b) Determine whether a third order image can be obtained with light of the same wavelength.
(c) State and explain a way in which the number of orders could be increased.
6. (a) Sketch the stress – strain curve for a specimen of rubber when it is loaded within the elastic limit and
then unloaded.
(b) Explain in molecular terms the shape of the graph during loading and unloading process, state clearly
the energy changes involved.
7. (i) Draw a block diagram for a radio transmitter and receiver.
(ii) Differentiate between FM and AM transmissions.
8. (a) State Coulomb’s law.
(b) In an experiment to determine the permittivity, 𝜀, of a medium, a series of values of force, F, and
corresponding separation, r, between similar charges Q each of value 3.8 x 10-3 C were obtained. A graph
1
of 𝑟 2 versus F was plotted as shown in figure 2.
<<ADVANCED LEVEL PHYSICS
Page 43
Figure 2
(i) From the graph obtain a value for the permittivity of the medium
(ii) From the value of the permittivity obtained from the graph, in what type of medium do you think the
charges were placed.
(b) Lighting occurred in a forest and a tree standing vertically in the forest provided path along which the
lightning passed.
Suppose 600 x 10-3 C of charge is conducted through the tree in 1.0 µs
(i) Calculate the average current
(ii) Sketch the temporal magnetic fields created by this current.
(iii) What will the magnetic field strength be 10 cm from the tree.
(d) (i) State Newton’s law of gravitation
(ii) Suppose a planet of mass m is moving in a circular orbit of radius r above the sun of mass M. prove
that the periodic time, T, of the planet round the sun is given by the expression.
𝑟3
𝑇 2 = 4𝜋 2
𝐺𝑀
(iii) If the universal gravitational constant is 6.67 x 10-11 Nm2kg-2, the radius of the moon’s orbit is 3.5 x
108 m and the mass of the earth is 6.0 x 1024 kg, calculate the period of rotation of the moon round the
earth.
<<ADVANCED LEVEL PHYSICS
Page 44
(e) Figure 3 shows how 𝑟 3 varies with 𝑇 2 for a planet of mass m. Use the graph in figure 3 to obtain the
mass, m of the planet given that G = 6.67 x 10-11 Nm2kg-2.
(f) Sketch a graph to show how the field strength, g, varies with distance from the centre of the earth to
some distance outside the earth assuming that the density of the earth is uniform.
9. A bullet is projected horizontally at a height above the ground with a velocity u of magnitude 5 ms-1.
(i) What is the trajectory of the ball?
(ii) Find the position and velocity of the ball after 0.5 s.
(iii) State how the velocity and acceleration of the ball vary with time.
(b) Figure 4 shows a framed picture of a body held up by two strings OP and OQ each at an angle of 600
to the vertical.
600
600
Calculate the weight of the framed picture.
P
Q
Figure 4
<<ADVANCED LEVEL PHYSICS
Page 45
(c) What is a collision?
Distinguish between elastic and inelastic collisions
(d) (i) State the principle of conservation energy. Give a mathematical form of this law, stating
clearly each term of the equation. (First law of thermodynamics may be used).
(ii) Give an example of a situation in which the principle of conservation of energy applies.
(e) (i) Differentiate between renewable and non renewable energy sources giving an example of each.
(ii) To harness tidal power a dam is built across the tidal region of water. Water is trapped at high tides
and released at low tides. Suppose the water trapped during high tides is in a basin of area 40 x 106 m2. If
the maximum height of water is 10 m.
(ii) (a) Calculate the gravitational potential energy change from high tides to low tides given that density
of water is 1000 kgm-3.
(ii)(b) Calculate the average power obtained if the flow from high to low tide took 6 hours
(f) How can
(i) Wind
(ii) Oil, as energy sources be derived from the sun.
10. (a) Describe an experiment to measure the specific charge, e/m of an electron
(b) The element uranium 238
92U undergoes radioactivity to give an alpha particle and the particle thorium
(Th)
(i) Explain the meaning of the underlined words
(ii) Write the equation of the decay
(c) A certain radioactive material contains 1010 atoms. The half life of the radioactive material is 20 days
(i) Calculate the number of disintegrations after one second
(ii) After how long will the material take to reduce to 104 radioactive atoms?
(d) (i) Define temperature coefficient of resistance
(ii) Describe an experiment to determine the temperature coefficient of resistance of a conductor.
(e) Distinguish between the following
(i) Ohmic and non – ohmic conductors
(ii) Potential difference and electromotive force
6V
3Ω
F
H
C
D
4V
A
2Ω
8Ω
B
Figure 5 shows a load of resistance R = 8Ω
connected across two cells in parallel. The
cells have internal resistances as shown in
figure 5.
(i) Determine the values of I1 and I2
(ii) Comment on the values obtained
Figure 5
<<ADVANCED LEVEL PHYSICS
Page 46
STUDENT’S PROPOSED ANSWERS TO JUNE 2008
JUNE 2009
1. (a) (i) State the principle on which the optical fibre operates
(ii)State any two uses of optical fibre
(iii) Draw a labeled diagram of an optical fibre and show on the diagram how a ray of light is transmitted
through the optical fibre.
1
2. (a) State the assumptions used in deriving the kinetic theory equation P = 𝜌𝑐̅̅̅2 where P is the pressure
3
exerted by the particles, 𝜌 is the density of the gas and ̅̅̅
𝑐 2 is the mean square speed.
3.
R1=3 Ω
E1=6 V
E2=9 V R2=3 Ω
E3=12 V
300
R3=3 Ω
Figure 1
plank
Figure 2
(a) Draw a diagram indicating the forces
Figure 1 is a circuit diagram showing how dc power sources
(E1,on
E2the
and
E3) are connected with loads of
acting
plank
Calculate
the
tension in the rope
resistances R1, R2 and R3. Calculate the currents through R(b)
,
R
and
R
.
1
2
3
4. A 25 kg plank is suspended horizontally by a rope as shown in the diagram in
5. Distinguish between solids and liquids using
(a) Intermolecular forces
(b) molecular motion (c) molecular arrangement
(d) Intermolecular spacing (e) bulk shape
6. (a) Draw two separate diagrams to show a p – n junction connected in the forward bias and reversed bias
(b) When a p – n junction diode is connected in a circuit and is reversed bias, there is a very small
leakage current across the junction. Explain the source of the current. How does the size of this current
depend on temperature of the diode?
7. A tennis player drives a ball at 60ms-1; 100 to the horizontal and 50 cm above a tennis court.
(a) Calculate the velocity at which the ball hits the court.
(b) Sketch the velocity – time graph for the velocity of the ball.
8. (a) Define simple harmonic motion
(b) Describe an experiment to measure the acceleration of free fall, using a simple pendulum. Your
description should include a diagram, procedure, precautions, observations and conclusion.
<<ADVANCED LEVEL PHYSICS
Page 47
(c) A small mass M is attached to the free end of a coiled spring on a smooth table and the other end of
the spring is fixed and the mass pulled through a distance of 8 mm and then released. If the spring
constant is 10 Nm-1
(i) Prove that the motion of the mass at the end of the spring is simple harmonic
(ii) If the mass oscillates at a frequency of 30 Hz. Calculate the value of M and the kinetic energy of the
body when the extension is 3.0 mm
(iii) State any assumption made in your calculations
(d) Define specific heat capacity
(e) Describe an experiment to determine the specific heat capacity of a metal. Your description should
include a diagram, procedure, precaution, observation and conclusions.
(f) An engine is used to raise an 800 kg of iron at a speed of 6.7 ms-1. 0.5 kg of glycerin initially at room
temperature of 230C is required every second to maintain the temperature of the engine bearing at θ. The
power developed by the engine is 1.0 x 105 W. If the specific heat capacity of glycerin is 2.5 x 103Jkg-1K1
. Calculate the value of θ.
(g) A well lagged aluminum calorimeter of mass 80 kg contains 150 g of water and 100 g of ice at 00C. A
heating coil rated 1.0 Kw is put in the calorimeter and the mixture stirred until its temperature is 330C.
Calculate how much ice is left after one minute. State any assumption you have made.
9. (a) Define the term resistivity
(b) The graph in figure 3 below shows how the resistance of a copper wire varies with length at 200C.
R/10-3 Ω120
100
80
60
40
20
0
0
1
2
3
4
5
l/10-3 m
6
The wire has a thickness of 1.00 mm. Use the graph to determine
(i) The resistivity of the wire.
<<ADVANCED LEVEL PHYSICS
Page 48
(ii) The conductivity of the copper wire. If the experiment were carried out at 300 C, how would this
affect the conductivity of the copper wire?
(iii) The length of the copper that has resistance of 56 mΩ.
(c) A milliammeter has a resistance of 10 Ω and a full scale deflection of 10 mA. How would you convert
it into?
(i) An Ammeter reading up to 10 A
(ii) A voltmeter reading up to 10 V
(d) State conditions that must be satisfied for a balanced to be obtained with a slide wire potentiometer.
6.0 V
l
X
R
Y
m
A
Figure 4
I/mA
50
0
-50
-100
-150
-200
0
10
20
30
<<ADVANCED LEVEL PHYSICS
40
50 4 60
figure
70
80
90
100
110
120
l/cm
Page 49
The circuit of figure 4 is used to obtain data from which a graph of current I against balanced length l, is
drawn as in figure 5. The internal resistance of the cell is negligible.
(i) Explain why for different values of l, the current can be positive, zero or negative
(ii) Using the graph, calculate the resistance of R. what assumption have you made in your calculations?
(iii) Hence, deduce the emf of the cell
(d) The flux density between the poles of a powerful electromagnet is 2.5 T. What is the force exerted on
15 mm of wire carrying a current of 3.0 A when the wire is
(i) At right angles to the field
(ii) Parallel to the field
(iii) At 300 to the field
10. (a) (i) Determine the dimensions of the universal gravitation constant G.
(b) Derive an expression for the acceleration, due to gravity at the earth’s surface in terms of G, the
radius of the earth R and it density, ρ.
(c) (i) A communication satellite revolves round the earth in a circular orbit at a height of 36.000 km
above the earth’s surface. Find the satellite’s period of revolution in hours. Comment on the result.
(ii) Distinguish between electric and gravitational fields.
(d) (i) Distinguish between photoelectric emission and thermionic emission
(ii) State four observations obtained from the experiment on photoelectric emission.
(iii) Choose any two of the observations and account for them in terms of the quantum theory of light.
(e) The 212
84𝑃𝑜 nucleus emits 𝛼 particle when it decays
(i) What is the significance of 212 and 84 in 212
84𝑃𝑜
(ii) Write out and complete the equation below representing this decay.
212
4
84𝑃𝑜 → 2𝛼 +
Calculate the energy that is emitted in the decay process of 212
84𝑃𝑜 in joules
Atomic mass: polonium = 211.9890 U, alpha particle = 4.0026 U, lead = 207.9767 U, where 1 U = 931
MeV.
STUDENT’S PROPOSED ANSWERS TO JUNE 2009
<<ADVANCED LEVEL PHYSICS
Page 50
JUNE 2010
1. The force per unit length, F, between two wires placed a distance, r, apart in a vacuum and varying
current, I1 and I2 in the same direction is given by 𝐹 =
𝜇𝑜 𝐼1 𝐼2
2𝜋𝑟
where 𝜇𝑜 is the permeability of a vacuum.
Show that the equation is homogeneous.
2. (a) When two semiconductor materials, A and B, are doped using pentavalent and trivalent materials
respectively and the two materials fused together, a p – n junction or a junction diode is created due to the
presence of the depletion layer.
(i) What is the depletion layer and how is it formed?
(ii) What is the barrier potential?
(iii) The p – n junction is connected to a power source (battery). Sketch the current pd curve in the
forward and reverse bias if it is a silicon diode.
0
3.
-0.54 eV
-0.85eV
-1.51 eV
Energy level in eV
-3.4 eV
-13.6 eV
Figure 1
Figure 1 above shows an energy level diagram for a certain atom
(i) What is meant by the ground state
(ii) What is the ionization energy of this atom
(iii) Determine the wavelengths λ1 and λ2 of the transitions from -3.4 eV and -1.5 eV to the ground
state respectively.
4.
<<ADVANCED LEVEL PHYSICS
Page 52
Load/N
6
5
4
3
2
1
0
0
5.
6.
7.
8.
1
figure 2
2
3
4
Extension/mm
5
Figure 2 shows some experimental results used to verify Hooke’s law. The length of the wire is 2.0 m
and its cross – sectional area is 0.10 m2. Use the graph to
(a) Calculate the external work done on the wire in extending it by 4.0 cm
(b) Calculate the Young’s modulus of the material of the wire
(a) How is the nuclide 238
92U altered by:
(i) The emission of one alpha particle (ii) the emission of one beta particle
(iii) The absorption of one neutron
(b) Calculate the energy released when 5 kg of 238
92U undergoes a fission according to the equation
238
141
92
1
1
92U + 0n → 56Ba + 36Kr + 3 0n
141
92
1
Mass of 238
92U = 235.04 U, 56Ba = 140.997 U, 36Kr = 91091 U, 3 0n = 1.01 U, 1U = 931 MeV
Na = 6.02 x 1023 mol-1.
A beam of parallel light is incident normally on a diffraction grating having 550 lines per mm. A
telescope is used to observe the second order in the spectrum. Calculate the angular separation in radians
of two spectral lines of wavelength 559 nm and 563 nm.
A faulty vehicle stands on a horizontal road. The vehicle is towed by means of a rope connected to
another vehicle. Draw a free body diagram showing the horizontal forces acting on
(a) The vehicle being towed
(b) The vehicle doing the towing
Identify the Newton Third law pair of forces
(a) (i) Define capacitance. State the factors that determine the capacitance of a parallel plate capacitor
<<ADVANCED LEVEL PHYSICS
Page 53
(ii) Describe an experiment to measure the permittivity of air. Your description should include a diagram,
procedure, precautions, observations, calculations and conclusion.
(b) (i) Calculate the escape velocity from the moon’s surface given that the radius of the moon is 1738
km.
(ii) Calculate and comment about the electrostatic and gravitational force between an electron and a
proton in an atom.
(c) (i) Define surface tension in terms of force. Show that the work done per unit area in changing the
area of a liquid surface under isothermal conditions is equivalent to the definition of surface tension in
terms of force.
(ii) Describe an experiment to show that surface tension of a liquid varies with temperature. Your account
should include a diagram, procedure, observations, precautions, calculations and conclusions.
(c) When a jet of steam is sent into a container having 1.8 x 103 kg of ice at 00C, the ice turns into water
at 50C. Find the smallest mass of steam that this requires if steam initially at 1000C ends up as water at
50C. State any assumptions in your calculations.
9. (a) (i) What is meant by modulus of elasticity and elastic limit.
(ii) Explain how strength differs from stiffness of a given material.
<<ADVANCED LEVEL PHYSICS
Page 54
10
U/eV
8
6
4
2
0
r/10-10 m
-2
-4
-6
-8
-10
0
1
2
3
4
5
6
7
8
(b) Figure 3 shows how the potential, U, between two neutral atoms varies with their distance apart.
(i) What is the minimum potential energy in joules?
(ii) With reference to the graph, explain the expansion of solids
(iii) Determine slopes at the points where the r values are 11 nm, 38 nm and 67 nm.
(iv) Plot a graph of slopes above against r, estimate the energy needed to separate the atoms completely.
(c) Explain how electromotive force is different from potential difference.
<<ADVANCED LEVEL PHYSICS
Page 55
V/V 100
90
80
70
60
50
40
30
20
10
0
0
1
2
3
4
5
6
7
figure 4
8
9
10
11
12
13
14
I/A
15
Figure 4 shows the way V, across a resistor varies with the current, I, through it.
(i) Sketch a possible circuit diagram from which such results were obtained. Briefly explain the set – up.
(ii) From the graph in figure 4, determine the values for the emf and internal resistance of the cell.
2V
2Ω
2V
2Ω
L
21Ω
Figure 4 shows three identical batteries each of
internal resistance 2Ω and emf 2 V. Calculate the
current and voltage of the device L
2V
2Ω
Figure 4
<<ADVANCED LEVEL PHYSICS
Page 56
10. (a) Distinguish between progressive waves and stationary waves in terms of energy, amplitude and phase
of vibrations.
(b) A beam of light is incident at an angle of 560 on a glass and the reflected light is completely plane
polarized.
(i) Explain what is meant by plane polarized beam?
(ii) Calculate the angle of refraction for the transmitted beam.
P
S1
S2
D
Q
Figure 6
Figure 6 shows an arrangement which in practice enables an interference pattern to be observed.
(i) What are the conditions necessary for an interference pattern to be observed?
(ii) Explain how S1 and S2 become sources of waves.
(iii) What are the approximate dimensions for D and d in a laboratory condition when used to
determine the wavelength of visible light?
(iv) What could be observed along PQ if S1 and S2 are illuminated by white light?
(d) (i) Distinguish clearly between elastic and inelastic collisions
(ii) What is meant by the term linear momentum?
(iii) State the law of conservation of linear momentum and explain how force is related to linear
momentum.
(e) A vehicle collides with a rain gate barrier when travelling at 27.7 ms-1 and is brought to rest in
0.075 s. If the mass of the vehicle and its occupants is 1000 kg, calculate the average force on the
vehicle. The driver used his seat belt which restricted his movement through a distance of 0.25 m
relative to the vehicle. What was the average force exerted by the belt on the driver if the driver’s
mass is 100 kg.
Pellet
plasticine
Toy gun
0.6 m
<<ADVANCED LEVEL PHYSICS
0.25 m
Page 57
X
Floor
(f) Figure 7 shows a set up designed by a student to determine the velocity of a pellet from a toy gun. A
piece of plasticine of mass 50 kg is balanced at the edge of a table such that it fails to fall off. A pellet of
mass 10 g is fired horizontally into the plasticine and remains embedded in it. As a result the plasticine
reaches the floor a horizontal distance of 0.25 m away from the edge of the table.
i.
ii.
iii.
What is the horizontal velocity of the plasticine given that the table surface is 0.6 m high?
What is the velocity of the pellet just before it hits the plasticine?
What is the velocity of the plasticiine just before impact with the floor at a horizontal distance of
0.25 m.
STUDENT’S PROPOSED ANSWERS TO JUNE 2010
<<ADVANCED LEVEL PHYSICS
Page 58
JUNE 2011
1. (a) Explain the fact that the homogeneity of a physical equation is not sufficient for the correct of the
equation.
(b) The electric field intensity, E acting on a point charge, q, placed from a test charge in a vacuum can
q
be expressed as E = 4πε r2. Where εo is the permittivity of a vacuum. Show that this equation is
o
homogeneous.
2.
6V
I2
Figure 1 shows how resistors and cells may
be connected in an electrical circuit.
15Ω
X
I1
10 Ω
Figure 1
(i) I1, I2 and I3 (ii) the Pd across XY
5Ω
I3
9V
Calculate values for
Y
3. Distinguish between elastic and inelastic collision
(ii) A ball of mass m, falls vertically from a height h1, to the ground and rebounces to a height h2.
Calculate the change in momentum in terms of m, h1 and h2.
4. A cathode ray oscilloscope has it Y – sensitivity set at 20 Vcm-1. A sinuosoidal input is suitably applied
to give a steady trace with time base so that the electron beam takes 10-2 s to traverse the screen. If the
trace has a peak to peak value of 4.0 cm and has 4 complete cycles. Estimate the value for
(i) Root mean square
(ii) Frequency of the input signal.
5.
S
P
600
30 N
T
R
Q
m
Figure 2
𝜃
= 𝜋𝑟 2
2.5 kg
Figure 2 shows a string PQRS. P and S are attached to a fixed support and mass m and 2.5 kg are
attached at the point Q and R respectively and the system is in equilibrium. Calculate
(i) the mass, m
(ii) the tension, T
(iii) the angle, 𝜃
6.
Figure 3
800
Figure 3 shows a cross section of an optical fibre used for telecommunication.
(a) State and explain two reasons why the optical fibre is preferred to the copper cables for this purpose.
(b) The speed of light in the core is 1.95 x 108 ms-1 while the smallest angle of incidence in the core is
800. Calculate the refractive indices for
<<ADVANCED LEVEL PHYSICS
Page 60
(i) The core
(ii) the cladding
7. (a) Explain why the specific heat capacity of a gas at constant pressure, Cp is greater than the specific
heat capacity at constant volume, Cv.
(b) Explain why a distinction between specific heat capacity at constant pressure and at constant volume
is important for gases but for solids.
8. (a) (i) Define surface tension
(ii) Describe an experiment to show how the surface tension of a liquid varies with temperature.
(b) (i) A soap bubble of radius 8.0 cm is blown on one end of a tube which is connected to a U – tube
containing water. Calculate the difference in the water levels.
(ii) If another soap bubble of radius of curvature 2.0 cm is now allowed to make contact with the first so
that the radius of curvature of the common surface tension for soap solution is 3.5 x 10-2 Nm-1, calculate
r.
(c) (i) The net force, F, between two particles in a solid varies with their separation , r, according to the
equation F =
8.0×10−20
r2
−
2.0 x 10−96
r10
.Calculate the equilibrium separation, ro
(iii) Sketch a graph showing the force, between two adjacent particles varies with their separation.
(d) (i) Define capacitance
(ii) Describe an experiment to show how the capacitance of a parallel plate capacitor varies with their
separation.
(e) A tiny pith ball of mass 5.0 x 10-4 kg is suspended by a light thread of negligible mass. The ball is
electrically charged and placed in a uniform horizontal electric field of strength 4.0 x 102 NC-1. Calculate
the charge q when it is deflected through an angle of 100.
(f) (i) Sketch a graph showing how the electric field strength E varies with distance, r, from the centre of
a uniform solid metal sphere of radius ro which is positively charged.
(ii) Explain the shape of your graph when r<r0 and when r>r0
9. (a) (i) Explain what is meant by the half – life of a radioactive nuclide.
(ii) Living wood has an activity of 16 counts per minute per gram which is due to the disintegration of
carbon – 14 atoms in the wood. The half – life of carbon – 14 is 5.6 x 103 years. Calculate the age of ship
with a sample of wood of mass 0.5 g from the ship whose activity is 6.5 counts min-1.
(b) Natural uranium contains 0.7 % U – 235. When U – 235 undergoes fission, 200 MeV of energy is
released. Calculate
(i) The number of U – 235 nuclei contained in 1 kg of natural uranium
(ii) The cost to be paid to AES SONEL at the rate of 60 francs per unit when the uranium – 235 conten in
1 kg completely undergoes fission.
(c) Sketch a block diagram of a nuclear reactor and explain the functions of
(i) The coolant (ii) the moderator (iii) the control rods
(d) (i) Explain what is meant by simple harmonic motion
(ii) Sketch graphs to show how the following quantities vary with period of oscillations for one complete
cycle.
(i) Kinetic energy (ii) potential energy (iii) total energy
(e) Pendulum of length 1.2 cm has a bob of mass 0.2 g. The bob is pulled aside a horizontal distance of
20. Cm and then released. Calculate
(i) The velocity of the bob at its lowest point.
(ii) The maximum kinetic energy of the bob.
(f) Mechanical systems may undergo free, damped and forced oscillations
<<ADVANCED LEVEL PHYSICS
Page 61
(i) Explain the meaning of the underlined words
(ii) A string has a length of 2.0 cm and a density of 8.0 x 103 kgm-3. When the string is vibrating in a
fundamental mode with a frequency of 200 Hz, the tension in the string produces a stain of 2 %..
Calculate the Young’s modulus for the string.
10. A student investigates the variation of potential difference, and the current, I, through a semiconductor
diode. The corresponding values of the potential difference and the current are displayed in table 1.
V/mV I/10-4 A
The equation relating I and V is 𝐼 = 𝐼 𝑒 𝐵𝑉 where I and B are
0
0
constants
255
0.004
315
0.016
345
0.036
(b) (i) Plot a suitable graph from which the values of I0 and B could be
obtained.
385
0.089
(ii) Determine the values of I0 and B
410
0.182
(c) Another equation linking I and V is 𝐼 = 𝐼0 (𝑒 𝐵𝑉 − 1)
455
0.552
475
0.903
What physical approximations could have been considered for 𝐼 =
𝐼0 𝑒 𝐵𝑉
495
1.400
505
1.820
515
2.230
530
3.100
(a) What is the physical significance of I0
Table 1
STUDENT’S PROPOSED ANSWERS TO JUNE 2011
<<ADVANCED LEVEL PHYSICS
Page 62
JUNE 2012
1. (i) When an alternating potential difference is applied to the primary of a transformer, why is an
alternating emf is produced (induced) in the secondary.
(ii) State with reasons the output of the secondary, if the alternating voltage of the primary were replaced
with direct current voltage.
(iii) Why are transformers coils wound on an iron core? State an important feature of such a core and the
function of the feature
2. Figure 1 shows a graph of stopping potential, V, plotted against frequency, for a certain metal, Z
Vest/
V
0
Figure 1 f/Hz
(i) Calculate the threshold frequency of the metal, given that the work
function is 3.8 eV.
(ii) To which region of the electromagnetic spectrum does the
wavelength calculated in (i) belong
(iii) Of what significance is the gradient of the graph in figure 1. Draw a set – up from which the results
above could have been obtained.
3. The voltage sensitivity, θ⁄V, of a moving coil meter is given by
θ⁄ = BAN
V
CR
Where B is the magnetic field strength, A is the area of the coil, N is the number of turns of the coil, C is
the torsional constant and R is the resistance of the coil. What are the units of C?
4. Figure 2 shows the loading and unloading curve for rubber
<<ADVANCED LEVEL PHYSICS
Page 64
load/N
1.2
A
1
0.8
0.6
0.4
0.2
0
0
10
20
30
figure 2
5.
6.
7.
8.
40
50
60
70
80
90
extension/mm
(a) Use the graph to estimate the gain in the internal energy by the strip of rubber when it is extended to
A and unloaded.
(a) Explain why the rubber can be stretched to about ten times its original length before its elastic limit is
reached.
Copper contains about 1029 free electrons per unit volume. Determine the average drift velocity of the
free electrons in a uniform copper wire of length 0.5 m when a p.d of 1.5 V is applied across it.
(Resistivity of copper = 1.7 x 10-8 Ωm)
The SONEL thermal plant in Limbe supplies SONARA with 1.0MW at a pd of 1.0 x 104 V. The
resistance between the power station and the factory is 0.5 Ω.
(a) What is the power output of the thermal plant
(b) Explain why the power station output voltage is always stepped up before transmission over a long
distance.
An observer is standing at the bank of a stream. At what speed will the observer see a boat sailing at 50
cms-1 relative to a stream which is flowing at 10 cms-1
(i) Downstream
(ii) perpendicular to the flow of the stream.
(a) (i) Differentiate between longitudinal and transverse waves
(ii) Describe an experiment to measure the speed of sound in free air from measurements of frequency
and wavelength using progressive wave. Your description should include a diagram, procedure and
precaution(s), observations, calculations and conclusion.
(b) Light is travelling in glass A with speed 1.9568 x 108 ms-1. It reaches an interface with a different
glass B, at an angle slightly greater than the critical angle of 87.600 and undergoes total internal
reflection.
(i) Explain with the aid of a diagram what is meant by critical and total internal reflection.
(ii) Calculate the speed of light in B.
<<ADVANCED LEVEL PHYSICS
Page 65
(c) An observer travelling with constant velocity of 25 ms-1 passes close to a stationary sound and noticed
that there is a change of frequency of 60 Hz as he passes the source. What is the frequency of the source?
(d) (i) Define the specific heat of vaporization
(ii) Describe an experiment to determine the specific heat of vaporization of water. Your description
should include a diagram, procedure and precautions, observations, calculations and concludions.
(e) An office uses a water tank containing 1.2 m3 of water as a thermal store. The water in the tank is
heated to 980 C in the night when there is less electrical energy consumption. In the morning when there
it is very cold , the water is pumped round the office to keep the office warm.
(i) Calculate the energy given out by the water on a day that the temperature drops from 980 C to 650 C.
(ii) The radiators in the office gives out an average power of 1.5 KW each. For how long can they operate
at this power before the water temperature drops to 650 C.
(iii) Explain why this heating system operates more efficiently in the morning than in the afternoon.
(f) State with reasons two thermometric properties.
9. (a) (i) Define capacitance
(ii) What are the physical factors on which the capacitance of a parallel plate capacitor depends?
(iii) How would you relate capacitance to permittivity?
(b) Figure 3 shows a circuit for charging and discharging of a capacitor using a two way switch.
G
R
C
Figure 3
9V
<<ADVANCED LEVEL PHYSICS
Page 66
Figure 4
The graph in figure shows how the current varies with time during the discharge. Use the graph to
(i) Calculate the value of the resistance R
(ii) Estimate the charge stored on the capacitor at the start of the discharge
(iii) Find the energy stored at the start of the discharge
(iv) Calculate the capacitance of the capacitor
(v) What would be the effect on the shape of the graph if the value of R were increased?
(c) A 5.0 µF capacitor and 8.0 µF capacitor are charged by a 12 V battery. The two capacitors are
then connected as shown in figure 5
5.0µF
Calculate the charge on the 5.0µF capacitor
as shown in figure 5
8.0µF
Figure 5
(d) Distinguish between intrinsic and extrinsic semiconductors
(e) The output characteristics in figure 7 are for the transistor circuit shown in figure 6
R1=74kΩ
C1
R3
C2
9V
Voutput (3 V)
Vin
R2
R4= 1kΩ
Figure 6
<<ADVANCED LEVEL PHYSICS
Page 67
8
IB = 80µA
IC /mA
7
IB = 60µA
6
5
IB = 40µA
4
3
IB = 2µA
2
1
0
-0.5
0.5
1.5
2.5
3.5
4.5
5.5VCE /V
(i) Construct a table of vaules which would enable you to draw the transfer characteristic at constant VCE
of 3.5 V. hence draw a graph from which the you could obtain the current gain hfe for the transistor. What
is the value of hfe.
(ii) What are the functions of R3, C1 and C2
(iii) Calculate R1 and R2
Given that the current through R4 = 1 mA, current through R3 = 1mA and VCE=0.6 V
10. (a) Explain what is meant by the emf and terminal pd of a battery. Why is the pd between the terminals of
a battery not always the same as the emf?
(b) The emf of the electricity supply to a rural farm house is 240 V. The resistance of the cables to the
farm house may be considered as the internal resistance of the supply. When an electric cooker is used in
the farm house the measured voltage across the cooker is 220. If the resistance of the cooker is 40 Ω,
(i) Calculate the power of the cooker.
(ii) Calculate the resistance of the cables to the farm house
(iii) Explain why the voltage measured at the cooker is less than the supply voltage when the cooker is in
use.
(iv) Suggest one disadvantage of this power supply.
(c) A semiconductor diode and a resistor of constant resistance are connected in some way inside the box
having two external terminals (figure 8). When a p.d of 4.0 V is applied across the terminals, the
ammeter reads 100 mA. If the same p.d is applied in the reverse direction, the ammeter reads 200 A
(i) What is the most likely arrangement of the resistor and
diode? Explain your deduction
(ii) Calculate the resistance of the resistor and the forward bias
resistance of the diode.
Ammeter
Figure 8
<<ADVANCED LEVEL PHYSICS
Page 68
(d) (i) State the observations obtained from the Rutherford α – scattering experiment with a thin gold foil.
What conclusions may be deduced from each of these observations?
(ii) Explain how and why the masses of compounds differ from the sum of the masses of their constituent
particles.
(e) Radium (Ra) decays to radon (Rn) by the reaction
226
222
4
89Ra → 87Rn + 2He + γ
(i) Estimate the energy (in joules) released when an atom of 226
89Ra decays
(ii) Estimate the wavelength of gamma photon emitted during this decay given that 4 % of the energy
turns to gamma radiations.
(iii) What happens to 96 % of the energy?
The atomic masses are radium = 3.7533 x 10-25 kg, radon = 3.686 x 10-25 kg, helium = 0.066 x 10-25 kg.
(f) An α – particle is accelerated to attained a kinetic energy of 1.34 x 10-15 KJ, collides head – on with a
gold nucleus. Calculate the upper limit of the radius of the gold nucleus. Proton number of gold is 79
STUDENT’S PROPOSED ANSWERS TO JUNE 2012
<<ADVANCED LEVEL PHYSICS
Page 69
JUNE 2013
1. The equation RT = Ro(1+αT+βT ), where α and β are constants, describes the variation of the resistance
of a wire with temperature T. This equation is homogeneous.
(i) Explain the meaning of the underlined word
(ii) What are the units of αT and βT2?
(iii) What are the units of α and β?
2. (a) What is meant by the term “moment of a force”?
(b) If three forces are in equilibrium they must be coplanar and concurrent. Explain the meaning of the
word “coplanar”
(c) A uniform metal bar of length 4.0 m and mass 80 kg rest with its upper end against a smooth vertical
wall and with its lower end on a rough surface of coefficient of friction 0.32. What is the maximum angle
made with the horizontal to which the bar can be inclined without sliding?
3. (a) Compare the image formed by a diverging lens and a converging lens, both of focal length 20 cm, if
an object is placed 12 cm from each.
(b) Why is a frequency modulated signal preferred to an amplitude modulated signal system in
communication?
4. A 1500 µF capacitor is fully charged using a 100 V d.c power supply. It is disconnected from the power
supply and connected to an unchanged 1000 µF capacitor.
(a) Calculate the p.d across the terminals of the capacitor
(b) Calculate the initial and final energy stored in the capacitors
(c) Why is there a loss in energy?
2
I3
5.
6V
2Ω
A
I1
6V
3Ω
4Ω
In figure 1, determine
(i) The current I1 and I2
(ii) The p.d between AB
6Ω
I2
B
Figure 1
6.
R2
R1
9.0 V
VBE
(i) What is a p – type semiconductor?
Figure 2 shows a transistor in the common emitter mode.
The transistor has the following characteristics VBE = 0.62 V,
hfe = 100. The input resistance R1=60 kΩ and the load
resistance R2 = 600 Ω
(ii) Calculate the current through the load
(iii) Calculate VCE
Figure 2
7. Distinguish between liquids and gases using
(i) Intermolecular force
(ii) The kinetic theory of matter
8. (a) (i) Explain what is meant by the thermometric property of a substance?
(ii) State two qualities which can make the thermometric property suitable for temperature measurements.
<<ADVANCED LEVEL PHYSICS
Page 70
(iii) The melting point of a metal is measured using a resistance thermometer and a constant volume gas
thermometer. Explain whether the values obtained would be the same or different.
(b) Describe an experiment to determine the specific latent heat of vaporization of water. Your
description should include a diagram, procedure, precautions, observations and conclusions.
(c) (i) A piece of metal block of mass 0.8 kg and specific heat capacity 455 Jkg-1K-1 is initially heated in a
furnace. The block is then immersed in 1.2 kg ice in an ice container and equilibrium temperature of 48oC
is obtained. Calculate the initial temperature of the block.
(ii) Explain whether all electrical insulators are necessarily good thermal conductors.
(d) (i) State the laws of electromagnetic induction.
(ii) An electron of charge, e, and mass, m, enters a uniform magnetic field B of value 2.0 x 10-3 T as
shown in figure 3 and moves with a speed, v.
Copy figure 3 into your paper and indicate the path of
the electron in the field.
(iii) Calculate the number of revolutions per second
made by the electrons.
v
Figure 3
(e) Describe an experiment to determine the specific charge of an electron. Your account should
include a diagram, procedure, observations and conclusions.
E
X
R
L
Y
Figure 4 shows two bulbs X and Y connected to a supply E. The
inductance of L is 6.0 x 10-3 H, the resistance of R is 2.0 Ω, while
the resistance of X and Y are each 2.0Ω.
(i) Calculate the current in Y when it is fully lighted.
(ii) Sketch, on the same axes, graphs to show how the p.d across
X and Y vary with time
Figure 4
9. (a) Define stopping potential
(b) Use Einstein photoelectric equation to explain
(i) Why for a particular metal, electrons are emitted only when the frequency of the incident radiation is
greater than a certain value.
(ii) Why the maximum speed of the emitted electrons is independent of the intensity of the incident
radiation.
(c) Figure shows how the frequency (f) of incident radiation on a metal surface varies with the energy of
the emitted photoelectrons
<<ADVANCED LEVEL PHYSICS
Page 71
Figure 5
i) From the graph, determine the threshold frequency and calculate the maximum wavelength of the
emitted
electrons.
(ii) Calculate values for: the plank constant and the work function
(d) An X – ray photon has wavelength of 3.0 x 10-10 m. Calculate the values for
(i) The momentum (ii) The energy
(iii) The mass of the particle associated with the proton which moves at the speed of light.
(e) (i) Define time constant.
Figure 6 shows how a resistor R and a capacitor may be connected in a circuit.
V
600 µF
R
The capacitor is fully charged and connected to the resistor R
and the reading on the voltmeter falls by half in 60 s.
(ii) Calculate the time constant and explain how its value could
be increased
Figure 6
(f) Figure 7 shows the displacement time graph for a vibrating system.
<<ADVANCED LEVEL PHYSICS
Page 72
displacement/cm
50
40
30
20
10
t/s
0
(i) Explain whether the
motion is simple harmonic or
not.
Use the graph to calculate
(ii) The amplitude and
frequency of oscillation
(iii) Write the wave equation
for the motion described in in
figure 7
-10
-20
-30
-40
-50
0
10
20
30
40
figure 7
50
60
70
80
90
100
(g) (i) Sketch a graph to show how the velocity changes with time for the motion above.
(ii) Compare nuclear fission and nuclear fusion as sources of energy.
10. The table below gives the force, F, between a pair of molecules in a solid at various separations, r.
Force, Separation,
F/10-7 r/10-10 m
N
8.8
0.1
5.6
0.26
0.8
0.34
-2.0
0.42
-5.0
0.52
-2.0
1.5
-0.8
1.8
-0.4
1.9
(a) Draw a graph of F against r for a pair of molecule
(b) (i)
From your
-8.0
0.8 graph, determine the molecular spacing for the molecules at equilibrium separation.
(ii) Calculate the energy used to separate the molecules completely
(iii) What
-6.6 is the
1.2physical significance of the energy calculated in (ii)
(c) How can your graph be used to explain that at some point
-4.0 law
1.34
(i) Hooke’s
is slightly obeyed
(ii) The vibration of the molecules is simple harmonic
<<ADVANCED LEVEL PHYSICS
Page 73
JUNE 2014
1
1. The resonant frequency, f, of an inductor – capacitor circuit (L –C) is given by the equation fr = 2π√LC
where L and C are respectively the inductance and capacitance of the circuit.
(i) Show that the equation is homogeneous.
(ii) Calculate the inductance of the circuit if the resonant frequency is 104 Hz, across a circuit of
capacitance 4.0 x 10-9 F.
2. Figure 1 shows how capacitors are connected in a circuit.
X
90 V
3 µF
Y
Figure 1
Calculate the
(i) Charge stored by the 4 µF capacitor.
(ii) Potential difference across YZ
4 µF
5 µF
Z
3. (a) Explain why the specific latent heat of vaporization of a substance is always larger than the specific
latent heat of fusion for the same substance.
(b) A mixture of 50 g of ice and 210 g of water at 00 C is passed in until all the ice just melts. Calculate
the mass of the water now in the container.
4. (a) Consider the acceleration of free fall on the moon’s surface to be 1.6 ms-2. Determine the length of a
simple pendulum which has a period of 1.0 s on the moon’s surface.
(b) A particle executing simple harmonic motion has 5 times the energy of another particle but their
masses and frequencies are equal. Calculate the ratio of the amplitudes for the two motions.
5. (a) Explain how the internal energy of a system is modified when it undergoes an isothermal change. In
one such change, 200 J of energy was added to a system. How much work was done on or by the system?
(b) Scientific analysis shows that light gases such as helium nuclei undergo fusion to release energy in
the sun. Estimate the root – mean – square speed of helium atom of mass 6.6 x 10-27 kg near the surface
of the sun where the temperature is about 6000 K.
6. (i) Draw a diagram of a tuning circuit of a radio.
(ii) Distinguish between A.M and F.M radio transmission systems.
7. (a) Figure 2 shows how the force, F, between two molecules varies with the separation, r.
<<ADVANCED LEVEL PHYSICS
Page 74
F/10-4
N
10
8
6
4
2
0
-2
r/10-10 m
-4
-6
0
1
2
3
4
5
6
Use the graph to calculate
(i) The energy needed to completely separate the two molecules initially at their equilibrium separation
(ii) Calculate the gradient of the graph around the linear region. What is the significance of the slope?
8. (a) (i) Define the term resistivity
(ii) Describe an experiment you will carryout to determine the electrical resistance of a specimen of
copper wire.
(b) A student designs an electrical heating element using a wire 5.0 m long of diameter 1.00mm so that it
dissipates 2 KW when connected to a 240 V mains. Calculate,
(i) The resistivity of the wire
(ii) The cost of using the element for 30 days if ENEO CAMEROON charges 60 frs per kilowatt – hour
and the coil is used for 6 hrs each day.
(c) Sketch on the same axis, graphs to show how the current across the following material vary with
potential difference across their ends.
(i) Copper wire (ii) Silicon
(iii) Filament bulb
(d) (i) Define temperature coefficient of resistance of a material.
(ii) Describe an experiment you can carryout to determine the temperature coefficient of resistance of a
metal wire.
(c) A surface of a metal is illuminated with light of wavelength 590 nm. A p.d of 0.15 V is applied
between the metal surface and collecting electrodes in order to prevent the collection of electrons.
Calculate
(i) The work function of the metal (ii) The work done against the most energetic photoelectrons.
(iii) The speed of the most energetic electron.
