PHYSICS
216
1. (a)
Length of air column L/mm
152.0
158.0
163.0
170.0
179.0
182.0
Temperature  /ºC
Temperature T / K
14.0
287.0
29.0
302.0
40.0
313.0
57.5
330.5
78.0
351.0
85.0
358.0
(b)
Graph of L/mm vs T/K
217
(c)
Slopes, S 
y2  y1
x2  x1
175  153
342  290
22

52
 0.42 mm K -1

(d)
The slope tells how the length of air column and effectively the volume of air
changes with absolute temperature.
(e)
Value of the length of the air column at 273 K  146 mm
(f)
Charles’ law states that the volume of a fixed mass of gas is directly proportional
to the absolute temperature provided the pressure remains constant.
(g)
Using
V1  V2
T1  T2
V2
2

 273  35  273  75

2. (a)
V
2
 2
308 348
2  348
V2 
308
 2.26 L
(i)
Quantity
Force
Formula
F  ma
Potential energy
E  mgh
Momentum
(b)
p  mv
Unit
N (kg ms-2 )
J  Nm or kg m 2 s -2 
kg ms-1
(ii)
For bodies undergoing collision, the total momentum before collision is
equal to the total momentum after collision, provided no external force is
acting.
(i)
The linear momentum is conserved in the crash. The total momentum
before collision is zero since the momentum of each truck is equal and
opposite (mv  mv) . After collision, the momentum is zero since the
velocity of each truck is zero.
218
(ii)
By the principle of conservation of momentum
Total momentum before impact  total momentum after impact
mb vb  mt vt   mb  mt  v
 0.1 vb    5.0  0    0.1  5.0  6.0
0.1vb  30.6
vb  306 ms -1
3. (a)
(b)
A longitudinal wave is one in which the vibration of the parties is parallel to the
direction of travel of the wave.
(i)
A wavelength – A to E (or (B to F, C to G etc.)
(ii)
Amplitude – B to W (or D to X etc.)
(i)
Using  
(c)
(d)

v
f
340
0.350 1000
  0.97 m
(ii)
The frequency remains unchanged in travelling from one medium to
another i.e. 0.350 KHz.
(iii)
Refractive index of water, nw 
water
air
1.29
0.97
nw  1.33

219
4. (a)
(b)
The features of the vacuum flask reduce the heat loss due to conduction,
convection and radiation.
 The vacuum between the double walls prevents the heat loss due to
conduction and convection.
 The silvered glass wall reduces heat loss due to radiation.
 The cork support and cork stopper reduces heat loss due to conduction and
convection.
(i)
Energy per day collected  Intensity  Area  Efficiency
 5  5  0.95
 23.75 kWh
(ii)
Energy per day to heat water  0.5  23.75
 11.88 kWh
(iii)
Energy per day available 
(iv)
Using E  mc
E

m
c
17.48  3.6 106

4 200   55  25 
80 92

 23.75
100 100
 17.48 kWh
 499 kg
5. (a)
The current and voltage are recorded each time the rheostat is varied. A graph of I
vs V is plotted.
220
(b)
(i)
R1 R2
R1  R2
1000 1000
 1000 
1000  1000
Total resistance, RT  R1 
RT  1500
(ii)
V
RT
110

1500
Current, I 
 0.073 A
(iii)
Using
P
V
1100

110
 10 A
I
 Fuse rating  10A  13A
6. (a)




The GM tube is first placed next to the source without any shielding and the count
rate observed.
A sheet of paper is then placed between the source and the GM tube. A significant
reduction in the count rate will confirm the presence of alpha particles.
The GM tube is then placed behind the aluminum sheet and a further reduction in
count rate indicates the presence of beta particles from the source.
When the GM tube is placed behind the lead sheet, a further reduction in count
rate confirms the presence of gamma rays.
221
(b)
(c)
14
16
C 
 147 N  1e
Mass of  21 H  21 H   2.014 0  2
 4.0280 u
Mass of 21 He  4.0026 u
Mass defect,
m  0.0254 u
 0.0255 1.66 10 27 kg
 4.216 4 10 29 kg
Energy released, E  mc2
 4.216 4 10 29   3 108 
2
 3.795 1012 J
222
1. (a)
Angle of incidence, iˆ /°
Angle of reflection, r̂ /°
sin iˆ
sin rˆ
30.0
23.5
0.500
0.398
40.0
30.5
0.643
0.508
50.0
38.0
0.766
0.616
60.0
43.7
0.866
0.691
70.0
48.5
0.940
0.749
(b)
Graph of sin iˆ vs sin rˆ
223
y2  y1
x2  x1
0.850  0.550