(f) Light of varying frequencies is incident on the surface of three different metals, X, Y, and Z. the work
function (W0) are in the order W0z< W0y< W0x. sketch on the same axis graphs to show how the
maximum kinetic energies of photoelectrons vary with frequency.
9. (a) State Newton’s laws of motion
Fixed
support
(b)
300
Pulle
F1
300 F2
Pull
<<ADVANCED
LEVEL
Bucket
of PHYSICS
sand
Figure
Figure 3 shows a bucket of sand which is pulled
upward at a building site at the instant shown, the
forces F1 and F2 have equal magnitudes of 600 N and
the bucket is moving with an acceleration of 2 ms-2.
Determine the mass of the bucket and its content.
Page 75
(c)
Fixed support
Bullet of mass 16 g
thread
Wood of mass 2kg
kg
Figure 4
A bullet is fired horizontally as shown in figure 4, so that it strikes the wood with a velocity of 18 ms-1. It
gets embedded in a block of wood suspended freely using a long thread. Calculate
(i) The magnitude of the momentum of the bullet just before it enters the block.
(ii) The magnitude of the initial velocity of the block immediately after the impact.
(iii) The kinetic energy of the block and bullet immediately after impact and use it to say whether the or
not the collision is elastic.
(iv) The maximum height attained above the equilibrium position by the block and the embedded bullet.
(d) (i) State the following laws: Newton’s law of universal gravitation, Coulomb’s law and Faraday’s law
of electromagnetic induction.
(ii)
+
Negatively charged
polythene
5.00 mm
180 V
-
Figure 5
Figure 5 shows a negatively charged polythene sphere of mass 3.5 x 10-15 kg held stationary between two
parallel plates. How many excess electrons are on the sphere?
(e) A space ship of mass 6.0 x106 kg is launched into space so that it orbits the earth at a height, H, above
the earth’s surface. Where H = RE is the mean radius of the earth.
(i) Explain why an astronaut moving about in the spacecraft at this height feels weightless.
(ii) Determine the minimum energy required to take the spaceship to the desired height. Explain why
more energy is needed in the practical situation than in the calculated value?
(ii) Calculate the period of the space ship in its orbit at this height and hence explain whether or not the
space ship is in a geo – stationary orbit.
<<ADVANCED LEVEL PHYSICS
Page 76
10. A student used the diffraction experiment to investigate the variation of nuclear radius, R, and nucleon
number, A, for several nuclear species. The corresponding value of R and A are recorded in the table
which follows.
R/10-15
m
4.4
4.7
5.0
5.3
5.7
6.0
6.2
6.5
6.8
7.0
A
25
50
75
100
125
150
175
200
225
250
R and A are related by an expression of the form 𝑅 = 𝑅0 𝐴𝑛 where, R0 and n are constants
(a) (i) Plot a suitable graph to determine the values of R0 and n
(ii) Hence determine the values of R0 and n
(b) (i) What is the physical significance of R0
(ii) State the relationship between R and A
STUDENT’S PROPOSED ANSWERS TO JUNE 2014
<<ADVANCED LEVEL PHYSICS
Page 77
JUNE 2015
1. (a) What is a homogeneous equation?
(b) The equation 𝑃 + ℎ𝜌𝑔 + 𝜌𝑣 2 = 𝐾 is homogeneous, where P is the pressure, h is the depth, g is the
acceleration due to gravity, ρ is the density, v is the velocity and K is a constant. Determine the base units
of K.
2. A girl swings a stone of mass 100 g at the end of a string in a horizontal circle of diameter 80.0 cm above
her head. Her friends observing the action sees a shadow of the motion of the stone on a nearby wall and
and notices that the stone makes 150 revolutions in 5 minutes. Calculate
(a) The frequency of the motion.
(b) The centripetal force on the stone
(c) Sketch a distance – time graph for this motion of the shadow for two cycles.
3. A quantity of steam at 1000 C is passed into a 1.5 kg of pure melting ice in a highly insulated calorimeter
so that the heat given out by the steam in condensing is just enough to melt the ice. Determine
(a) The quantity of steam passed into the ice to melt it completely.
(b) The equilibrium temperature attained by the mixture
4. Figure 1 show a small spherical charged metal bob 50 g which initially hangs vertically between two
conducting plates. When a potential difference of 12.0 V is maintained across the plates the thread makes
an inclination of 300 to the vertical.
(a) Draw a free body diagram for the bob when the p.d
is applied.
(b) Determine the charge on the bob
300
-
5.0 cm
+
Figure 1
5. Figure 2 shows how resistors may be connected in an electrical circuit. The bridge circuit is balanced
when the voltmeter M3 reads 3.0 V
M
2 kΩ
2
R1
M
Determine the
(a) Reading of M1 and M2
(b) Resistance of R1 and R2
-4
I = 5.0 x 10 A
12 kΩ
𝜀
6 kΩ
M
R2
Figure 2
6. (a) (i) What is meant by a coherence light source?
<<ADVANCED LEVEL PHYSICS
Page 78
(ii) Describe how you can determine the wavelength of monochromatic light using Young’s Double slits.
Your account should include a diagram, observations, precautions and how you would use the
observations to reach a conclusion.
(b) A car sounding an alarm at a frequency of 512 Hz is approaching a stationary listener at a speed of 8
ms-1.
(i) Explain why the listener has the impression that the frequency of the sound heard is varying.
(ii) What is the apparent frequency of the horn as perceived by the listener?
(c) An object is placed 20. Cm away from a thin convex lens and then a thin concave lens each of focal
length 10.0 cm and perpendicularly to the principal axis in each case. Use either ray diagrams or
otherwise distinguish between the images obtained in each case when the object is brightly illuminated
such that light rays from the object reach the lens parallel to the principal axis.
(d) (i) Explain what is meant by a material is elastic?
(ii) Describe an experiment to determine Young’s modulus for a copper wire. Your description should
include a diagram, procedure, precautions observations and conclusions.
(e) A toy having a plastic head resting on one end of a light spring is stretched as shown in figure 3
8
F/N
E
6
4
2
0
0
10
20
30
40
extension/mm
(i) Explain why there is a change in the gradient at the point E
(ii) Calculate the maximum elastic potential energy of the toy – spring system
(iii) Explain what happens to the energy of the system when the load is removed and the spring regains it
original length.
(f) The kinetic theory of matter describes the behavior of a gas in terms of the properties of its molecules.
Use this theory to explain
(i) Why a gas in a container at room temperature exerts pressure on the walls of the container.
(ii) Why the pressure increases when more of the same gas is introduced into the same container at the
same temperature.
7. An experiment was performed to investigate how the resistance of a material wire varies with the
temperature. The following data was recorded.
Resistance/Ω Temperature
330.0
10
340.0
20
350.0
30
360.0
40
Theory suggests that the resistance of the wire is related to its
temperature by the expression R = R 0 (1 + αθ) where R 0 is the
resistance at a temperature 00C, and α is a constant.
(a) Plot a suitable graph from which R 0 and α could be obtained.
(b) Use the graph to obtain the values of R 0 and α
(c) Say whether this material is a conductor or semiconductor.
<<ADVANCED LEVEL PHYSICS
Page 79
370.0
50
380.0
60
389.0
70
400.0
80
410.0
85
OPTIONS
OPTION 1: ENERGY RESOURCES AND ENVIRONMENTAL PHYSICS
8. (a) Sate two advantages of nuclear fusion over nuclear fission as sources of energy
(b) Biomass, solar energy and hydroelectricity are some of the sources of energy from which functional
energy could be obtained.
(i) What is meant by functional energy?
(ii) Choose any two of the sources and briefly explain how functional energy could be obtained from
them.
(c) The power derived from a windmill is given by the equationP =
ρAv2
2
, where ρ is the average air
density and, A is the area of the blade and v is the average wind speed. One such aero – generator has a
blade of diameter 6.0 m. given that the efficiency of the system is 25 % at a wind speed of 13.5 ms-1,
(i) Calculate the power output of the aero – generator. Assume the average density of the air to be 1.2
kgm-3
(ii) Why is the efficiency of the system less than 100 %
(d) (i) Name a substance which is responsible for the depletion of ozone layer.
(ii) State and explain the impact of the depletion of ozone layer on the environment.
OPTION 2: COMMUNICATION
9. (a) A radio station uses a carrier frequency of 200 kHz to transmit an amplitude – modulated wave.
The transmission consists of audio signals within the frequency range 50 kHz – 9 kHz.
(i) Explain the meaning of the bolded phrases
(ii) Calculate the minimum and the maximum frequency sidebands and the bandwidth.
(b) Figure 4 shows a simple tuning radio circuit.
Antenna
Output decoder
Inductor
Variable capacitor
(i) Explain how the tuning circuit functions
(ii) Given that the coil used has an inductance of 4.0
MH, calculate the value for the capacitor required to
tune into the broadcast described in 9(a)
(iii) What is the use of the decoder in this circuit?
(c) (i) State three advantages which digital transmission has over analogue transmission.
(ii) Explain how several telephones conversation can be transmitted at the same time along a single
optical fibre.
<<ADVANCED LEVEL PHYSICS
Page 80
OPTION 3: ELECTRONICS
10. (a) (i) What is meant by thermionic emission?
(ii) Distinguish between n – type and p – type semiconductors
(b) You are given two circuits consisting of:
(i) A resistor of 500Ω and a capacitor connected in series to 9.0 V d.c supply.
(ii) An inductor and a resistor of 500 Ω connected in series to 9.0 V dc supply.
Sketch current – time graphs for these circuits and explain the difference between them.
(c) (i) Explain how a transistor is used as a switch.
(ii) State in words and in the form of a truth table, the action of an OR logic gate with two inputs.
OPTION 4: MEDICAL PHYSICS
11. (a) (i) Draw a simple diagram of the human eye, showing clearly the parts which enable the eye to form
an image of an object.
(ii) Name any two eye defects, explaining how each defect manifests and explain how each defect may be
corrected.
(b) X – rays and ultrasound are two techniques used for imaging of parts of the human body.
(i) State one part of the body where each of the techniques would be more suitable than the other.
(ii) Explain why ultrasound is not likely to replace X – rays completely for medical diagnosis.
(c) Explain how the magnetic Resonance (MR) scanner produces a visual image of a cross – section of a
part of the body of a patient.
<<ADVANCED LEVEL PHYSICS
Page 81
JUNE 2016
1.
(a) Explain why the homogeneity of a physical equation is not a sufficient condition for the correctness
of the physical equation?
𝑑𝐼
(b) Faraday’s law may be stated in the form𝐸 = −𝐿 𝑑𝑡, where, E, is the induced emf, L is the inductance
𝑑𝐼
of the coil and 𝑑𝑡 is the rate of change of current. Determine the base units of L, if the equation is
homogeneous.
2. A simple pendulum of lengthℓ, has a period T, on the surface of the earth. The simple pendulum is
carried in a space craft to a height of 2R, above the earth’s surface where R is the radius of the earth.
Explain where the period of the pendulum at this height would increase or decrease.
3. (a)Distinguish between thermionic emission and photo-electric effect.
(b) An electromagnetic radiation of wavelength 6.3 x 10-14m falls on a clean metal surface which has a
work function of 2.25 x 10-14 J. Explain whether photo – electrons would be emitted or not.
4. (a) A transformer cannot be used to run a 230 V, 100 V mains lamp directly from a 12 V d.c car batter
Suggest in terms of fields and energy why the system cannot work
(b) (i) Discuss how the system can be adapted to function
(ii) What type of transformer does figure 1 represents?
5. Figure 2 shows how resistors and cells may be connected in an electric circuit.
60 Ω
6.0 V
50 Ω
12.0 V
40 Ω
Figure 2
20 Ω
3.0 V
Calculate the:
(i) Current flowing through the 40 Ω resistors
(ii) Voltage drop across the 80 Ω resistor
80 Ω
6. (a) (i) Distinguish between transverse and longitudinal waves
(ii) Describe an experiment to determine the speed of sound in air. Your account should include a
diagram, procedure, precautions, observation and conclusion
(b) A source of sound whose frequency is 51.6 Hz is placed in front of a flat vertical smooth wall, if a
microphone is moved from the source directly towards the wall a series of minimum and maximum
values in its output are observed at equally spaced intervals. The speed of sound at room temperature is
330 ms-1.
(i) Explain how these minimum positions are formed
(ii) Calculate the separation of these minimum points
<<ADVANCED LEVEL PHYSICS
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(iii) What can be done to increase the separation calculated in (ii) above.
(d)Explain why the specific heat capacities of gases are either measured at constant pressure or at
constant volume while this is not requiring for solids or liquids.
(e) Describe an experiment to determine the specific heat capacity of a liquid. Your account should
include a diagram, procedure, precautions, and conclusion.
(f) In terms of molecular behavior explain,
(i) How liquids are similar to gases but different from gases.
(g) A highly lagged compound bar 25.0 cm long is made from a copper 15.0 cm long joined to aluminum
bar of equal cross – sectional area. The free end of the copper is maintained at 1000C while that of
aluminum is maintained at 00C. Calculate the temperature gradient for each of the bars under steady
states, given that the ratio of thermal conductivities of copper to aluminum is 15:7
7.Table 1 shows the force, F, between two charged particles in a substance. The force is given by the
𝑄2
equation 𝐹 = 𝑟 2 4𝜋𝜀. In order to confirm this relationship the following data was recorded for various
values of F and r, the distance between the charged particles. Q = 4.4 x 10-6 C
F/N
1.0
1.5
2.0
2.5
3.0
4.0
4.5
r/nm
355.1
297.5
258.2
230.6
210.8
182.6
172.0
5.0
6.0
163.3
149.0
Table 1
(a) Plot a suitable graph from which 𝜀 could be determine
(b) (i) Find the slope S of the graph
(ii) What does S represent? (iii) Calculate the value of 𝜀
(c) What would the nature of the forces if the experiment was conducted in a medium of higher dielectric
constant?
OPTIONS
OPTION 1: ENERGY RESOURCES AND ENVIROMENTAL PHYSICS
8. (a) (i) What do you understand by finite and renewable energy resources?
(ii) Given that the mean distance of the earth from the sun is 1.5 x 1011 m and the power output of the
sun is 4 x 1026 W, calculate a value for the solar constant. State the assumption that you have made in
your calculation.
(b) Describe the processes by which electrical energy could be obtained from the following sources of
energy.
- Geothermal energy
- Wind energy
(c) (i) Discuss the consequences on humanity of the destruction of the ionosphere layer
(ii) Explain ways by which the ionosphere can be protected from destruction.
OPTION 2: COMMUNICATION
9. (a) (i) Draw a basic block diagram of a mobile phone handset
(ii) Compare the use of optical fibres and copper cables in the transmission of information in terms of:
Security, Noise, and Signal attenuation
(b) What is the full meaning of the following abbreviations?
- SIM
- SMS
(c) Explain how a radio receiver works
OPTION 3: ELECTRONICS
10. (a) Explain why a piece of pure silicon may not conduct electricity at 100C but would conduct at 800C
<<ADVANCED LEVEL PHYSICS
Page 83
(b) A Capacitor, an ammeter and an a.c power source are connected in series and the readings on the
ammeter noted. The capacitor and the ammeter are disconnected and connected to a d.c power source.
The reading is also noted will the ammeter readings in the two cases be similar or different? Explain.
Figure 3 is an amplifier circuit using NPN transistor in the common emitter mode. The base current is
25 µF when the output voltage V0 is 6.0 V for a current gain of 60.
+9.0 V
RL
Rb
Ib
0V
Calculate
(i) The base current
Cout
Vout
Figure 3
(ii) The value of RL
(iii) Explain the use of the capacitor Cout
OPTION 4: MEDICAL PHYSICS
11. (a) (i) Draw a simple structure of the ear and describe how the ear functions.
(b) (i) Name two light – sensitive receptors in the human eye.
(ii) By reference to refraction at the cornea and the lens, draw a diagram showing how the rays from a
distant object form a blurred image in the eye.
(iii) A patient suffering from long sight has a near point which is 1.5 m from his eyes. Determine the type
of lens that the patient should use to correct this defect.
(c) Explain the principle of operation for obtaining the ECG waveform. How is it useful in diagnosing the
heart problems?
<<ADVANCED LEVEL PHYSICS
Page 84
JUNE 2017
1. The energy stored in an air filled parallel plate capacitor whose area of overlap is “A” is given by the
equation 𝐸 =
𝑉 2 𝐴∈0
2𝑑
, where d is the separation between the plates, V, is the potential difference across the
plates and ∈0 , is the permittivity of free space. Show that this equation is homogeneous.
2. An object is placed 30.0 cm from a converging lens of focal length 15.0 cm and 30.0 cm from a diverging
lens of the same focal length as the converging lens. Calculate the magnification of the lenses and
describe the images formed by the:
(i) converging lens
(ii) diverging lens
3. Figure 1 shows how capacitors, X, Y, and Z are connected to a battery in an electric circuit.
400 µF
400 µF
Y
X
Z
600 µF
12.0 V
Figure 1
Calculate
a) Voltage across the capacitor Y
b) Energy stored in the capacitor X
4. (a) Explain what is meant by the wave – particle duality.
(b) Estimate the de Broglie wave length for an electron emitted by thermonic emission into vacuum from
a hot cathode and accelerated by a pd of 3.0 x 104 V.
5. (i) Forces may generally be classified as contact forces or action at a distance forces. Explain the
meaning of the phrases in bold
(ii) Give an example of each type in 5(i).
6. (a) Distinguish between an ideal gas and a real gas
(b) One form of the equations for an ideal gas is 𝑃𝑉 = 𝑛𝑅𝑇 … … … … … … … … … . (𝑖)
Where P, is the pressure of the gas, V is the volume occupied by the molecules, n is the number of moles,
R is the molar gas constant and T is the absolute temperature of the gas.
1
Another expression relating pressure for an ideal gas is 𝑃 = 𝜌𝑐̅̅̅2 … … … … … … … … . (𝑖𝑖)
3
(i) State the four assumptions of the kinetic theory of matter used to derive equation (i)
(ii) By considering the equations for the ideal gas, show how the average kinetic energy is related to
the absolute temperature
(c) (i) State the second law of thermodynamics.
(ii) A system delivers an amount of heat Qh to an engine which does mechanical work, W, and
releases Q0 to the atmosphere. Considering the first law of thermodynamics, use the symbols Qh, Q0 and
W to work out the efficiency of the engine.
(iii) Explain why Efficiency is less than 100 %
(d) (i) Compare a moving coil instrument and an oscilloscope as current measuring devices
(ii) Why is a.c system preferred to d.c system for long distance transmission of electrical energy?
<<ADVANCED LEVEL PHYSICS
Page 86
(e) A thin wire of mass 5.0 g is wound round a dry piece of wood to produce several turns and the ends
joined together. When a powerful bar magnet is swiftly moved over the coil in an interval of 0.3 s the
wire burns out.
(i) Explain why the wire burns out.
(ii) State and explain what can be done to reduce the time for the wire to burn
(iii) A wire burns out when a minimum current of 5.0 A flows in at a pd of 150 V. If the temperature of
the wire changes by 800 C, estimate the specific heat capacity of the wire.
(f) A coil of inductance 50 H, a capacitor of 200 µF and a resistor of 1 kΩ are connected in series to a
signal generator.
(i) Write down the equation for the impedance of the circuit
(ii) Determine the frequency of the a.c signal which allows for a maximum potential difference across
the resistor
(iii)Explain why the potential difference across the resistor is maximum at this frequency.
7. In an experiment to determine the properties of a acr battery, various loads were connected to its
terminals in a closed circuit. A record of the different potential difference, V, across each load and the
corresponding current, through the battery was recorded as didplayed in the table 1 below.
V/V
16.6
15.2
12.0
8.6
7.8
6.6
5.2
3.8
I/mA
4.0
8.0
16.8
26.4
28.8
32.0
36.0
40.0
Table 1
Theory holds that the potential difference, V, and the current, I, vary according to the equation
𝜀 = 𝑉 + 𝐼𝑟. Where 𝜀 and r are the electromotive force and internal resistance of the battery respectively.
a) Plot a suitable graph from which 𝜀 and r could be obtained
b) Determine the value of 𝜀 and r
c) Would you expect this source to deliver power more efficiently when connected to a 200 Ω or
2000 Ω resistor? Explain
OPTIONS
OPTION I: ENERGY RESOURCES AN ENVIRONMENTAL PHYSICS
8. (a) (i) Distinguish between renewable and non renewable energy sources giving an example of each
(ii) A solar panel delivers power of 2.0 kW when the rays of the sun fall normally on it. If the solar
constant is 1.2 x 103 Wm-2 and its efficiency is only 40 %, calculate the area of the solar panel.
(b) (i) Explain why Cameroon cannot rely completely on the solar energy for its energy needs.
(ii) Draw an energy flow diagram for an energy scheme in which wood is burnt to produce electrical
energy.
OPTION 2: COMMUNICATION
9. (a) (i) Draw a block diagram of a radio system
(ii) A station is broadcasting on a frequency of 92.5 MHz. Determine the capacitance of the capacitor for
which should be associated with an inductor of 1.25 x 10-9 H to receive this station
(b) (i) Compare analogue and digital systems as means of transmitting information
(ii) Discuss the problems and the advantages for Cameroon changing from analogue to digital
transmission in the near future.
OPTION 3: ELECTRONICS
10. (a) State in words and in the form of a truth table the actions of the following logic gates
(i) AND (ii) OR (iii) NAND
<<ADVANCED LEVEL PHYSICS
Page 87
(b) Figure 2 shows a transistor circuit operating in the common emitter mode with a current gain of 60
and VBE of 0.7 V
200 Ω
4.5 V
10 𝑘Ω
V0
2.7 V
Figure 2
Calculate the output voltage, V0
OPTION 4: MEDICAL PHYSICS
11. (a) Explain using ray diagrams how a normal eye focuses an image of an object on the retina
(b) A doctor notices that one of her patients can see clearly some text when it is near but will see the
same text appearing blurred when moved further away. Explain how such a defect can be corrected using
a named lens.
(c) Select a non – ionizing imaging technique and explain how it is used in medical diagnosis
<<ADVANCED LEVEL PHYSICS
Page 88
JUNE 2018
1. (a) State Newton’s law of gravitation and Coulomb’s law.
(b) Bring out two ways in which the force experienced in a gravitational field may differ from the force
experienced in an electric field.
2. A simple oscillating pendulum has as amplitude 0.05 m and period 2.0 s.
(a) Calculate the velocity of the pendulum as it passes through the equilibrium position.
(b) The expression for the displacement of the pendulum is 𝑦 = 𝐴𝑠𝑖𝑛𝜔𝑡. Sketch a graph of acceleration,
a, against time, t, for a complete oscillation. Note 𝑎 = −𝜔2 𝑦
3. Figure 1 shows an electric circuit with two power sources connected to resistors.
E
15 V
3Ω
6Ω
R
3A
1A
Determine
(a) The current through R
(b) The value of R
(c) The emf, E.
4. (a) What is meant by photoelectric effect?
(b) Briefly outline observations of the photoelectric effect which cannot be explained by classical physics.
5. (a) State the conditions necessary for a body to be in equilibrium on a plane
Uniform wooden bar
(b)
l/4
800 N
Pivot
Figure 2
200 N
Figure 2 shows a uniform wooden bar length, l, in equilibrium.
Determine the weight of the bar.
6. (a) (i) Use examples to explain why and how waves are classified.
(b) Describe an experiment to measure the speed of sound in free air. Your account should include a
diagram, procedure, observations, precautions and how the observations are used to obtain a conclusion.
(b) As an ambulance sounding a siren approaches a control point, the frequency of the siren is measured
to be 360 Hz and as it passes and moves away the frequency is presumed to be 320 Hz. Explain why
<<ADVANCED LEVEL PHYSICS
Page 89
there is this difference and calculate the speed of the ambulance given that the speed of sound in air is
340 ms-1.
(d) From the kinetic theory of ideal gases, the pressure, P, of a fixed mass of an ideal gas trapped in a
1
container is given by 𝑃 = 𝜌𝑐̅̅̅2 where ̅̅̅
𝑐 2 is the mean square speed of the gas particles.
3
(i) State the assumptions used to derive the equation
(ii) Hence derive the equation.
(e) Describe an experiment to determine the specific latent heat of vaporization of water. Your account
should include diagram, procedure, observations, precautions and how the observations are used to obtain
a conclusion.
(f) 8.0 kg of hot liquid A initially at 900 are mixed with 3.0 kg of water initially at 220 C in an insulated
container. If the specific heat capacity of the liquid A is half that of water:
Determine
(i) The equilibrium temperature reached by the system
(ii) The ratio of the change of temperature of the liquid A to that of water when equilibrium is reached.
𝜌𝑔
7. The equation 𝑃 = 𝑃𝑜 𝑒 −𝑘ℎ is called the law of atmospheres. K is a constant given by 𝑘 = 𝑃 where 𝜌 is
0
the density of air at stp and g is the gravitational field strength. Table 1 which follows gives some values
of height above sea level with their corresponding atmospheric pressure P.
h/km
P/104 Nm-2
9.6
4.346
11.7
2.691
18.0
1.094
28.1
0.221
34.9
0.0993
40.0
0.0602
44.8
0.173
51.0
0.0138
(a) Plot a suitable graph from which values of K and P0 can be determined
(b) Determine the values of K, P0 and hence 𝜌
(c) What is the pressure at a height of 65 km above sea level?
8. (a) (i) A small container ship called “THE ACHUKA” needs to enter and berth a port in Cameroon. What
does the captain of the ship need to know about the weather in this locality in which the port is located?
(ii) Explain how weather forecast can be done from a distance.
(b) Below is an example of a solar panel. Study the diagram carefully and answer the questions that
follow.
<<ADVANCED LEVEL PHYSICS
Page 90
Explain why the panel should have the following features
(i) The pipes are blackened (ii) The top of the panel is covered with glass
(iii) The walls are highly insulated and blackened.
c) (i) Draw a cross section of a hydro – electric power plant
(ii) Explain the energy changes that take place in the plant.
9. (a) (i) Name four functions of a mobile
(ii) State the meaning of the following words SMS, MMS and SIM
(b) Figure 4 shows a simple circuit receiver. Using A to E identify the components that best fits with the
following
A
D
C
D
Figure 4
(i) Which part of the circuit has all the transmitted signal?
(ii) Which component is used to select a particular signal?
(iii) Which component is used to remove the radio signal?
(iv) The component that generates input signals into the radio receiver
c) (i) Define the term bandwidth
(ii) Distinguish between analogue and digital transmission stating clearly the advantages of one over the
other.
10.
<<ADVANCED LEVEL PHYSICS
Page 91
(a) (i) With the help of the band theory only, distinguish between an insulator and a semiconductor?
(ii) Use the graph in the figure above to plot another graph of collector current, IC against the base
current, IB.
Obtain the current gain for the transistor.
(iii) Explain the term thermal runway.
D
Input
R
Output
Figure 6
<<ADVANCED LEVEL PHYSICS
Page 92
The input into figure 6 is an alternating voltage. Draw the output voltage for the circuit.
11. (a) (i) Draw and describe the basic structure of the human ear.
(ii) Consider the ear to be a pipe closed at one end, and that the length of the human auditory canal is
approximately 28 mm. If the velocity of sound in air is 340 m/s then what is the frequency of the
fundamental note in the ear?
(iii) The ossicles in the ear act as a lever. What does this mean?
(b) Optical fibres are considered to be the major breakthrough in many fields of study including
medicine.
(i) State the concept under which optical fibres is very useful.
(ii) Draw a diagram under which optical fibre is very useful.
(iii) Describe an application of optical fibres in medicine.
<<ADVANCED LEVEL PHYSICS
Page 93
SOLUTONS
JUNE 2001
1.Heat lost by calorimeter and water = heat absorbed by ice to change state plus heat used in warming the
melted ice
i.e mc cc (θR − θ) + mw cw (θR − θ) = mi lf + mi cw (θ − 273), with 𝜃 = 282.7𝐾
Simplifying the above expression gives
mi lf + mi cw (θ − 273)
0.05(3.25 × 105 ) + (0.05)(4200)(282.7 − 273)
θR =
+ θ ⇒ θR =
+282.7
mc cc + mw cw
0.1(380) + 0.3(4200)
⇒ θR = 14.088 + 282.7 = 296.8K
2. This question has a problem with the formulation. Now the question says “the source S supplies 300 W to
the resistors R1, R2 and 100 Ω respectively”. This means the total power supply of the cell is 900 W
𝑡𝑜𝑡𝑎𝑙 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑠𝑢𝑝𝑝𝑙𝑦 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑐𝑒𝑙𝑙 = 1.5𝐴 + 1.5 𝐴 = 3.0 𝐴
(i) 𝑃 = 𝐼𝑉 ⟹ 𝑉 =
900
3
= 300𝑉. This is the emf of the cell. Since S, R1, 100Ω and R2 are in parallel,
𝑉𝑅1 = 𝑉𝑅2 = 𝑉100Ω = 300𝑉
(ii) 𝑅2 =
𝑉𝑅2
𝐼𝑅2
=
300
1.5
= 200Ω
3.The solution of this problem is beyond the scope of advanced level physics
1
1
1
v
v
v
v
4.(a) From the lens equation f = v + u ⇒ f = 1 + u ⟹ f = 1 + m ⟹ m = f − 1, so a graph of m against v
1
gives slope f . From the graph, slope =
2.5
13
1
= 0.19cm−1 ⟹ f = 0.19 = 5.3cm
(b) Advantages of optical fibres over copper wires
- optical fibres have a larger bandwidth ie is they have a large information carrying capacity
- Transmission security is more enhanced since it is impossible to tap information from the fibre
without breaking it.
- The glass of which the optical fibres are made is considerably cheaper than copper cables.
- Optical fibres are free from electrical interference
- Fewer amplifiers are required
- There is little or no risk (electrical accidents) with optical fibre since they do not carry electricity.
5.(a)
Stres
s
Q
Loading
R
Hysteresis loop
Unloading
P
O
Strain
During loading, energy is supplied to the wire and
released during unloading. The area under ORQ
represents the work done per unit volume during the
loading process. The area under the curve OPQ
represents the energy per unit volume released during
the unloading process. The area of the loop OPQRO
represents the energy lost in the form of heat during the
loading and unloading process.
(b) This is because the car tyre should not get hot faster when the car is in motion and get exploded. The
small area of hysteresis means that less mechanical energy is converted to heat energy and as such fuel
consumption is also minimized.
<<ADVANCED LEVEL PHYSICS
Page 94
Stress
Iron
(c)
Glass
Strain
0
−t
6.Q = AC(1 − eBC )
Units of AC = units of Q ⟹ units of A =
units of Q
units of C
units of V × units ofC
= units of V = volt
units of C
units of t
units ofQ⁄units of I
units of V
units of BC = units of t ⟹ unitsof B =
=
=
= ohm
units of C units of Q ⁄units of V units of I
⟹ units of A =
7.
9V
3 kΩ
IC
50 kΩ
IB
Vin
ac
V0
VCE = V0
Vin
VBE
0V
(i) Applying Kirchhoff’s voltage law, Vin − IB R B − VBe = 0 ⇒ IB =
Vin −VBe
RB
, assuming the transistor is a
2.0−0.6
made from silicon, VBe = 0.6 V ⟹ IB = 50×103 ⟹ IB = 2.8 × 10−5 A
I
(ii) β = I c ⟹ IC = βIB = 60 × 2.8 × 10−5 = 1.68 × 10−3 A
B
(iii) Applying Kirchhoff’s voltage law, VCE = Vcc − IC R L = 9 − (0.00168 × 3000) = 3.96V
Output voltage
V/V
Input voltage
t/s
Since the transistor is a current amplifier, the
amplitude of the output is greater the input. Also, the
transistor is an inverter so the input and output are
1800 out of phase.
8. (a) (i) Coulomb’s law states that the magnitude of the force between two electrically charged bodies is
directly proportional to the product of the charges and inversely proportional to the square of their
separation.
<<ADVANCED LEVEL PHYSICS
Page 95
Q2
1
Q2
(ii) By coulomb’s law, F= 4πεr2 = r2 . 4πε , since the charges are of the same magnitudes. From the
Q2
1
Q2
equation, a graph of F against r2 gives a straight line with slope= 4πε ⟹ ε = 4π×slope
From the graph, slope =
4−0
(3.0−0)10−5
ε
5
(4.4×10−3 )
2
2
= 1.333 × 10 Nm ⟹ ε = 4π×1.33×105 = 1.15 × 10−11 Fm−1
1.33×10−11
(iii) Dielectric constant, εr = ε = 8.85×10−12 = 1.5
0
(b) (i)
4 µC
+
-
⃗⃗⃗⃗
𝐸1 ⃗⃗⃗⃗
𝐸2
Q1
50 cm
-3 µC
Q2
P
d
(ii) Between the two charges, the fields created by both charges are in the same direction. Hence there is
no point between the two charges that the resultant field can be zero. Since Q1>Q2, the resultant field
can only be zero to the right of Q2, say at the point P, a distance d from Q2.
Q1
Q2
At the point P, ⃗⃗⃗⃗
E1 + ⃗⃗⃗⃗
E2=o
⃗ ⟹ E1 = E2 ⟹ 4πε(0.5+d)
2 = 4πεd2 ⟹ d =
Q
0.5√ 2
Q1
Q
1−√ 2
⟹d=
4
3
0.5√
4
3
1−√
Q1
⟹ d = 0.32 m
(c) The bird is on an equipotential surface. Hence the potential difference between the legs is zero, no
current flows and the bird remain unelectricuted.
(d) (i) Newton’s law of gravitation states the magnitude of the force of attraction between any two
particles in
the universe is directly proportional to the product of the masses and inversely
proportional to the square of their separation.
(ii) For a planet of mass m, circling the sun with mass M, the centripetal force is provided by the
gravitational attraction between the sun and the planet. i.e
Centripetal force = gravitational attraction.
GMm
r2
= mω2 r ⇒
GM
ω2
= r 3 , but ω =
2π
T
⇒ r3 =
GM
2π 2
( )
T
GM
⇒ r 3 = 4π2 T 2 as required to demonstrate
4π2 r3
(iii) From the above relation, T = √
GM
(e) From r 3 = 4π2 T 2 ⇒ T 2 =
through the origin with slope
4π2
GM
GM
4π2 (3.5×108 )3
⇒ T = √1.67×10−11 ×6.0×1024 ⇒ T = 2.1 × 106 s
r 3 . This implies that a graph of T 2 against r 3 is a straight line passing
4π2
GM
(1.2−0.2)1012
From the graph, slope = (1.4−0.1)1027 = 1.692 × 10−16 ⟹
4π2
GM
4π2
= 1.692 × 10−16 ⟹ G = 1.692×10−16 M
4π2
⟹ G = 1.692×10−16 ×7.0×1026 = 7.33 × 10−11 kg −1 s−2 m3
3.5x105 m
(f)
M
𝑔𝑚
⃗⃗⃗⃗⃗
𝑔𝑒
⃗⃗⃗⃗
P
Me
d
<<ADVANCED LEVEL PHYSICS
Let the gravitational field strength be zero at P,
⃗ at the point P i.e g m = g e
then ⃗⃗⃗⃗
g e +g⃗⃗⃗⃗⃗m =O
GM
⟹ (3.5×10m
5 −d)2 =
GMe
d2
Solving the above equation gives d= 3.1 x 105 m
Page 96
9.(a) (i) When the ball is thrown, the vertical component of the velocity decreases to zero at the top of the
motion and thereafter it increases until the ball strikes the ground. Since there is no acceleration in the
horizontal direction, the horizontal component of the velocity stays constant throughout the motion.
(ii) Air resistance will reduce the maximum horizontal range of the ball, since air resistance opposes
motion.
(b) (i)
u
y
uy
𝑢𝑦 = 𝑢𝑠𝑖𝑛𝜃,𝑢𝑥 = 𝑢𝑐𝑜𝑠𝜃, 𝜃 = 300, u=300 ms-1. The
vertical displacement of the bullet is given by
300
O
x
1
𝑦 = 𝑢𝑡𝑠𝑖𝑛𝜃 − 𝑔𝑡 2
2
ux
100 m
Ground
When the ball strikes the ground, y = -100 m
1
⟹ −100 = 300tsin30o − gt 2 ⟹ −100 = 150 − 4.9t 2 ⟹ 4.9t 2 − 150t − 100 = 0
2
Solving the above equation gives t =-0.65 s or t=31.3 s. Since time is never negative, we take t=31.3 s
and discard t = -0.65s.
Distance of bullet from the cliff, x = ux t ⟹ x = (300cos30o )(31.3) ⟹ x = 8132 m
(ii) vx = ucosθ ⟹ vx = 300cos30 ⟹ vx = 259.8 ms-1 , vy = usinθ − gt
⟹ vy = 300sin30 − 9.8 × 31.3 ⟹ vy = −156.74 ms-1 ⟹ v = √vy2 + vx2 ⟹ v = 300 m-1
vy
Direction, α = tan−1 v ⟹ α =31.10. This angle is below the horizontal
x
R
(c) (i)
Kv
4.8 N
2
(ii) By Newton’s second law,
4.8 – 1.5 – Kv2 = 0 ⟹ 𝑣 =
√
Fr = 1.5 N
3.3
6.0×10−2
⟹ 𝑣 = 7.4 ms-1
W
(d) (i) The zeroth law of thermodynamics states that if bodies X and Y are separately in thermal
equilibrium
with body Z then X and Y are in thermal equilibrium with each other.
(ii) Temperature tells us in which way heat will flow if two bodies are brought in contact with each other.
(e) (i) Primary sources of energy are those which used in form in which the occur naturally e.g wood,
coal,
etc.
(ii) These are sources of energy which are obtained by converting other forms of energy into the required
form.
GPE
(f) (i)P = time =
h
2
mg( )
time
1
=2
ρVh
t
1
=2
ρAh2
t
1
=2
×1100×4.0×107 ×102
12×3600
= 5.0 × 1012 W
(ii) The variation of water level with seasons will lead to fluctuations in the output. Also, the cost of
trapping water is very high.
10. (a) Consult your notebooks
(b) (i) Radioisotopes are radioactive isotopes produced as a result of bombardment of nuclei with fast
moving particles such as alpha particles, neutron, beta particles etc.
The half life is the time taken for half the number particles present in a given sample to decay.
<<ADVANCED LEVEL PHYSICS
Page 97
−
𝑙𝑛2
𝑙𝑛2×12
𝑡
(ii) 𝑁 = 𝑁0 𝑒 −𝜆𝑡 = 𝑁0 𝑒 𝑇1⁄2 = 8.0 𝑚𝑔𝑒 − 30 = 6.1 𝑚𝑔
(c) (i) mass defect, ∆m = (3.345 + 5.008) × 10−27 − (6.647 + 1.675) × 10−27 = 0.031 × 10−27
Energy released, ∆E = ∆mc 2 = 0.031 × 10−27 × (3.0 × 108 )2 = 2.79 × 10−12 J
1
(ii) Number of moles of 1 kg of deuterium, 𝑛 = 0.002 = 500
Number of atoms = 500 × 6.02 × 1023 = 3.01 × 1026
Thus the energy released per kilogram =3.01 × 1026 × 2.79 × 10−12 J = 8.40 × 1014 J
(d) Consult your notebooks
(e) (i) An ohmic conductor is one whose potential difference across its ends at constant temperature is
directly proportional to the current flowing through it. While a non – ohmic conductor is one whose p.d
across its does not vary linearly with the current flowing through it.
(ii) Electromotive force is the work done per unit charge to convert other forms of energy (chemical or
mechanical) to electrical energy, while potential difference is the work done per unit charge to convert
electrical energy into other forms of energy such as heat.
(f)
V15Ω
V20Ω
(i) KCL:𝐼1 = 𝐼2 + 𝐼3 - - - - - - - -- - -- - -- -- (1)
(i)
A
I2
20Ω
15Ω
6.0 V
24.0 V
2
I1
5Ω
1
V10Ω
10Ω
I3
V5Ω
KVL loop 1: 10𝐼2 − 6 − 15𝐼3 − 5𝐼3 = 0
⟹ 10𝐼2 − 20𝐼3 = 6 ⟹ 5𝐼2 − 10𝐼3 = 3 - - - - -(2)
KVL loop 2: 6 − 10𝐼2 + 24 − 20𝐼1 = 0
⟹ 2𝐼1 + 𝐼2 = 3 - - - - - -- - - -- - - -- - - -- - -(1)
Solving the above equations,
𝐼1 = 1.05 𝐴, 𝐼2 = 0.9 𝐴, 𝐼3 = 0.15𝐴
B
(ii) 𝑉𝐴𝐵 = 6.0 − 10𝐼2 = 6.0 − 10(0.9) = −3.0 𝑉. From these calculations, it implies A is at a higher
potential than B i.e the potential difference between A and B is simply
1.(i) Optical fibres are thin flexible glass rods with diameter almost equal to that of a human hair used to
transmit light using the principle of total internal reflection.
(ii) Total internal reflection
(b) In medicine, optical fibres are used in endoscopy to transmit and guide light to some spots in the
human body.
2.Binding energy is the energy required to free all the nucleons in the nucleus of an atom.
(b)
<<ADVANCED LEVEL PHYSICS
Page 98
3.By the second law of thermodynamics, ∆Q = ∆U + ∆W, where ∆W = P∆V
In the process ab, ∆V = 0 ⟹ ∆W = 0 ⟹ ∆U = ∆Q ⟹ ∆U = 150 J
(ii) In the process abd, ∆Qabd = ∆Uabd + ∆Wabd ⟹ ∆Uabd = ∆Qabd − ∆Wabd
⟹ ∆Uabd = ∆Qab + ∆Qbd − ∆Wab − ∆Wbd
∆Uabd = 150J + 600J − 0 − (8.0 × 104 )(5.0 − 2.0)10−3 J = 510J
4.(i) F is the magnitude of the force of attraction between m1 and m2, G is the gravitational constant, m1 and
m2 are masses and r is the mean distance between the centers of gravity of m1 and m2
(ii) F =
Gm1 m2
r2
Fr2
⟹G=m
1 m2
⟹ [G] =
[F][r]2
[m]2
=
Nm2
kg2
=
kgms−2 m2
kg2
= kg −1 m3 s −2
(b) (i) A physical equation is said to be homogeneous when all the terms in the equation have the same
base units or dimensions. For terms to be added, subtracted or equated, they must have the same base
units or dimensions. If an equation is not homogeneous, it means the terms in the equation do not have
the same base units or dimensions and hence the equation must be wrong.
1
(ii) Some physical equations have dimensionless constants e.g K. e = 2 mv 2 . So homogeneity with
respect to units or dimensions cannot be used to test the correctness of a physical equation. It is only
valid for equations with no dimensionless constants.
5.(i) Scattering refers to the irregular reflection or refraction of waves
(ii) Signal attenuation is the loss in the intensity of the signal due to the absorption of the signal energy
by the transmitting medium. Signal attenuation is taken care of by the use of booster or repeaters.
1
4
λ
850 4
(b) (i) A ∝ λ4 ⟹ A1 λ14 = A2 λ42 ⟹ A2 = (λ1 ) A1 ⟹ A2 = (1500) (2.0) = 0.21 dBkm−1
2
(ii) The decibel is the measuring unit of intensity level
6. Molecules of liquid and gases undergo translational motion whereas those of solid undergo vibrational
motion.
(ii) Intermolecular forces of attraction are negligible in gases whereas in liquids and solids, the
intermolecular force of attraction is strong
(b) (i) and (ii) Describe any observation of Brownian motion.
7. The charged particle (e.g electron) enters the field at right angles to the field and experiences a maximum
force on it. From Fleming left hand rule, the force experienced by the electron is always perpendicular to
its direction of motion in the field. Since the field strength is constant, therefore the force is constant (F =
Bev) .The speed is constant but the direction of motion changes which implies the velocity changes. If
<<ADVANCED LEVEL PHYSICS
Page 99
the velocity changes, then the electron accelerates. But there is no tangential acceleration. Therefore the
existing acceleration is centripetal, giving rise to a centripetal force. Hence the path of the electron is
circular.
(b) (i) Centripetal force = magnetic force ⟹ Fc = Bev = 0.5 × 1.6 × 10−19 × 3.0 × 105 = 2.4 ×
10−14 N
(ii) The electron will spiral inwards.
8.(i) Specific latent heat of fusion is the thermal energy required to change the state of a unit mass of a
substance from solid to liquid at the melting point of the substance
(ii) Consult your textbooks or note books (Diagram, procedure, observation, calculations and conclusion)
m c
b(i) Qw = Qa ⟹ mw cw ∆θw = ma ca ∆θa ⟹ ∆θw = (m a ca ) ∆θa but ma = ma = m and cw = 2ca
w w
mca
1
⟹ ∆θw = m(2c ) = 2 ∆θa. Thus the temperature change for the water is half that of alcohol.
a
(ii) Heat lost by molten lead in melting + heat lost by lead in cooling = heat used to melt ice + heat used
in warming the melted ice.
i. e mp lp + mp cp (327 − θ) = mi li + mi cw (θ − 0)
⟹ 10 × 2.4 × 103 + 10(1.28 × 103 )(327 − θ) = 1 × 3.34 × 105 + 1 × 4200θ
⟹ 24500 + 12800(327 − θ) = 334000 + 4200θ
⟹ 4210100 − 12800θ = 334000 + 4200θ ⟹ 3876100 = 17000θ ⟹ θ =
3876100
= 228. 0o C
17000
3 )(327
Heat lost by lead, Q = ml lp + mp cp (327 − θ) = 10 × 2.4 × 103 + 10(1.28 × 10
− 228.0)
⟹ Q = 24500 + 1267200 = 1291700J
(c) The match handle is made of wood which is a poor conductor. Also, the gap between the flame and
the fingertips is air which is also a poor conductor of heat. Hence the energy reaching the fingertips is
small.
(d) (i) Young’s modulus is defined as the ratio of the tensile stress to the tensile strain
(ii) Consult your note books (Diagram, procedure, observation, calculations and conclusion)
(e) The stress is maximum at A. This is because when the string is loaded, the maximum force is at A.
f) (i) Stress =
Force
=
102×9.8
(ii) strain = original length =
Stress
= 1.0 × 108 Pa
area
0.1×10−4
extension
2.2×10−3
2.0
= 1.1 × 10−3
1.0 ×108
(ii) E= Strain = 1.1×10−3 = 9.1 × 1010 Pa
9.(a) (i) Q = CV ⟹ C =
Q
⟹C=
V
3.5×10−3
6.0
= 5.8 × 10−4 F
(ii) Initial current, Io = slope of graph at t = 0 ⟹ Io =
∆Q
∆t
=
(3.5−1.0)10−3
0−100
= −2.5 × 10−5 A, the negative
sign shows that the current is decreasing with time.
V
6.0
R = R = 2.5×10−5 = 2.4 × 105 Ω
time constant, τ = RC = 5.8 × 10−4 × 2.4 × 105 = 139.2 s
(iii)
V/V
1
6.0
b) (i) fr = 2π√LC =
1
2π√9×10−12 ×2.8×10−3
= 1.0 × 106 Hz = 1.0 MHz
VC
VR
(ii) 𝑄𝑚𝑎𝑥 = 𝐶𝑉𝑚𝑎𝑥 = 9 × 10−12 × 12 = 1.08 × 10−10 𝐶
<<ADVANCED LEVEL PHYSICS
t/s
Page 100
Imax = ωQmax = 2πfQmax = 2π × 1.0 × 106 × 1.08 × 10−10 = 6.79 × 10−4 A
(c) (i) Resultant force = slope of momentum – time graph
FAB =
(20−10)103
(10−5)(60)
(0−20)103
= 33.3N, FBC = 0 N, FCD = (20−15)(60) = −66.7N
Area under graph
(ii) Total displacement =
mass
= (S1) + (S2 ) + (S3 )
1
S1 = 2 (
(15 − 10)(60)(20 × 103 )
10 × 60
) (20 × 103 ) = 6.0 × 103 m, S2 = (
) = 6.0 × 103 m,
1000
1000
(20−15)(60)(20×103 )
N.B
= 3.0 × 103 m,  Between A and B, the car is accelerating, meaning the
1000
of4 m
the displacement – time graph should increase
total displacement = (6.0 + 6.0 + 3.0)103 = gradient
1.5 × 10
with time.total displacement is given by
OR since the area enclosed by the graph is a trapezium,
S3 = 12
s/103 S =
15.0
3
1 (5+20)(60)(20×10 )
2
1000
= 1.5 × 104 m
12.
 Between B and C, the car is moving at constant speed,
meaning the gradient should be constant. Hence a straight
line
 Between C and D car is decelerating. Hence the slope of the
graph should decrease with time
6.0
0
1
1
20
t/mi
N
 B
e
t
w
(b) (i) By the principle of conservation of linear momentum,
e Pmax 20×103
mc
mc vmax = (mc + mv )v ⟹ v = (m +m
v
,
but
v
= m = 1000 = 20 ms −1
max
max
e
c
v)
1000
1000
n
⟹ v = (1000+1500) 20 = (2500) 20 = 8 ms −1
A
1
5
2
(
)(
)2
Initial kinetic energy just before collision, Keinitial = 12mv
a max = 2 1000 20 = 2.0 × 10 J
Kinetic energy of the interlocked, K. efinal = 12(1000+1500)(8)
n 2 =8.0×104 J
d and sound while some is used in deformation.
Some of the initial kinetic energy of the car is lost as heat
10. (a) (i) A line emission spectrum is the spectrum of light Bradiated by individual atoms in a hot gas when
the electrons in the atom jump from higher energy levels, to a lower energy levels. The spectrum consists
t
of colored lines on a dark background.
h
A line absorption spectrum is produced when a beam of white is passed through a cool gas, such
e
that that the photons whose energies are equal to the excitation energies of the gas atoms are absorbed.
c
An absorption spectrum consists of dark lines on a colored
background.
a
r
i
<<ADVANCED LEVEL PHYSICS
Page 101
s
a
c
The table below summarizes the differences between line absorption spectrum and line emission
spectrum
Line emission spectrum
Line absorption spectrum
A series of colored lines
Series of dark lines
The radiation has not passed through an
absorption medium
Radiation has passed through an absorption
medium
A result of excited electrons falling from
higher to lower energy level
Results from an electron taking energy to go
from a lower to a higher energy level
An example is the light from a hydrogen or
sodium vapour
An example is white light passed through a
hydrogen or sodium vapour
b) (i) The ground state of an atom is its lowest energy level available. Electrons have the least energy in
the ground state, so this is the case for a normal atom. The highest energy level is given the value zero
and the lower values are therefore negative.
(ii) The ground state energy is E1 = -13.6 eV, and the energy at infinity is E∞ = 0. The energy
difference E∞ − E0 = 0 − −13.6 eV = 13.6 eV , which is the ionization energy.
(iii) ∆E = hf ⟹ E5 − E1 = hf ⟹ f =
(−0.54—13.6)(1.6×10−19 )
6.6×10−34
= 3.2 × 1015 Hz, which lies in the
ultraviolet region of the electromagnetic spectrum.
(iv) The maximum wavelength will correspond to the minimum energy difference i.e E3 – E2
⟹ Emin = E3 − E2 = λ
hc
max
⟹ λmax = E
hc
3 −E2
6.6×10−34 ×3.0×108
= (−1.5−−3.4)(1.6×10−19 ) = 6.5 × 10−7 m
This wavelength lies in the visible region of the electromagnetic spectrum.
(c) (i) The fact that the α – particles were reflected back suggests that an atom has a nucleus. Since only
few were reflected back through angles greater than 900, it means the nucleus is very massive and small.
(ii) The fact that some of the α – particles were reflected through angles greater than 900 also suggests
that the nucleus is massive and positively charged.
(d) Electromagnetic waves can travel through a material medium as well as a vacuum at very high speeds
while mechanical waves require a material medium for their propagation. Examples of electromagnetic
waves include: x – rays, gamma rays, infra – red, visible light, etc. examples of mechanical waves
include: sound waves, water waves, waves produced in a spring, seismic waves, etc
(e) Differences between stationary and progressive waves
<<ADVANCED LEVEL PHYSICS
Page 102
Stationary Wave
Progressive Wave
Energy
No net transfer of energy from one point to
another. Energy is confined within the wave
and there is interchange of K.E. and P.E.
Energy is transported in
the direction of travel of
the wave travel.
Phase
All particles between two adjacent nodes are
in phase. Particles on opposite sides of a
node will be in antiphase.
All particles within one
wavelength vibrate with
different phase.
Amplitude
Varies from zero at nodes to maximum at
antinode.
Same amplitude for all
particles in the wave
Wavelength 2 x distance between adjacent nodes or
antinodes.
Distance between adjacent
particles which are in
phase.
Frequency
All particles vibrate in SHM with same
frequency except at nodes.
All particles vibrate in
SHM with same
frequency.
Waveform
Does not advance.
Advances in the direction
of velocity of wave.
A stationary wave is produced when two progressive waves have the same amplitude, wavelength and
frequency and travelling in opposite directions interfere.
(f) (i) Harmonics are frequencies which are integral multiples of the fundamental frequency
(ii) Wavelength = twice distance between antinodes = 2x 10 =20 m; Amplitude = 10 cm
(iii) 𝑣 = 𝑓𝜆 = 600 × 0.2 = 120 𝑚𝑠 −1
In the fundamental mode of vibration, the whole string is half the wavelength (see diagram below)
𝑙 = 60 𝑐𝑚 = 0.6𝑚 ⇒
𝜆
2
𝜆
120
= 0.6 = 1.2 𝑚 ⇒ 𝑓0 =
= 100 𝐻𝑧
2
1.2
JUNE 2003
1.(a) (i) In terms of medium of transmission, we have mechanical and electromagnetic waves
(ii) In terms of the mode of propagation we have transverse and longitudinal waves.
<<ADVANCED LEVEL PHYSICS
Page 103
 Examples of mechanical waves are: water waves, sound waves, waves produced in a slinky spring, shock
waves etc..
 Examples of electromagnetic waves include: visible light, x-rays, gamma says, radio waves, etc
 Examples of longitudinal waves are: sound waves, waves produced in a spring when displaced along its
axis etc
 Examples of transverse waves include: all electromagnetic waves, waves produced in a spring when
displaced perpendicularly along its axis.
2.(i)
V
A
Black box
R
E
S
(ii) I/A
I/A
I/A
Junction diode
Copper
V/V
3.(i)
Filament lamp
V/V
V/V
220
212
0
X→
Y + 2 ( β) + 2 42He
86
84
−
(ii) A = A0 e
−λt
1
T1
A0
A
2
⟹ ln( ) = λt ⟹ t = λ ln ( A ) ⟹ t = ln2
ln ( A0 ) ⟹ t =
A0
A
6000
ln2
19
ln ( 7 ) = 8.64 × 106 yr
1
4.(i) dsinθ = mλ ⟹ N sinθ = mλ ⟹ sinθ = Nmλ ⟹ θ = sin−1(Nmλ) ⟹ θ = sin−1(1 × 600 × 103 ×
5.9 × 10−7 ) ⟹ θ = sin−1 (0.354) = 20o
(ii) For the third order image, m = 3 ⟹ sinθ = 3 × 600 × 103 × 5.9 × 10−7 = 1.062, which is greater
than one. Hence the third order image is not possible.
1
From dsinθ = mλ ⟹ m ∝ λ. Thus if the wavelength is increased, the number of orders will decrease.
5
1
3P
3P
3×1.0×10
5.(a) P = ρc̅2 ⟹ c 2 = ⟹ crms = √ ⟹ √
= 339.77 ms −1
3
ρ
ρ
2.6
(b) The speed of sound in air at S.T.P which is 330mS-1 is less than the root mean square of nitrogen
molecules at S.T.P. This is as a result of random notion.
6.(i)
Projecting
Tension
20o
T
Vertically, Tcos20 = mg ........(1)
Horizontally, Tsin20 = ma .....(2)
<<ADVANCED LEVEL PHYSICS
Weight
W
Solving (1) and (2) simultaneously,
a = 3.6 ms-2
Page 104
(ii) At constant speed, the bob oscillates about the equilibrium position. That is the bob describes simple
harmonic motion about the equilibrium.
7.(a) Crystalline Solid are materials which have a basic repeating structure called a unit cell, possess
cleavage planes, have definite melting points and show long range order. An example is sodium chloride.
Amorphous materials have no particular order arrangement of atoms, no definite melting points and
posses no cleavage planes. An example is glass.
Polymeric materials are materials made up of long chains built up from simpler units called monomers.
They can be stretched to about ten times their original length. An example is rubber.
NB: A unit cell is the fundamental unit from which the entire crystal may be constructed by purity
translation (like bricks in a wall).
8.(i) Specific latent heat of vaporization is the heat required to change the state of a unit mass of a substance
from liquid to gas at constant temperature, While latent heat of vaporization is the heat required to change
the state of a substance from liquid to gas at constant temperature.
ii) Consult your text books or note books
b) Electrical energy lost by heater = heat gained by water + heat gained by kettle
ie Pt = mc∆θ + C∆θ ⇒ t = (
Mc+C
P
) ∆θ =
(15×103 ×4200+400)(100−20)
2000
= 18.52 s
ii) Time taken to convert water of mass m to steam is t=5(60)-18.52= 281.48s
⇒ Pt = mlv ⇒ m =
Pt
lv
=
2000×281.48
2.0×106
= 0.21 kg
This result is not realistic as the mass of water evaporated is greater than the initial mass of the water.
c) Pt = mlv ⇒ t =
mlv
P
=
15×10−3 ×2.0×106
2000
= 15 s.
Assumption: All the electrical energy supplied goes to boil and evaporate all the water. No heat is
absorbed by the kettle and no energy is lost to the surrounding.
(d) (i) Consult your notes
(ii) Consult your textbooks
(e) (i) By the principle of conservation of linear momentum:
Momentum before collision = Momentum after collision.
⇒ muA + mB uB = mA vA + mB vB ⇒ vB =
1
1
muA + mB uB −mA vA
mB
1
=
(2m)(5)+(m)(2)−(2m)(3)
m
⇒ VB =60ms −1
1
ii) Initial K.e=2 mA vA2 +2 mB u2B =2 (2m)(5)2+2 (m)(2)2 = 27m
1
1
1
1
Final kinetic energy = 2 mA vA2 + 2 mB vA2 = 2 (2m)(32 ) + 2 (m)(62 ) = 27m
Since initial k.e is equal to the final k.e, the collision is elastic
Or coefficient of restitution =
velocity of separation
velocity of approach
v −v
3−6
−3
= uA−uB = 2−5 = −3=1
B
A
Since the coefficient of restitution is unity, the collision is elastic.
(f)
9.(i) Photoelectric effect is the ejection of electrons from the surface of a metal when radiation of sufficient
frequency falls on it.
ii) Electrons are emitted only when the frequency of the incident radiation above some threshold value, no
matter how intense the light is.
- The maximum kinetic energy of the emitted electrons depends on the frequency of the incident radiation.
- Photoelectrons are emitted almost at once when radiation of a sufficiently high frequency strikes the metal.
<<ADVANCED LEVEL PHYSICS
Page 105
- The number of photoelectrons ejected is proportional to the intensity of the incident radiation.
iii) The classical theory agrees with the number of photoelectrons with intensity, but cannot explain why
this increase in intensity cannot lead to an increase in the kinetic energy of the ejected electrons. The
classical theory cannot explain why the maximum kinetic energy of the emitted electrons is frequency
dependent but independent of the intensity. According to the classical theory, the kinetic energy of the
electrons could be increased with any frequency of radiation simply by making the light more intense.
Classical theory cannot explain why the emission of electrons is instantaneous. The expectation of classical
physics is that photoelectrons will absorb energy over period of time as the radiation continues to fall,
eventually gaining energy to be ejected.
h
1
(b) (i) eVs = hf − Φ ⟹ Vs = (e) f − (e) Φ
0.75−0.25
Slope of graph = (6.6−5.2)1014 = 3.57 × 10−15 Vs ⟹
h
⟹ h = 3.57 × 10−15 × 1.6 × 10
h
= 3.57 × 10−15
e
−15
= 5.71 × 10−34 Js
1
(ii) Vs = (e) f − (e) Φ, choosing the intercept (5.2 × 1014 , 0.25);
5.71×10−34
Φ
⟹ 0.25 = ( 1.6×10−19 ) (5.2 × 1014 ) − 1.6×10−19
⟹ 0.25 × 1.6 × 10−19 = 5.71 × 10−34 × 5.2 × 1014 − Φ ⟹ Φ = 2.57 × 10−19 J = 1.61 eV
(c) (i) Thermionic emission is the emission of free electrons from the surface of metals when it is
sufficiently heated electrically the metal must be of high melting point such as tungsten. In this process,
the work function of the metal must be overcomed.
ii) See notes for the structure of the electron gun.
d)
Solids
Force
Kinetic energy
Strong intermolecular forces of attraction
Lowest kinetic energy
Liquids Weak intermolecular forces of attraction
Small kinetic energy but intermediate
between phase of solids and gazes
Creases Negligible intermolecular forces of
attraction
Largest kinetic energy compare
(e)(i) There exist two kinds of forces between the molecules-repulsive and attractive forces. Below the
equilibrium, face is repulsive (i-e in the range (0 < 𝑟 ≤ 𝑂. 95𝑥10−10 m) the strength of the repulsive force
decreases towards the equilibrium position. Force is attractive for separation greater than the equilibrium
position at the equilibrium the net force is zero.
ii) At and near the equilibrium separation (0.95x10−10 m, the force extension is linear Skagit line). The
linear nature means the displacement of the molecules is proportional to the force, provided the force is not
strong of molecules, which is hook’s law.
(iii)
U/J
0
0.95
x/10-10 m
<<ADVANCED LEVEL PHYSICS
Page 106
iv) Work done in separation molecules is the area between the graph and separation axis from equilibrium
separation to infinity (try to do it). Count the number of squares between the curve and the separation axis
from equilibrium separation, and then multiply it by the area of one square.
v) The work represents the latent heat of vaporization of the material.
10. (a) i) Copper conducts by the movement of free mobile electrons. When the temperature of the copper
increases the amplitude of vibration of the atoms increases leading to more collision by free mobile
electrons with the atoms and hence the resistance increases hence the conductivity, of the copper drops.
Silicon is a pure semiconductor, so conduction is by free mobile electrons and holes under an applied pd..
At OK, all the electrons in pure silicon are found in the valence. As temperature increases, electrons leave
the valence band and enter the conduction band and the conductivity increases.
ii) An n-type semi conductor is produced by “dopping” a pure semi conductor (e.g silicon) with a
pentavalent element, such that the materials conductivity is dominated by electrons. The pentavalent
element donates conduction electron to the silicon.
Unpaired electron
available for
conduction
Si
Si
Si
P
Phosphorous has five electrons in the outermost shell.
When covalent bonds are formed between an atom of
phosphorous and four silicon atoms, the central atom
(phosphorous) has an extra unpaired electron. The
unpaired electron is available for conduction and moves
to the conduction band. This lead to the formation of an
n – type semiconductor since the dominated charge
carriers are electrons.
Si
Formation of a p – n junction:A p – n junction is produced when a crystal of a pure semiconductor is
doped such that one half is p – type and the other half is n – type. Immediately the junction is created,
electrons from the n – type side migrate to fill some of the dominant holes on the p – type side.
Conversely, holes move to the n –type side to be captured by electrons.
The information given in this question is incomplete. So the solution will be in terms of unknowns
(b) (i) By KVL, VCC − IC R C − VCE = 0. At saturation, VCE = 0 ⟹ IC =
VCC
RB
By KVL, VCC − IB R B − IB R B = 0, but VCC = IC R C ⟹ VBE = IC R C − IB R B = IB (βR C − R B )
(ii) R B =
VBE
IE
= IE R BE = IB (βR C − R B ) ⟹ R B = βIC +
(c)(i)
ii
IE RBE
IB
, with IE = IB + IC
IC/mA
Load
VCE/V
<<ADVANCED LEVEL PHYSICS
Page 107
(d) (i) The tuning circuit selects only one station because the capacitance is not variable (has only one
value) and so the corresponds to one
resonant frequency hence selects only one station.
(ii) The demodulator extracts the information signals from the carrier waves.
The amplifier boosts up the strength of the signals
(ii) The function of the demodulator is to extract the information from the carrier wave. The amplifier
boosts up the strength of the signal
(e) (i) Amplitude modulation is the process is a process whereby the audio signal is superimpose on the
carrier wave signal by varying the amplitude, whereas frequency modulation is the process whereby the
audio
Amplitude modulation
Frequency modulation
signal is
Circuit
Simple
Complex
Commercial Cheaper
Aspects
Expensive
Bandwidth
Greater bandwidth
Smaller bandwidth
Range
Longer range
Shorter range
superimpose unto the carrier signal by varying the frequency.
The differences between AM and FM can be summarized in the table below
(ii) 𝑓𝑟 =
1
1
⟹ 𝐿 = (2𝜋𝑓 )2 𝐶 ⟹ 𝐿 = 1.27 × 10−8 H
2𝜋√𝐿𝐶
𝑟
(iii) By the use of a variable capacitor.
(f) (i) If the satellite is to circle the plane of the equator, then the centripetal force is provided by the
gravitational attraction between satellite and the earth.
<<ADVANCED LEVEL PHYSICS
Page 108
𝑐𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑜𝑛 𝑠𝑎𝑡𝑒𝑙𝑙𝑖𝑡𝑒 = 𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑡𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑠𝑎𝑡𝑒𝑙𝑙𝑖𝑡𝑒 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑒𝑎𝑟𝑡ℎ
𝑖. 𝑒
𝑚𝑣 2
𝑟
=
𝐺𝑀𝑒 𝑚
𝑟2
⟹𝑣=√
𝐺𝑀𝑒
𝑟
= √𝑟𝑔 = √6.4 × 106 × 9.8 = 7.9 × 103 𝑚𝑠 −1 = 7.9 𝑘𝑚𝑠 −1 .
Therefore, the satellite must be provided with a velocity of about 7.9 kms-1 so that it can attain a required
circular orbit. To do this, the satellite is carried by a rocket to the required height of the orbit, and then
projected along a tangent to the orbit.
(ii) Suppose the direction of rotation the satellite is the same as that of the earth and the satellite stays in a
fixed position above the surface of earth, then it is in a geostationary orbit and hence has a period of 24 hrs.
Centripetal force on satellite = gravitational attraction between the earth and the satellite
𝑚𝜔2 𝑟 =
𝐺𝑀𝑒 𝑚
𝑟2
⟹ 𝜔2 =
𝐺𝑀𝑒
𝑟3
⟹ 𝑇2 =
4𝜋 2 𝑟 3
𝐺𝑀𝑒
, 𝑤𝑖𝑡ℎ 𝑟 = 𝑅𝑒 + ℎ, where Re is the radius of the earth and h is
the height of the orbit above the satellite above the earth surface.
3
𝑇 2 𝐺𝑀𝑒
⟹ 𝑅𝑒 + ℎ = √
4𝜋 2
3
2 2
3
(24×60×60)2 ×(6.4×106 )2 ×9.8
𝑇 𝑅 𝑔
⟹ ℎ = √ 4𝜋𝑒2 − 𝑅𝑒 = √
4𝜋 2
− 6.4 × 106 = 3.7 × 107 𝑚
JUNE 2004
E(r)
1.(i)
+
(ii)
+
Electrons
+
+
+
+
R0
r
N.B: When the negative charge is placed at the centre of the sphere, it induces positive charges on the
walls of the sphere and negative charges on the outside. Since the sphere is earth, electrons on the surface
will now flow to the earth, leaving the outside surface with a zero net charge.
2.
300
600
25 N
Resolving
Vertically, 40𝑐𝑜𝑠60° + 25𝑐𝑜𝑠30° − 𝑚𝑔 = 𝑚𝑎𝑦
⟹ 4ay = 40 × 0.5 + 25 × 0.866 − 4 × 9.8 = 2.45
⟹ ay = 0.613ms−2 - - - - - - - - -- -(1)
Horizontally, 40𝑠𝑖𝑛60° − 25𝑠𝑖𝑛30° = 𝑚𝑎𝑥 ⟹ 4ax = 22.14
⟹ ax = 5.535 ms−2 - - - - - - - - - - -- - (2)
40 N
4
𝑎 = (5.535𝑖̂ + 0.613𝑗̂)𝑚𝑠 −2 ⟹ a = √5.5352 + 0.6132 = 5.6ms−2
mg
3.
A
A
R
R
<<ADVANCED LEVEL PHYSICS
Page 109
V
V
Figure 3
y
(i) It can be seen in figure 3 (x) that the voltmeter measures a voltage V but the actual voltage drop across
the unknown resistance R is VR, given by the expression VR =IR, where I is the current through the
resistance R. The indicated value of the voltage drop, V is given by = 𝐼𝑅 + 𝐼𝑅𝑎 , where, Ra is the internal
resistance of the ammeter. The internal resistance of the ammeter is usually very small and so its p.d
drop. If the resistance under measurement is large enough, then VR will be much greater than Va, hence
keeping the error in the measurement very low.
This method is therefore recommended for the measurement of large values of resistances. This
method in electrotechnics is called the upset connection.
(ii) In figure 3 (y), the voltmeter measures a voltage V which is the actual voltage drop across the across
the unknown resistance R, since the voltmeter is in parallel with R alone. Note here that the ammeter
reads a current I not the current through the unknown resistance R. if the resistance of R is much smaller
than the resistance of the voltmeter, then the greater part of the current will flow through the unknown
resistance R, hence reducing the error in the measurement of R.
This method is therefore acceptable when the value of the resistance is much smaller than the internal
resistance of the voltmeter. In electrotechnics, this method is called the down set connection.
4.For a pipe closed at one end, the end is a node (See diagram below for the fundamental mode of vibration)
𝑣
𝑇
𝑇
279
𝑣 <∝ √𝑇 ⟹ 𝑣1 = √𝑇1 ⟹ 𝑣2 = 𝑣1 √𝑇2 = 340√273.15 = 354.53 𝑚/𝑠
𝜆
𝑙=
4
2
2
1
Thus the speed of sound in air at 0℃ is 354.53 𝑚/𝑠
𝑣
𝑣
𝑣
354.53
The fundamental frequency, 𝑓0 = 𝜆 = 4𝑙 ⟹ 𝑙 = 4𝑓 = 4×512 = 0.173 𝑚 = 17.3 𝑐𝑚
0
5.(i) By the conservation of mechanical energy, p.e lost = k.e gained
1
𝑖. 𝑒 𝑚𝑔ℎ = 2 𝑚𝑣 2 ⟹ 𝑣 = √2𝑔ℎ = √2 × 9.8 × 1.5 × 10−2 = 0.54 𝑚𝑠 −1
(ii) By the work energy principle, k.e lost = work done against friction
1
⟹ 2 𝑚𝑣 2 = 𝐹. 𝑑 ⟹ 𝐹 =
6.
Reservoir
Dam
𝑚𝑣 2
2𝑑
=
50×10−3 ×(2×9.8×1.5×10−2 )
2×0.5
Penstock
Turbine
= 1.47 × 10−2 𝑁
Generator
Transformer
 The reservoir stores the reserved water and provides the potential energy of water that is
converted into kinetic energy of the turbine
 The dam backs up large quantity of water
 The penstock determines the amount of water admitted to the generating unit and hence the
power output.
 The turbine converts the gravitational potential energy of water to rotational kinetic energy and
provides the emf that produces electricity as it is linked to the generator
 The generator converts the rotational kinetic energy of the rotor blades to electrical energy
 The transformer steps up the voltage
(b) The environmental hazards are:
 The dam serves as breeding grounds for mosquitoes
<<ADVANCED LEVEL PHYSICS
Page 110
 Dams can cause floods which leads to erosion and destruction of habitats
 The dam blocks large amounts of water which often submerge farm lands, ecosystems and may
even require the displacement of settlements
7.(a) See June 2001 Q10
𝑙𝑛2
(b) 𝐴 = 𝐴0 𝑒 −𝜆𝑡 , 𝑤𝑖𝑡ℎ 𝜆 = 𝑡
1⁄2
𝑙𝑛2
𝑙𝑛2
= 5730 𝑦𝑟 −1 ⟹ 𝐴 = (3 × 104 )𝑒 −(5730𝑦𝑟
−1 )(25000 𝑦𝑟)
= 1.46 ×
103 Counts per minutes. This is the number of count per minutes in 2 kg.
Therefore the count rate in 0.3 kg will be
0.3
2
× 1.46 × 103 = 2.19 × 102 𝑐𝑜𝑢𝑛𝑡𝑠 𝑝𝑒𝑟 𝑚𝑖𝑛𝑢𝑡𝑒
8. (a) Contact forces are those which come into play when bodies are in physical contact i.e when the bodies
are touching each other. Examples include: friction, upthrust, etc. whereas action at a distances results
when bodies are not in physical contact i.e not touching. Examples include: electrostatic force,
gravitational force, magnetic force, etc.
(b) Consult your notebook
(c) Let the car driver take time t to catch up with the truck driver. At that instant, both of them must have
travelled equal distances.
N.B: The time taken by the truck driver is (t+0.7)s
1
1
Distance travelled by car, 𝑠𝑐 = 2 𝑎𝑡 2 = 2 × 2𝑡 2 = 𝑡 2
36×1000
Distance travelled by truck, 𝑠𝑡 = 𝑣𝑡 = (
60×60
) (𝑡 + 0.7) = 10(𝑡 + 0.7)
𝑠𝑐 = 𝑠𝑡 ⟹ 𝑡 2 = 10(𝑡 + 0.7) ⟹ 𝑡 2 − 10𝑡 − 7 = 0 ⟹ 𝑡 = 5 ± √32 ⟹ 𝑡 = 10.7 𝑠𝑜𝑟 𝑡 = −0.7 𝑠
But 𝑡 > 𝑜. Thus the time taken by the car to catch up with the truck driver is 10.7 s.
(d) See June 2002 Q8
(e) Consult your notebook
(f) Amplitude, 𝑎 = (
14−4
) × 10 𝑐𝑚 = 50 𝑐𝑚
2
1
1
Frequency, 𝑓 = 𝑇 = 31 𝑑𝑎𝑦𝑠 = 0.032𝑑𝑎𝑦 −1
9.(a) (i) Self induction is the phenomenon whereby changing current in a coil and the resulting changing
magnetic flux induces an emf in the same coil with the emf opposing the change producing it. That is the
overall effect of self induced emf in a coil is acts against the voltage supply. While mutual induction is a
phenomenon whereby the changing current in one coil (primary) induces an emf in another adjacent coil
(secondary).
(ii)
(iii) 29 min
(iv) If a resistor replaces the inductor, the current through inductor will be constant with time. Hence the
graph will simple to a horizontal straight line (see diagram below)
I/A
t/s
<<ADVANCED LEVEL PHYSICS
Page 111
(b)
F= tension force in the insulated thread
W=weight (i.e the gravitational pull of the earth on the
charge)
Fe = electrostatic force
Since the charge is in equilibrium,
Resolving horizontally, 𝐹𝑐𝑜𝑠60° = 𝐹𝑒 - - - - - - -(1)
Resolving vertically, 𝐹𝑠𝑖𝑛60° = 𝑚𝑔 - - - - - - - -(2)
𝑚𝑔
𝑚𝑔
(1)⁄(2) ⟹ 𝑡𝑎𝑛60° = 𝐹 ⟹ 𝐹𝑒 = 𝑡𝑎𝑛60° - - - - -(3)
F
Fe
600
W=mg
𝑒
But 𝐹𝑒 = 𝑞𝐸 =
𝑞𝑉
𝑑
⟹
𝑞𝑉
𝑑
𝑚𝑔
𝑑𝑚𝑔
= 𝑡𝑎𝑛60° ⟹ 𝑞 = 𝑉𝑡𝑎𝑛60° =
(c)(i) See June 2001Q10
(ii) See your notebook for the setup
(iii) 𝑉 = −𝑟𝐼 + 𝐸 . From the graph, 𝐸 = 3.0 𝑉
𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑔𝑟𝑎𝑝ℎ
0.05×0.05×9.8
600𝑡𝑎𝑛60°
= 3.36 × 10−5 𝐶
(𝑏) 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑟 =
3−0
𝑠𝑙𝑜𝑝𝑒 = 0−4 = −0.75 Ω ⟹ 𝑟 = − − 0.75 = 0.75Ω
(d) 𝜌 =
𝑅𝐴
𝑙
⟹𝑙=
𝑅𝐴
,𝑃=
𝜌
𝑉2
𝑅
⟹𝑅=
𝑉2
𝑃
⟹𝑙=
𝑉2𝐴
𝑃𝜌
=
202 ×1.0×10−7
10×1.0×10−6
= 4.0 𝑚
(e) Electrical energy consumed, 𝑊 = 𝑝𝑡 = 0.01𝑘𝑊 × 30 × 24 = 7.2 𝑘𝑊 ⟹ 𝑐𝑜𝑠𝑡 = 7.2 × 60 =
432 𝑓𝑟𝑠
10.
+6 V (a) (i) Applying KVL, 𝑉 + 𝑉 + 𝑉 − 𝑉 = 0
𝐵𝐸
𝐶𝐵
𝑅
𝐶𝐶
Neglecting the voltage across the base – collector junction,
𝑉𝑅 = 𝑉𝐶𝐶 − 𝑉𝐵𝐸 = 6 − 0.6 = 5.4 𝑉
A
VR
R
⟹ 𝐼𝐶 =
VCB
𝑉𝑅
𝑅
=
5.4
𝑅
. If the value of R is known then IC can be
found, which is the reading of the ammeter
𝑄
VCE =V
3.2 𝜇𝐶
(ii) input voltage, 𝑉𝑖𝑛 = 𝐶 = 4.7 𝜇𝐹 = 0.68 𝑉
V
Voltage gain, 𝐴 =
VBE
0V
𝑉0𝑢𝑡
𝑉𝑖𝑛
=
𝑉𝐶𝐸
𝑉𝑖𝑛
⟹ 𝑉𝐶𝐸 = 𝐴𝑉𝑖𝑛 = 20 × 0.68
⟹ 𝑉𝐶𝐸 = 13.6𝑉. Thus the voltmeter reads 13.6 V
This value obtained the reading of the voltmeter is greater than the applied voltage, meaning that the
transistor must have become saturated.
(b)
12 V
10mA
12
Forward bias resistance of LED,𝑅𝑓 = 0.01 = 1200Ω
P.d across LED = P.d across R = 12 V
R
⟹ 𝐼𝑅 𝑅 = 12 ⟹ 𝑅 =
12
𝐼𝑅
12
= 𝐼−0.01 , where I is the current delivered by the battery
(c) (i) 𝑄 = 𝐶𝑉 = 0.33 × 10−6 × 2.4 = 7.92 × 10−7 𝐶
1
1
(ii) 𝑊 = 2 𝑄𝑉 = 2 × 7.92 × 10−7 × 2.4 = 9.50 × 10−7 𝐽
<<ADVANCED LEVEL PHYSICS
Page 112
(d)(i) The diode ensures that current flows only in one direction (forward bias direction).
(ii) 𝑄 = 𝐶∆𝑉 = 0.33 × 10−6 × (4 − 1) = 4.62 × 10−7 𝐶 (iii) 𝑄 = 𝐼∆𝑡 ⟹ ∆𝑡 =
(e)(i) Molar volume =
Atomic volume =
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝑑𝑒𝑛𝑠𝑖𝑡𝑦
𝑚𝑎𝑙𝑜𝑟𝑎 𝑣𝑜𝑙𝑢𝑚𝑒
𝑁𝑎
=
=
6.4×10−2
8.9×103
𝑄
𝐼
=
4.62×10−7
1.0×10−6
= 0.462 𝑠
= 7.19 × 10−6 𝑚3 𝑚𝑜𝑙 −1
7.19×10−6 𝑚3 𝑚𝑜𝑙−1
6.02×1023 𝑚𝑜𝑙−1
= 1.195 × 10−29 𝑚3
3
Assuming that atoms are cubes of sides, d, then 𝑑 = √1.195 × 10−29 = 2.3 × 10−10 𝑚
4
𝑑 3
N.B: if we assume that the atoms are spherical, 𝑉 = 3 𝜋 ( 2 ) =
𝜋𝑑3
6
6 × 1.195 × 10−29
√
⟹𝑑=
= 2.8 × 10−10 𝑚
𝜋
3
F
d=2.3 x 10-10 m
0
x
d is called the equilibrium
position, where the net force is
zero
(f) The pressure difference in the smaller droplet is greater than in the bigger one. Hence the
smaller droplet breaks and enters the bigger one as shown in the diagram below. This is because surface
tension is greater in small droplet than a big droplet.
+
(g) (i) See June 2006 Q8
(ii) The separation of molecules in a liquid is slightly greater than the equilibrium separation. Hence there
is a
net attractive force between each pair of neighboring molecules. The molecules in the bulk (or in
the body) of the liquid experience a net zero force as each neighboring molecule is being pulled equally
in all directions. The molecules at the surface of the liquid are without neighbors above and
are
more spaced out. Hence there is now a net force on each surface molecule, directed into the bulk of
the liquid. Therefore the surface of a liquid is under a force due to attraction from its neighbors. So
making molecules to resist being stretched or compressed from a widely spaced distance.
<<ADVANCED LEVEL PHYSICS
Page 113
(iii) Molecules in the bulk of a liquid experience a net zero force as they are been pulled equally in all
directions. While at the surface, molecules are without neighbors above and hence experience a net
attractive
force
1
1
2
1
2𝛾
1
1
2×7.0×10−2
2
1
9.8×1000
(h) ℎ𝜌∆ℎ = 2𝛾 (𝑟 − 𝑟 ) ⟹ ∆ℎ = 𝑔𝜌 (𝑟 − 𝑟 ) =
<<ADVANCED LEVEL PHYSICS
1
1
(10−3 − 2×10−3 ) = 0.007𝑚 = 7 𝑚𝑚
Page 114
JUNE 2005
1.(a) An equation is homogeneous if all the terms in the equation have the same base units or dimensions.
Q2
Q2
(b) c 2 μ0 εo = 1, from F = 4πε
From 𝐹 =
𝜇0 𝐼 2 𝑙
2𝜋𝑟
2
0r
[r][F]
[Q]2
⟹ ε0 = 4πFr2 ⟹ [ε0 ] = [F][r]2 = kg −1 m−3 s 4 A2
⟹ [μ0 ] = [I]2 [l] ⟹ [μ0 ] = kgms −2 A−2
Units of LHS =(ms−1 )2 (kgms −2 A−2 )(kg −1 m−3 s4 A2 ) = 1. Hence the equation is homogeneous.
l
l
1
g
1
2.(i) For a simple pendulum, T = 2π√g ⟹ T 2 = (2π)2 g , but 𝑓 = 𝑇 ⟹ f 2 = (4π2 ) . l . Thus a graph of 𝑓 2
1
𝑔
against 𝑙 is straight line passing through the origin with slope 4𝜋2
From the graph, 𝑠𝑙𝑜𝑝𝑒 =
1.5−0.5
6−2
g
𝑔
= 0.25𝑚𝑠 −2 ⟹ 4𝜋2 = 0.25 ⟹ g = 0.25 × 4π2 = 9.87 ms −2
1
g
1
9.87
(ii) We know that f 2 = (4π2 ) . l ⟹ l = (4π2 ) . f2 = 4π2 (202 ) = 6.25 × 10−4 = 0.625mm
3.By the law of conservation of linear momentum,
- Horizontally, mA vA = mA vA cos60 + mB mB cosθ , but mA = mB = m ⟹ 5.5 = 2.5cos60 + vB cosθ
⟹ vB cosθ = 4.25 − − − − − − − − − −(1)
- Vertically, 0 = mvA sin60 − mvB sinθ ⟹ vB sinθ =
√3
(2.5)
2
⟹ vB sinθ = 1.25√3 − − − − − − − − − (2)
1.25√3
Solving (1) and (2) simultaneously, gives vB = 4.77 ms −1 and θ = tan−1 (
c
4.(a) vp = n =
p
3.0×108
1.40
c
= 2.14 × 108 ms −1 and vQ = n =
Q
3.0×108
1.45
4.25
) = 27.0°
= 2.07 × 108 ms −1
SincevP > vQ , the light will emerge first from P
(b) Optical path of X, dx = nP t = 1.40 × 20 × 10−6 = 2.8 × 10−5 m
Optical path of Y, dY = nQ t = 1.45 × 20 × 10−6 = 2.9 × 10−5 m
Optical path difference, ∆x = dy − dx = 1.0 × 10−6
Phase difference, ∆φ =
2π∆x
λ
=
2π×1.0×10−6
450×10−9
= 13.96 rad
5.
f=+20cm
f=-30cm
I/
I
10 cm
d
25 cm
For the convex lens in position, the image is formed at I. when the concave lens is now placed between
the screen and the convex lens, it diverges the rays from the convex lens slightly so that they now
converge at the point I/. Therefore the screen must be shifted through a distance d as indicated on the
diagram
The image formed by the convex lens acts like a virtual object to the concave lens. Thus for the concave
1
1
1
𝑢𝑓
lens, 𝑓 = −30.0 𝑐𝑚, 𝑢 = −15 𝑐𝑚 ⟹ 𝑓 = 𝑣 + 𝑢 ⟹ 𝑣 = 𝑢−𝑓 =
(−30)(−15)
−15−−30
⟹ 𝑣 = 30 𝑐𝑚
Therefore the screen must be shifted a distance d = 30 cm – 15 cm = 15 cm backwards.
<<ADVANCED LEVEL PHYSICS
Page 116
6.
I1
12 V
I3
10 Ω
V1
6V
1
I2
V3
5Ω 2
V2
(i) Applying KVL
Loop 1, 𝑉2 + 𝑉1 − 6 = 0
⟹ 5𝐼3 + 10𝐼1 − 6 = 0 ⟹ 5𝐼3 − 10𝐼1 = 6 − − − (1)In
loop 2,−𝑉3 + 12 − 𝑉2 = 0
⟹ −5𝐼2 + 12 − 5𝐼3 = 0 ⟹ 𝐼2 + 𝐼3 = 2.4 − − − (2)
By KCL, 𝐼1 + 𝐼2 − 𝐼3 = 0 − − − − − − − − − −(3)
Solving the above equations simultaneously, gives
𝐼2 = 𝐼3 = 1.2 𝐴, 𝐼1 = 0
(ii) 𝑉𝑥𝑦 = 5𝐼3 = 5 × 1.2 = 6.0 𝑉
5Ω
I3
Figure 4
Y
FBP
7.(i)
A
B
P
d1
𝐾𝑄𝐴 𝑄𝐵
2
1 +𝑑2 )
𝐹𝐴𝑃 = (𝑑
𝐹𝐵𝑃 =
𝐾𝑄𝐵 𝑄𝑃
𝑑22
=
=
FAP
d2
9.0×109 ×4.0×10−8 ×2×10−8
(𝑑1 +𝑑2 )2
9.0×109 ×1.0×10−8 ×2×10−8
𝑑22
7.2×10−6
2
1 +𝑑2 )
= (𝑑
=
1.8×10−5
𝑑22
The net force on P due to A and B, F = FBP − FAP =
18×10−6
d22
7.2×10−6
2
1 +d2 )
− (d
If the distances are known then the magnitude of F can be gotten.
(ii) Since the charge P is positive, B is negative and A is positive, if P is placed between A and B, the
force
on P due to A is to the right and that due to B is also to the right. Hence the net force cannot be
zero.
8.(a) (i) For the Newton’s laws of motion, see your textbooks.
(ii) Consider two bodies A and B with masses mA and mB respectively. If initially A and B have speeds
uA and uB respectively and collide such that after the collision, the have respective speeds vA and vB, then
By Newton’s second law, the forces exerted on A and B are
𝑚𝐴 (𝑣𝐴 − 𝑣𝐵 )
𝑚𝐴 (𝑣𝐵 − 𝑣𝐵 )
𝐹𝐴 =
𝑎𝑛𝑑 𝐹𝐵 =
∆𝑡
∆𝑡
𝑚𝐴 (𝑣𝐴 −𝑣𝐵 )
𝑚𝐴 (𝑣𝐵 −𝑣𝐵 )
By Newton’s third law, 𝐹𝐴 = −𝐹𝐵 ⟹
=−
⟹ 𝑚𝐴 𝑣𝐴 − 𝑚𝐴 𝑢𝐴 = −𝑚𝐵 𝑣𝐵 − 𝑚𝐵 𝑢𝐵
∆𝑡
∆𝑡
⟹ 𝑚𝐴 𝑢𝐴 + 𝑚𝐵 𝑢𝐵 = 𝑚𝐵 𝑣𝐵 + 𝑚𝐴 𝑣𝐴 , which is the principle of conservation of linear momentum.
(b) See your textbooks
(c) A conservative force is one in which the total work done under its influence in a closed path is zero.
Examples are: gravitational force, electrostatic force. A non – conservative force is one in which the total
work done in a closed path under it influence is not zero. Examples are: friction, drag force, upthrust etc.
d) (i)
Kirchhoff’s current law states that the total current entering a junction equal the total current
leaving the junction. The law is essentially a law of conservation of charge because current is the rate of
flow of charge with time. If current entering a junction equal current leaving the junction, it therefore
implies that the quantity of charge arriving the junction per unit time equal the quantity of charge leaving
the junction per unit time hence charge is conserved.
<<ADVANCED LEVEL PHYSICS
Page 117
Kirchhoff’s voltage law states that in a closed loop, the algebraic sum of the p.d drops across loads
equal algebraic sum of emf. This law is the law of conservation of energy in that the total work done
per unit charge in converting other forms of energy to electrical energy is equal to the work done per unit
charge in converting electrical energy to other forms of energy. Hence from the work energy principle,
energy is conserved.
(e) See you notebooks
(f)
8Ω
1
6V
I
V
2
6Ω
I2
KCL, 𝐼 = 𝐼1 + 𝐼2 − − − − − −(1)
KVL, loop (1): 12 − 8𝐼 − 6 − 6𝐼2 = 0
⟹ 4𝐼 + 3𝐼2 = 3 − − − − − − − (2)
KVL, loop (2): 6𝐼2 − 12𝐼1 = 0
⟹ 𝐼2 = 2𝐼1 − − − − − −(3)
Solving, 𝐼 = 0.5 𝐴, 𝐼1 = 0.167 𝐴, 𝐼2 = 0.33 𝐴
𝑉6𝛺 = 𝑉 = 𝐼2 𝑅 = 0.333 × 6 = 2.0 𝑉
12 V
I1
12 Ω
Figure 6
9.
 Molecules of a particular gas are identical
 Collision between the molecules and with the walls of the container is perfectly elastic.
 The molecules exert no force on each other except during impact which may be assumed to be of
negligible duration.
 There are sufficiently large numbers of molecules for the laws of statistic to be meaningfully
applied.
 The sizes of molecules are negligible compare to their separation.
Wall
X
Consider a gas molecule enclosed in a cube of sides L. let each molecule of the gas have
mass m. consider a single molecule with x – component of speed, u1 moving towards the wall X. the x –
component of the momentum of the molecules is mu1 towards the wall. The molecule will reverse its
direction after colliding with the wall. Since the collision is perfectly elastic, its x – component in the
reverse direction will be –mu1. The change in the x – component of the momentum is mu1 - - mu1 = 2mu1.
The molecule has travelled a distance 2L (to and fro). The time for the molecule to move to and from the
wall is is 2L/u1
2mu
The rate of change of momentum of the molecule due to the collision is 2L⁄u1 =
1
mu21
L
. By Newton’s second
law, the rate of change of momentum is equal to the force exerted on the wall X
i.e Force on wall, F =
mu21
L
Therefore force per unit area, pressure P =
mu21 ⁄L
L2
=
mu21
L3
(area side x side = L2)
if there are N molecules in the container with x – components of velocity u1, u2, u3, . . ., uN, then the total
pressure exerted on the wall will be given by
<<ADVANCED LEVEL PHYSICS
Page 118
m
P = L3 (u12 + u22 + ⋯ + u2N )
m ̅̅̅
2 , where ̅̅̅
Therefore, P = Nu
u2 is the mean square velocity in the x – direction.
L3
The total mass of all the molecules is mN. Therefore the density of the gas is given by ρ =
̅̅̅2 - - - - - - - - (+)
P = ρu
mN
L3
. Thus
If the c is the resultant speed of a molecule whose x - , y – and z – components of velocity are
𝑢, 𝜔 𝑎𝑛𝑑 𝜗 respectively, then c 2 = u2 + ω2 + ϑ2 ⟹ c̅2 = ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
u2 + ω2 + ϑ2 ⟹ c̅2 = ̅̅̅
u2 + ̅̅̅̅
ω2 + ̅̅̅
ϑ2
1
Since there are large number of molecules and are moving randomly, ̅̅̅
u2 = ̅̅̅̅
ω2 = ̅̅̅
ϑ2 ⟹ ̅̅̅
u2 = c̅2 .
3
1
Therefore,(+) becomes P = 3 ρc̅2
1
1
(b) (i) P = 3 ρc̅2 ⟹ slope of graph = 3 c̅2 . From the graph, slope =
(1.3−0)105
1.5−0
= 8.87 × 104 m2 s −2
1
⟹ 3 c̅2 = 8.87 × 104 ⟹ c̅2 = 3 × 8.87 × 104 ⟹ crms = √3 × 8.87 × 104 = 510 ms −1
(ii) 𝑇 < 300 𝐾 Because slope of graph at the temperature T is less than the slope at the temperature 300 K.
Or by calculations,
Slope at temperature T is given by
(1.0−0)×105
1.85−0
= 5.4 × 104 𝑚2 𝑠 −2
⟹ crms = √3 × 5.4 × 104 =
𝑐
𝑇
𝑐
2
402.7ms −1
402.7 2
Now, 𝑐1 = √𝑇1 ⟹ T2 = (𝑐1 ) T1 ⟹ T2 = ( 510 ) × 300 = 187.0 K . Thus T = 187.0 K < 300 K.
2
2
2
(i)
N(v)
300 K
600 K
c0 cm crms
C0 = most probable speed
Cm = mean speed
Crms = root mean square speed
v/ms-1
(d) See June 2003
(e)
Stress
Strain
f) (i) Maximum energy = area between the curve and the extension axis Area of one square 0.5x1000
500 J .Approximate number of squares under the graph =16.5
Total area = 16.5 x 500J = 8250 J. therefore maximum energy = 8250 J
<<ADVANCED LEVEL PHYSICS
Page 119
𝑠𝑡𝑟𝑒𝑠𝑠
(ii) Young’s modulus, = 𝑠𝑡𝑟𝑎𝑖𝑛 =
𝐹 ⁄𝐴
𝑒 ⁄𝑙
𝐹
𝑙
𝑙
𝑙
𝑙
= 𝐴 × 𝑒 ⟹ 𝑒 = 𝐸𝐴 . 𝐹 ⟹ 𝑠𝑙𝑜𝑝𝑒 = 𝐸𝐴 ⟹ 𝑠𝑙𝑜𝑝𝑒×𝐴 . where the
𝑑 2
material is calculated as follows 𝐴 = 𝜋 ( 2 ) , where d is the diameter.
area of the
Intensity
c) (i)
3𝜆/𝑑
−3𝜆/𝑑 −2𝜆/𝑑 −𝜆/𝑑 0 𝜆/𝑑
2𝜆/𝑑
separation
(ii) From the diagram above,
the width of the central
bright fringe is Angular
given by
𝜆
𝜆
2𝜆
𝑑
𝑑
𝑑
𝐷 = −− =
=
2×5.5×10−7
0.15×10−3
= 7.33 × 10−3 𝑟𝑎𝑑
(d) (i) (ii) See June 2001
(e) (i)
 The effect electrostatics force can be shielded while that of gravitational force cannot be shielded
 Electrostatic force is attractive or repulsive while gravitational force is only attractive.
 When elementary particles are considered, the electrostatic force is stronger than the gravitational force.
(ii)
 Both of them obey the inverse square law
 Both are action at a distance forces
 Both field strength equal their negative potential gradient
𝐺𝑀𝑚
(f) 𝑈 = (𝑅+ℎ)
(i) On the surface of the earth, the potential of the earth is given by 𝑉 =
the
−𝐺𝑀
𝑅
, while at a height h above
−𝐺𝑀
surface, the potential is given by 𝑉 = (𝑅+ℎ). The potential energy gained in lifting a mass m to a height h
above the surface of the earth is given by
−𝐺𝑀
𝑈 = 𝑚∆𝑉 = 𝑚 ( 𝑅+ℎ − −
the
following relation 𝑈 =
𝐺𝑀
𝑅
𝐺𝑀𝑚ℎ
𝑅2
1
1
𝐺𝑀𝑚ℎ
) = 𝐺𝑀𝑚 (𝑅 − 𝑅+ℎ) = 𝑅(𝑅+ℎ) =
, but 𝑔 =
𝐺𝑀
𝑅2
𝐺𝑀𝑚ℎ
ℎ
𝑅 2 (1+ )
𝑅
. Asℎ ≪ 𝑅,
ℎ
𝑅
≈ 0, and one has
⟹ 𝑈 = 𝑔𝑚ℎ = 𝑚𝑔ℎ hence.
(ii) If the space craft is to leave the earth completely, the initial kinetic energy should be equal to the
potential
energy needed to take the space craft to infinity.
𝑖. 𝑒, 𝐾. 𝑒 = 0 − −
𝐺𝑀𝑚
𝑅
=
𝐺𝑀𝑚
𝑅
6
, but𝑔𝑅 2 = 𝐺𝑀 ⟹ 𝐾. 𝑒 =
𝑔𝑅 2 𝑚
𝑅
= 𝑔𝑅𝑚
⟹ 𝐾. 𝑒 = 100 × 9.8 × 6.4 = 6.3 × 109 𝐽
(iii) If the energy is less, the space craft will move in an elliptical path. If the energy is is more, the space
craft will escape from the earth, and the path will describe a hyperbola
<<ADVANCED LEVEL PHYSICS
Page 120
JUNE 2006
1.(a) 𝐹 = 𝐵𝐼𝑙𝑠𝑖𝑛𝜃
One condition for an equation to be physically correct is that it should be homogeneous.
Units of LHS =units of force = kgms-2
Units of right hand side = (units of B)x(units of I)x(units of l)=kgA-1s-2.A.m= kgms-2
Since the base units on both sides of the equation are the same, it means the equation is homogeneous
and may be physically correct.
𝑄2
(b) 𝐹 = 4𝜋𝜀
2.
𝜃/℃
0
𝑄2
𝑟2
[𝑄]2
𝐴2 𝑠2
⟹ 𝜀0 = 4𝜋𝐹𝑟 2 ⟹ [𝜀0 ] = [𝐹][𝑟]2 = 𝑘𝑔𝑚𝑠−2 𝑚2 = 𝑘𝑔−1 𝑚−3 𝐴2 𝑠 4
unlagged
Lagged
Length/m
3.(i) Emax = hf − ϕ ⟹ ϕ = hf − Emax =
hc
λ
− Emax ⟹ ϕ =
6.63×10−3 ×3.0×108
248×10−9
− 8.6 × 10−20 J
7.2×10−19
⟹ ϕ = 7.2 × 10−19 = 1.6×10−19 = 4.5 eV
(ii) ϕ = hf0 ⟹ f0 =
ϕ
h
7.2×10−19
= 6.63×10−34 = 1.09 × 1015 Hz
4.See June 2002 Q10
5.(i) Let ss be the distance travelled by superman sb be the distance travelled by the baby.
Let superman take time t to catch the baby.
1
1
⟹ ss = sb ⟹ vs t = 2 g(t + 2)2 ⟹ 39.2t = 2 (9.8)(t + 2)2 ⟹ 8t = t 2 + 4t + 4 ⟹ t 2 − 4t − 4 = 0
(t − 2)2 = 0 ⟹ t = 2s .Thus the time taken by superman to catch the child is t=2 s.
(ii) ss = vst = 39.2x2 = 78.4 m.
4×4
6.𝐶𝑃 = 3𝜇𝐹 + 1𝜇𝐹 = 4𝜇𝐹, 𝐶𝑇 = 4+4 = 2𝜇𝐹
(i) 𝑄 = 𝐶𝑉 = 2 × 10−6 × 12 = 24 𝜇𝐶. Thus the charge on the 4 𝜇𝐹 is 24 𝜇𝐶
(ii) If S1 is opened and S2, the capacitors will discharge through the resistance, R, and the initial
maximum current that will flow is 𝐼0 =
7.
𝑉
𝑅
=
12
5
= 2.4 𝐴
T
mv 2
r
𝑇 − 𝑚𝑔 =
mg
mv 2
v2
⟹ T = m (g + )
r
r
⟹ T = 45 (9.8 +
32
) = 495 N
7.5
8.(a) A uniform magnetic field can be produced in the laboratory by
Using a solenoid.This is achieved by passing a steady d.c through a long coil and a uniform magnetic
field is created at the centre.
Using Helmholtz coils. To do this, the same d.c is passed through two Helmholtz coils connected in
series and a uniform magnetic field is produced between them.
<<ADVANCED LEVEL PHYSICS
Page 122
(b) As the electron enters the field, magnetic force = centripetal force
⟹ Beϑ =
mϑ2
r
e
⟹ ϑ = Br (m) . . . . . . . . . . . . . . . . . . . . . . . . . . .(1)
Electrical energy lost = gained in kinetic energy
1
Ber 2
1
e
2V
⟹ eV = 2 mϑ2 ⟹ eV = 2 m ( m ) ⟹ m = B2 r2 as required to demonstrate.
(c) Consult your textbooks
1
2eV
(d) (i) eV = 2 mϑ2 ⟹ ϑ = √
m
= √2 × 1.76 × 1011 × 1400 = 1.2 × 105 ms−1
(ii) For no deflection, magnetic force equal electric force
⟹ Beϑ = eE ⟹ E = Bϑ = 0.4 × 1.2 × 105 = 4.8 × 104 NC−1
(e) Surface tension is the force per unit length acting normally to one side of an imaginary line drawn in
the surface of the liquid.
(f)
Large drop
Small drop
The small mercury drop is spherical because the surface tension force is greater than the weight of the
drop but the large one tend to flatten because the weight of the drop is greater than the the surface
tension force
N.B 𝛄 ∝ 𝐀 𝐚𝐧𝐝 𝐰𝐞𝐢𝐠𝐡𝐭 ∝ 𝐯𝐨𝐥𝐮𝐦𝐞
100 cm
(g)
5cm
F
45cm
m
20cm
W
30cm
0.049 N
When the ring is balanced by the 5 g mass, its weight w is given by
W(0.45) = (0.049)(0.1) ⟹W = 0.0109 N
When the beaker is placed and the 5 g mass moves to the 70 cm mark,
By the principle of moment (𝑤 + 𝐹)(0.45) = (0.049)(0.2) ⟹ 𝑊 + 𝐹 = 0.0218 𝑁
𝐹
0.0109
⟹ 𝐹 = 0.0218 𝑁 − 0.0109 𝑁 = 0.0109 𝑁, but 𝛾 = 2𝜋𝑟 = 2𝜋×0.01 = 0.18 𝑁
9.(a) (i)The resistivity of a material is the resistance per unit length of a material of unit cross.
ρ
Insulator
<<ADVANCED LEVEL PHYSICS
Page 123
A
(ii)
b(i)
Conductor
Semiconductor
(ii)From 𝑉 = 𝐼𝑅 ⟹ 𝑠𝑙𝑜𝑝𝑒 =
⟹𝜌=
𝑅𝐴
𝑙
=
𝜋𝑑2 𝑅
4𝑙
∆𝑉
6.4−2.0
4.4
= 𝑅 ⟹ 𝑅 = 1.00−0.32 = 0.68 = 6.47 Ω
∆𝐼
2
=
𝜋×(5×10−3 ) ×6.47
4×80×10−2
= 1.6 × 10−4 Ωm
c) 𝑊 = 𝐼𝑉𝑡 = 60 × 12 × 60 × 60 = 2.59 × 106 𝐽
d) (i) A material is elastic if it has the ability to regain its original length after the removal of a
deforming force.
(ii)
F/N
Loading
(e) (i)
V
Unloading
O
e/m
Fixed load
Energy dissipated equal area of the loop
(ii) 𝐸 =
𝐹 ⁄𝐴
𝑒 ⁄𝑙
𝐴𝐸
⟹ 𝐹 = ( 𝑙 ) 𝑒 ⟹ 𝑠𝑙𝑜𝑝𝑒 =
𝑠𝑙𝑜𝑝𝑒 =
𝐴𝐸
𝑙
⟹𝐸=
𝑙×𝑠𝑙𝑜𝑝𝑒
𝐴
Variable load
. From the graph,
(6 − 0)103
0.8 × 2.143 × 103
= 2.143 × 103 𝑁𝑚−1 ⟹ 𝐸 =
= 9.7 × 107 𝑃𝑎
2.8 − 0
𝜋(7.5 × 10−3 )2
1
(iii) W area under graph, 𝑊 = 2 × 6.6 × 1000 × 3 = 9900 𝑁
(f) 𝐹 = 𝑚𝑔 = (450 + 750) × 9.8 = 11760 𝑁
𝐹
𝐹𝑜𝑟𝑐𝑒
11760
𝑠𝑡𝑟𝑒𝑠𝑠 = 𝐴 ⟹ 𝐴 = 𝑠𝑡𝑟𝑒𝑠𝑠 = 4.0×108 = 2.94 × 10−5 𝑚2 .
2.94×10−5
𝐴
But 𝐴 = 𝜋𝑟 2 ⟹ 𝑟 = √𝜋 = √
𝜋
= 3.06 𝑚𝑚
10. (a) (i), (ii) See June 2002 Q10
(b) (i)
(ii) The presence of different types of elements could be identified by comparing their spectra with those
of known element until they correspond
1
1
(iii) Energy difference, ∆𝐸 = 𝐸1 − 𝐸2 = (𝜆 − 𝜆 ) ℎ𝑐
1
1
1
⟹ ∆𝐸 = (589.0×10−9 − 589.6×10−9 ) (6.67 × 10
2
−34 )(3.0
× 108 ) = 3.46 × 1022 𝐽
(c) (i) Nuclear fission is the splitting of a larger nuclei into smaller ones accompanied with the release of
energy.
<<ADVANCED LEVEL PHYSICS
Page 124
A controlled chain reaction is a reaction in which the product one reaction initiates a further reaction,
giving rise to more products which in turn cause further reactions.
Nuclear energy
k.e of fission
products
Internal energy
of coolant
Electrical energy
in generator
Internal energy
of water
k.e of spinning
rotor
k.e of steam
k.e of spinning
turbines
(d) See June 2003 Q10
(e) (i) Temperature increases, the more electrons leave the valence band and enter the conduction band
and conductivity of the intrinsic semiconductor increases.
(ii) The depletion layer in a pn junction is fairly free from majority charge carriers. The width of the
depletion layer is smaller than that of the p – and n – type regions.
(f)(i) When temperature rises, the resistance of the thermistor decreases, leading to a drop in the input
p.d, which is unable to turn on a large collector current. As a result, the alarm does not ring. When
temperature falls, the resistance of the thermistor increases, leading to an increase in the input voltage,
resulting in a large collector current, which enables the alarm to ring.
N.B: if the alarm is to ring when temperature rises, then the thermistor should be placed in position R.
(ii) In the dark, the resistance of the LDR increases. The input or base voltage increases, leading to an
increase in the collector current, which causes the alarm to ring. As daylight comes, the resistance of the
LDR falls, and the input voltage also drops, resulting in a fall in the collector current and the alarm goes
off.
<<ADVANCED LEVEL PHYSICS
Page 125
JUNE 2007
1.c =
1
√μ0 ε0
.
Q2
Q2
From F = 4πε
From F =
μ0
2
0r
2
I l
2πr
[Q]2
A 2 s2
⟹ ε0 = 4πFr2 ⟹ [ε0 ] = [F][r]2 = kgms−2 m2 = kg −1 m−3 A2 s 4
⟹ μ0 =
2πrF
I2 l
⟹ [μ0 ] =
mkgms2
A2 m
= kgms −2 A−2
Units of LHS= units of 𝑐 = 𝑚𝑠 −1
Units of RHS =
1
1
√kgms−2 A−2 .kg−1 m−3 A2 s2
= √𝑚−2
𝑠2
= 𝑚𝑠 −1
Since the LHS and the RHS have the same base units, it means the equation is homogeneous.
1.5 A
2.
E
1
3Ω
0.5 A
I
R
I = 1.5A – 0.5 A = 1.0 A
KVL in loop (1), 4(1) + 3(1.5) − 𝐸 = 0
⟹ E = 4 + 4.5 = 8.5 A
KVL in loop (2), 4(1) − 𝑅(0.5) = 0 ⟹ R = 8.0 Ω
4Ω
2
3.Stress – strain is preferable because it gives the same shape for any length and area while the force –
extension graphs for the same materials of different length and area have the same shape.
(ii) Work done = area of hysteresis loop OABCO
Area of one square = 1x1000x0.25x10-3 = 0.25 J
Approximate number of squares = 35
Therefore work done = 0.25x35 =8.75 J
4.(i) Electrical energy supplied, W = IVt = 1.5x240x60 =21600 J
0.75x21600 = mc∆θ ⟹ m =
16200
c∆θ
16200
⟹ m = 460×20 = 1.7 kg
(ii) Assumption: Mass of the piece of iron is constant
5.
IC/mA
IC/mA
IB/mA
IB/mA
Transfer
VCE/V
VBE/V
Input
Output
(b) The current gain of the amplifier is slope of the transfer characteristic curve.
6.(i) Critical angle is the angle of incidence in the dense medium for which the angle of refraction in the less
dense medium is 900.
(ii) n1>n2 to enable total internal reflection occur at the core cladding boundary.
n
(iii) Considering figure 3, n1 sinθc = n2 sin90° ⟹ sinθc = n2. Assuming light leaves glass into air, the
1
𝑛𝑎
1
critical angle for glass is given by sinC = 𝑛 ⟹ C = sin−1 (1.5) = 42°
𝑔
7.(i) This is because the constant volume gas thermometer is a standard on which other thermometers are
calibrated and besides the constant volume gas thermometer is calibrated on the thermodynamic scale
<<ADVANCED LEVEL PHYSICS
Page 126
while the mercury in glass thermometer is calibrated on the Celsius scale. Since the Celsius scale is
defined from the thermodynamic scale, it therefore implies both can give readings in degree Celsius.
Note that a temperature rise of one Kelvin on the thermodynamic scale is equivalent to a temperature rise
of one degree Celsius on the Celsius scale.
(ii) The variation of the length of the mercury thread with temperature is not the same as the variation of
gas pressure with temperature.
8.(a) See your notebooks or textbooks.
b(i) Elastic collision means kinetic energy is conserved.
(ii) By the law of conservation of linear momentum,
𝑚𝑢𝑥 = 𝑚𝑣𝑥 + 𝑚𝑣𝑦 ⟹ ux = vx + vy . . . . . . . . . . . . . . . . . . . . . . . (1)
Conservation of kinetic energy gives
1
2
1
1
𝑚𝑢𝑥2 = 2 𝑚𝑣𝑥2 + 2 𝑚𝑣𝑦2 ⟹ 𝑢𝑥2 = 𝑣𝑥2 + 𝑣𝑦2 . . . . . . . . . . . . . . . . . . .. .(2)
Solving (1) and (2) simultaneously, gives𝑣𝑥 = 0 𝑎𝑛𝑑 𝑣𝑦 = 2.5 𝑚𝑠 −1. This means that X comes to rest
and Y continues to move with the initial velocity of X. That is X transfers all it momentum and kinetic
energy to Y as a result of the collision.
(iii) 𝐹 =
∆𝑃
∆𝑡
𝑣−𝑢
= 𝑚(
𝑡
0−2.5
) = 0.4 (60×10−6 ) = −1.67 × 104 𝑁. Thus the average force =1.67 × 104 𝑁
P
t/s
(d) See your textbooks
(e) Seee June 2005 Q9
9.(a) (i) To be radioactive means the substance is capable of spontaneously and randomly emitting radiation.
This can be as a result of low binding energy per nucleon.
(ii) The use of cloud chambers: In a cloud chamber, alpha tracks are thick and straight while beta tracks
are thin and crooked.
The use of electric field: In a uniform electric field, alpha particles are deflected toward the positive
plate while beta particles are deflected toward the positive plate
b(i) See your notes
7.0 − 5.0
−2
2 −1
lnA = lnA0 − λt ⟹ lnA = −λt + lnA0 , slope = −λ ⟹ −λ =
=
⟹λ=
s
(0 − 6.4)60 384
384
But 𝑇1⁄2 =
𝑙𝑛2
λ
=
384ln2
2
= 133.08 s = 2.22 min
(ii) 𝑙𝑛𝐴0 = 7 ⟹ A0 = e7 = 1.1 × 103 min−1
(d) A capacitor is a device used to store charge
(iia) Both diode and capacitor allows d.c to flow through them
(iib) A capacitor allows only a.c to flow through it whereas a diode allows both a.c and d.c in one
direction only.
e(i) Consult your notebook
(ii) Initial charge, 𝑄0 = area under graph
<<ADVANCED LEVEL PHYSICS
Page 127
Area of one square = 50x10-3 = 0.5 C
Total number of squares under the graph = 25.5
Q0 = 25.5 x 0.5 12.75 C
Q
𝑄 = 𝐶𝑉 ⟹ C = V =
12.75
10
= 1.275 F
(iii) Time constant, 𝜏 = 𝑅𝐶 = 1.275 × 100 = 127.5𝑠
10. (a) (i) Wave speed is the distance travelled by wave energy per unit time.
(ii) Displacement of a wave is the distance travelled by wave energy in a particular direction.
(b) See June 2002 Q10
(c) 𝑥 =
λD
𝑑
.Fringe separation, 𝑥 =
12.5
10
= 1.25 𝑚𝑚 ⟹ λ =
xd
D
=
1.25×10−3 ×0.05×10−3
1.0
= 6.25 × 10−7 m
(d) (i)For electrons to be emitted from the surface of a metal, the frequency of the incident radiation must
be greater than the threshold frequency of the metal. If electrons are emitted with uv light it means the the
frequency of uv is greater than the threshold frequency of the metal, whereas for visible light, the
frequency is less than the threshold frequency. That is why electrons are not emitted with visible light.
(ii) The work function for zinc is greater than that of potassium. Hence the maximum kinetic energy of
electrons ejected from the surface of zinc will be greater than that for those ejected from the surface of
potassium.
(e)(iii) The number of electrons ejected per second is proportional to the intensity of the incident
radiation. A single can only eject a single electron. This happens when the frequency of the incident
radiation is greater than the threshold frequency of the metal. This implies that an increase in frequency
will result to an increase in the maximum kinetic energy of the ejected electrons. Increase in intensity
only result to an increase in the number of electrons ejected per second from the metal surface. We
observe that electrons are only ejected when the incident frequency is greater than the threshold
frequency, kinetic energy is frequency dependent while electrons ejected per second is intensity
dependent. Hence doubling the intensity doubles the number of electrons ejected per second
(iv) From the Einstein photoelectric equation, 𝐸𝑚𝑎𝑥 = ℎ𝑓 − ℎ𝑓0 = ℎ(𝑓 − 𝑓0 ). Since h is constant, it
therefore implies that 𝐸𝑚𝑎𝑥 ∝ (𝑓 − 𝑓0 )
(v) Gamma photons have a shorter wavelength than infrared; hence they carry much energy than infrared
radiation. Due to the shorter wavelength, they have a high penetration power than infrared and as a result
cause more destruction.
(iv) ∆𝐸 =
ℎ𝑐
λ
hc
6.67×10−34 ×3.0×108
⟹ λ = ∆E = (−1.5—13.6)1.6×10−19 = 1.0 × 10−7 m = 100 nm
<<ADVANCED LEVEL PHYSICS
Page 128
JUNE 2008
1.(i) Homogeneous means units or dimensions of all the terms in the equation are the same.
4
(ii) 𝜂𝜋𝑎𝜐 = 3 𝜋𝑎3 (𝜌 − 𝜌′)𝑔
Units of LHS = units of( 𝜂𝑎𝜐) = 𝑘𝑔𝑚−1 𝑠 −1 . 𝑚. 𝑚𝑠 −1 = 𝑘𝑔𝑚𝑠 −2
Units of RHS = units of (𝑎3 (𝜌 − 𝜌′)𝑔) = 𝑚3 𝑘𝑔𝑚−3 𝑚𝑠 −2 = 𝑘𝑔𝑚𝑠 −2
Since the base units on both sides are the same, the equation is homogeneous.
1
1
1
𝑣
𝑣
𝑣
25.5−22.0
2.𝑓 = 𝑣 + 𝑢 ⟹ 𝑓 = 1 + 𝑢 ⟹ 𝑓 = 1 + 𝑚 ⟹ 𝑣 = 𝑚𝑓 + 𝑓. Slope of graph = 𝑓 ⟹ 𝑓 = 1.50−1.20 = 11.67𝑐𝑚
b) See June 2001 Q4
3.(i) When 𝑄 =
𝑄0
2
, 𝑡 = 𝑇1⁄2 ⟹
𝑄
(ii) When 𝑄 = 2𝑒0 , 𝑡 = 𝜏 ⟹
𝑄0
2
𝑄0
𝑒
= 𝑄0 𝑒 −
𝑇1⁄2
𝑅𝐶
⟹ 𝑇1⁄2 = 𝑅𝐶𝑙𝑛2
𝜏
= 𝑄0 𝑒 −𝑅𝐶 ⟹ 𝜏 = 𝑅𝐶
(iii) 𝑇1⁄2 = 𝜏𝑙𝑛2
4.Simple harmonic motion is the motion in which the acceleration is directly proportional to the
displacement from a fixed point.
𝑦 = 𝑟𝑠𝑖𝑛𝜔𝑡 Where
r is the amplitude (maximum displacement from equilibrium position)
𝜔 is the pulsation (also called angular frequency)
𝑡 is time interval during which the motion takes place.
1
5.(a) 𝑑𝑠𝑖𝑛𝜃 = 𝑚𝜆 ⟹ 𝑁 𝑠𝑖𝑛𝜃 = 𝑚𝜆 ⟹ 𝜆 =
𝑠𝑖𝑛𝜃
𝑚𝑁
𝑠𝑖𝑛36°
= 2×(5000×102 ) = 5.88 × 10−7 𝑚
(b) For m = 3, 𝑠𝑖𝑛𝜃 = 3 × 5.88 × 10−7 × 5000 × 102 = 0.881 < 0. Hence a order image is possible
(c) From𝑑𝑠𝑖𝑛𝜃 = 𝑚𝜆 ⟹ 𝑚 =
𝑑𝑠𝑖𝑛𝜃
𝜆
1
⟹ 𝑚 ∝ 𝜆. Thus to increase the number of orders, the wavelength of
the light should be reduced. The number of orders can also be increased by increasing the spacing of the
grating.
6.See June 2001 Q5
7.(i)
Information
source
Input
transmitter
Modulator
Transmitter
Carrier producer
(ii) See June 2004 Q8
8.(a) See June 2001 Q8
(b) (i) 𝐼 =
𝑄
𝑡
=
600×10−3
1.0×10−6
= 6.0 × 105 𝐴
(iii) 𝐵 =
(ii)
𝜇0 𝐼
2𝜋𝑟
=
2𝜋×10−7 ×6.0×105
2𝜋×0.1
= 0.6 𝑇
(d) See June 2001 Q8
<<ADVANCED LEVEL PHYSICS
Page 130
(f)
g(r)
𝑔(𝑟) ∝ 𝑟
1
𝑔(𝑟) ∝ 𝑟2
r
R
9.(a) (i) The path (trajectory) is a parabola
u = 5 ms-1
x
1
vx
r
vy
(ii) Displacement
x = ut = 5 × 0.5 = 2.5 m
v
y
1
y = 2 gt 2 = 2 × 9.8 × 0.52 = 1.23m
Thus displacement, r = (2.5î + 1.23ĵ)m
vx = u = 5ms −1 rand vy = gt = 9.8 × 0.5 = 4.9ms−1
⟹v
⃗ = (5î + 4.9ĵ)ms−1
(iii) Vertical component of acceleration is constant while horizontal component is zero. For the velocity,
the vertical component increases with time while the horizontal component is constant
(b)
6cos60
6N 6N
6cos60
W = 6cos60 + 6cos60 = 12 N
(c) A collision is a situation in which bodies exert relatively strong forces on each other in a relatively
short time interval.
An elastic collision is one in which kinetic energy is conserved while an inelastic collision is one in
which kinetic energy is not conserved
(d) (i) The principle of conservation of energy states that energy can neither be created nor destroyed but
can only be transformed from one form to another.
From the first law of thermodynamics, ∆Q = ∆U + ∆W
Where, ∆Q is the energy supplied to the system
∆U is the change in the internal energy of the system
∆W is the work done on or by the system
(ii) A ball falling under gravity, the loss in gravitational potential energy is equal to the gain in kinetic
energy.
In a battery, when connected to a load, chemical energy is converted to electrical energy.
(e) Renewable energy sources are sources be continually be replenished naturally and make use of
processes that are part of our natural environment. They will thus not get exhausted as they are exploited
for energy production. They are also called infinite energy sources. Examples include: biomass, sun, wind,
flowing streams, etc. While non – renewable sources are those which get exhausted as they are being
<<ADVANCED LEVEL PHYSICS
Page 131
used.
They are also called finite energy sources. Examples include: fossil fuels like coal, oil, natural gas
etc.
(iia) mass of water, 𝑚 = 𝜌𝑉 = 𝜌𝐴ℎ
ℎ
ℎ
1
1
𝐺. 𝑃. 𝑒 = 𝑚𝑔 ( ) = 𝜌𝐴ℎ ( ) = 𝜌𝐴ℎ2 = × 1000 × 40 × 106 × 9.8 × 102 = 1.95 × 1013 𝐽
2
2
2
2
(iib) 𝑃 =
𝐺.𝑝.𝑒
𝑡
=
1.95×1013
6×60×60
= 9.02 × 108 𝑊
(f) When the heat from the sun heats up large air masses, it results to air movement (wind). The kinetic
energy of the wind can then be converted to electrical energy by the wind mills.
(ii) Living organisms (plants) manufacture their food by the process of photosynthesis using sunlight.
When they die and are buried, they decay to form fossil fuels after so many years e.g oil.
10. (a) Consult your textbooks
(b) (i) Radioactivity is the random and spontaneous decay (or disintegration) of unstable nuclei to form
more stable ones by emitting nuclear radiations.
An alpha particle is an excited helium nucleus that has lost two of its orbital electrons
234
4
(ii) 238
92U ⟶ 90Thn + 2α + energy
(c) (i)
𝑑𝑁
𝑑𝑡
= 𝑁𝜆 =
𝑁𝑙𝑛2
𝑇1⁄2
1010 𝑙𝑛2
= 20×24×60×60 = 4.0 × 103 𝐵𝑞
1
𝑁
(ii) 𝑁 = 𝑁0 𝑒 −𝜆𝑡 ⟹ 𝑡 = 𝜆 𝑙𝑛 ( 𝑁0 ) =
𝑇1⁄2
𝑙𝑛2
𝑁
𝑙𝑛 ( 𝑁0 ) =
20 𝑑𝑎𝑦𝑠
𝑙𝑛2
1010
𝑙𝑛 ( 104 ) = 398.6 𝑑𝑎𝑦𝑠 ≈ 399 𝑑𝑎𝑦𝑠
(d) (i) Temperature coefficient of resistance of a material is the fractional increase in the resistance of the
material at zero degree per degree rise in temperature.
(ii) See June 2001 Q10
(f)
6V
F
3Ω
I1
2
I2
C
I3
1
4V
2Ω
8Ω
A
<<ADVANCED LEVEL PHYSICS
KCL: 𝐼3 = 𝐼1 + 𝐼2 . . . . . . . . . . . . . .(1)
H KVL in loop (2), −3𝐼1 − 6 + 4 + 2𝐼2 = 0
⟹ −3𝐼1 + 2𝐼2 = 2 . . . . . . . . . . . . . . .(2)
D KVL in loop (1); −4 − 8𝐼3 − 2𝐼2 = 0
⟹ 𝐼2 + 4𝐼3 = −2 . . . . . . . . . . . . . . .(3)
Solving the above three equations,
𝐼1 = −0.609 𝐴; 𝐼2 = 0.087 𝐴, 𝐼3 = −0.522 𝐴
B
Page 132
JUNE 2009
1.See June 2001
2.See June 2005
3.
I1 R1=3 Ω
1 E =9 V
2
R2=3 Ω
I2
I3
KCL: 𝐼1 + 𝐼2 = 𝐼3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(1)
KVL loop1: 6 − 3𝐼1 − 9 + 3𝐼2 = 0 ⟹ 𝐼2 − 𝐼1 = 1 . . . . . .(2)
KVL loop 2: 3𝐼1 − 6 + 3𝐼3 − 12 = 0 ⟹ 𝐼1 + 𝐼3 = 6 . . . . .(3)
Solving the above equations simultaneously, gives:
𝐼1 = 1.67 𝐴; 𝐼2 = 2.67 𝐴; 𝐼3 = 4.33 𝐴
E1=6 V
E3=12 V
2
R3=3 Ω
Figure 1
4.
T
R
30
α
Taking moment about the point A,
(𝑇𝑠𝑖𝑛30°)(2𝐿) − 𝑊𝐿 = 0
0
𝑊
A
𝑚𝑔
⟹ 𝑇 = 2𝑠𝑖𝑛30° = 2×0.5 = 𝑚𝑔 = 25 × 9.8 =245 N
B
Question: Determine the reaction R
W
5.See June 2003Q9
6.(a)
Forward bias
P
Reverse bias
n
P
n
(b) The leakage current originates from the minority charge carriers which are thermally generated. As
temperature increase, the size of the leakage current increases also because more electron holes pairs will
be
generated.
7.(a)
y
u= 60ms-1
100
0.5m
Tennis court
vx
θ
vy
x
1
1
𝑦 = 𝑢𝑦 𝑡 − 𝑔𝑡 2 = 60𝑡𝑠𝑖𝑛10° − × 9.8𝑡 2
2
2
When the strikes the ground y=-0.5 m
⟹ −0.5 = 60𝑡𝑠𝑖𝑛10° − 4.9𝑡 2
⟹ 4.9𝑡 − 60𝑠𝑖𝑛10° − 0.5 = 0
Solving t =2.173 s or t = -0.047 s
But t > 0⟹t = 2.173 s
𝑣𝑥 = 𝑢𝑐𝑜𝑠10° = 60𝑐𝑜𝑠10° = 59.088 𝑚𝑠 −1
𝑣𝑦 = 𝑢𝑠𝑖𝑛10° − 𝑔𝑡
⟹ 𝑣𝑦 = 60𝑠𝑖𝑛10° − 9.8 × 2.173 = −10.88𝑚𝑠 −1
Thus 𝑣 = (59.088𝑖̂ − 10.88𝑗̂)𝑚𝑠 −1
v
<<ADVANCED LEVEL PHYSICS
Page 134
𝑣𝑦
10.88
OR 𝑣 = √59.0882 + (−10.88)2 = 60.1𝑚𝑠 −1 at angle 𝜃 = tan−1 (𝑣 ) = tan−1 (59.088) = 9.69°
𝑥
(b)
Vy/ms-1
Vx/ms-1
t/s
t/s
8.(a) See June 2008 Q4
(b) Consult your textbooks or notebooks
(c) (i)
By Newton’s second law, ∑ ⃗Fext = ma⃗
⟹ −kx = ma ⟹ a = −(mk )x
⟹ a ∝ x. Since the acceleration is
proportional to the displacement from a
fixed point, the motion is simple harmonic
k
1
1
1
(ii) ω2 = m ⟹ 2 mω2 (r 2 − x 2 ) = 2 k(r 2 − x 2 ) = 2 (10)(0.0082 − 0.0032 ) = 2.75 × 10−4 J
(iii) Assumption: Mass of spring is negligible
(d) Specific heat capacity is the heat required to raise the temperature of a unit mass of a substance by 1 K.
(e) Consult your notebooks
f) Energy supplied by engine = work done to raise block + heat gained by glycerin
mg
h
⟹ Pt = mb gh + mg c∆θ ⟹ P = mb g ( t ) + (
m
⟹ P = mb gv + ( tg)c∆θ ⟹ ∆θ =
P−mb gv
mg
( t )c
=
h
) c∆θ, but v = and
t
t
1.0×105 −(800)(9.8)(6.7)
0.5×2.5×103
𝑚𝑔
𝑡
=mass of glycerine per second.
= 37.98℃ .
But∆𝜃 = 𝜃 − 23 ⟹ θ = 61.0℃
(g) Electrical energy lost = thermal energy gained
𝑃𝑡 = 𝑚𝑐 𝑐𝐶 ∆𝜃 + 𝑚𝑤 𝑐𝑤 ∆𝜃 + 𝑚𝑖 𝑙𝑓 + 𝑚𝑖 𝑐𝑤 ∆𝜃 ⟹ 𝑚𝑖 =
1000×60−(80×100+0.15×4200)(33−0)
3.4×105 +4200×(33−0)
𝑃𝑡−(𝑚𝑐 𝑐𝑐 +𝑚𝑤 𝑐𝑤 )∆𝜃
𝑙𝑓 +𝑐𝑤 ∆𝜃
= 0.18 𝑘𝑔 , which is more than the initial mass of ice
Assumption: the heat supplied goes to melt the ice.
9.(a) See June 2006
b) (i) 𝜌 =
𝑅𝐴
𝑙
ρ
ρ
⟹ R = (A) l ⟹ slope = A ⟹ ρ = A × slope
(80−0)10−3
⟹ ρ = π(0.5 × 10−3 ) × (4.3−0)10−5 = 1.46 × 103 Ωm
1
1
(ii) 𝜎 = 𝜌 = 1.46×103 = 6.84 × 10−4 Ω−1 𝑚−1
If the experiment was carried out at 300C, the conductivity will reduce since resistivity increases with an
increase in temperature.
(iii) 𝑙 =
𝑅𝐴
𝜌
=
56×10−3 ×𝜋(0.5×10−3 )
1.46×103
2
= 3.01 × 10−11 𝑚
<<ADVANCED LEVEL PHYSICS
Page 135
(c) (i)
10Ω
0.01 A
10
0.01x10=R(9.99) ⟹ R=0.01Ω
10 A
R
9.99 A
10Ω
0.01 A
R
10
(R+10)(0.01)=10⟹R=990Ω
V
(di) Emf of driver cell must be greater than the emf of the test cell
Positive terminal of the driver cell must be connected to positive terminal of the test cell
(ii) If the emf of the driver cell is greater than that of the test cell, the current will be positive
If the emf of the driver cell is equal to that of the test cell, the current will be zero
(iii) Circuit diagram has a problem (see diagram below)
6.0 V
l
X
R
Y
Short circuit
m
A
Figure 4
(e) 𝐹 = 𝐵𝐼𝑙𝑆𝑠𝑖𝑛𝜃
(i) 𝐹 = 2.5 × 3.0 × 15 × 10−3 𝑠𝑖𝑛90° = 0.113𝑁 (ii) = 2.5 × 3.0 × 15 × 10−3 𝑠𝑖𝑛0° = 0 𝑁
(iii) = 2.5 × 3.0 × 15 × 10−3 𝑠𝑖𝑛30° = 0.056 𝑁
10. (a) 𝐹 =
(b) 𝑔 =
𝐺𝑀𝑚
𝑟3
𝐺𝑀
𝑅2
𝐹𝑟 2
⟹ 𝐺 = 𝑀𝑚 ⟹ [𝐺] =
=
𝐺𝜌𝑉
𝑅2
=
𝐺𝜌(43𝜋𝑅3 )
𝑅2
[𝐹][𝑟]2
[𝑚]2
=
𝑀𝐿𝑇 −2 𝐿2
𝑀2
= 𝑀−1 𝐿3 𝑇 −2
4
= 3 𝐺𝜌𝜋𝑅
(c) Centripetal force equal gravitational attraction
𝐺𝑀𝑚
2𝜋 2
𝐺𝑀
(𝑅+ℎ)3
⟹ 𝑚𝜔2 (𝑅 + ℎ) = (𝑅+ℎ)2 ⟹ ( 𝑇 ) = (𝑅+ℎ)3 ⟹ 𝑇 = 2𝜋√
⟹ 𝑇 = 2𝜋√
(36000×103 +6400×103 )3
9.8(6400×103 )
𝐺𝑀
(𝑅+ℎ)3
= 2𝜋√
𝐺𝑅 2
= 8.658 × 104 𝑠 = 24.05 ℎ𝑟𝑠 ≈ 24 ℎ𝑟𝑠
<<ADVANCED LEVEL PHYSICS
Page 136
Comment: The period of rotation of the satellite is same as that of the earth. Hence it will always remain
in one position above the earth’s surface as the earth rotates. i.e the satellite is in a geostationary orbit.
(c) See June 2005
(d) See June 2003
(ei) 212 = nucleon number (proton + neutron)
208
4
(ii) 212
84𝑃𝑜 → 2𝛼 + 82𝑃𝑏 + ∆𝐸
∆𝐸 = 211.9890𝑈 − (4.0026 + 207.9767)𝑈 = 0.0097 𝑈
⟹ ∆𝐸 = 0.097 × 931 × 106 × 1.6 × 10−19 = 1.44 × 10−12 𝐽
<<ADVANCED LEVEL PHYSICS
Page 137
JUNE 2010
1.𝐹 =
𝜇0 𝐼1 𝐼2
2𝜋𝑟
Units of LHS =𝑁𝑚−1 = 𝑘𝑔𝑚𝑠 −2 𝑚−1 = 𝑘𝑔𝑠 −2
Units of RHS=
(𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝜇0 )(𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝐼)2
𝜇0 𝐼
From 𝐵 = 2𝜋𝑟 ⟹ 𝜇0 =
⟹ 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑅𝐻𝑆 =
2𝜋𝑟
2𝜋𝑟𝐵
⟹ 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝜇0 =
𝐼
𝑘𝑔𝑚𝑆 −2 𝐴−2 .𝐴2
𝑚
𝑚.𝑁𝐴−1 𝑚−1
𝐴
=
𝑚.𝑘𝑔𝑚𝑠 −2 𝐴−1 𝑚−1
𝐴
= 𝑘𝑔𝑚𝑠 −2 𝐴−2
= 𝑘𝑔𝑠 −2
Since base units on both sides are the same, the equation is homogeneous
2.(a) (i) The depletion layer is the region between the n – and p – type materials that is fairly free from
majority charge carriers. Immediately the junction is formed, electrons diffuses into the p –type while the
holes diffuses into the n – type. When the diffusion of electrons stop at saturation, a depletion layer is set
up.
3.(i) Ground state energy is the lowest available energy level
(ii) 𝐸𝑖𝑜𝑛𝑖𝑠𝑎𝑡𝑖𝑜𝑛 = 𝐸∞ − 𝐸0 = 0 − −13.6𝑒𝑉 = 13.6 𝑒𝑉
6.63×1034 ×3.0×108
ℎ𝑐
(iii) 𝜆1 = Δ𝐸 = (−3.4−−13.6)×1.6×10−9 = 1.2 × 10−7 𝑚
ℎ𝑐
6.63×10−34 ×3×108
𝜆2 = ∆𝐸 = (−1.5—13.6)×1.6×10−19 = 1.09 × 10−7 𝑚
1
4.(a) Work done = area under force extension curve ⟹ 𝑊 = 2 × 5.2 × 4 × 10−3 = 1.04 × 10−2 𝐽
𝑠𝑡𝑟𝑒𝑠𝑠
(b) From 𝐸 = 𝑠𝑡𝑟𝑎𝑖𝑛 =
𝑙×𝑠𝑙𝑜𝑝𝑒
𝐹 ⁄𝐴
⟹
𝑒 ⁄𝑙
5.2−0
𝐸𝐴
𝐹 = ( 𝑙 ) 𝑒 ⟹ 𝑠𝑙𝑜𝑝𝑒 =
𝐸𝐴
𝑙
1
= 2.0 (4×10−3 −0) × 0.01 = 2.6 × 109 𝑁𝑚−1
𝐴
234
238
234
4
5.(ai) 238
92𝑈 → 90𝑇ℎ𝑛 + 2𝛼 . Hence 92𝑈 changes to 90𝑇ℎ𝑛
238
0
(ii) 238
92𝑈 → 92𝑁𝑝 + −1𝛽 + ∆𝐸
141
92
1
1
(iii) 238
92𝑈 + 0𝑛 → 56𝐵𝑎 + 56𝐾𝑟 + 3( 0𝑛) + ∆𝐸
⟹𝐸=
(b) ∆𝐸 = (235.04𝑈 + 1.0𝑈) − (140.997𝑈 + 91.91𝑈 + 3(1.01𝑈)) = 0.2000𝑈
⟹ ∆𝐸 = 0.200 × 931𝑀𝑒𝑉 = 0.200 × 931 × 106 × 1.6 × 10−19 = 2.99 × 10−11 𝐽
235.04 × 10−3 𝑘𝑔 contains 6.02 × 1023 𝑎𝑡𝑜𝑚𝑠
5
∴ 5𝑘𝑔 will contain 235.04×10−3 × 6.0 × 1023 = 1.28 × 1025 𝑎𝑡𝑜𝑚𝑠
⟹ 𝐸𝑛𝑒𝑟𝑔𝑦 𝑟𝑒𝑙𝑒𝑎𝑠𝑒𝑑 𝑏𝑦 5𝑔 = 1.28 × 1025 × 2.99 × 10−11 = 3.83 × 1014 𝑘𝑔
6.𝑑𝑠𝑖𝑛𝜃 = 𝑚𝜆 ⟹ 𝑠𝑖𝑛𝜃 =
𝑚𝜆
𝑑
= 𝑁𝑚𝜆
𝑠𝑖𝑛𝜃1 = 550 × 103 × 2 × 559 × 10−9 = 0.61490 ⟹ 𝜃1 = sin−1(0.61490) = 37.9°
𝑠𝑖𝑛𝜃2 = 550 × 103 × 2 × 563 × 10−9 = 0.61930 ⟹ 𝜃1 = sin−1(0.61930) = 38.2°
⟹ 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 = 𝜃2 − 𝜃1 38.2° − 37.9° = 0.3°
7.
(a)
Pulling force
(b)
Friction
Tension force
Pulling force
Friction
T
Friction
T
Friction
<<ADVANCED LEVEL PHYSICS
Pulling force
Page 138
By Newton’s third law, the force exerted on the string by the vehicle (tension) is the same action while
the force exerted by the vehicle being towed on the on the string is the reaction.
8.(ai) Capacitance is the ability of a capacitor to store charges or it is the charge stored by a capacitor when
the potential difference across its terminals is 1 V.
(ii) Consult your textbooks
(bi) Initial kinetic energy = work done to take the object to infinity
1
⟹ 2 𝑚𝑣𝑒2 = (0 − −
𝐺𝑀
𝑅
𝐺𝑀
) 𝑚 ⟹ 𝑣𝑒 = √ 𝑅 . The mass of the moon or the gravitation field strength on the
surface of the moon are not given.
(ii) 𝐹𝑔 =
𝐺𝑚𝑝 𝑚𝑒
𝑟2
𝑒2
1
=
6.7×10−11 ×9.1×10−31 ×1.67×10−31
(10−10 )2
𝐹𝑒 = 4𝜋𝜀 . 𝑟 2 = 9 × 109 ×
0
𝐹𝑒
𝐹𝑔
1.6×10−19
(10−10 )2
= 1.018 × 10−47 𝑁
= 2.3 × 10−8 𝑁
2.3×10−8
= 1.08×10−47 = 2.26 × 1039 ⟹ 𝐹𝑒 = 2.26 × 1039 𝐹𝑔 . Thus when elementary particles are considered,
the force of gravity is very weak.
(ci) Surface tension in terms of force is defined as the force per unit length acting in the surface of a
liquid and perpendicular to one side of an imaginary line drawn in the surface.
If AB is moved to a distance δx to A’B’, then work has to
be done against this force
Provided AB is moved isothermally to A’B’, the force on
AB will be constant and therefore
Work done, δW=force, F x distance, δx
Therefore, δW=2𝛾Lδx
Increase in surface area = 2Lδx
δx
L
A
A’
B
B’
2𝛾𝐿
Liquid film
Movable wire
Rigid frame Work done per unit area = 2𝜎 =
𝛿𝑊
𝛿𝐴
=
2𝛾Lδx
2Lδx
=𝛾
Thus the surface tension is the work done in isothermally increasing the surface area of a liquid by unit area
(ii) Consult your textbooks.
(d) Assuming no heat lost to the surroundings,
Heat lost by steam =heat gained by ice to melt +heat gained to warm the melted ice.
ms lv + ms cw (100 − 5) = mi lf + mi cw (5 − 0) ⟹ ms = mi (
lf +5cw
lv +95cw
)
3.34×105 +5×4200
⟹ ms = 1.8 × 103 (2.26×106+95×4200) = 240.3 kg
9.(ai) Modulus of elasticity the ratio of stress to strain within the elastic limit, while elastic limit is the
maximum force beyond which a material will no regain its original shape or size.
(ii) Strength relates to the maximum force which can be applied on material without it breaking, while
stiffness relates to the resistance which a material offers to have its size and/or shape changed.
(bi) 𝑈𝑚𝑖𝑛 = −7.2 𝑒𝑉 = −7.2 × 1.6 × 10−19 = 1.15 × 10−18 𝐽
(ii) When temperature increases, the separation of the atoms increases i.e separation becomes greater than
the equilibrium separation and the slope of the graph is positive.
<<ADVANCED LEVEL PHYSICS
Page 139
U
r0
Separation greater than r0
gradient of graph is
positive
𝑑𝑈
(iii) Slope of graph is given by 𝐹 = − 𝑑𝑟
𝑠𝑙𝑜𝑝𝑒 𝑎𝑡 11 𝑛𝑚 =
(7.5−4.0)×1.6×10−19
(0.5−1)10−10
𝑠𝑙𝑜𝑝𝑒 𝑎𝑡 38 𝑛𝑚 =
𝑠𝑙𝑜𝑝𝑒 𝑎𝑡 67 𝑛𝑚 =
= −7.0 × 10−9 𝑁
(−4.0—7.5)×1.6×10−19
(5.0−3.25)10−10
(−2.0—3.5)×1.6×10−19
(5.0−3.0)×10−10
r/10-10 m
F/10-9 N
1.1
7.0
3.8
-3.2
6.7
-1.2
= 3.2 × 10−9 𝑁
= 1.2 × 10−9 𝑁
Energy needed to separate atoms
completely = area under graph
Area of one square =
1.0 × 10−10 × 1.0 × 10−9 = 1.0 × 10−19 𝐽
Approximate number of squares =10.5
Energy = 10.5 × 1.0 × 10−19 𝐽 = 1.05 ×
10−18 𝐽
(c) See June 2008
(di) See June 2008
(ii) Assuming that the graph shows the variation of the p.d across the cell, then
90−37
𝑉 = −𝑟𝐼 + 𝐸 , then the slope=-r ⟹ 𝑟 = − ( 0−10 ) = 5.3 Ω, and E=90.0 V
2×2
(e) Total resistance, 𝑅𝑇 = 21Ω + 2Ω + 2+2 = 24Ω, total emf, E = 2V+2V = 4 V
Let the current in the circuit be I, then by Kirchhoff’s voltage law,
<<ADVANCED LEVEL PHYSICS
Page 140
𝑑𝐼
𝐸 − 𝐼𝑅𝑇 − 𝐿 𝑑𝑡 = 0
𝑑𝐼
𝐸
4
When the current is steady, 𝑑𝑡 = 0 ⟹ 𝐼 = 𝑅 = 24 = 0.17𝐴
𝑇
VL = 0.17 x 21 = 3.5 V
10. (a) (i) See June 2007 Q10
(bi) A plane polarized wave is one in which all the vibrations are in a single plane containing the
direction of propagation of the wave.
(iii) The angle of refraction for the transmitted beam is when the reflected beam and refracted beam
are perpendicular.
i
Completely polarized beam
𝑖 + 𝑟 = 90° ⟹ 𝑟 = 90° − 𝑖
i
⟹ 𝑟 = 90° − 56° = 34°
Partially polarized beam
r
(ci) For an interference pattern to be observed, the light from the two slits must be coherent, i.e having the
same wavelength, frequency and should maintain a constant phase difference.
(ii) The source slit S diffracts the light that falls on it and as a result, illuminates S1 and S2. Diffraction
takes
place again at S1 and S2 and where the light from slits S1 and S2 overlap, an interference pattern
occurs there.
Since the light that emerge from S1 and S2 come from the main source, S, then S1 and S2
now become coherent sources.
(iii) 0.1 𝑚𝑚 ≤ 𝑑 ≤ 1.0 𝑚𝑚, 0.5 ≤ 𝐷 ≤ 2.0𝑚
(iv) If white light is used, the central bright fringe will be white and the fringes on either side are colored.
Blue is the color near the central bright and red is furthest away.
(di) See June 2008
(ii) See June 2005
(e) 𝐹 = 𝑚𝑎 = 𝑚 (
𝑣−𝑢
𝑡
0−27.5
) = 1000 ( 0.075 ) = −3.7 × 105 𝑁
From the equation of motion 𝑣 2 = 𝑢2 + 2𝑎𝑠 ⟹ 𝑎 =
𝑣 2 −𝑢2
2𝑠
=
02 −27.52
2×0.25
5
= −1540 𝑚𝑠 −2
By Newton’s second law, 𝐹 = 𝑚𝑎 = 1000(1540) = 1.54 × 10 𝑁
1
2𝑦
2×0.6
(f i) 𝑦 = 𝑔𝑡 2 ⟹ 𝑡 = √ = √
2
𝑔
9.8
𝑥
0.25
𝑡
0.35
= 0.35 𝑠 ⟹ 𝑣𝑥 = =
= 0.71 𝑚𝑠 −1
(ii) By the law of conservation of momentum, 𝑚𝑝𝑒𝑙𝑙𝑒𝑡 𝑣𝑝𝑒𝑙𝑙𝑒𝑡 = (𝑚𝑝𝑒𝑙𝑙𝑒𝑡 + 𝑚𝑝𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑛𝑒 )𝑣
⟹ 𝑣𝑝𝑒𝑙𝑙𝑒𝑡 =
(iv)
(𝑚𝑝𝑒𝑙𝑙𝑒𝑡 +𝑚𝑝𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑛𝑒 )𝑣
𝑚𝑝𝑒𝑙𝑙𝑒𝑡
=
(0.01+0.05)0.71
0.01
= 4.26 𝑚𝑠 −1
1
By the principle of conservation of energy, 𝑚𝑔ℎ = 2 𝑚𝑣 2
⟹ 𝑣 = √2𝑔ℎ = √2 × 9.8 × 0.6 = 3.43 𝑚𝑠 −1
3.43
⟹ 𝑣 = (0.71𝑖̂ + 3.43𝑗̂)𝑚𝑠 −1 OR |𝑣| = √0.712 + 3.432 = 3.5 𝑚𝑠 −1 at tan−1 0.71 = 78.2° below the
horizontal
<<ADVANCED LEVEL PHYSICS
Page 141
JUNE 2011
1.(a) For an equation to be homogeneous, all the terms in equation should have the same base units or
dimensions. If a dimensionless constant which was suppose to be in that equation is absent or if a
dimensionless constant which was not suppose to be in that equation is present, that equation will be
homogeneous be physically wrong.
(b) See previous examples.
2.
6V
1
X
10 Ω
2
9V
I2
KVL:𝐼1 + 𝐼2 − 𝐼3 = 0 - - - - - - - - - - - - - - - - - - -- - - -(1)
KVL loop 1: 6 − 10𝐼1 + 15𝐼2 = 0
⟹ 10𝐼1 − 15𝐼2 = 0 - - - - - - - - - - - - - - - - - - - - - - - -(2)
15Ω
KVL loop 2: 10𝐼1 − 9 + 5𝐼3 = 0
Y
⟹ 10𝐼1 + 5𝐼3 = 9 - - - - - - - - - - - - - - - - -- - - - - - -- (3)
I1
Solving the above equations, we have
5Ω
𝐼1 = 𝐼3 = 0.6 𝐴 𝑎𝑛𝑑 𝐼2 = 0 𝐴
I3
𝑉𝑋𝑌 = 10𝐼1 = 10 × 0.6 = 6 𝑉
3.(i) See June 2008 Q9
(ii) By conservation of energy, loss in p.e = k.e just before the ball hits the ground
⟹ 𝑚𝑔ℎ1 = 12𝑚𝑣12 ⟹ 𝑣1 = √2𝑔ℎ1
When the ball renounces back to a height h2, then by conservation of energy,
1
2
𝑚𝑣22 = 𝑚𝑔ℎ2 ⟹ 𝑣2 = √2𝑔ℎ2
Change in momentum, ∆𝑃 = 𝑚𝑣2 − 𝑚𝑣1 = √2𝑔ℎ2 − √2𝑔ℎ1
4.(i) 𝑉0 =
20.0×4.0
(ii) 𝑓 =
2
𝑛
𝑡
= 40 𝑉 ⟹ 𝑉𝑟𝑚𝑠 =
1
𝑉
√2 0
=
40
√2
= 28.3 𝑉
4
= 10−2 = 400 𝐻𝑧
5.(i) At the point Q, 30𝑐𝑜𝑠60° = 𝑚𝑔 ⟹ 𝑚 =
30×0.5
9.8
= 1.53 𝑘𝑔
(ii)
6.(a) See June 2001 Q4
(bi) 𝑛𝑐𝑜𝑟𝑒 = 𝑣
3.0×108
𝑐
𝑐𝑜𝑟𝑒
= 1.95×108 = 1.54
(ii) 𝑛𝑐𝑜𝑟𝑒 𝑠𝑖𝑛𝜃𝑖 = 𝑛𝑐𝑙𝑎𝑑𝑑𝑖𝑛𝑔 𝑠𝑖𝑛𝜃𝑟
The minimum angle of incidence = critical angle and occurs when 𝜃𝑟 = 90°
⟹ 𝑛𝑐𝑜𝑟𝑒 𝑠𝑖𝑛𝐶 = 𝑛𝑐𝑙𝑎𝑑𝑑𝑖𝑛𝑔 𝑠𝑖𝑛90° ⟹ 𝑛𝑐𝑙𝑎𝑑𝑖𝑖𝑛𝑔 = 𝑛𝑐𝑜𝑟𝑒 𝑠𝑖𝑛80° = 1.52
7.(a) The specific heat capacity at constant pressure is greater than that at constant volume because at
constant pressure, the energy supplied goes to increase the internal energy and also do work against the
pressure, while at constant volume, the energy supplied only goes to increase the internal energy.
(b) The change in volume for gases is large and varies from gas to gas. From ∆𝑊 = 𝑃∆𝑉, it means that
the work done is also large. For solids and liquids, the change in volume is very very small
(approximately zero). Hence the work done either at constant pressure or at constant volume is
negligible.
8.(a) See June 2010
<<ADVANCED LEVEL PHYSICS
Page 142
(bi) ∆𝑃 =
1
4𝛾
𝑟
=
1
1
1
2
∆ℎ.𝜌.𝑔
𝑟
1
4×3.5×10−2
4𝛾
⟹ ∆ℎ = 𝜌𝑔𝑟 = 103 ×9.8×0.08 = 1.8 × 10−4 𝑚
1
8
(ii) 𝑟 = 𝑟 − 𝑟 = 2 − 8 ⟹ 𝑟 = 3 = 2.7𝑐𝑚
8.0×10−20
(ci) 𝐹 =
𝑟2
8.0×10−20
⟹
𝑟02
−
−
2.0×10−96
𝑟 10
2.0×10−96
𝑟010
at equilibrium separation, F = 0 and r=r0
8
= 0 ⟹ 𝑟0 = √2.5 × 10−77 = 2.6 × 10−10 𝑚
(ii) See June 2010 Q9
(d) See June 2010 Q8
(e)
Resolving
Vertically, F𝑐𝑜𝑠10° = 𝑚𝑔 - - - - - -- - - --(1)
Horizontally, 𝐹𝑠𝑖𝑛10° = 𝐹𝑒 - - - - - - -- - - (2)
10°
(1)𝑎𝑛𝑑 (2) ⟹
F
Fe
𝐹𝑒
𝑚𝑔
= 𝑡𝑎𝑛10° ⟹ 𝐹𝑒 = 𝑚𝑔𝑡𝑎𝑛10°
𝐸𝑞 = 𝑚𝑔𝑡𝑎𝑛10° ⟹ 𝑞 =
⟹𝑞=
5×10−4 ×9.8×𝑡𝑎𝑛10°
4.0×10−2
𝑚𝑔𝑡𝑎𝑛10°
𝐸
= 2.2 × 10−6 = 2.2𝜇𝐶
mg
(fi)
E
0
E
For a solid metallic sphere we
have
r0
For a hollow metallic sphere
E=0
0
r
r0
r
N.B: For a hollow metallic conductor, the net field inside the conductor is zero. This is as a result of the
superposition of field inside the conductor, giving rise to a zero field inside. For r>r0, the spherical hollow
conductor behaves like a point charge, producing a radial field.
9.(ai) See June 2001 Q10
1
𝐴
(ii) 𝐴 = 𝐴0 𝑒 −𝜆𝑡 ⟹ 𝑡 = 𝜆 𝑙𝑛 ( 𝐴0 ) =
𝑇1⁄2
𝑙𝑛2
𝐴
𝑙𝑛 ( 𝐴0 ) =
5.6×103
𝑙𝑛2
16
𝑙𝑛 (13) = 1677.5 𝑦𝑟𝑠
0.7
(bi) mass of uranium 100 × 1𝑘𝑔 = 0.007 𝑘𝑔
0.007
⟹Number of nuclei = 235×10−3 × 6.02 × 1023 = 1.8 × 1022 𝑛𝑢𝑐𝑙𝑒𝑖
(ii) Each nucleus releases 200 MeV =200x106x1.6x10-19 = 3.2 x 10-11 J of energy
There 1.8 x 1022 nuclei will release 1.8 x 1022 x 3.2 x 10-11 = 5.72 x 1011 J
At SONEL, one unit = 1KWh = 1x1000x60x60J = 3.6 x 106 J
Therefore number of units consumed =
5.72×1011
3.6×106
= 1.59 × 105 𝑢𝑛𝑖𝑡𝑠
Cost of consumption 1.59 x 105 x 60 = 9546667 Frs
(c)
Core of reactor
Heat exchangers
<<ADVANCED LEVEL PHYSICS
Turbine
Generator
Page 143
(i) Coolant: Carries away heat generated in the reactor core
(ii) The moderator: To slow down the neutrons so as to enable cause fission on colliding with uranium
nuclei
(iii) Control rods: To absorb some of the neutrons so that the reaction does not go out of control
(di) See June 2008
(ii)
Energy
K.E
P.E
Time
(ei) 𝑣 2 = 𝜔2 (𝑟 2 − 𝑥 2 ) at equilibrium position, x = 0⟹ 𝑣 = 𝜔𝑟 =
𝑔
2𝜋𝑟
𝑇
𝑙
but 𝑇 = 2𝜋√𝑔
9.8
⟹ 𝑣 = 𝑟√ 𝑙 = 0.2√1.2×10−2 = 5.7 𝑚𝑠 −1
1
1
(ii) 𝐾. 𝐸𝑚𝑎𝑥 = 2 𝑚𝑣 2 = 2 × 0.2 × 10−3 × 5.722 = 3.27 × 10−3 𝐽
f(i) Free oscillation: This occurs when a body is set into motion allowed to oscillate on its own. In this
type of oscillation, the total energy is constant.
Damped oscillation : this is a type of oscillation in which the amplitude progressively decreases with
time. The decrease in amplitude is as a result of energy loss.
Forced oscillation: this is a type of oscillation in which a body is set in motion as a result of an external
vibrating body
1
𝑇
(ii) 𝑓0 = 2𝑙 √𝜇but 𝜇 =
𝑚
𝑙
=
𝜌𝑉
𝑙
=
𝜌𝐴𝑙
𝑇
𝑙
= 𝜌𝐴 ⟹ 𝑇 =
𝑠𝑡𝑟𝑒𝑠𝑠
⟹ 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝐴 = 4𝑙 2 𝑓02 𝜌 ⟹ 𝐸 = 𝑠𝑡𝑟𝑎𝑖𝑛 =
4𝑙2 𝑓02 𝜌
𝑠𝑡𝑟𝑎𝑖𝑛
4𝑙 2 𝑓02 𝜌𝐴
⟹𝐸=
4×22 ×2002 ×8.0×103
0.02
= 5.56 × 1011 𝑃𝑎
10. (a) I0 is the reverse saturation current also called leakage current
(bi) 𝐼 = 𝐼0 𝑒 𝐵𝑉 ⟹ 𝑙𝑛𝐼 = 𝐵𝑉 + 𝑙𝑛𝐼0 .Thus a plot of lnI against V gives a straight line with slope = B and
intercept on lnI = 𝑙𝑛𝐼0
V/mV
I/10-4A Ln(I/A)
255
0.004
-14.73
315
0.016
-13.35
345
0.036
-12.53
385
0.089
-11.63
410
0.182
-10.91
455
0.552
-9.8
475
0.903
-9.31
495
1.4
-8.87
<<ADVANCED LEVEL PHYSICS
Page 144
505
1.82
-8.61
515
2.23
-8.41
530
3.1
-8.08
∆𝑉
(ii) 𝑠𝑙𝑜𝑝𝑒 = ∆𝑙𝑛𝐼 =24.5 V-1 (do the calculations)
Intercept =-21.00 (do the calculations) ⟹ 𝑙𝑛𝐼0 = −21.00 ⟹ 𝐼0 = 𝑒 −21.00 = 7.58 × 10−10 𝐴
(c) The physical constant, B depends on temperature
<<ADVANCED LEVEL PHYSICS
Page 145
JUNE 2012
1.(i) The changing current in the primary induces a changing magnetic field in the secondary, which by
Faraday’s law of electromagnetic, the changing magnetic field in the secondary will induce an emf.
(ii) If the input in the primary is d.c, the output in the primary will be zero.
(iii) Iron is a soft magnetic material which ensures that all the flux from the primary are linked to the
secondary. Such a core reduces energy losses due to magnetic reversals, since iron can easily be
magnetized and demagnetized.
ℎ𝑐
hc
0
Φ
2.(i) Φ = 𝜆 ⟹ 𝜆0 =
=
6.63×10−34 ×3.0×108
3.8×1.6×10−19
= 3.27 × 10−7 𝑚
(ii) Ultraviolet
(iii) 𝑒𝑉𝑠 = ℎ𝑓 − Φ .Thus the slope of the graph gives the plank constant. For the set up, see your notes
𝐵𝐴𝑁
𝐵𝐴𝑁
𝑁𝑚−1 𝐴−1 .𝑚2
3. 𝜃⁄𝑉 = 𝐶𝑅 ⟹ 𝐶 = 𝜃 ⟹ 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝐶 = 𝑉𝐴−1 𝑟𝑎𝑑𝑉 −1 = 𝑘𝑔𝑚2 𝑠 −2 𝑟𝑎𝑑 −1
𝑅( )
𝑅
4.(𝑎) Energy supplied= area of hysteresis loop
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 = 5.5
Area of one square = 20 x 10-3 x 0.2 x = 4.0 x 10-3 J
Total energy = 5.5 x 4 x 10-3 = 2.2 x 10-2 J
(b) This is because rubber has a long chain coiled molecules
5. I = nevA - - - - - - - - -- - - -- - - - - -- - - - - - (1)
𝜌=
𝑅𝐴
𝑙
; 𝑎𝑛𝑑 𝑅 =
𝑉
𝐼
⟹𝐼=
𝑉𝐴
From (1) and (2) ⟹ 𝑛𝑒𝑣𝐴 =
𝑃
- - - - -- - - -(2)
𝜌𝑙
𝑉𝐴
𝜌𝑙
𝑉
1.5
⟹ 𝑣 = 𝜌𝑙𝑛𝑒 = 1.7×108 ×0.5×1029 ×1.6×10−19 = 1.1 × 10−2 𝑚𝑠 −1
1.0×106
6.(a) 𝐼 = 𝑉 = 1.0×104 = 100 𝐴
Power lost along transmission line due to resistance of cables = 𝐼 2 𝑅 = 1002 × 0.5 = 5000 𝑊
Power output = 106 W + 5000W = 1.005 x 106 W
(b) By stepping up the voltage, the current is reduced. So energy lost due to resistance of cables is
minimized
7.(a) Resultant speed = 10 cms-1 + 50 cms-1 = 60 cms-1
50 cms-1
10 cms-1
(b)
10 cms-1
50 cms-1
𝑣 = √502 + 102 = 50.99 ≅ 51.0𝑐𝑚𝑠 −1
𝐼2 𝑅
8.(ai) See June 2004
(ii) See notes
(bi)
C
Critical angle
<<ADVANCED LEVEL PHYSICS
i>C
i>C
Total internal reflection
Page 146
c
c
A
B
v
A
(ii) nA sinC = nB sin90° ⟹ v sinC = v ⟹ vB = sinC
=
𝑣+𝑢0
(c) Approaching frequency, 𝑓𝑎 = (
𝑣−𝑢0
Receding frequency, 𝑓𝑟 = (
365
𝑣
350
315
𝑣
340+25
) 𝑓𝑠 = (
340−25
) 𝑓𝑠 = (
340
1.9568×108
) 𝑓𝑠 =
340
315
sin87.7°
365
𝑓
340 𝑠
= 1.9585 × 108 ms−1
) fs = 340 fs
350
𝑓𝑎 − 𝑓𝑟 = (340 − 340) 𝑓𝑠 = 340 𝑓𝑠 ⟹ 340 𝑓𝑠 = 60 ⟹ fs = 408 Hz
(di) See June 2003 Q8
(ii) See notes
(ei) 𝑄 = 𝑚𝑐∆𝜃 = 𝜌𝑉𝑐∆𝜃 = 1000 × 1.2 × 4200 × (98 − 65) = 1.66 × 108 𝐽
𝑄
(ii) 𝑡 = 𝑃 =
1.66×108
1.5×103
= 1.13 × 105 𝑠
(iii) By Newton’s law of cooling, rate of heat loss is proportional to the excess temperature (temperature
change). In the morning, the excess temperature is high, which implies rate of heat loss is also high
(f) Consult your notes
9.(a) (i) (ii) See June 2010
(iii)
𝐶
𝐶0
=
(bi) 𝑅 =
𝜀
𝜀0
𝑉0
𝐼0
= 𝐶 ⟹ C = C0 εr
9
= 45×10−6 = 2.0 × 105 Ω
(ii) Charge stored = area under graph = 18.5 × 100 × 5 × 10−6 = 9.25 × 10−3
1
1
(iii) Energy stored 2 𝑄𝑉 = 2 × 9.25 × 10−3 × 9 = 4.16 × 10−2
𝑄
(iv) 𝐶 = 𝑉 =
9.25×10−3
9
= 1.03 × 10−3 𝐽
(v) If R is increased, the capacitor will take a longer time to discharge.
(c) Since both of the charged from a 12 V supply, the common p.d across their terminals when connected
together is 12 V.
⟹ charge stored on the 5μF = 12 × 5 μC = 60 μC
(e) (i)
IB/µA
20
40
60
80
IC/mA
2.3
4.3
6.2
8.3
<<ADVANCED LEVEL PHYSICS
Page 147
9
I C /mA
y = 0.0995x + 0.3
8
7
6
5
4
3
2
1
0
0
10
20
30
40
50
60
70
80
90
IB /µA
ℎ𝑓𝑒 = 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑔𝑟𝑎𝑝ℎ = 99.0 (Do the calculations yourself)
C1 stops the d.c component in the input from entering the circuit
C2 ensures that only a.c appears at the output.
(iii) VCC − IC R 3 − VCE = 0 ⟹ R 3 =
VCC −VCE
IC
9−3
= 1.0×10−3 = 6.0 × 103 Ω
R 2 and R1 form a potential divider with VCC ⟹ VR2 = R
R2
1 +R2
VCC
Also, VR4 + VBe − VR2 = 0 ⟹ VR2 = VR4 + VBe = 10−3 × 103 + 0.6 = 1.6 V
⟹ 1.6 = (R
R2
1 +R2
) 9 ⟹ 1.6R1 + 1.6R 2 = 9R 2 ⟹ R 2 =
1.6R1
7.4
=
1.6×74000
7.4
= 16 𝑘Ω
10. (a) Emf is the work done per unit charge to drive current through the whole circuit (i.e including the
external and internal resistance), while terminal p.d is the work done per unit charge to drive current
through a load only.
The p.d across the terminals of a battery is not always equal to the emf because when a battery is
connected to a load such that it supplies current, the chemical energy inside the cell is gradually been
converted to electrical energy. Hence the terminal p.d of the battery gradually reduces as the battery is
being used up.
From 𝐸 = 𝑟𝐼 + 𝑉 ⟹ E > 𝑉
(bi) 𝑃 =
𝑉2
𝑅
=
2202
40
= 1210 𝑊
(ii) 𝐸 − 𝑉 − 𝐼𝑟 = 0 ⟹ r =
E−V
I
𝑉
. But 𝐼 = 𝑅 =
220
40
= 5.5 𝐴 ⟹ r =
240−220
5.5
= 3.6 Ω
(iii) When the cooker is in use, the voltmeter measures only the p.d across the cooker. Since current is
flowing in the conducting wires, it means there exist a p.d drop across the wires which is not taken into
account by the voltmeter. Hence the measured p.d across the cooker is less than the supply p.d.
(iii) Much energy is lost along the transmission since the electricity is transmitted at low p.d.
(di)
<<ADVANCED LEVEL PHYSICS
Page 148
 Most of the alpha particles went through the thin gold foil undeviated. Thus, most of the atom is
empty and all its mass is concentrated in the nucleus.
 Some of the particles were deflected through small angles. Thus the atom has a core which is
positively charged
 A small number of the alpha particles were deflected large angles. Thus, the tiny core or the
nucleus is massive or dense
(ei) Mass defect, ∆𝑚 = 3.7533 × 10−25 − (3.686 + 0.066) × 10−25 = 1 × 10−28 𝑘𝑔
Energy released, ∆𝐸 = ∆𝑚𝑐 2 = 1.0 × 10−28 × (3.0 × 108 )2 = 9.0 × 10−12 𝐽
4
(ii) Energy of gamma radiation = 100 × 9.0 × 10−12 𝐽 = 3.6 × 10−13
ℎ𝑐
𝜆
hc
= 3.6 × 10−13 ⟹ λ = 3.6×10−13 =
6.63×10−34 ×3.0×108
3.6×10−13
= 5.5 × 10−13 m
(iii) 96 % is converted to kinetic energy of the daughter nuclei
(f) At the distance of closest approach, all the kinetic energy of the alpha particle will be converted to
potential energy.
𝐾. 𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑃. 𝐸𝑓𝑖𝑛𝑎𝑙 ⟹ K. E =
(Qα )(QGold )
4πεr
⟹r=
(Qα )(QGold )
4πεK.E
=
(2e)(79e)
4πεK.E
158e2
= 4πεK.E
2
158×(1.6×10−19 )
⟹ r = 4π×8.85×10−12 ×1.34×10−12 = 2.7 × 10−14 m
<<ADVANCED LEVEL PHYSICS
Page 149
JUNE 2013
1.(i) See Previous years
(ii) αT and αT2 have no units
(iii) Unit of α = K-1, while units of 𝛽 = 𝐾 −2
2.(a) The moment of a force is defined as the product of force and perpendicular distance from the line of
action of the force.
⃗ and the sum
⃗ =0
Since the system is in equilibrium, ∑ F
(b) Coplanar means the forces acting in
of moment about any point should also give zero
the same plane.
N2
Resolving:
N1
R
2sinα
µN1
O 2cosα
α
W
𝜇𝑁1 = 𝑁2 - - - - - - - - - - (1)
𝑁1 = 𝑊 - - - - - - - - - - - --(1)
Taking moment about O;
2𝑊𝑐𝑜𝑠𝛼 + 4𝑁2 𝑠𝑖𝑛𝛼 − 4𝑁1 𝑐𝑜𝑠𝛼 = 0
⟹ 2𝑊𝑐𝑜𝑠𝛼 + 4𝜇𝑊𝑠𝑖𝑛𝛼 − 4𝑊𝑐𝑜𝑠𝛼 = 0
1
1
⟹ 𝑡𝑎𝑛𝛼 =
⟹ 𝛼 = tan−1 (
) = 57.4°
2𝜇
2 × 0.32
3. For the convex lens, f>0, and from the thin lens equation we have
1
𝑓
1
1
1
1
1
𝑣
30
= 𝑣 + 𝑢 ⟹ 𝑣 = 20 − 12 ⟹ 𝑣 = −30 𝑐𝑚; 𝑚 = 𝑢 = 12 = 2.5
Thus the image is magnified, virtual and upright
For the concave lens, f<0;
1
𝑓
1
1
1
1
1
= 𝑣 + 𝑢 ⟹ 𝑣 = −20 − 12 ⟹ 𝑣 = −7.5 𝑐𝑚; 𝑚 =
7.5
12
= 0.63
Thus the image is virtual, diminished and upright
The images formed in each case can be represented as shown.
(b) See June 2003
4. (a)
1500 µF
Concave lens
1000 µF
1500
µF lens
Convex
1000 V
V
Initial charge on the 1500 µF, 𝑄𝑖 = 1500 × 10−6 × 1000 = 0.15 𝐶
By conservation of charge, after the connection
.015
𝑄𝑖 = (1500 + 100) × 10−6 𝑉 ⟹ 𝑉 = 2500×10−6 = 60 𝑉
1
1
(b) Initial energy stored, 𝑊𝑖 = 2 𝐶𝑉 2 = 2 × 15 × 10−6 × 10002 = 7.5 𝐽
1
1
Final energy stored, 𝑊𝑓 = 2 𝐶𝑉 2 = 2 (2500 × 10−6 ) × 602 = 4.5 𝐽
<<ADVANCED LEVEL PHYSICS
Page 150
(c) Energy is lost in the conducting wire due to it resistance
I3
I1
A
A
I1
5. (i)
6V
6V
2Ω
6V
3Ω
6Ω
4Ω
6V
2Ω
I2
B
2 2.4Ω
1 3Ω
I2
Figure 1
KCL: I1 = I2 + I3 - - - - - - - - - - - (1)
𝐾𝑉𝐿; 𝑙𝑜𝑜𝑝 1: 3𝐼2 − 6 + 6 − 2𝐼3 = 0 ⟹ 3𝐼2 − 2𝐼3 = 0 - - - - - - - - - -(2)
𝐾𝑉𝐿; 𝑙𝑜𝑜𝑝 2: −2.4𝐼1 + 6 − 3𝐼2 = 0 ⟹ 3𝐼2 + 2.4𝐼1 = 6 - - - - - - - - - (3)
Solving the above equations simultaneously, 𝐼1 = 1.67𝐴, 𝐼2 = 0.67 𝐴, 𝐼3 = 1.01 𝐴
(ii) 𝑉𝐴𝐵 = 2.4𝐼1 = 2.4 × 1.67 = 4.0 𝑉 or ⟹ 𝑉𝐴𝐵 = 6 − 3𝐼2 = 6 − 3 × 0.67 = 4.0 𝑉
6. (i) A p – type semiconductor is an impure semiconductor in which the majority charge carriers are holes.
It is produced by doping a pure semiconductor with a trivalent element.
(ii)VCC − IB R1 − VBE = 0 ⟹ 𝐼𝐵 =
R1
R2
VR1
9.0 V
IC
IB
VCE
VBE
VCC
⟹ 𝐼𝐵 =
9.0−0.62
60×10−3
VCC −VBE
R1
= 1.4 × 10−4 𝐴
𝐼𝐶 = 𝛽𝐼𝐵 = 100 × 1.4 × 10−4 = 1.4 × 10−2 𝐴
(iii) VCC − IC R 2 − VCE = 0 ⟹ VCE = VCC − IC R 2 − VCE
⟹ VCE = 9.0 − 1.4 × 10−2 × 600 = 0.62 V
7. See June 2003 Q9
8. (a) (i) It is a physical quantity that changes linearly or uniformly with temperature change
(ii) Accurate, sensitive and reproducible
(iii) The values would be slightly different. This is because the variation of temperature with pressure is
not the same as the variation of temperature with resistance.
(b) (i) Consult your notebooks
(ci) Heat lost by metal block = heat gained by melting ice + heat required to warm the melted ice
𝑚𝑏 𝑐𝑏 (𝜃 − 48) = 𝑚𝑖 𝑙𝑓 + 𝑚𝑖 𝑐𝑤 (48 − 0) ⟹ 𝜃 =
⟹𝜃=
1.2×3.34×105 +48×1.2×4200
0.8×455
𝑚𝑖 𝑙𝑓 +48𝑚𝑖 𝑐𝑤
𝑚𝑏 𝑐𝑏
= 1765.7℃
(ii) At the same temperature, all electrical insulators are good thermal insulators. This is because in both
cases, conduction is by vibration of atoms.
(di) Faraday’s law of electromagnetic induction states that the magnitude of the induced current in a
circuit id directly proportional to the rate of flux linkage with time.
Lenz’s law states that the direction of the induced emf in a circuit is in such a way as to oppose the
effect causing it.
(ii)
𝑒
𝑒
⟹ 𝐹𝐵 = 𝐹𝐶 ⟹ 𝑚𝜔2 𝑟 = 𝐵𝑒𝜔𝑟 ⟹ 𝜔 = 𝐵 ( ) ⟹ 2𝜋𝑓 = 𝐵 ( )
𝑚
𝑚
<<ADVANCED LEVEL PHYSICS 𝐵(𝑚𝑒 ) 2.0×10−3 ×1.67×1011
Page 151
⟹ 𝑓 = 2𝜋 =
= 5.3 × 107 𝑟𝑒𝑣𝑠 −1
2𝜋
(e) See Your notes
I
𝐸−𝐿
E
V
Applying Kirchhoff voltage law,
L
X
𝑑𝐼
𝑑𝑡
𝑑𝐼
𝑑𝑡
Y
X
Y is fully lighted, the current will
become steady and
R
Y
− 𝐼𝑅𝑌 = 0 When the lamp
𝐼=
= 0. Hence
t
𝐸
𝑅𝑌
(e) See your notes
9. (a) The stopping potential is the least negative potential to stop all the electrons emitted from the cathode
in a photoelectric cell
(bi) From the Einstein photoelectric equation, we have 𝐸𝑚𝑎𝑥 = ℎ𝑓 − ℎ𝑓0 = ℎ(ℎ − 𝑓0 ) . Since𝐸𝑚𝑎𝑥 ≥ 0,
it therefore implies that𝑓 > 𝑓0 . Thus there exists a minimum frequency needed for the ejection of
electrons
2(ℎ𝑓−ℎ𝑓0 )
(ii) 𝐸𝑚𝑎𝑥 = ℎ𝑓 − ℎ𝑓0 ⟹ 𝑣𝑚𝑎𝑥 = √
, from here we observe that the maximum speed is only
𝑚
frequency dependent not intensity dependent.
(ci) ℎ𝑓 = ℎ𝑓 − ℎ𝑓0 ⟹ 𝑓 =
𝐸𝑚𝑎𝑥
ℎ
𝐸𝑚𝑎𝑥
+ 𝑓0 ⟹
ℎ
ℎ𝑐
1
ℎ𝑐
+ 𝜆 , slope = ℎ and intercept on the f – axis = 𝜆
0
0
𝑐
3.0×108
From the graph, intercept = 3.5 × 1014 𝐻𝑧 ⟹ 𝑓0 = 3.5 × 1014 𝐻𝑧 ⟹ 𝜆0 = 𝑓 = 3.5×1014 = 8.6 × 10−7 𝑚
0
Thus the maximum wavelength, 𝜆0 = 8.6 × 10−7 𝑚
(10.2−3.5)×1014
𝑠𝑙𝑜𝑝𝑒 = (2.5−0)×1.6×10−19 = 1.675 × 1033 𝐽−1 𝑠 −1
1
1
⟹ ℎ = 𝑠𝑙𝑜𝑝𝑒 = 1.675×1033 𝐽−1 𝑠−1 = 6.0 × 10−34 𝐽𝑠
𝑊𝑜𝑟𝑘 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛, Φ = ℎ𝑓0 = 6.0 × 10−34 × 3.5 × 1014 = 2.1 × 10−19 𝐽 = 1.31 𝑒𝑉
(d)(i) By the de Broglie’s hypothesis, 𝑃 =
(ii) 𝐸 = ℎ𝑓 =
(iii) 𝑚 =
𝑃
𝑐
=
ℎ𝑐
ℎ
𝜆
=
6.63×10−34
3.0×10−10
8
= 2.2 × 10−24 𝑘𝑔 𝑚𝑠 −1
= 𝑃𝑐 = 2.2 × 10−24 × 3.0 × 10 = 6.6 × 10−16 𝐽 = 4143.75 𝑒𝑉
𝜆
2.2×10−24
3.0×108
= 7.33 × 10−31 𝑘𝑔
(e)(i) Time constant is the time taken for the charge on a discharging capacitor or the p.d across the
capacitor to decrease to 𝑒 −1 of it initial value. OR it is the time taken for the charge on a charging
capacitor or the p.d across the capacitor to increase to (1-e-1) of the value that it wil have when fully
charged.
(ii) 𝑇1⁄2 = 𝑅𝐶𝑙𝑛2 = 𝜏𝑙𝑛2 ⟹ 𝜏 =
𝑇1⁄2
𝑙𝑛2
=
60
𝑙𝑛
= 86.6𝑠
The value of the time constant could be increased by increasing the capacitance and using a resistor of
higher resistance.
<<ADVANCED LEVEL PHYSICS
Page 152
(f) (i) The motion is simple harmonic because the variation of the displacement with time is sinusoidal.
1
1
(ii) Amplitude = 0.4 0 m, 𝑓 = 𝑇 = 40 = 0.025 𝐻𝑧
(iii) Generally, 𝑦 = 𝑎𝑠𝑖𝑛(𝜔𝑡 + 𝜑), where 𝜔 is the pulsation and 𝜑 is the phase.
The initial conditions (as seen on the graph) are that: at t = 0, y =0.4 m
⟹ 0.4 = 0.4𝑠𝑖𝑛(0 + 𝜑) ⟹ 𝑠𝑖𝑛𝜑 = 1 ⟹ 𝜑 = 90° ⟹ 𝑦 = 0.4𝑠𝑖𝑛(𝜔𝑡 + 90°) = 0.4cosωt, since the
sine and cosine function are 90° out of phase
𝜔 = 2𝜋𝑓 = 2𝜋 × 0.025 = 0.05𝜋 ⟹ 𝑦 = 0.4𝑐𝑜𝑠(0.05𝜋𝑡)
(g)(i) 𝑣 =
𝑑𝑦
𝑑𝑡
=
𝑑(0.4𝑐𝑜𝑠(0.05𝜋𝑡))
𝑑𝑡
= −0.02𝜋𝑠𝑖𝑛(0.05𝜋𝑡)
v/ms-1
0.02𝜋
80 t/s
40
10. (a)
F /10-7N
-0.02𝜋
10
8
6
4
2
r/10 -11 m
0
-2
-4
-6
-8
-10
0
2
4
6
8
10
12
14
16
18
20
(b) (i) Equilibrium separation, r0 = 3.6 x 10-11 m
(ii) Energy needed to separate molecules, E = area between curve and separation axis
Number of squares between curve and separation axis = 18
<<ADVANCED LEVEL PHYSICS
Page 153
Area of one square 2 × 10−7 × 2 × 10−11 = 4 × 10−18 ⟹ 𝐸 = 18 × 4 × 10−18 = 7.2 × 10−17 𝐽
(iii) They energy calculated above is the binding energy for the pair of molecules.
(ii) At and near the equilibrium separation, the force separation is linear, meaning the displacement of the
molecules is directly proportional to the intermolecular, which is Hooke’s law.
(iii) To be checked latter.
<<ADVANCED LEVEL PHYSICS
Page 154
JUNE 2014
1
1. (i) 𝑓𝑟 = 2𝜋√𝐿𝐶
Units of LHS =s-1
1
Units of RHS = 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 (
√𝐿𝐶
𝑢𝑛𝑖𝑡𝑡𝑠 𝑜𝑓 𝑄
But units of C=
)
𝐴𝑠
𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑉
𝑑𝐼
𝑉
𝑉𝐿 = 𝐿 𝑑𝑡 ⟹ 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝐿 = 𝐴𝑠−1 =
Thus units of RHS =
𝐴𝑠
= 𝐽𝐴−1 𝑠−1 = 𝐾𝑔𝑚𝑠−2 𝑚𝐴−1 𝑠−1 = 𝑘𝑔−1 𝑚−2 𝐴2 𝑠 4
𝐾𝑔𝑚𝑠 −2 𝑚𝐴−1 𝑠 −1
𝐴𝑠−1
1
= 𝑘𝑔𝑚2 𝑠 −2 𝐴−2
1
√𝑘𝑔−1 𝑚−2 𝐴2 𝑠4 𝑘𝑔𝑚2 𝑠 −2 𝐴−2 .
= √𝑠2 = 𝑠 −1
Since units of RHS = units of LHS, the equation is homogeneous
1
1
(ii) From the give equation, 𝐿 = (2𝜋𝑓 )2 𝐶 = (2𝜋×104 )2 ×4.0×10−9 = 0.063 𝐻
𝑟
2.
X
𝐶𝑇 =
90 V
𝑉𝑋𝑌 =
Y
Z
5 µF
Figure 1
=
27
𝜇𝐹
12
−6
𝑄 = 𝐶𝑉 = 2.25 × 10
4 µF
3 µF
3×(4+5)
𝜇𝐹
3+(4+5)
𝑄
𝐶𝑋𝑌
=
2.025×10−4
3.0×10−6
= 2.25𝜇𝐹
× 90 = 2.025 × 10−4 𝐶
= 67.5 𝑉
𝑉𝑌𝑍 = 90 − 67.5 = 22.5𝑉
⟹ 𝑄4𝜇𝐹 = 4 × 10−6 × 22.5 = 9.0 × 10−5 𝑉
3. (a) This is because during vaporization, there is a large increase in volume; as a result the amount of work
done against the atmosphere is great. Whereas during fusion, the increase in volume is very small, so
little or no work is done against the surrounding atmosphere. The energy supplied during fusion only
goes in to break the bonds of the molecules to form a liquid. Thus the energy supplied per unit mass to
cause vaporization is greater than the energy supplied per unit mass to cause fusion.
(b) Thermal energy supplied by steam in condensing = thermal energy gained by ice in melting
𝑖. 𝑒 𝑚𝑠 𝑙𝑣 = 𝑚𝑖 𝑙𝑓 ⟹ 𝑚𝑠 =
𝑚𝑖 𝑙𝑓
𝑙𝑣
=
50𝑔×3.34×105
2.26×106
= 7.39𝑔
Thus final mass of water=mass of melted ice + mass of condensed steam + initial mass of water
⟹ 𝑚 = 50𝑔 + 7.39𝑔 + 210𝑔 = 267.39 𝑔
𝑇2𝑔
𝑙
4. (a) 𝑇 = 2𝜋√𝑔 ⟹ 𝑙 = 4𝜋2 =
1
1.02 ×1.6
4𝜋 2
= 4.1 × 10−2 = 4.1 𝑐𝑚
1
1
1
(b) 𝐸1 = 2 𝑚𝑣 2 = 2 𝑚(2𝜋𝑓𝑟1 )2 and 𝐸2 = 2 𝑚𝑣 2 = 2 𝑚(2𝜋𝑓𝑟2 )2
𝐸1
𝐸2
𝑟
2
= (𝑟1 ) ⟹
2
5𝐸2
𝐸2
𝑟
2
𝑟
= (𝑟1 ) ⟹ 𝑟1 = √5 ⟹ 𝑟1 : 𝑟2 = √5
2
2
5. (a) For an isothermal change, temperature is constant implying ∆𝑇 = 0, ∆𝑈 = 𝐶∆𝑇 ⟹ ∆𝑈 = 0
By the first law of thermodynamics, ∆𝑄 = ∆𝑈 + ∆𝑊 ⟹ ∆𝑊 = ∆𝑄 − ∆𝑈 = 200 𝐽 − 0 = 200𝐽
3𝐾𝑇
(b) 𝑐𝑟𝑚𝑠 = √
6. (i)
𝑚
3×1.38×10−23 ×6000
=√
6.6×10−27
= 6.13 × 103 𝑚/𝑠
Antenna
Output decoder
Inductor
(ii) See June 2003 Q10
Variable capacitor
7. (a) (i) Energy needed = area between graph and the separation axis (Try to do it yourself)
<<ADVANCED LEVEL PHYSICS
Page 155
(ii) Try to calculate the gradient. The gradient is force constant (𝐹 ∝ 𝑥 ⟹ 𝐹 = −𝑘𝑥)
8. (a) (i) The resistivity of a material is the resistance per unit length of a material of unit cross sectional
area.
(ii) Consult your notebook
(b) (i) 𝜌 =
𝑅𝐴
𝑙
,𝑅 =
𝑉
𝐼
=
𝑉2
𝑃
⟹𝜌=
𝑉2𝐴
𝑃𝑙
=
2402 ×𝜋(0.5×10−3 )
2000×5.0
2
= 4.52 × 10−6 Ω𝑚
(ii) Electrical energy consumed = 𝑃𝑡 = 2𝐾𝑊 × 6ℎ𝑟 × 30 = 360 𝐾𝑊ℎ
Electric bill = 360 x 60 =21600 frs
I/A
Copper
Filament bulb
Silicon
V/V
(d) (i) See june 2006 Q9
(ii) Consult your notebooks
c
(c) (i) 𝑤𝑜𝑟𝑘 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛, Φ = ℎ𝑓 − 𝐸𝑚𝑎𝑥 , but and 𝑒 𝐸𝑚𝑎𝑥 = 𝑒𝑉𝑠 ⟹ Φ = h λ − 𝑒𝑉𝑠
⟹Φ=
6.63×10−34 ×3.0×108
590×10−9
− 1.6 × 10−19 × 0.15 = 3.1 × 10−19 𝐽 = 1.9375𝑒𝑉
(ii) Work done against the most energetic electrons = gained in kinetic energy
⟹ W = eV = 1.6 × 10−19 × 0.15 = 2.4 × 10−20 = 0.15eV
2×2.4×10−20
1
2
(iii) 2 𝑚𝑣𝑚𝑎𝑥
= 2.4 × 10−20 ⟹ vmax = √ 9.31×10−31 = 2.27 × 105 m/s
(f)
Emax
Z
Y
X
f/Hz
9. (a) Consult previous years or see your notes
F1
F2
300 300
By Newton’s second law, ∑ 𝐹 = 𝑚𝑎
Resolving vertically; 𝐹1 𝑐𝑜𝑠30° + 𝐹2 𝑐𝑜𝑠30° − 𝑚𝑔 = 𝑚𝑎
⟹ 600𝑐𝑜𝑠30° + 600𝑐𝑜𝑠30° = 𝑚(𝑎 + 𝑔)
⟹ 1200𝑐𝑜𝑠30° = 𝑚(𝑎 + 𝑔)
1200𝑐𝑜𝑠30° 1200𝑐𝑜𝑠30°
⟹𝑚=
=
= 88.1 𝑘𝑔
𝑎+𝑔
2 + 9.8
mg
<<ADVANCED LEVEL PHYSICS
Page 156
(c) (i) Momentum of bullet before collision = mu =0.016 x 180 = 2.88 kgms-1
𝑚𝑏 𝑢𝑏
(ii) 𝑚𝑏 𝑢𝑏 = (𝑚𝑏 + 𝑚𝑏𝑙 )𝑣 ⟹ 𝑣 = (𝑚
2.88
𝑏 +𝑚𝑏𝑙 )
= 0.016+2 = 1.43 𝑚/𝑠
1
1
(iii) Kinetic energy of bullet before collision, 𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 2 𝑚𝑏 𝑢𝑏2 = 2 × 0.016 × 1802 = 2.59 × 103 𝐽
1
Kinetic energy of bullet and block after the collision 𝐸𝑓𝑖𝑛𝑎𝑙 = 2 (𝑚𝑏 + 𝑚𝑏𝑙 )𝑣 2
1
⟹ 𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 2 (0.016 + 2)1.432 = 2.1𝐽
Since there is a loss in kinetic energy due to the collision, the collision is inelastic.
𝑣2
1
(iv) By the law of conservation of mechanical energy, 2 (𝑚𝑏 + 𝑚𝑏𝑙 )𝑣 2 = (𝑚𝑏 + 𝑚𝑏𝑙 )𝑔ℎ ⟹ ℎ = 2𝑔
1.432
⟹ ℎ = 2×9.8 = 0.11 𝑚
(d) (i) See June 2001 and 2013
(ii)
Since the sphere is stationary, by Newton’s second law,
𝐹𝑒 − 𝑚𝑔 = 0 ⟹ 𝐸𝑞 = 𝑚𝑔 ⟹ 𝑞 =
+
Fe
5.00 mm
-
⟹𝑞=
180 V
mg
𝑚𝑔𝑑
𝑉
=
⟹ 𝑛𝑒 = 𝑞 ⟹
𝑚𝑔
𝐸
3.5×10−15 ×9.8×5×10−3
= 9.53 × 10−19 𝐶
180
𝑞
9.53×10−19
𝑛= =
= 5.95 ≅ 6
𝑒
1.6×10−19
(e)(i) this is because the spaceship and the astronaut have the same acceleration towards the earth’s
centre. As a result, the reaction of the floor of the astronaut on the spaceship is zero and the astronaut
feels weightless.
(ii)Required energy = increase in potential
𝐺𝑀
𝑖. 𝑒 𝑈 = 𝑚∆𝑉 = (− 𝑅+𝐻 − −
But 𝐺𝑀 = 𝑔𝑅 2 ⟹ 𝑈 =
𝑔𝑅𝑚
2
𝐺𝑀
𝑅
=
1
1
𝐺𝑀𝑚𝐻
) 𝑚 ⟹ 𝑈 = 𝐺𝑀𝑚 (𝑅 − 𝑅+𝐻) = 𝑅(𝑅+𝐻) =
9.8×6.4×106 ×6.0×106
2
𝐺𝑀𝑚𝑅
2𝑅 2
=
𝐺𝑀𝑚
2𝑅
’ since R = H
= 1.88 × 1014 𝐽
More energy is needed in practice because work has to be done against friction (air resistance) in
the earth’s atmosphere.
(iii) Centripetal force = gravitational attraction
𝑖. 𝑒 𝑚𝜔2 𝑟 =
𝐺𝑀𝑚
𝑟2
8𝑅 3
2𝜋 2
⟹ (𝑇) =
8𝑅
𝐺𝑀
𝑟3
𝑟3
⟹ 𝑇 = 2𝜋√𝐺𝑀, but r = 2R
8×6.4×106
⟹ 𝑇 = 2𝜋√ 𝐺𝑀 = 2𝜋√ 𝑔 = 2𝜋√
9.8
= 1.44 × 104 𝑠 = 4.00 ℎ𝑟𝑠
Since the period of rotation of the spaceship is different from that of the earth, the spaceship is not in a
geostationary orbit
10. (a) (i) 𝐴 = 𝑅0 𝐴𝑛 ⟹ 𝑙𝑛𝑅 = 𝑙𝑛𝑅0 + 𝑛𝑙𝑛𝐴 ⟹ 𝑙𝑛𝑅 = 𝑛𝑙𝑛𝐴 + 𝑙𝑛𝑅0
Thus a graph of lnR against lnA is a straight with slope =n and intercept on the lnR axis = 𝑙𝑛𝑅0
<<ADVANCED LEVEL PHYSICS
Page 157
30
40
-3250
50
ln(A)/10-1
60
-3260
-3270
-3280
ln(R/m)/10-2
-3290
-3300
-3310
(ii) From the graph, slope, 𝑛 = 0.21 (try to do it yourself)
Intercept, 𝑙𝑛𝑅0 = −3380 × 10−2 ⟹ 𝑅0 = 𝑒 −33.80 = 4.66 × 10−15 𝑚
(b)(i) Observing from the above equation, when A=1, the corresponding element is hydrogen and the
corresponding
radius R=R0 is the radius of a hydrogen atom.
(ii) R and and A are related by as follows 𝑅 = (4.66 × 10−15 𝑚)𝐴0.2
<<ADVANCED LEVEL PHYSICS
Page 158
JUNE 2015
1. (a) An equation is homogeneous if all the terms in the equation have the same base units or dimensions
(b) 𝑃 + ℎ𝜌𝑔 + 𝜌𝑣 2 = 𝐾
Since the equation is homogeneous, all the terms have the same units or dimension
Units of 𝐾 = 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝑁𝑚−2 = 𝑘𝑔𝑚−1 𝑠 −1
2. (a) 𝑓 =
𝑛
𝑡
150
= 15×60 = 05 𝑟𝑒𝑣𝑠 −1
(b) 𝐹 = 𝑚𝜔2 𝑟 = 𝑚(2𝜋𝑓)2 𝑟 = 0.1 × (2𝜋 × 0.5)2 × 0.4 = 0.39 𝑁
(c)
Distance/m
4.0
2.0
Time/s
3. (a) Heat lost by steam on condensing = heat absorbed by ice in melting.
𝑖. 𝑒
𝑚𝑠 𝑙𝑣 = 𝑚𝑖 𝑙𝑓 ⟹ 𝑚𝑠 =
𝑚𝑖 𝑙𝑓
𝑙𝑓
=
1.5×3.34×105
2.26×106
= 0.22 𝑘𝑔
ℎ𝑒𝑎𝑡 𝑙𝑜𝑠𝑡 𝑏𝑦 𝑠𝑡𝑒𝑎𝑚 𝑖𝑛 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝑓𝑟𝑜𝑚 100℃
ℎ𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑚𝑒𝑙𝑡𝑒𝑑 𝑖𝑐𝑒 𝑖𝑛 𝑤𝑎𝑟𝑚𝑖𝑛𝑔
(b) (
)= (
)
𝑓𝑟𝑜𝑚 0℃ 𝑡𝑜 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒, 𝜃
𝑡𝑜 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑡𝑒𝑚𝑝, 𝜃
100𝑚𝑠
⟹ 𝑚𝑠 𝑐𝑤 (100 − 𝜃) = 𝑚𝑖 𝑐𝑤 (𝜃 − 0) ⟹ (𝑚𝑠 + 𝑚𝑖 )𝜃 = 100𝑚𝑠 𝑐𝑤 ⟹ 𝜃 = 𝑚
𝑠 +𝑚𝑖
=
100×0.22
0.22+1.5
= 12.8℃
4. (a)
𝐹 = 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑡𝑟𝑖𝑛𝑔
30°
𝐹
𝐹
⃗⃗⃗
𝐹𝑒
⃗⃗⃗
𝐹𝑒
⃗⃗⃗
𝐹𝑒 = 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑠𝑡𝑎𝑡𝑖𝑐 𝑓𝑜𝑟𝑐𝑒
⃗⃗⃗ = 𝑤𝑒𝑖𝑔ℎ𝑡
𝑊
⃗⃗⃗
𝑊
⃗⃗⃗
𝑊
Free body diagram
(b) Since the bob is in equilibrium, by Newton’s second law, ∑ ⃗⃗⃗⃗⃗⃗⃗
𝐹𝑒𝑥𝑡 = ⃗0
Resolving vertically, 𝐹𝑐𝑜𝑠30° = 𝑚𝑔 - - - - - - - -- - -(1)
Resolving horizontally, 𝐹𝑠𝑖𝑛30° = 𝐹𝑒 - - - - - - - - - -(2)
𝐹
𝑒
From (1) and (2), 𝑡𝑎𝑛30° = 𝑚𝑔
⟹ 𝐹𝑒 = 𝑚𝑔𝑡𝑎𝑛30°; but 𝐹𝑒 = 𝐸𝑞 =
⟹𝑞=
𝑚𝑔𝑑𝑡𝑎𝑛30°
𝑉
=
0.05×9.8×0.05×𝑡𝑎𝑛30°
12.0
𝑉𝑞
𝑑
⟹
𝑉𝑞
𝑑
= 𝑚𝑔𝑡𝑎𝑛30°
= 1.7 × 10−3 𝑘𝑔
5.
<<ADVANCED LEVEL PHYSICS
Page 159
Q
M
2 kΩ
At balanced, 𝑉𝑄 = 𝑉𝑆 ⟹ 𝑉𝑄𝑆=0 , meaning
no current flows through the meter M1
(a) M1 reads 0 A
𝑉𝑃𝑆 = 𝑉𝑃𝑄
⟹ (5.0 × 10−4 × 12000) = 2000𝐼1
⟹ 𝐼1 = 3.0 × 10−3 𝐴
Thus M2 reads 3.0 × 10−3 𝐴
R1
I1
P
M
R
-4
I = 5.0 x 10 A
𝜀
12 kΩ
6 kΩ
M
S
R2
Figure 2
(b) 𝑉𝑄𝑅 = 𝑉𝑅𝑆 ⟹ 6000𝐼 = 𝐼1 𝑅1 ⟹ 𝑅1 =
6000𝐼
𝐼1
=
6000×5.0×10−4
3.0×10−3
= 1000Ω = 1 𝑘Ω
3.0
𝑅2 (𝐼 + 𝐼1 ) = 3.0 ⟹ 𝑅2 = 3.0×10−3 +5.0×10−4 = 8.6 × 103 Ω
By KVL, 𝜀 − 𝐼(12000 + 6000) − 3 = 0 ⟹ 𝜀 = 3 + (5.0 × 10−4 )(18000) = 12 𝑉
6. (a) (i) Coherent sources of light have the same frequency, wavelength and consequently maintain a
constant phase different.
(ii) Consult your notebook
(b)(i) This is due to Doppler effect of sound. As the source moves towards the observer, the wave fronts
are compressed leading to a decrease in wavelength and hence a gradual increase in the frequency of the
sound.
𝑣
340
(ii) 𝑓 ′ = (𝑣−𝑢 ) 𝑓𝑠 = (340−8) × 512 = 525.3𝐻𝑧
𝑠
1
1
1
1
1
1
𝑢𝑓
(c) From the thin lens equation, 𝑓 = 𝑣 + 𝑢 ⟹ 𝑣 = 𝑓 − 𝑢 ⟹ 𝑣 = 𝑢−𝑓
For a converging lens, 𝑓 > 0 ⟹ 𝑣 =
For the diverging lens, 𝑓 < 0 ⟹ 𝑣 =
20×10
= 20 𝑐𝑚, magnification, 𝑚 =
20−10
20×(−10)
20−−10
𝑣
𝑢
=
20
=1
20
𝑣
= −6.67𝑐𝑚, magnification, 𝑚 = 𝑢 =
6.67
20
= 0.33
From the above calculations, we can bring out the following differences
 The image produced by the converging lens is same as the object, while that produced by the
diverging lens id diminished.
 The image produced by the converging lens is real while that produced by the diverging lens is
virtual
 The image produced by the converging lens is inverted while that produced by the diverging lens is
upright
 The image produced by the converging lens is on the opposite side of the lens while that produced
by the diverging lens is on the same side as the object.
(d) (i) A material is elastic if it has the ability to return to its original shape or size after the removal of
the deforming force. The deforming must not exceed the elastic limit of the material.
(ii) Consult your notebooks
(e)(i) Beyond E, the elastic limit of the toy has been exceeded, so the applied force is longer proportional
to the extension
(ii) Maximum elastic potential energy = Area between force – extension curve and the extension axis
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑢𝑛𝑑𝑒𝑟 𝑔𝑟𝑎𝑝ℎ = 24.5 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 = 1 × 5 × 10−3 = 5 × 10−3 𝐽
Total energy, 𝑈 = 24.5 × 5 × 10−3 = 0.12𝐽
<<ADVANCED LEVEL PHYSICS
Page 160
(iii) When the load is removed, the elastic potential energy of the system is converted to the kinetic
energy of the toy.
(f)(i) The collision of the gas molecules with the walls of the container results to a change in momentum.
The rate of change of momentum of the gas molecules with time equals the force exerted on the walls of
container. The force exerted per unit area on the walls of the container is the pressure.
(ii) Introducing more of the same gas will result to an increase in the number of collisions per unit time,
leading to an increasing in the force exerted on the walls of the container, and hence the pressure.
7. (a) 𝑅 = 𝑅0 (1 + 𝛼𝜃) ⟹ 𝑅 = (𝑅0 𝛼)𝜃 + 𝑅0 thus a graph of R against 𝜃 will give a straight line with slope
equals to (𝑅0 𝛼) and intercept on the R axis equals to 𝑅0
(See graph below)
(b) From the graph, intercept, 𝑅0 = 319.0 𝛺 (Try to verify)
1.028
Gradient of graph, 𝑅0 𝛼 = 1.028 𝛺℃−1 ⟹ 𝛼 = 319.0 = 4.7 × 10−3 ℃−1 (Verify this)
(c) The material is a conductor because the temperature coefficient of resistance is positive.
450
R/Ω
440
430
420
410
y = 1.028x + 319.06
400
390
380
370
360
350
340
330
320
0
10
20
30
40
50
60
70
80
90
Temp./0C
8. (OPTION: Energy Resources and Environmental Physics)
(a) Advantages of nuclear fusion over nuclear fission
 The raw material for nuclear fusion is abundant while that for nuclear fission is scarce
<<ADVANCED LEVEL PHYSICS
Page 161

Waste products of nuclear fusion are not radioactive while those from nuclear fission are
radioactive
 Nuclear fusion produces greater energy yield per unit mass of fuel used than nuclear fission
(b) (i) Functional energy refers to energy in its final form
(ii)
SOLAR ENERGY: the use of solar panels or solar cells to convert solar energy to electrical energy
which can be used for lighting and running of engines.
BIOMASS: Biomass such as wood can be burnt and the energy obtained can be used for cooking
HYDROELECTRICITY: Electrical energy generated from the hydroelectric power plant can be
distributed to industries to power engines or to homes for lighting.
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝜀 =
(c)(i)
𝑃𝑖𝑛 =
𝜌𝐴𝑣 2
𝑃
2
=
6 2
1.2×𝜋( ) ×13.53
2
2
𝑝𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡,𝑃0
𝑝𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡,𝑃𝑖𝑛
= 4.17 × 103 𝑊
25
𝜀 = 𝑃 0 ⟹ 𝑃0 = 𝜀𝑃𝑖𝑛 = 100 × 4.17 × 104 𝑊 = 1.04 × 104 𝑊
𝑖𝑛
(ii) The efficiency of the system is less than 100% because part of the energy input is use overcome
friction at the hinge of the blades as it rotates.
(d) (i) Substances responsible for the depletion of the ozone layer include chlorofluorocarbons, a family
of chlorine containing gases.
(ii) The depletion of the ozone layer will result to ultraviolet radiations from outer space reaching the
earth which may result to skin burns and damages to the eyes.
9. OPTION: Communication
(a)(i) Carrier frequency is the frequency used to transport information using electromagnetic waves.
The channel in operation is defined by the carrier frequency.
Amplitude modulation refers to a situation in which the amplitude of the waves varies in response to the
amplitude of the information signal, with no effect on the frequency.
(ii) Minimum frequency = fc – fmax = 200KHz – 50 KHz = 191 KHz
Maximum frequency = fc – fmax = 200KHz + 50 KHz = 209 KHz
Signal bandwidth = maximum frequency – minimum frequency = 209 – 191 = 18 KHz
b(i) The capacitance of the varied capacitor is varied until the resonance occur i.e the natural frequency
of the of the L- C – R circuit matches with one of the incoming radio waves and a particular radio station
is selected.
(ii) This correspond to the capacitance at resonance
1
1
1
Resonant frequency, fr = 2π√LC ⟹ C = (2πf)2 L = (2π×200×103 )2 ×4.0×10−3 = 1.6 × 10−10 F
(iii) The function of the decoder is to remove (extract) the original information which has been
transmitted from the received signal.
(c)(i) Advantages of digital signal over analogue
- Digital signals are easier to process, store and retrieve
- Digital systems are cheaper than analogue systems. They are easier to be designed and
constructed using integrated circuits
- Digital signals can be encrypted to ensure privacy
- Digital transmission systems are less prone to interference or noise
- Digital signals are easily transmitted. In this case, the transmitter and the receiver need only to tell
the difference between on (1) and off (o).
<<ADVANCED LEVEL PHYSICS
Page 162
(ii) Simultaneous transmission of several telephone conversation can be done through multiplexing
(multiplexing is a method by which multiple analog or digital signals are combined into one signal
over a shared medium). We time – division multiplexing and frequency – division multiplexing.
In time – division multiplexing, each conversation is divided into bursts of digitally encoded
information. The bursts of information are then rooted alternately through the same channel, usually
of low band width.
In frequency – division multiplexing, the channel bandwidth of the fibre is divided into blocks eachof
about 4 kH, corresponding to the signal bandwidth for the telephone conversation. Different
telephone conversations are tooted through the different slices of the channel bandwidth.
10. a) (i) Thermionic emission is the process by which free electrons are liberated from the surface of a metal
when it absorbs thermal energy.
(ii)
Aspect considered
n-type semiconductor
p-type semiconductor
Method of production
Produced by doping a pure
semiconductor with a
pentavalent element such as
phosphorus
Produced by doping a pure
semiconductor with a
trivalent element such as
aluminium
Majority charge carriers
Majority charge carriers are
electrons having a negative
charge
Majority charge carriers are
holes having a positive
charge
I/A
I/A
Imax
(b)
Imax
Current in the C-Rcircuit
I/A
<<ADVANCED LEVEL PHYSICS
Current in the L-R circuit
I/A
Page 163
Current in C-R circuit
Current in L-R circuit
The current in the capacitor decreases
exponentially with time until it becomes
zero.
The current increases exponentially with
time, approaching a steady maximum
value of 18 mA.
As the capacitor is energized by the d.c
supply, the p.d between its plates
increases from zero to the emf of the
source, when the capacitor is fully
charged. As a result of this, the current
decreases exponentially with time.
As the switch is closed, a back emf is
induced in the inductor due to its self
inductance. The back emf opposes the
growth of current in the circuit so that the
current increases slowly to its final steady
value.
t
E RC
I e
R
R
t
E
I  (1  e L )
R
(c) (i) Transistor as a switch
+Vcc
IC

RL

IB

V
0V
V0
When the input voltage Vi is small, IB =0
and consequently, IC = o and the transistor
is off.
When the input voltage is large, IB is
maximum and hence IC and the transistor
become saturated (V0=0 V)
Thus the transistor can be made to function
between two states depending on the input
voltage.
(ii) The output of and OR is high if either one of the input is high or both inputs are high. An algebraic
expression for an OR gate is given by S = P+Q, where P and Q are the inputs and S is the output. The
truth table of an OR gate is shown below.
Input
output
P
Q
S
0
0
0
0
1
1
<<ADVANCED LEVEL PHYSICS
Page 164
1
0
1
1
1
1
11). (a) (i) Draw a simple diagram of the human eye showing the parts which enable it to form image
(cornea, lens etc)
Eye defect
How it manifests
Correction
Short – sightedness (Myopia)
Astigmatism
Long – sightedness
(hypermetropia)
Only nearby objects can be seen
clearly. Distant objects cannot
be brought to focus on the retina
The eye has two focal lengths in
two different planes, due to the
cornea being slightly barel
shaped rather than the true
spherical shape
Only distant objects are seen
clearly
Using a diverging lens placed in
front of the eye
Using cylindrical lens
Using a converging lens
b(i) X-rays are suitable in producing images of dense tissues like the bones, while ultrasound is suitable
in producing images of less dense tissues like the liver, heart and kidney.
(ii) Ultrasound are not likely to replace X-rays in medical diagnosis because they cannot be used to image
hard tissues like the bones and deep – lying tissues of the body.
(c) How the MRI produces an image
The human body is mostly water. The water molecules contain hydrogen nuclei (protons), which become
aligned in a magnetic field. An MRI applies a very strong magnetic field (about 0.2 T to 3 T), generated
by superconducting magnets which aligns the proton "spins." The scanner also produces a radio
frequency current that creates a varying magnetic field. The protons absorb the energy from the variable
field and flip their spins. When the field is turned off, the protons gradually return to their normal spin, a
process called precession. The return process produces a radio signal that can be measured by receivers in
the scanner and made into an image.
<<ADVANCED LEVEL PHYSICS
Page 165
JUNE 2016
1. (a) Homogeneity is a necessary but insufficient condition for the physical correctness of an equation
because in the presence or absence of a dimensionless constant in an equation, the equation will still be
homogeneous but physically wrong. For example the period of oscillation of a simple pendulum is given
ℓ
by the equation 𝑇 = 2𝜋√𝑔. In the absence of the constant2𝜋, the equation will still be homogeneous but
physically wrong.
𝑑𝐼
𝑑𝐼
(b) 𝐸 = 𝐿 𝑑𝑡 ⟹ 𝐿 = 𝐸 ⁄𝑑𝑡 ⟹ 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝐿 =
⟹ 𝑢𝑛𝑖𝑡𝑡𝑠 𝑜𝑓 𝐿 =
𝐾𝑔𝑚𝑠 −2 𝑚.𝐴−1 𝑠−1
𝐴𝑠 −1
𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝐸
𝑢𝑛𝑖𝑡𝑠 𝑜𝑓
𝑑𝐼 =
𝑑𝑡
𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑊𝑜𝑟𝑘
⁄𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑐ℎ𝑎𝑟𝑔𝑒
𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
⁄𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑡𝑖𝑚𝑒
𝐽𝐶 −1
= 𝐴𝑠−1
= 𝑘𝑔𝑚2 𝐴−2 𝑠 −2
2. Let g be the gravitational field strength on the earth surface, then the period of the simple on the earth is
given by
ℓ
𝑇 = 2𝜋√𝑔 . . . . . . . .. . . . . . . . . .. . . . . .(1)
The gravitational field strength at a height 2R above the surface of the earth is given by
𝑟
2
2
𝑅
1
𝑔/ = ( 𝑟𝐸 ) 𝑔 = (𝑅+2𝑅) 𝑔 = 9 . 𝑔 . . . . . . . . . .. . . . . .(2)
The period of oscillation of the simple pendulum at a height 2R above the surface of the earth surface is
given by
ℓ
ℓ
9ℓ
ℓ
𝑇 / = 2𝜋√𝑔, = 2𝜋√𝑔 = 2𝜋√ 𝑔 = 3 × 2𝜋√𝑔 = 3𝑇 ⟹ 𝑇 / = 3𝑇
⁄9
Thus the period of oscillation will increase by a factor of 3
3. (a)Thermionic emission is emission of free electrons from the surface of metals when sufficiently heated
electrically. The metal must be of high melting point. While photo – electric effect is the ejection of
electrons from the surface of a metal when radiation of a sufficiently high frequency falls on it.
(b) From the Einstein photoelectric equation, 𝐸𝑚𝑎𝑥 = ℎ𝑓 − Φ. For photoemission to occur,
𝐸𝑚𝑎𝑥 > 0 ⟹ ℎ𝑓 − Φ > 0.
Now,𝐸𝑚𝑎𝑥 = ℎ𝑓 − Φ =
hc
λ
−Φ=
6.63×10−34 ×3.0×108
6.3×10−14
− 2.25 × 10−14 = 3.13 × 10−12 > 0
Since the photon energy is greater than the work function, it therefore implies that photoemission is
possible.
4. (a) The input voltage is d.c hence the resulting magnetic field created is constant and as a result, no emf is
induced in the secondary coil and so the system cannot work.
The number of turns in the primary of the transformer equal to the number of turns in the secondary,
meaning for the system to function, the input voltage should be equal to the output voltage. This is not
the case with this transformer. The input voltage (primary voltage) is by far less than the expected voltage
output, meaning the input power is by far less than the power output. Consequently, the energy input is
less than the expected energy output and so the system cannot work.
(b)(i) The system can be adapted to function by
- Using an a.c across the primary
- Increasing the number of turns in secondary to step up the voltage.
(ii) Figure 1 represents an isolation transformer
<<ADVANCED LEVEL PHYSICS
Page 166
5.
60 Ω
KVL loop 1: −50𝐼1 − 60𝐼1 + 6 − 40𝐼2 + 12 = 0
⟹ 110𝐼1 + 40𝐼2 = 18 . . . . . . . . . . . . . . . . [1]
KVL loop2: −50𝐼1 − 60𝐼1 + 6 − 20𝐼3 − 3 + 80𝐼3 = 0
50 Ω
⟹ 110𝐼1 + 100𝐼3 = 3 . . . . . . . . . . . . . . . . .[2]
KCL: 𝐼1 = 𝐼2 + 𝐼3 . . . . . . . . . . . . . . . [3]
Solving [1],[2] and [3] simultaneously,
𝐼1 = 0.087𝐴; 𝐼2 = 0.212𝐴; 𝐼3 = −0.125 𝐴
(i) Current through 40Ω, 𝐼2 = 0.212𝐴
(ii) 𝑉80Ω = 80𝐼3 = 80 × 0.125 = 10.0 V
6.0 V
I1
I2
2
40 Ω
1
12.0 V
I3
20 Ω
Figure 2
80 Ω
3.0 V
6. (i) Transverse waves are waves in which the oscillations are perpendicular to the direction of motion
while longitudinal waves are waves in which the oscillations are parallel to the direction of oscillations.
(ii) Consult your note books.
(b) (i) Minimum positions are obtained when the incident wave and its echo interfere destructively i.e out
of phase. This means crest meets with trough of a reflected wave and the amplitude drops due to
cancellation.
𝑣
300
(ii) From the wave equation: 𝑣 = 𝑓𝜆 ⟹ 𝜆 = 𝑓 = 51.6 = 6.4 𝑚
𝜆
6.4
But for a stationary wave, the distance between two adjacent minimum position = 2 = 2 = 3.2 𝑚
(iii) Use a sound source of lower frequency
(d) For gases, heat changes are usually accompanied by changes in volume. The heat needed to change
the temperature at constant volume is different from that needed at constant pressure. However, for
liquids and solids, the changes in volume are little and can be neglected.
(e) Consult your not book: Diagram, procedure, calculations, result and precautions.
(f) (i) Molecules of liquids and gases can actually move and interchange their positions due to weak
forces of attraction but solids only vibrate about their fixed positions due to a strong force between their
molecules.
(ii) Molecules of solids and liquids have relatively stronger forces of attraction than those of gases. As a
result solids and liquids have fixed volumes but gases do not.
(g)
100℃
𝜃
0℃
Cu
Al
10.0 cm
15.0 cm
At steady state, the rate of heat flow through the copper equal rate of heat flow through aluminum
𝑑𝑄
𝑑𝑄
( 𝑑𝑡 )
𝐶𝑢
= ( 𝑑𝑡 )
𝐴𝑙
⟹
𝐾𝐶𝑢 𝐴(100−𝜃)
𝑙𝐶𝑢
=
𝐾𝐴𝑙 𝐴(𝜃−0)
𝑙𝐶𝑢
⟹
𝐾𝐶𝑢 (100−𝜃)
𝐾𝐴𝑙
.
𝑙𝐶𝑢
=𝑙
𝜃
𝐶𝑢
⟹
15 100−𝜃
7
.
0.15
𝜃
= 0.1 ⟹ 𝜃 = 58.82℃
Thus the junction temperature, 𝜃 is 58.82℃.
∆𝜃
Temperature gradient for Cu, (∆𝑥 )
𝐶𝑢
∆𝜃
=
Temperature gradient for Al, , (∆𝑥 )
𝐴𝑙
100−58.82
=
0.15
58.82−0
𝑄2
0.1
= 274.5℃𝑚−1
= 588.2℃𝑚−1
1
𝑄2
7. (a) From the equation 𝐹 = 4𝜋𝜀𝑟 2,drawing a graph of F against 𝑟 2, gives a straight line with slope𝑆 = 4𝜋𝜀
<<ADVANCED LEVEL PHYSICS
Page 167
F/N
1.0
r/nm
355.1 297.5
𝟏
𝒓𝟐
1.5
/1013 m-2 0.793 1.13
2.0
2.5
3.0
4.0
4.5
5.0
6.0
258.2
230.6
210.8
182.6
172.0
163.3
149.0
1.50
1.88
2.25
3.00
3.38
3.75
4.50
7
y = 1.3413x - 0.0283
F/N
6
5
4
3
2
1
0
0
1
2
3
4
(1/r2)/1013 m5-2
(b) (i) From the graph, slope, S = 1.34 x 10-13 Nm-2
𝑄2
(ii) Slope of graph represents 4𝜋𝜀
(iii) From the equation 𝑆 =
𝑄2
4𝜋𝜀
⟹𝜀=
𝑄2
(4.4×10−6 )
2
4𝜋×1.34×10−13
𝜀
given by 𝜀𝑟 = 𝜀
4𝜋𝑆
(c) By definition, the dielectric constant is
=
0
= 11.50 𝑁 −1 𝑚2 𝐶 2
⟹ 𝜀 = 𝜀0 𝜀𝑟 . Thus, the greater the dielectric
constant, the greater the permittivity of the medium. Also, the force is inversely proportional to the
permittivity of the medium. Hence the greater the dielectric constant the smaller the force.
8. (a) (i) Finite energy sources are sources which cannot be replenished once used. Examples include fossil
fuels, coal and natural gas. While renewable sources of energy are those sources which can continually be
replenished naturally and make use of processes that are part of our natural environment. Examples
include sun, wind, waves, biomass, flowing streams etc.
(ii) Power output = solar constant, K x area spanned by the radius
<<ADVANCED LEVEL PHYSICS
Page 168
4×1026
𝑃
𝑃0 = 𝐾 × 𝐴 = 𝐾 × 4𝜋𝑅 2 ⟹ 𝐾 = 4𝜋𝑅0 2 = 4𝜋×(1.5×1011)2 = 1.42 × 103 𝑊𝑚−2
Assumption: All the energy that leaves the sun reaches the earth. This is not actually true because clouds,
water droplets and many other particles in the atmosphere absorb a lot of the energy.
(b) Geothermal: To obtain electrical energy from geothermal energy, two holes are drilled several
kilometers down into dry rocks. Cold water is then sent down through one of the holes and comes back
through the other hole as steam. The steam is used to drive turbines to generate electricity.
Wind: The kinetic energy of moving air can be transformed to electrical energy by the use of windmills
or aero – generators.
(c) (i) Destruction of the ionosphere leads to gene mutation because cosmic rays from the sun directly
heat the surface of the earth.
(ii) Prohibition of the use of chlorofluorocarbons, planting of more trees to absorb carbon dioxide which
is a green house gas.
9. (a) (i)
DISPLAY
KEYPAD
ON/OFF
ANTENNA
SIM
ROM
CPU
RF
RAM
BATTERY
SPEAKER
MICROPHONE
Block Diagram of a Cell phone
(ii)
ssAspects
considered
Security
Optical fibers
Noise
Less noise due to no electrical interference
Signal attenuation
Information transmitted in the form of light
by simple reflection. Hence very little lose
of signal strength
-
More security
Less cross talk
Copper cable
-
Less secured
There is the possible of
cross talk
Large (high) NOISE
Much loss of signal strength
(b) (i) SIM stands for Subscribers Identification module
SMS stands for Short Message Service
Homework: Write down the full meaning of the following acronyms
MMS, ENS, MIN, SID, MTSO, SMTP
<<ADVANCED LEVEL PHYSICS
Page 169
(c) A radio receiver has an antenna which receives radio waves from many transmitters and changes them
into electrical signals. The signals pass to the tuning and amplifying circuit with the aid of inductors and
capacitors. The signal from the required station is picked up and sent to the demodulator and finally to
the speaker.
10. (a) At 0℃ most of the electrons are found in the valence band. So just few electrons are found in the
conduction band and as a result the semiconductor may not conductor. At 80℃ almost all the electrons in
the valence band gain thermal energy and are promoted to the valence to the conduction band and are
available as conduction electrons. As a result, the semiconductor can conduct due to increase in the
conductivity.
A
A
(b)
Circuit 2: ammeter does not display
Circuit 1: ammeter displays
In circuit 1, the ammeter displays a value since a capacitor allows a.c to pass through it. In circuit the
ammeter does not display any value since a capacitor does not allow d.c to pass through it. In circuit, the
capacitors acts like an opened switch.
+9.0 V
Ic
VL
RL
Vout
Rb
Ib
VR
VB
VCE = Vout
Ie
E
0V
N.B: In d.c analysis of a transistor, the capacitor Cout is short circuited
(i) Applying KVL; 𝑉𝐵𝐸 + 𝑉𝑅𝑏 − 9 = 0 ⟹ 𝑉𝑅𝑏 = 9 − 𝑉𝐵𝐸 ⟹ 𝐼𝑏 𝑅𝑏 = 9 − 𝑉𝐵𝐸 ⟹ 𝑅𝑏 =
9−𝑉𝐵𝐸
𝐼𝑏
, assuming
9−0.6
VBE = 0.6V⟹ 𝑅𝑏 = 25×10−6 = 336000Ω = 336 𝑘Ω
(ii) 𝐼𝐶 = 𝛽𝐼𝐵 = 60 × 25 × 10−6 = 1.5 × 10−3 𝐴
Applying KVL, 𝑉𝑜𝑢𝑡 + 𝐼𝐶 𝑅𝐿 − 9 = 0 ⟹ 𝑅𝐿 =
9−𝑉𝑜𝑢𝑡
𝐼𝐶
9−6
= 1.5×10−3 = 2000 Ω = 2 kΩ
(iii) The capacitor Cout is to remove the d.c component in the output.
11. (a)(i)
<<ADVANCED LEVEL PHYSICS
Page 170
Blurred image
formed before the
retina Retina
(b) (i) The rods and the cones
(ii)
Cornea
Lens
For the eye to focus a distant object,
- The ciliary muscle is relaxed
- This causes the lens to flatten, increasing its focal length
For an object at infinity, the focal length of the eye is equal to the fixed distance between lens
and retina
(iii) The type of lens is a convex lens
(c) Muscular actions of the heart produces voltages that set up as pulses, conducted through body fluids
to the outer surface of the body and recorded on an electrograph. This portrays heat beat rate and rhythm
which aids in highlighting heart problems like lack of blood through the heart, fast or slow.
<<ADVANCED LEVEL PHYSICS
Page 171
JUNE 2017
1. For an equation to be homogeneous, all the terms on both sides of the equation must have the same units
or dimensions
Units of LHS = units of energy i.e units of E = units of energy, but E = force x distance
𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝐸 = 𝑘𝑔𝑚𝑠 −2 × 𝑚 = 𝑘𝑔𝑚2 𝑠 −2
Unit of RHS
𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑤𝑜𝑟𝑘
𝑈𝑛𝑖𝑡𝑠 𝑜𝑓 𝑉 = 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑐ℎ𝑎𝑟𝑔𝑒 =
𝑄 𝑄2
𝐹𝑟𝑜𝑚 𝐹 = 4𝜋𝜀1
0𝑟
𝑘𝑔𝑚2 𝑠 −2
= 𝑘𝑔𝑚2 𝐴−1 𝑠 −3 ; 𝑈𝑛𝑖𝑡𝑠 𝑜𝑓 𝐴 = 𝑚2 ; 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑑 = 𝑚
𝐴𝑠
𝐴2 𝑠2
𝑄 𝑄
2
1 2
−1 −3 4 2
⟹ 𝜀0 = 4𝜋𝐹𝑟
𝑚 𝑠 𝐴
2 ⟹ 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝜀0 = 𝑚2 𝑘𝑔𝑚𝑠 −2 = 𝑘𝑔
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑅𝐻𝑆 =
𝑘𝑔2 𝑚4 𝐴−2 𝑠−6 × 𝑚2 ×𝑘𝑔−1 𝑚−3 𝑠4 𝐴2
𝑚
= 𝑘𝑔𝑚2 𝑠 −2 . Since both sides of the
equation have the same base units, then the equation is homogeneous.
2. (i) For the converging lens, u = 30.0cm, f = +15.0 cm
1
1
1
𝑢𝑓
𝑓
𝑢
𝑣
𝑢−𝑓
From the thin lens equation, = + ⟹ 𝑣 =
=
15×30
30−15
𝑣
= 30; 𝑚 =
𝑢
=
30
30
=1
The magnification is 1, meaning the image has the same size as the object. The image distance is
positive, hence the image is real.
𝑢𝑓
(ii) For the diverging lens, u = 30.0 cm, f = -15.0 cm, 𝑣 = 𝑢−𝑓 =
𝑚=
−10
30
−15×30
30—15
=
−15×30
45
= −10.0 𝑐𝑚;
1
= − 3 < 0 . The magnification is negative and its magnitude is less than 1, meaning the image
is diminished and virtual
400 µF
3.
Y
400 µF
X
Z
600 µF
12.0 V
For the capacitors X and Y in series, 𝐶
1
1
𝑋𝑌
1
𝐶 𝐶
400×400
= 𝐶 + 𝐶 ⟹ 𝐶𝑋𝑌 = 𝐶 𝑋+𝐶𝑌 = 400+400 = 200 𝜇𝐹
𝑋
𝑌
𝑋
𝑌
(a) 𝑉𝑋𝑌 = 12.0 𝑉; 𝑄𝑋𝑌 = 𝐶𝑋𝑌 𝑉𝑋𝑌 = 200 𝜇𝐹 × 12.0 𝑉 = 2400 𝜇𝐶
For capacitors in series, the same charge flow through them. Thus the same amount of charge flows
through X and Y 𝑉𝑦 =
𝑄𝑌
𝐶𝑌
2400 𝜇𝐶
=
= 6.0 𝑉
400 𝜇𝐹
1
𝐶 𝑉 2
2 𝑋 𝑋
(b) 𝐸𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑜𝑟𝑒𝑑 𝑖𝑛 𝑋 =
1
= 2 × 400 × 10−6 × 62 = 7.2 × 10−3 𝐽
4. (a) Wave particle duality: Matter can have wave properties and waves can behave as particles.
Reflection, diffraction and polarization of light show that light is a wave. Electron diffraction show that
matter can behave like wave. Electromagnetic radiation has momentum and hence can be considered as a
particle.
(b) By the law of conservation of energy, lost in electrical energy = gained in kinetic
1
2𝑒𝑉
𝑒𝑉 = 2 𝑚𝑒 𝑣𝑒 2 ⟹ 𝑣𝑒 = √ 𝑚 ⟹ 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛, 𝑃 = 𝑚𝑒 𝑣𝑒 = √2𝑒𝑉𝑚𝑒
𝑒
ℎ
6.6×10−34
The associated de Broglie wavelength is 𝜆 = 𝑝 = √2×1.6×10−19
×3.0×104 ×9.1×10−31
= 7.06 × 10−12 𝑚
5. See June 2008 question 8(a)
<<ADVANCED LEVEL PHYSICS
Page 172
6. (a) A real gas is one which obeys the gas laws and cannot liquefy no matter the condition of temperature
and pressure. They deviate from the gas laws when subjected to high pressure
(b) (i) See June 2005 question 9(a)
1
1
(ii) From 𝑃 = 𝜌𝑐̅̅̅2 , 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑏𝑦 𝑉, 𝑃𝑉 = 𝑉𝜌𝑐̅̅̅2 , 𝑏𝑢𝑡 𝜌𝑉 = 𝑚𝑎𝑠𝑠, 𝑀 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑠
3
3
1
𝑃𝑉 = 𝑀𝑐̅̅̅2 , 𝑖𝑓 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 = 𝑁 𝑎𝑛𝑑 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑜𝑛𝑒 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 = 𝑚 ⟹ 𝑀 = 𝑚𝑁
3
1
2 1
2 1
1
3𝑛
⟹ 𝑃𝑉 = 3 𝑁𝑚𝑐̅̅̅2 = 3 (2 𝑚𝑐̅̅̅2 ) 𝑁, 𝑏𝑢𝑡 𝑃𝑉 = 𝑛𝑅𝑇 ⟹ 𝑛𝑅𝑇 = 3 (2 𝑚𝑐̅̅̅2 ) 𝑁 ⟹ 2 𝑚𝑐̅̅̅2 = 2 𝑁 𝑅𝑇
𝑁
1
3 𝑅
𝑏𝑢𝑡 𝑛 = 𝑁 , 𝑤𝑖𝑡ℎ 𝑁𝐴 = 𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ⟹ 2 𝑚𝑐̅̅̅2 = 2 𝑁 𝑇; 𝑤𝑖𝑡ℎ
𝐴
𝐴
1
3
Boltzmann constant ⟹ 2 𝑚𝑐̅̅̅2 = 2 𝐾𝑇. But
3
2
𝑅
𝑁𝐴
= 𝐾 where K is the
1
𝐾 is a constant and 2 𝑚𝑐̅̅̅2 is the average translational k.e i.e
𝐾. 𝑒 ∝ 𝑇
(c)(i) The second law of thermodynamics states that energy does not transfer spontaneously by heat from
a cold object to a hot object. (Summary of the Clausius and the Kelvin – Planck statements)
(ii) By the first law of thermodynamic, 𝑄ℎ = 𝑄0 + 𝑊 ⟹ 𝑊 = 𝑄ℎ − 𝑄0 ;
𝐵𝑢𝑡 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
ℎ𝑒𝑎𝑡 𝑡𝑎𝑘𝑒𝑛 𝑖𝑛
⟹𝜂=
𝑊
=
𝑄ℎ
𝑄ℎ −𝑄0
𝑄ℎ
𝑄
= 1 − | 0|
𝑄ℎ
(iii) Efficiency is less than 100 % because the engine rejects some of the heat at a lower temperature.
(d) (i) A moving coil instrument has a pointer which deflects along a scale to indicate the magnitude of a
physical quantity. An oscilloscope displays in the form of a wave the magnitude of a physical quantity. A
moving coil instrument may directly read the value of a quantity but the oscilloscope is used to compare
a value with a standard.
(ii) With an a.c system, it is possible to step up voltages before transmission enabling low power losses.
D.c voltages cannot be stepped up or down
(e) (i) Moving the magnet over the coil leads to changing fields, hence induced emf and hence an induce
current on the wire. The resistance of the wire is high leading to the production of heat.
𝑑𝜙
(ii) From Faraday’s law of electromagnetic induction, induced emf, 𝐸 = −𝐿 𝑑𝑡 ; Thus increasing the
time(i.e moving the magnet slowly) will reduce the rate of change of the flux linked with the wire hence
low emf and low current.
(iii) By conservation of energy, Δ𝑄 = 𝑚𝑐Δ𝜃 ⟹ 𝐼𝑉𝑡 = 𝑚𝑐Δ𝜃
𝐼𝑉𝑡
⟹ 𝑐 = 𝑚𝑐Δ𝜃 =
(f)(i) 𝑍 =
𝑉0
𝐼0
5.0×150×0.3
5×10−3 ×80
= 562.5 𝐽𝑘𝑔−1 𝐾 −1
= √(𝑋𝐿 − 𝑋𝐶 )2 + 𝑅 2 , 𝑤𝑖𝑡ℎ 𝑋𝐿 = 2𝜋𝑓𝐿 𝑏𝑒𝑖𝑛𝑔 𝑡ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟
1
𝑎𝑛𝑑 𝑋𝐶 = 2𝜋𝑓𝐶 𝑏𝑒𝑖𝑛𝑔 𝑡ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟
(ii) For maximum potential difference across the resistor, resonance must occur and XL = XC and the
frequency, f becomes the resonant frequency i.e f = f0
1
𝑋𝐿 = 𝑋𝐶 ⟹ 𝑓0 = 2𝜋√𝐿𝐶 =
1
2𝜋√50×200×10−6
= 1.59 𝐻𝑧
(iii) For series circuits, current is same across all the components. The potentials across the capacitor and
inductor oppose each other. At resonant, both potentials balance and cancel out. Hence all p.d is across
resistor and hence maximum.
7. 𝜀 = 𝑉 + 𝐼𝑟 ⟹ 𝑉 = −𝑟𝐼 + 𝜀. Hence a graph of V against I is a straight line with gradient –r and intercept
on the V – axis as 𝜀.
<<ADVANCED LEVEL PHYSICS
Page 173
V/V
16.6
15.2
12.0
8.6
7.8
6.6
5.2
3.8
I/mA
4.0
8.0
16.8
26.4
28.8
32.0
36.0
40.0
(a) See graph below
(b) From the graph, intercept on V – axis is 18.0 V. Thus 𝜀 = 18.0 𝑉
Slope of graph is – 356.0 VA-1. Thus r = 356 Ω (Try to determine the slope by yourself.
8. (a) (i) Renewable energy sources are those sources which cannot be exhausted completely i.e they are
continually replenished after usage e.g wood, solar energy, e.t.c. Non renewable energy sources can be
exhausted after usage e.g uranium, fossil fuel, coal, natural gas etc.
(ii) 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝜂 =
𝑝𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡,𝑃𝑜𝑢𝑡
𝑃𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡,𝑃𝑖𝑛
× 100 %
Total power supplied to the panel Pin ⟹ 𝑃𝑖𝑛 =
𝑃
𝑃
𝑃𝑜𝑢𝑡 ×100
,𝜂
=
2000×100
40
= 5000 𝑊 = 5.0 𝑘𝑊
5000
𝑠𝑜𝑙𝑎𝑟 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝑆𝐶 = 𝐴 ⟹ 𝐴 = 𝑆 = 1.2×103 = 4.2 𝑚2
𝐶
(b) (i) Cameroon cannot solely depend on solar energy for its needs for the following reasons
 The total amount of energy delivered varies from one region to another in Cameroon
<<ADVANCED LEVEL PHYSICS
Page 174



Storage of solar energy after conversion is very costly
Construction of solar panel is costly and occupies too much space
Seasonal variations leads to variation in the intensity of solar energy delivered to the panel.
(ii)
Chemical
energy in wood
Heat energy
in water
Heat energy
Kinetic energy
of steam
Rotational kinetic
energy of turbine
Electrical energy
9. (a) (i) Block diagram of a radio receiver
Information
source
Input
transducer
Modulator
Receiver
Transmitter
Carrier wave
Received
information
Output
transducer
Demodulator
(ii) The station is received at the resonant frequency i.e
1
1
1
𝑓0 = 2𝜋√𝐿𝐶 ⟹ 𝐶 = 4𝜋2 𝑓 2 𝐿 = 4𝜋2 ×(92.5×106 )2 ×1.25×10−9 = 2.37 × 10−9 𝐹 = 2.37 𝑛𝐹
0
(b) With analogue transmission, information is transmitted as continuously varying physical quantity i.e
electrical signals vary in the same way as the physical quantity producing them. With digital systems,
signals are coded i.e information is converted to electric pulses with only two levels of signal either low
or high.
(ii) Some of the problems Cameroon may face in changing from analogue to digital include:
 Problem of reinforcement or recycling of personnels
 They need to purchase new equipment which are very costly
Some advantages include:
 Easy transmission since transmitter and receiver need only to tell the difference between on and
off.
 Digital signals will be less prone to interference and noise
 Digital systems are more reliable and cheap to design.
10. (a)
(i) An AND gate is a logic gate whose output is high only if all inputs are high. The table below shows
the truth table of an AND gate with two inputs.
Input
Output
A
B
C
N.B: 𝐶 = 𝐴 × 𝐵
<<ADVANCED LEVEL PHYSICS
Page 175
0
0
0
0
1
0
1
0
0
1
1
1
(ii) An OR gate: The output is high if either inputs or both are high.
Input
A
Output
B
C
0
0
0
0
1
1
1
0
1
1
1
1
N.B: 𝐶 = 𝐴 + 𝐵
(iii) NAND gate: The output is high if either or both inputs are low
Input
A
Output
B
C
0
0
1
0
1
1
1
0
1
1
1
0
(b)
N.B: 𝐶 = ̅̅̅̅̅̅̅̅
𝐴×𝐵
200 Ω
4.5 V
10 𝑘Ω
V0
2.7 V
VBe
From the circuit above (input), 2.7 − 10000𝐼𝐵 − 𝑉𝐵𝑒 = 0 ⟹ 𝐼𝐵 =
𝐼𝐶 = 𝛽𝐼𝐵 = 60 × 2 × 10−4 = 1.2 × 10−2 𝐴 = 12 𝑚𝐴
<<ADVANCED LEVEL PHYSICS
2.7−𝑉𝐵𝑒
10000
=
2.7−0.7
10000
= 2.4 × 10−4 𝐴
Page 176
From the above circuit (output),
𝑉𝑜 + 200𝐼𝐶 − 4.5 = 0 ⟹ 𝑉0 = 4.5 − 200𝐼𝐶 = 4.5 − 200 × 1.2 × 10−2 = 2.1 𝑉
11. (a) When rays of light leave an object, they approach the eye almost parallel to the principal axis. The
combined action of the aqueous humour and the eye lens lead to refraction of the rays, passing through
the focus and forming an inverted image on the retina (as shown on the diagram below).
(b) This patient is suffering from shortsightedness or myopia. Distant objects are not seen clearly and the
image is formed in front of the retina (see diagram below)
This type of eye defect can be corrected by the use of a diverging lens of appropriate focal length. This
will diverge the rays and bring the image to focus on the retina as shown in the diagram below.

(c) Some non – ionizing imaging techniques include: MRI (magnetic resonant imaging) and ultrasound.
MRI: Magnetic Resonance Imaging (MRI) works by measuring the way that hydrogen atoms absorb and
then relax and re-emit electromagnetic energy. Most of the human body is made up of water molecules,
which consist of only hydrogen and oxygen atoms and fat, which also contains hydrogen atoms. You are
made up of about 60% hydrogen atoms! The nucleus of a hydrogen atom is a proton, and protons are
very sensitive to magnetic fields. When the proton spins it generates a magnetic field, therefore the
nucleus of a hydrogen atom is like a tiny magnet. When your body is in a strong magnetic field all of
your hydrogen nuclei line - just like a row of compass needles lining up with a magnetic field.
MRI scanners use powerful magnets. When the powerful magnets that are used in magnetic
resonance imaging (MRI) are switched on, all the protons in your body are pulled so that they spin in the
same direction, in the same way that a magnet can pull the needle of a compass. The scanner contains
several electric coils. This produces variations in the strength of the magnetic field at different points in
your body. This variation means that each hydrogen nucleus experiences a slightly different magnetic
field strength. This is important for detecting the position of a particular hydrogen nucleus.
The frequency of these waves depends on the strength of the magnetic field where each nucleus is
and this means that the scanner can work out the location of each nucleus. The MRI scanner sends a
pulse of radio signals to certain areas of the body which ‘snaps’ the protons out of position. The pulse
gives enough energy to the hydrogen nuclei in that area to change direction. When the pulse of energy
ends the nuclei snap back to their original orientation and each nucleus gives off energy in the form of a
<<ADVANCED LEVEL PHYSICS
Page 177

radio wave. When this happens, each proton transmits a radio signal that provides information about its
exact location in the body.
Ultrasound: A very high frequency sound (ultrasound) is sent into a particular organ of a body
(particularly soft tissues e.g the heart, kidney, breast etc). The ultrasound pulse travels into the body and,
when it reaches an interface or boundary between different tissues in the body, some of the sound energy
is reflected back towards the transducer where it causes an electrical signal to be generated which can
then be transformed into images for studies.
<<ADVANCED LEVEL PHYSICS
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June 2018
1. (a)
-
Newton’s law of gravitation states that “the magnitude of the force of attraction between any two
particles in the universe is directly proportional to product of their masses and inversely
proportional to the square of their separation.”
Coulomb’s law states that “the magnitude of the force (attractive or repulsive) between two
charged particles is directly proportional to the product of the charges and inversely proportional
to the square of the distance between the charges.”
(b)
-
The effect of the electrostatic force can be shielded while that of a gravitational force cannot be
shielded.
- The electrostatic force can either be attractive or repulsive while the gravitational force is only
attractive.
2. (a) At the equilibrium separation, the displacement, x is zero (x = 0) and the velocity is said to be
maximum, vmax.
𝐹𝑟𝑜𝑚, 𝑣 = 𝜔√𝑟 2 − 𝑥 2 , when x = 0, 𝑣 = 𝑣𝑚𝑎𝑥 = 𝜔𝑟 =
(b) 𝑦 = 𝐴𝑠𝑖𝑛𝜔𝑡, 𝑎 = −𝜔2 𝑦 = −𝜔2 𝐴𝑠𝑖𝑛𝜔𝑡
2𝜋
𝑇
𝑟=
2𝜋(0.05)
2.0
= 0.16 m/s
a/ms-2
0.49
2.0
t/s
-0.49
E
3.
15 V
I
3Ω
6Ω
2
1
R
3A
1A
(a) Let the current through R be I, then by Kirchhoff’s current law, 𝐼 + 1 = 3 ⟹ 𝐼 = 2 𝐴
6
(b) Apply KVL in loop 1, −𝐼𝑅 + 15 − 3 × 3 = 0 ⟹ −𝐼𝑅 + 15 − 9 = 0 ⟹ 𝐼𝑅 = 6 ⟹ 𝑅 = 2 = 3Ω
c) Applying KVL in loop 2, 𝐸 − (3)(3) − (6)(1) = 0 ⟹ 𝐸 − 9 − 6 = 0 ⟹ 𝐸 = 15𝑉
4. (a) Photoelectric effect is the ejection of electron from the surface of a metal when light of a sufficiently
high frequency falls on it.
(b)
- The maximum frequency of the ejected electrons depends on the frequency of the incident
radiation but not on the intensity.
- The emission of electrons commences immediately the surface is irradiated with radiation of a
sufficiently high frequency.
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-
Emission of electrons occurs only if the frequency of the incident radiation is above the a certain
maximum value (called the threshold frequency) for a particular metal no matter the intensity.
5. (a) Conditions for a body to be in equilibrium
- The vector sum of all the forces acting on the body should be zero. i.e the forces in all the
directions must balance.
- The resultant moment about any axis must be zero. .i.e clockwise moment equal anti – clockwise
moment about any axis.
𝑙 /4
𝑙 /4
Pivot
800 N
𝑙 /2
200 N
W
Since the bar is in equilibrium, anti – clockwise moment about the pivot = clockwise moment
𝑙
𝑙
3𝑙
800 × = 𝑊 × + 200 × ⟹ 𝑊 = 200 𝑁
4
4
4
Where W is the weight of the bar.
6. (a) (i) Waves can be classified based on the mode of propagation or medium of transmission.
Based on the mode of propagation we have longitudinal and transverse waves.
Longitudinal waves are waves in which the oscillations of the particle of the medium are parallel to the
direction of propagation. Examples of longitudinal waves include: sound waves, waves produced on a
spring when plucked along the axis etc.
Transverse waves are waves in which the vibrations of the particles are perpendicular to the direction of
propagation. Examples include: All electromagnetic waves, water ripples etc.
Based on the medium of transmission, we have mechanical and electromagnetic waves
Mechanical waves require a material medium for their transmission. Examples include: sound waves,
water waves, waves produced on a spring etc.
Electromagnetic waves can travel through a material medium and even a vacuum. Examples include:
visible light, ultraviolet radiation, gamma radiation etc.
Waves could also be classified as progressive or stationary (see differences between stationary and
progressive waves)
(b) See notes
c) There is a difference in the frequencies perceived by the observer because there is relative motion
between the observer and the source.
𝑣
340
Approaching frequency, 𝑓 / = 𝑣−𝑢 𝑓𝑠 ⟹ 360 = 340−𝑢 𝑓𝑠 ⟹ 360(340 − 𝑢𝑠 ) = 340𝑓𝑠 … … … … … (1)
𝑠
Receding frequency, 𝑓
//
𝑣
𝑠
340
= 𝑣+𝑢 𝑓𝑠 ⟹ 320 = 340+𝑢 𝑓𝑠 ⟹ 320(340 + 𝑢𝑠 ) = 340𝑓𝑠 ……………(2)
𝑠
𝑠
(2) = (1) ⟹ 360(340 − 𝑢𝑠 ) = 320(340 + 𝑢𝑠 ) ⟹ 12240 − 36𝑢𝑠 = 10880 + 32𝑢𝑠
⟹ 1360 = 68𝑢𝑠 ⟹ 𝑢𝑠 = 20 𝑚𝑠 −1
(d) See previous years for the assumptions and derivation.
(e) Se notes
(f) (i) Heat lost by liquid A = heat gained by water
<<ADVANCED LEVEL PHYSICS
Page 180
𝑐𝑤
) (90 − 𝜃𝑓 ) = (3)(𝑐𝑤 )(𝜃𝑓 − 22)
2
⟹ 360 − 4𝜃𝑓 = 3𝜃𝑓 − 66 ⟹ 𝜃𝑓 = 60.90 𝐶
𝑚𝐴 𝑐𝐴 (90 − 𝜃𝑓 ) = 𝑚𝑤 𝑐𝑤 (𝜃𝑓 − 22) ⟹ (8) (
∆𝜃
90−60.9
29.1
(ii) ∆𝜃𝐴 = 60.9−22 = 38.9 = 0.75
𝑤
7. (a) 𝑃 = 𝑃0 𝑒 −𝐾ℎ ⟹ 𝑙𝑛𝑃 = 𝑙𝑛𝑃0 − 𝐾ℎ
A graph of lnP against h gives a straight line with gradient –K and intercept on the lnP axis as lnP0.
h/km P/104 Nm- ln(P/Nm-2
2
9.6
4.346
10.68
11.7
2.691
10.20
18
1.094
9.30
28.1
0.221
7.70
34.9
0.0993
6.90
40
0.0602
6.40
44.8
0.173
7.46
51
0.0138
4.93
<<ADVANCED LEVEL PHYSICS
Page 181
(b) From the graph, gradient , - K = - 1.21 x 10-4 m-1 i.e K = 1.21 x 10-4 m-1
Intercept, 𝑙𝑛𝑃0 = 11.55 ⟹ 𝑃0 = 𝑒 11.55 = 1.04 × 105 𝑁𝑚−2
𝜌𝑔
𝐾 × 𝑃0 1.21 × 10−4 × 1.04 × 105
𝐾=
⟹𝜌=
=
= 1.49 𝑘𝑔𝑚−3
𝑃0
𝑔
9.8
8. (a) (i) The captain needs to know:
- The wind speed and the wind direction
- Wave periodicity and height
- The precipitation
- The cloud coverage.
(ii) Weather forecast can be done using satellites which have radiometers. Information about the
weather at the moment is made in the form of electrical signals (voltages) and transmitted to receiving
stations. From the information, the state of the atmosphere can then be predicted for a period of time
either by persistent forecasting, statistical forecasting or computer forecasting.
(b) (i) Black surfaces are good absorbers of radiant heat. As such, the water that flows the pipes is
heated .
(ii) Glass is transparent hence short wavelength radiations from the sun can penetrate to heat up the
water but long wavelength radiations will no longer move out (green house effect).
(ii) Insulated and blackened walls prevent heat from escaping through the walls.
<<ADVANCED LEVEL PHYSICS
Page 182
c) (i) Check from your textbooks.
(ii) Potential energy of water in the damp is converted to k.e as water runs down the penstock. The
kinetic energy of the water in the penstock is converted to k.e of the turbine and hence the rotor. The
k.e of the rotor is converted to electrical energy.
9. (a) (i) Functions of the mobile phone
- To make and receive calls
- To send short messages
- To carryout internet services like emails
- Mobiles phones have short range wireless communication like infrared or Bluetooth .
(ii) SMS: Short message Service
MMS: Multimedia Message Service
SIM: Subscriber Identification Module
(b) (i) A, (ii) C (iii) D/E (v) A
c) (i) Bandwidth is a range of frequencies which can be sent through a particular channel.
(ii) With analogue transmission, the electrical signals vary in the same way as the physical quantity
producing them. While in digital transmission, the information is converted into electrical pulses or
codes.
10. (a) The energy needed for electrons to move from the valence band to the conduction is very large for
insulators i.e the forbidden energy gap for insulators is larger than for semiconductors.
(b) (i)
Q
I
Y
X
P
I
B
V
I
(ii)
IB/mA
IC/A
50
2.5
100
40
200
5.5
<<ADVANCED LEVEL PHYSICS
Page 183
400
7.0
500
9.5
∆𝐼
9.0−1.5
7.5
Current gain 𝛽 = ∆𝐼𝐶 = (440−70)×10−3 = 3.7×10−3 = 20
𝐵
(iii) Thermal runway is the increase in temperature of transistors which lead to an increase in collector
current.
c)
V/V
t/s
11. (a) (i)
<<ADVANCED LEVEL PHYSICS
Page 184
-
The external or outer ear helps to collect sound and channels it towards the auditory canal. The
ear has the eardrum which vibrates when sound wave is incident on it.
The middle ear has the system of three bones which receive the vibrations from the eardrum and
transfer the energy to the oval window to develop high pressure in the inner ear.
The inner ear has the cochlea which picks up the pressure and sends impulses to the brain.
𝜆
𝑣
(ii) for a closed pipe, the fundamental occurs at 𝑙 = 4 ⟹ 𝜆 = 4𝑙 , from 𝑣 = 𝑓𝜆 ⟹ 𝑓0 = 4𝑙
340
⟹ 𝑓0 =
= 3.04 × 103 𝐻𝑧
4 × 28 × 10−3
(iii) They receive vibrations from the eardrum and in turn transmit a higher pressure to the inner ear.
(b) (i) Optical fibres function on total internal reflection of light. Signals are incident at angles greater
than the critical angle and hence are totally internally reflected.
(ii) The core receives the signal which is converted to light pulses.
The cladding of lower refractive index helps to totally internally reflect the signals.
(iii) They are used to illuminate round bends and inaccessible regions e.g in endoscopy to view internal
organs of the body.
<<ADVANCED LEVEL PHYSICS
Page 185
1.
2.
3.
4.
MULTIPLE CHOICES (with proposed answers in square brackets)
At temperatures close to 0 K, the specific heat capacity of a particular solid is given by c = b T3, where T
is the thermodynamic temperature and b is a constant characteristic of the solid. What are the SI base
units of b?
A. kg m2 s-2 K-4
B. m2 s-2 K-3
C. m2s-2 K-4 D. kg m2 s-2 K-3
[C]
Four physical quantities P, Q, R and S are related by the equation PQ2= 8Q + RS. Which statement must
be correct for the equation to be homogeneous?
A. P, Q, R and S all have the same units.
B. P, Q, R and S are all scalar quantities
C. The product PQ2 is numerically equal to (8Q+RS). [D]
D. The product PQ2 has the same units as Q, and the product RS.
Which of the following experimental techniques reduces the systematic error of the quantity being
investigated?
A. Timing a large number of oscillations to find a period.
B. Measuring the diameter of a wire repeatedly and calculating the average.
C. Adjusting an ammeter to remove its zero error before measuring a current
D. Plotting a series of voltages and current readings for an ohmic device on a graph and using its gradient
to find resistance.
[C]
What is the reading shown on this milliammeter?
A. 2.35 mA
B. 2.7 mA
C. 3.4 mA
D. 3.7 Ma [C]
5. A micrometer screw gauge is used to measure the diameter of a copper wire. The reading with the wire in
position is shown in diagram 1. The wire is removed and the jaws of the micrometer are closed. The new
reading is shown in diagram 2.
What is the diameter of the wire?
A. 1.90 mm
B. 2.45 mm
C. 2.59 mm
D. 2.73 mm
6. Vernier calipers, reading to 0.1 mm, are used to find the internal diameter (10.0 ± 0.1 mm) and the
external diameter (12.0 ± 0.1 mm) of a length of glass tubing. The mean wall thickness would be quoted,
therefore, as
A. 1.0 ± 0.1 mm
B. 1.0 ± 0.2 mm
C. 2.0 ± 0.1 mm
D. 2.0 ± 0.0 mm
[C]
<<ADVANCED LEVEL PHYSICS
Page 186
7. What is the ratio 1 µm/1 nm?
A. 103
B. 10-9
C. 10-12
D. 10-15 [A]
8. Four STUDENT’S each made a series of measurements of the acceleration of free fall g (in ms-2). The
table shows the results obtained.
Student A
9.81
9.79
9.82
9.83
Student B
9.21
10.32
10.13
9.59
Student C
9.45
9.21
8.99
8.76
displacement/m
Which student obtained a set of results that
Student D 8.45
8.46
8.50
8.41
could
be described as accurate but not precise? [A]
9. A car at rest in a traffic queue moves forward in a straight line and then comes to rest again. The graph
shows the variation with time of its displacement.
time/s
0
0
20
40
60
80 100 120 140
What is its speed while it is moving?
A. 0.70 ms–1
B. 0.80 ms–1
C. 1.25 ms–1
D. 1.40 ms–1 [B]
10. Which of the following statements is correct?
A. An object can have a constant velocity even though its speed is changing.
B. An object moving with constant acceleration can reverse its velocity.
C. An object cannot have a constant speed if its acceleration is not zero.
D. An object cannot have an instantaneous zero velocity if it is moving with a constant acceleration. [D]
11. A motorcycle stunt-rider moving horizontally takes off from a point 1.25 m above the ground, landing 10
m away as shown.
What was the speed at take-off?
A. 5 m s–1
B. 10 m s–1
C. 15 m s–1
D. 20 m s–1 [D]
12. A stone is projected horizontally in a vacuum and moves along a path as shown. X is a point on this path.
XV and XH are vertical and horizontal lines respectively through X. XT are the tangent to the path at X.
<<ADVANCED LEVEL PHYSICS
Page 187
Path of stone
X
H
V
T
Along which direction or directions is the velocity constant?
A. XV
B. XH C. Xv and XH
D. XT [B]
13. A bomb is released from an aircraft which is travelling horizontally at a constant velocity of 200 m s-1 at
a height of 1200 m above the ground. What is the speed of the bomb when it hits the ground?
A. 150 m s-1
B. 200 m s-1
C. 250 m s-1
D. 300 m s-1 [C]
14. Which of the following physical quantities has the same base units as energy?
A. moment of a force
B. specific heat capacity
C. impulse
D. voltage [A]
15. A particle which moves from rest is acted upon by two forces : a constant forward force and a retarding
force which is directly proportional to its velocity. Which one of the following statements about the
subsequent motion of the particle is true?
A. Its velocity increases from zero to a maximum.
B. Its acceleration increases from zero to a maximum and then decreases
C. Its acceleration increases from zero to a maximum.
D. Its velocity increases from zero to a maximum and then decreases. [A]
16. The diagram shows four forces applied to a cylinder.
30 N
20 N
20 N
30 N
Which of the following describes the resultant force and resultant torque on the object?
A. zero resultant force, zero resultant torque B. zero resultant force, non-zero resultant torque
C. non-zero resultant force, zero resultant torque
D. non-zero resultant force, non-zero resultant torque
17. A uniform rod of weight 10 N is freely hinged to a wall at X as shown in the diagram. The rod is
supported at the other end by a cable perpendicular to the rod. The rod makes an angle of 60o to the wall.
<<ADVANCED LEVEL PHYSICS
Page 188
If a load W of weight 18 N is suspended at the other end of the rod, what is the tension in the cable?
A. 12 N
B. 16 N
C. 20 N
D. 32 N
[C]
18. A ball is thrown vertically upwards. Neglecting air resistance, which statement is correct?
A. The kinetic energy of the ball is greatest at the greatest height attained.
B. By the principle of conservation of energy, the total energy of the ball is constant throughout its
motion.
C. By the principle of conservation of momentum, the momentum of the ball is constant throughout its
motion
D. The potential energy of the ball increases uniformly with time during the ascent.
[B]
19. Car X is travelling at half the speed of car Y. Car X has twice the mass of car Y. Which statement is
correct?
A. Car X has half the kinetic energy of car Y. B. Car X has one quarter of the kinetic energy of car Y.
C. Car X has twice the kinetic energy of car Y. D. The two cars have the same kinetic energy [B]
20. A small electric motor is used to raise a weight of 2.0 N through a vertical height of 80 cm in 4.0 s. The
efficiency of the motor is 20 %. What is the electrical power supplied to the motor?
A. 0.080 W
B. 0.80 W
C. 2.0 W D. 200 W [C]
21. A car of mass 1200 kg travels along a horizontal road at a speed of 10 m s-1. At the time it begins to
accelerate at 0.2 m s-2, the total resistive force acting on the car is 160 N. What is the total output power
developed by the car as it begins the acceleration?
A. 0.80 kW
B. 1.6 kW
C. 2.4 kW
D. 4.0 Kw [D]
22. The graph shows the behaviour of a sample of a metal when it is stretched until it starts to undergo plastic
deformation.
What is the estimated work done in stretching the sample from zero extension to 12.0 mm?
A. 3.30 J
B. 3.55 J
C. 3.60 J
D. 6.60 J [C]
<<ADVANCED LEVEL PHYSICS
Page 189
23. A 1.00 × 103 kg car is driven clockwise around a flat circular track of radius 25.0 m. The speed of the car
is a constant 5.00 m s-1.
What minimum friction force must exist between the tires and the road to prevent the car from skidding as it
rounds the curve?
A. 1.25 × 105 N B. 9.80 × 104 N C. 5.00 × 103 N D. 1.00 × 103 N
[D]
24. A bucket of water is swung in a vertical circle at arm’s length of 0.70 m. The minimum number of
revolutions per second it must be swung to keep the water from spilling out of the bucket is
A. 3.74
B. 2.62
C. 1.68
D. 0.60
[D]
25. A small object of mass m is released at the rim of a smooth semi-spherical bowl of radius r , as shown in
the diagram.
What is the magnitude of the contact force acting on the object when it passes the bottom of the bowl?
A. 0.5 mg B. mg C. 2 mg D. 3 mg
[D]
26. A ball of mass 0.10 kg is attached to a string and swung in a vertical circle of radius 0.50 m as shown
below. At the top of the circle, the tension in the string is 6.2 N.
What is the speed v of the ball at this instant?
A. 5.1 m s-1 B. 6.0 m s-1 C. 36 m s-1 D. 72 m s-1 [B]
27. An aircraft is moving in a horizontal plane at a constant speed. It banks at an angle of 50o to the vertical
in order to make a turn as shown in the diagram below. The only forces acting on the aircraft are lift, L
and weight, W.
What is the ratio of the centripetal force to the weight of the aircraft?
A. o.643
B. 0.766
C. 0.839
D. 1.19 [D]
28. A toy car, of mass 0.10 kg, is travelling along a track which contains a vertical circular loop of radius
0.10 m as shown in the diagram.
<<ADVANCED LEVEL PHYSICS
Page 190
What is the minimum entry speed v of the car required to prevent it from falling off the track at the top of
the loop?
A. 0.3 m s-1 B. 1.0 m s-1 C. 1.4 m s-1 D. 2.2 m s-1 [D]
29. A solid X is in thermal equilibrium with a solid Y, which is at the same temperature as a third solid Z.
The three bodies are of different materials and masses. Which one of the following statements is certainly
correct?
A. X and Y have the same heat capacity B. X and Y have the same internal energy
C. It is not necessary that Y should be in thermal equilibrium with Z
D. There is no net transfer of energy if X is placed in thermal contact with Z [D]
30. Hot water tank of heat capacity 5000 JK-1 contains 10 kg of water at 25 °C. What is the time taken to
raise the temperature of the water to 45 °C using a heater coil of power of 3.0 kW, given that the specific
heat capacity of water is 4200 Jkg-1 K-1?
A. 61 s
B. 280 s
C. 310 s
D. 610 s [C is the most correct 313.3]
31. A student tries to determine the specific latent heat of vaporisation of a liquid by an electrical method. A
heater is used to boil the liquid and when the liquid is boiling, the mass of liquid vaporised per second is
measured at two different powers for the heater. At the power of 40 W and 80 W, the liquid is vaporised
at the rate of 0.0417 kg s-1 and 0.0893 kg s-1 respectively. What is the best estimate of the specific latent
heat of vaporisation of the liquid?
A. 840 J kg-1 B. 896 J kg-1 C. 928 J kg-1 D. 959 J kg-1
[A]
32. A gas cylinder is fitted with a safety valve which releases a gas when the pressure inside the cylinder
reaches 2.0 x 106 Pa. Given the maximum mass of this gas that the cylinder can hold at 10 °C is 15 kg,
what would be the maximum mass at 30 °C?
A. 5.0 kg B. 14 kg C. 20 kg D. 45 kg
[B]
33. 2 bulbs X and Y are filled with an ideal gas and connected by a capillary tube. The volume of bulb X is
three times that of bulb Y. The number of gas molecules in X and Y are 2N and N respectively. If the
temperature of the gas in Y is 300 K, what is the temperature of the gas in X?
A. 150 K
B. 200 K
C. 450 K
D. 600 K
[B]
34. An ideal gas undergoes a cycle of processes as shown in the p-V diagram below.
Which statement correctly describes the situation?
A. The internal energy of the gas increases over one complete cycle.
B. Over the entire cycle, work is done by the gas.
C. The gas absorbs more heat than it releases heat over the whole cycle
D. The gas gives out more heat than it absorbs
over the whole cycle.
35. A fixed mass of an ideal gas slowly releases 1500 J of heat and as a result contracts slowly, at a constant
pressure of 2.0 x 104 Pa, from a volume of 0.050 m3 to 0.025 m3 . What is the effect on the internal
energy of the gas?
[D]
A. It decreases by 2000 J B. It decreases by 1000 J C. It is unchanged D. It increases by 1000 J.
36. A loudspeaker produces sound waves in air of wavelength 0.68 m and speed 340 ms-1. How many cycles
of vibration does the loudspeaker diaphragm make in 10 ms?
A. 5
B. 10
C. 50
D. 100
[B]
37. In a transverse progressive wave of frequency 400 Hz, the least distance between two adjacent points
which have a phase difference of π/2 rad is 0.40 m. What is the speed, in ms-1 of the wave?
A. 160
B. 320
C. 640
D. 1280
[C]
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38. A microwave transmitter is used to direct microwaves of wavelength 30 mm along a line XY. A metal
plate is positioned at right angles to XY with its mid-point on the line, as shown.
When a detector is moved gradually along XY, its reading alternates between maxima and minima.
Which one of the following statements is not correct?
A. The distance between two minima could be 15 mm
B. The distance between two maxima could be 30 mm.
C. The distance between a minimum and a maximum could be 30 mm
D. The distance between a minimum and a maximum could be 37.5 mm.
39. Which one of the following statements about stationary waves is true?
A. Particles between adjacent nodes all have the same amplitude
B. Particles between adjacent nodes are out of phase with each other.
C. Particles immediately on either side of a node are moving in opposite directions
D. There is a minimum disturbance of the medium at an antinode [C]
40. The frequency of the fundamental note produced by an organ pipe that is open at both ends is 300 Hz.
The frequency of the next harmonic is
A. 400
B. 600 C. 800
D. 900
[B]
41. The volume of a hot-air balloon is 600 m3 and the density of the surrounding air is 1.25 kg m-3. The
balloon just hovers clear of the ground. What is the total mass of the balloon, including the hot air inside?
A. 48 kg B. 75 kg C. 480 kg
D. 750 kg [D]
42. A helicopter is hovering in the air. Its rotor blades propel 2500 kg of air vertically downwards every
second. The air, initially at rest, is accelerated to a speed of 15 m s-1. What is the mass of the helicopter?
A. 1640 kg C. 2500 kg C. 3000 kg D. 3820 kg
[D]
43. A ball is bounced on the ground as shown in the diagram.
The change in momentum of the ground is
A. zero.
B. half the magnitude of the change in momentum of the ball
C. the same magnitude as the change in momentum of the ball, but in the opposite direction
D. the same magnitude as the change in momentum of the ball, and in the same direction
44. A cart with mass 200 g moving on a frictionless linear air track at an initial speed of 1.2 m s-1 undergoes
an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart
continues in its original direction at 1.00 m s-1. What is the mass of the second cart?
A. 18 g
B. 36 g
C. 100 g D. 200 g
[A]
45. Four identical railway trucks, each of mass m, are coupled together and are at rest on a smooth horizontal
track. A fifth truck of mass m and moving at 5.00 m s-1 collides and couples with the stationary trucks.
What is the speed of the trucks after the impact?
[C]
-1
-1
-1
-1
A. 0.67 m s
B. 0.83 m s C. 1.00 m s
D. 1.67 m s
46. Which of the following pairs of forces is an action-reaction pair?
A. The force a ladder leaning on a smooth wall exerts on the wall and the normal reaction force from the
wall.
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B. The force a ladder leaning on a smooth wall exerts on the rough floor and the normal reaction from the
floor.
C. Weight of a parachutist and the pull of the parachute on him when he is moving with terminal
velocity.
D. Weight of a floating object and the upthrust
acting on it.
[C]
47. A radioactive source consists of a mixture of two isotopes P and Q
P has a half-life of 60 minutes and Q has a halflife of 30 minutes. The initial activity recorded
–1
by a suitable counter is 800 min . After 120
minutes the counter registers an activity of 80
–1
min .What was the initial contribution of P to
the count rate?
–1
–1
–1
A. 160 min B. 240 min
C. 270 min
D.1480 min–1
48. The binding energy per nucleon may be used as a measure of the stability of a nucleus. This quantity
A. increases uniformly throughout the Periodic Table.
B. is higher for the daughter nuclide in a nuclear fission reaction.
C. is a minimum for nuclides in the middle of the Periodic Table.
D. is directly proportional to the neutron/proton
ratio of the nuclide.
49. A radioactive nuclide emits an α-particle and two β-particles. Compared with the original nuclide, the
resulting nuclide will have
A. the same nucleon number
B. a higher proton number
C. a lower proton number D. the same proton number [D]
50. The equation
shows the fission of a Uranium−235 nuclide by a slow-moving neutron into a Rhodium-121 nuclide, a
Silver-113 nuclide and two neutrons.
binding energy per nucleon of Uranium−235 = 7.59 MeV
binding energy per nucleon of Rhodium-121= 8.26 MeV
binding energy per nucleon of Silver-113 = 8.52 MeV
What is the energy change during this fission process?
A. 73.9 MeV of energy is released B. 73.9 MeV of energy is absorbed.
C. 178 Mev of energy is released D. 178 Mev of energy is absorbed.
51. The radioactive isotope of potassium-24, with half-life of 12 hours, is being used in the localization of
brain tumours. A patient is given a dose containing 96 µCi of potassium-24 . After 2 days of ingestion of
potassium-24, the activity from the brain region is monitored and found to be 3 µCi.
What was the percentage of the ingested potassium-24 that was concentrated in the brain region at the
end of the 2-day period?
A. 3.13 % B. 25 %
C. 50 % D. 75 %
[A]
52. In a photoelectric experiment, electromagnetic radiation of wavelength 240 nm and intensity 8.2 x 103 W
m-2 is incident normally on a metal surface of area 2.0 x 10-4 m2. What is the number of photons incident
per second?
A. 2.0 x 1015
B. 2.0 x 1018 C. 2.0 x 1021 D. 2.0 x 1024 [B]
53. Which of the following is true when photoelectric effect occur?
A. The maximum speed of the photoelectrons is proportional to the intensity of the incident light.
B. The number of electrons emitted per second is proportional to the intensity of the incident light.
C. The maximum energy of the photoelectrons increases with the wavelength of the incident light
D. The wavelength of the incident light is greater than the threshold value. [B]
54. In a photoelectric emission experiment, photoelectrons are produced when an electromagnetic radiation is
incident on a metal surface. Both the intensity and the wavelength of the electromagnetic radiation are
then reduced, resulting in
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A. a reduction in the maximum kinetic energy of the electrons but an increase in their rate of emission
B. a reduction in both the maximum kinetic energy of the electrons and their rate of production
C. an increase in both the maximum kinetic energy of the electrons and their rate of production
D. an increase in the maximum kinetic energy of the electrons but a reduction in their rate of emission
[D]
55. In a series of photoelectric emission experiments, metals with different work functions Φ were
illuminated with light of different frequencies f and intensities I. The maximum kinetic energy of
photoelectrons in each experiment depends on
A. Φ but not on f and I B. Φ and f but not I
C. Φ and I but not f D. Φ, f and I [B]
56. The maximum kinetic energy Ek of emitted electrons is measured in photoelectric experiments using
light of particular intensity.
Which of the following is a possible graph showing how Ek varies with the wavelength λ of the light?
[A]
57. The electron energy levels in a certain atom are represented by the given diagram.
If the atom in the ground state E1 is bombarded with an electron of energy 4.0 eV, which is/are possible
transition/s that can take place?
A. Only E1 to E3
B. Only E1 to E2 or E1 to E3
C. Only E1 to E2 or E1 to E3 or E1 to E4
D. Any transition between the four energy levels is possible.
[A]
58. The de Broglie wavelength of a particle that has kinetic energy is Ek. The wavelength λ is proportional to
A. Ek
B. 1/Ek
C. 1/√Ek
D. 1/Ek2
[C]
59. Through what minimum potential difference must an electron in an X-ray tube be accelerated so that it
can produce X-rays with a wavelength of 0.100 nm?
A. 1.24 V B. 124 V C. 1.24 x 104 V D. 1.24 x 106 V
[C]
60. Which one of the following provides evidence for the existence of atomic energy levels?
A. The photoelectric effect B. Characteristic X-ray spectra
C. Matter waves C. Alpha particle scattering
[B]
61. A wire lies perpendicularly across a horizontal uniform magnetic field of flux density 2.0 × 10–2 T so that
0.30 m of the wire is effectively subjected to the field.
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If the force exerted on this length of wire due to a current in it is 3.0 × 10–2 N downward, what is the
current in the wire?
62. A. 0.45 A from P to Q B. 0.45 A from Q to P C. 5.0 A from P to Q D. 5.0 A from Q to P [D]
The diagram shows a vertical square coil whose plane is at right angles to a horizontal uniform magnetic
field B. A current, I, flows in the coil, which can rotate about a vertical axis OO’.
Which one of the following statements is correct?
A. The forces on the two vertical sides of the coil are equal and opposite.
B. A couple acts on the coil.
C. No forces act on the horizontal sides of the coil.
D. If the coil is turned through a small angle about OO', it will remain in position. [A]
63. An particle and a β particle both enter the same uniform magnetic field, which is perpendicular to their
direction of motion. If the β particle has a speed 15 times that of the α particle, what is the value of the
ratio magnitude of the force on the β particle to magnitude of the force on the α particle ?
A. 3.7 B. 7.5
C. 60.0
D. 112.5
[A]
64. An electron travelling at constant speed enters a uniform electric field at right angles to the field. While
the electron is in the field it accelerates in a direction which is
65. A. in the same direction as the electric field B. in the opposite direction to the electric field.
C. in the same direction as the motion of the electron C. in the opposite direction to the motion of the
electron. [B]
66. Three parallel wires, on the same plane, carry currents of equal magnitude, X, Y and Z, in the direction
shown in the diagram. Wires X and Z are x distance to the left and right of wire Y respectively.
The resultant force experienced by Y due to currents in X and Z is
A. Perpendicular to the plane of the paper B. to the left C. to the right D. zero
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[D]
Page 195
67. The figure shows the cross-sections of three long vertical wires that pass through the corners of an
equilateral triangle PQR. They carry equal currents into or out of the paper as shown.
The resultant force on wire R is
A. downward B. upwards C. to the left D. to the right
[C]
68. An a.c. source is connected to a resistance R . If the rms voltage output of the a.c. source is doubled, the
power dissipated in the resistor increases by a factor of
A. 1 B. √ 2
C. 2
D. 4
[D]
69. A resistor is connected in series with an alternating current supply of negligible internal resistance. The
peak value of the supply voltage is Vo and the peak value of the current in the resistor is Io. The average
power dissipation in the resistor is
A. VoIo/2 B. VoIo/√2
C. VoIo D. 2VoIo
[A]
70. From Young’s double-slit experiment, we can deduce that light
A. is electromagnetic in nature B. behaves like a wave
C. consists of transverse oscillations D. consists of photons [B]
71. In a Young.s double slit interference experiment, monochromatic light placed behind a single slit
illuminates two narrow slits and the interference pattern is observed on a screen placed some distance
away from the slits. Which one of the following decreases the separation of the fringes?
A. increasing the width of the single slit
B. decreasing the separation of the double slits
C. increasing the distance between the double slits and the screen
D. using monochromatic light of higher
frequency
[B]
72. Light of wavelength 700 nm is incident on a pair of slits, forming fringes 3.0 mm apart on a screen. What
is the fringe spacing when light of wavelength 350 nm is used and the slit separation is doubled?
A. 0.75 mm
B. 1.2 mm C. 3.0 mm D. 6.0 mm [A]
73. A diffraction grating has 500 lines per millimetre. When a parallel beam of light is incident normally on
the grating, the angular separation between the two second order maxima is 60o. What is the wavelength
of the incident beam?
A. 200 nm B. 400 nm C. 500 nm
D. 600 nm [B]
74. A narrow beam of monochromatic light is incident normally on a diffraction grating. Third-order
diffracted beams are formed at angles of 45o to the original direction. What is the highest order of
diffracted beam produced by this grating?
A. 3rd
B. 4th
C. 5th D. 6th [B]
75. Two monochromatic radiations X and Y are incident normally on a diffraction grating. The second order
intensity maximum for X coincides with the third order intensity maximum for Y. What is the ratio
wavelength of X : wavelength of Y?
A. 1/2
B. 2/3
C. 3/2
D. 2/1
[B]
76. The figure below show the side view of a conducting strip which consists of two strips of equal length x
but different diameters. The diameter of the narrow section is half that of the wider section. A current is
flowing through the strip as shown.
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Which of the following statements is correct?
A. The resistance of the narrow section is more than that of the wide section.
B. The resistance per unit length of the narrow section is twice that of the wide section
C. The resistivity of the narrow section is greater than that of the wide section
D. The current in the narrow section is less than
that in the wide section. [A]
77. A cylindrical piece of a soft, electrically-conducting material has resistance R. It is rolled out so that its
length is doubled but its volume stays constant.
What is its new resistance?
A. R/2 B. R
C. 2R
D. 4R [D]
78. A p.d. of 12 V is connected between P and Q.
What is the p.d. between X and Y?
A. 0 V B. 4 V
C. 6 V
D. 8 V
[B]
79. The diagram shows an arrangement of resistors.
What is the total electrical resistance between X and Y?
A. less than 1 Ω B. 7.5 Ω C. between 10 Ω and 30 Ω D. 40 Ω [B]
80. A potential divider consists of a fixed resistor R and a light-dependent resistor (LDR).
What happens to the voltmeter reading, and why does it happen, when the intensity of light on the LDR
increases?
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Page 197
A. The voltmeter reading decreases because the LDR resistance decreases
B. The voltmeter reading decreases because the LDR resistance increases
C. The voltmeter reading increases because the LDR resistance decreases
D. The voltmeter reading increases because the LDR resistance increases.
[D]
81. The diagram shows a potentiometer and a fixed resistor connected across a 12 V battery of negligible
internal resistance.
The fixed resistor and the potentiometer each have resistance 20 Ω. The circuit is designed to provide a
variable output voltage.
What is the range of output voltages?
A. 0 – 6 V B. 0 – 12 V C. 6 – 12 V D. 12 – 20 V [A]
82. The diagram shows a potentiometer circuit.
The contact T is placed on the wire and moved along the wire until the galvanometer reading is zero. The
length XT is then noted.
In order to calculate the potential difference per unit length on the wire XY, which value must also be
known?
A. the e.m.f. of the cell E1
B. the e.m.f. of the cell E2
C. the resistance of resistor R
D. the resistance of the wire XY
83. In the potentiometer circuit below, the moveable contact is placed at N on the bare wire XY, such that the
galvanometer shows zero deflection.
The resistance of the variable resistor is now increased.
What is the effect of this increase?
A. The p.d. across XY is increased and the balanced length XN is increased
B. The p.d. across XY is increased and the balanced length XN is decreased
C. The p.d. across XY is decreased and the balanced length XN is increased
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D. The p.d. across XY is decreased and the balanced length XN is decreased
84. The equation relating the pressure and volume of a gas with its thermodynamic temperature T is given by
pV = KT, where k is a constant. What are the base units of K?
A. no units, because k is a constant B. kg m-2 s-2 K-1 C. kg m2 s-2 K-1 D. kg m s-2 K-1 [C]
85. A bucket of water of mass 1.5 kg is whirled in a vertical circle of radius 0.50 m.
What is the minimum speed, in m s-1 of the bucket at the highest point of the circle so that the water does
not spill out?
A. 1.6
B. 2.2
C. 2.5
D. 4.9 [B]
86. The smallest distance between two points of a progressive transverse wave which have a phase difference
of π/3 rad is 0.050 m. If the frequency of the wave is 500 Hz, what is the speed of the wave?
A. 25 m s-1 B. 75 m s-1 C. 150 m s-1 D. 1700 m s-1 [C]
87. A lamp emits light of two distinct wavelengths, x and y. When the lamp is used as a source for a
diffraction grating experiment, it is found that the second order maximum for wavelength x occurs at
precisely the same angle as the third order maximum for wavelength y. The ratio x to y is therefore
A. 2
B. 3/2
C. 2/3
D. ½ [B]
88. A particle oscillates with undamped simple harmonic motion. Which one of the following statements
about the acceleration of the oscillating particle is true?
A. It is least when the speed is greatest. B. It is
always in the opposite direction to its velocity.
C. It is proportional to the frequency. D. It decreases as the potential energy increases [C]
89. In an X-ray tube, the accelerating potential difference across the tube determines
A. the maximum frequency of the X-rays B. the maximum wavelength of the X-rays
C. the wavelengths of the characteristic spectra
D. the maximum intensity of the X-ray beam
90. A rectangular tank, with vertical sides, contains water to a depth of 30 cm. On a cold day, the water is
initially 0 0C. The top 3.0 cm of it freezes into ice at 0 0C.
Assume that half of the latent heat given out by the ice goes to heating the remainder of the water. What
is now the temperature of the water below the ice?
(The specific latent heat of fusion of water is 330 kJ kg–1 and the specific heat capacity
of water is 4.2 kJ kg–1K–1)
A. 3.9 oC B. 4.4 oC C. 7.9 oC D 8.7 oC
91. A cell of e.m.f. 2.0 V and negligible internal resistance is connected to the network of resistors shown.
Which of the following statement is correct?
A. Electric potential at P is higher than Q by 0.20 V
B. Electric potential at P is lower than Q by 0.20
V
C. Electric potential at P is higher than Q by 0.50 V
D. Electric potential at P is lower than Q by 0.50
V [A]
92. A 2 kg ball is attached to a 0.80 m string and whirled in a horizontal circle at a constant speed of 6 m/s.
The work done on the ball during each revolution is:
(A) 90 J
(B) 72 J
(D) 16 J
(D) zero [D]
93. A football is kicked off the ground a distance of 50 m downfield. Neglecting air resistance, which of the
following statements would be INCORRECT when the football reaches the highest point?
A. all of the balls original kinetic energy has been changed into potential energy
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B. the ball’s horizontal velocity is the same as when it left the kickers foot
C. the ball will have been in the air one-half of its total flight time
D. the vertical component of the velocity is equal to zero [B]
94. What is the kinetic energy of a satellite of mass m that orbits the Earth, of mass M, in a circular orbit of
radius R?
A 1 GMm B. 1 GMm
2
R
4
R
C. 1 GMm
D. GMm
2
2
2 R
R
[A]
95. A spring-loaded gun can fire a projectile to a height h if it is fired straight up. If the same gun is pointed
at an angle of 45° from the vertical, what maximum height can now be reached by the projectile?
h
h
h
h
A.
B.
C.
D.
[C]
2
2
4
2 2
96. For a particle moving in a circle with uniform speed, which one of the following statements is correct?
A . The kinetic energy of the particle is constant.
B. The force on the particle is in the same direction as the direction of motion of the particle.
C .The momentum of the particle is constant.
D. The displacement of the particle is in the direction of the force. [A]
97. A mass on the end of a string is whirled round in a horizontal circle at increasing speed until
the string breaks. The subsequent path taken by the mass is
A. A straight line along a radius of the circle.
B. A horizontal circle.
C. A parabola in a horizontal plane.
D. A parabola in a vertical plane. [D]
98. Which of the following statements is true when an object is performing simple harmonic motion about a
central point O
A. The acceleration is always away from O
B. The acceleration and the velocity are always in opposite directions.
C. The acceleration and displacement from O are always in the same direction
D. The graph of acceleration against time is a straight line
[A]
99. A plannet has a radius of half the earth’s radius and a mass a quater of the earth’s mass. What is the
approximate gravitational field strength at the surface of the planet?
A. 1.6 Nkg-1 B. 5.0 Nkg-1
C. 10 Nkg-1
D. 20 Nkg-1
[C]
100.
An electron and a proton are 1.0 × 10–10m apart. In the absence of any other charges, what is the
electric potential energy of the electron?
A
+2.3 × 10–18 J B
–2.3 × 10–18 J
C
+2.3 × 10–18 J D
–2.3 × 10–18 J [
101.
An ion carrying a charge of +4.8 × 10–19C travels horizontally at a speed of 8.0 × 105 ms–1. It enters a
uniform vertical electric field of strength 4200 V m–1, which is directed downwards and acts over a
horizontal distance of 0.16m. Which one of the following statements is not correct?
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Page 200
Positive ion
Uniform electric
field of 4200 Vm-1
0.16 cm
A. The ion passes through the field in
2.0 ×10–7s.
B. The force on the ion acts vertically downwards at all points in the field.
C. The magnitude of the force exerted on the ion by the field is 1.6 × 10–9N.
D The horizontal component of the velocity of the ion is unaffected by the electric field.
102.
The electric potential at a distance r from a positive point charge is 45 V. The potential increases
to 50 V when the distance from the charge decreases by 1.5 m. What is the value of r?
A
1.3 m
B
1.5 m
C
7.9 m
D
15 m
[C]
103.
A 400 μF capacitor is charged so that the voltage across its plates rises at a constant rate from 0 V
to 4.0 V in 20 s. What current is being used to charge the capacitor?
A
5 μΑ
B
20 μΑ
C
40 μΑ
D
80 μΑ
[D]
104.
In experiments to pass a very high current through a gas, a bank of capacitors of total
Capacitance 50 μF is charged to 30 kV. If the bank of capacitors could be discharged completely in 5.0
ms, what would be the mean power delivered?
A
22 kW B
110 kW
C
4.5 MW D
9.0 MW [C]
105.
A 10 mF capacitor is charged to 10 V and then discharged completely through a small motor.
During the process, the motor lifts a weight of mass 0.10 kg. If 10% of the energy stored in the capacitor
is used to lift the weight, through what approximate height will the weight be lifted?
A
0 05 m B
0.10 m
C
0.50 m D
1.00 m
[A]
106.
A capacitor of capacitance C discharges through a resistor of resistance R. Which one of the
following statements is not true?
A. The time constant will decrease if C is increased.
B. The time constant will increase if R is increased.
C. After charging to the same voltage, the initial discharge current will increase if R is
decreased.
D. After charging to the same voltage, the initial discharge current will be unaffected if C is increased.
[A]
107.
An uncharged capacitor is connected to a battery.Which graph shows the variation of charge with
potential difference across the capacitor?
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Page 201
Q/C
Q/C
A
B
V/V
Q/C
V/V
Q/C
C
C
D
V/V
V/V
108.
Figure (i) shows a vertical plane square coil of 50 turns, carrying a current of 3.0 A. The length of
each side of the coil is 4.0 cm. Figure (ii) shows a view of this coil from above within a horizontal
magnetic field of flux density 0.20 T.
P 3.0 A Q
Q
300 0.2 T
4.0 cm
P
S
R
Figure (ii)
Figure (i)
The force on side QS is
A 120 N B 60 N C1.2 N D 0.60 N
109.
Take the acceleration due to gravity, gE, as 10 m s–2 on the surface of the Earth. The acceleration
𝑔
due to gravity on the surface of the Moon is 6𝐸 .An object whose weight on Earth is 5.0 N is dropped from
rest above the Moon’s surface. What is its momentum after falling for 3.0s?
A 2.5 kg m s–1 B 6.2 kg m s–1
C 15 kg m s–1
D 25 kg m s–1
[A]
110.
If the potential difference between a pair of identical, parallel, conducting plates is known, what is
the only additional knowledge required to determine the electric field strength between the plates?
A. The permittivity of the medium between the plates
B. The separation and area of the plates
C. The separation and area of the plates and the permittivity of the medium between the
plates
D. The separation of the plates [D]
111.
Which line, A to D, gives correct units for both magnetic flux and magnetic flux density?
x Magnetic flux Magnetic flux density
A Wbm-2
Wb
B Wb
T
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B
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C Wbm-2
Tm-2
D Tm-2
Wbm-2
112.
A girl of mass 40 kg stands on a roundabout 2.0 m from the vertical axis as the roundabout
rotates uniformly with a period of 3.0 s. The horizontal force acting on the girl is approximately
A zero.
B 3.5 × 102 N.
2
C 7.2 × 10 N. D 2.8 × 104 N.
113.
Which one of the following gives the phase difference between the particle velocity and the
particle displacement in simple harmonic motion?
𝜋
𝜋
A. 4 𝑟𝑎𝑑
B. 2 𝑟𝑎𝑑
3𝜋
C. 4 𝑟𝑎𝑑
D. 2𝜋 𝑟𝑎𝑑
[B]
114.
A capacitor of capacitance 15 µF is fully charged and the potential difference across its plates is
8.0V. It is then connected into the circuit as shown.
The switch S is closed at time t = 0. Which one of the following statements is correct?
A The time constant of the circuit is 6.0 ms.
B The initial charge on the capacitor is 12 µC.
C After a time equal to twice the time constant, the charge remaining on the capacitor is
Q0e2, where Q0 is the charge at time t = 0.
D After a time equal to the time constant, the potential difference across the capacitor is 2.9 V. [D]
115.
Two charges, P and Q, are 100 mm apart.
+4 µF
P
+6 µF
Q
100 mm
X is a point on the line between P and Q. If the potential at X is 0 V, what is the distance from P to X?
A. 40 mm B. 45 mm
C. 50 mm
D. 60 mm
[B]
116.
X
ZZ
Uniform
magnetic
field
Y
The diagram shows a square coil with its
plane parallel to a uniform magnetic field.
of the following would induce an emf in the coil?
A. movement of the coil slightly to the left
B. movement of the coil slightly downwards
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Which one
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C. rotation of the coil about an axis through XY
D. rotation of the coil about an axis perpendicular to the plane of the coil through Z
117.
Two isolated point charges are separated by 0.04 m and attract each other with a force of 20 µN.
If the distance between them is increased by 0.04 m, what is the new force of attraction?
A 40 µN B 20 µN
C 10 µN D 5 µN
[D]
118.
The diagram shows a uniform electric field of strength 10 V m-1
A charge of 4 µC is moved from P to Q and then from Q to R. If the distance PQ is 2 m and QR
is3 m, what is the change in potential energy of the charge when it is moved from P to R?
A 40 µJ B 50 µJ
C 120 µJ D 200 µJ
[C]
119.
An electron travelling at constant speed enters a uniform electric field at right angles to the field.
While the electron is in the field it accelerates in a direction which is
A In the same direction as the electric field.
B In the opposite direction to the electric field.
C In the same direction as the motion of the electron.
D In the opposite direction to the motion of the electron. [B]
120.
Protons, each of mass m and charge e, follow a circular path when travelling perpendicular to a
magnetic field of uniform flux density B. What is the time taken for one complete orbit?
𝟐𝝅𝒆𝑩
𝒎
𝒆𝑩
𝟐𝝅𝒎
A. 𝒎
B. 𝟐𝝅𝒆𝑩 C. 𝟐𝝅𝒎 D. 𝒆𝑩
[D]
121.
The primary winding of a perfectly efficient transformer has 200 turns and the secondary has
1000 turns. When a sinusoidal pd of rms value 10 V is applied to the input, there is a primary current of
rms value 0.10 A rms. Which line in the following table, A to D, gives correct rms output values
obtainable from the secondary when the primary is supplied in this way?
rms output Rms output current/A
emf/V
A 50
0.10
B 50
0.02
C 10
0.10
D 10
0.02
Questions 123 – 125 relate to a heat engine which uses heat to do 50 J of work, and then exhausts 100 J
of energy into a cold reservoir
122.
The heat added to the heat engine is
A. 150 J B. 100 J C. 50 J D 2 J
[A]
123.
The efficiency of the heat engine is
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Page 204
A. 25% B. 33% C. 50% D. 75%
[B]
124.
An ideal gas is enclosed in a container which has a fixed volume. If the temperature of the gas is
increased, which of the following will also increase?
I.
The pressure against the walls of the container.
II.
The average kinetic energy of the gas molecules.
III.
The number of moles of gas in the container.
A. I only
B. I and II only
C. II and III only
D. II only
[B]
2
1
3
4V
4
Questions 125 – 127 consider the capacitors in
the circuit
above. Each capacitor has a
capacitance of 2 μF.
125.
The total capacitance of the circuit is
A. 8 μF B. 4 μF C. 2/3 μF D. 8/3 μF
[D]
126.
The charge on capacitor 1 is
A. 8 μC B. 4 μC C. 2/3 μC D. 4/3 μC
[A]
127.
The charge on capacitor 2 is
A.8 μC B. 8/3 μC C. 4 μC D. 2/3 μC
[B]
128.
A positive charge moves through a magnetic field of magnitude 10-2 T with a speed of 106 m/s
when it experiences a force of 10-14 N. The magnitude of the charge is
A. 10-22 C B. 10-18 C C. 10-10 C D.10-6 C
[B]
129.
I
P
A wire on the y – axis of a coordinate system
has a current I in the +y direction as shown
above. What is the
direction of the
magnetic field due to the wire at point
P?
A. to the left B. to the right
C. down into the page and perpendicular to the page
D. up out of the page and perpendicular to the page
[C]
130.
Green light is passed through two narrow slits and forms a pattern of bright and dark lines on a
screen. The phenomena primarily responsible for this pattern is
A. refraction B. reflection
C. polarization D. interference
[D]
131.
Which of the graphs below represent the energy of a photon vs. its frequency?
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A
C
B
D
[B]
132.
Which of the following statements is true?
A. The binding energy of the nucleus is equal to the mass of the nucleus in atomic mass units.
B. The binding energy of the nucleus is equal to the mass of the nucleus in MeV
C. The atomic mass of a nucleus is greater than the sum of the masses of the individual nucleons.
D. The atomic mass of a nucleus is less than the sum of the mmasses of the individual nucleons
[D]
133.
The second law of thermodynamics (law of entropy) explains which of the following?
A. The heat lost by one object must be gained
by another object.
B. Heat flows naturally from a hotter body to a
cooler body.
C. Celsius degrees and Kelvin degrees are equivalent.
D. Heat can be transformed into work.
[B]
134.
The water in a river is running due west with a speed of 4 m/s. A boy in a boat tries to cross the
river by rowing due south at 3 m/s. The velocity of the boat relative to the shore is
A.12 m/s SW B. 7 m/s SW
C. 5 m/s SW D. 4/3 m/s SW
[C]
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