0.680  0.440
0.300

0.240
 1.25
(c)
Gradient 
(d)
(i)
The incident ray, refracted ray and normal at the point of incident are all
on the same plane.
(ii)
For a wave travelling from one medium to another the value of
sin iˆ
is a
sin rˆ
constant called refractive index, n.
(e)
Refractive index, n  gradient  1.25
sin iˆ
Using n 
sin rˆ
sin 90
1.25 
sin rˆ
sin 90
 sin rˆ 
1.25
1.00

1.25
 0.800
rˆ  sin 1 0.800
 53.1
(f)
Refractive index of coating, n  nL

nL  n2
 1.252
 1.56
2. (a)
(i)
Velocity is the rate of change of displacement.
(ii)
Acceleration is the rate of change of velocity.
(iii)
Linear momentum is the product of mass and linear velocity.
224
(b)
Total distance
Total time
60

6.5
 9.23 ms -1
(i)
Average speed 
(ii)
Using distance  Average speed  Time
 vu 
s
t
 2 
v0
60  
t
 2 
2  60
v
t
2  60

6.5
 18.46 ms-1
v u
t
18.46  0

6.5
 2.84 ms -1
(iii)
Acceleration, a 
(iv)
a)
He possessed kinetic energy.
b)
Kinetic energy, k.e. 
1 2
mv
2
1
  86  18.462
2
 14 650 J
 14.65 kJ
3. (a)
(i)
225
(ii)
(iii)
(b)
(iv)
Both (ii) and (iii) represent direct current (dc).
(i)
The truth table represents a NAND gate.
(ii)
A
0
0
1
1
4. (a)
B
0
1
0
1
C
0
1
1
1
D
1
0
1
0
E
0
0
1
0
(i)
As the switch is closed the current flows through the brushes and through
the coil.
From Fleming’s left hand rule, the magnetic acting on AB causes a
downward force and on CD an upward force.
The momentum of rotation of the coil allows it to cross the vertical
position after which the current in the coil is reversed to produce
continuous rotation.
(ii)
The purpose of the commutator is to reverse the current in the coil every
half turn thus allowing continuous flow of current and continuous rotation.
226
(b)
Work done
Time
mgh

t
25 10  30

5
 1500 W
(i)
Power provided by the motor, P 
(ii)
Using P  IV
1500  I  24
1500

I
24
 62.5 A
(iii)
Since P  I , greater power will increase the current.
5. (a)




Determine the mass of the metal block using a balance after the holes were
made for the heater and thermometer.
The initial temperature of the block was measured before heating.
Turn on the switch and allow the block to be heated for a measured period
of time.
Ensure that the current and voltage are constant during the heating. Adjust
the thermostat if necessary. Record the readings of the ammeter, voltmeter
and final temperature.
Calculation:
Provided no heat loss,
Heat supplied by heater  Heat gained by metal block
IVt  mc
IVt
 Specific heat capacity, c 
m
(b)
(i)
Heat supplied  Heat gained by liquid
E  mc
13.6 1000  0.1 c   50  25 
13600

c
0.1 25
 5440 J kg-1 °C -1
(ii)
There will be no change.
227
(iii)
The specific heat capacity is fixed for a particular substance and its value
remains constant.
6. (a)
(b)
Radiation
Range in air
Alpha (α)
About 5cm of air
Gamma (γ)
Travels much further
Behaviour in an
electric field
Deflected opposite to
direction of electric
field
No deflection
(i)
Number of neutrons in Xenon (Xe)  143  54
 89
(ii)
Mass defect, m  1.00867  235.04393 
Type of track in a
cloud chamber
Bold and straight
Short faint tracks
 142.93489  89.907 30  3 1.00867  u
 236.052 60  235.868 20
 0.184 40 u
m , in kg  0.18440 1.66 10 27
 3.06104 10 28 kg
Energy released, E  mc2
 3.06104 10 28   3 108 
2
 2.75 1011 J
(iii)
The preferred method would be the artificial decay.
(iv)
This is because it produces much more energy  2.75 1011 J  , than
natural decay  9.98 1013 J  .
228
1. (a)
Velocity is the rate of change of displacement.
Acceleration is the rate of change of velocity.
(b)
Graph of velocity vs time
229
y2  y1
x2  x1
7.5  2.5

6.0  2.0
 1.25 ms -2
(c)
Slope 
(d)
Slope of a velocity vs time graph is equal to the acceleration.
(e)
Resultant force, F  ma
 60  1.25
 75.0 N
(f)
Distance covered after 10 seconds  Area under graph
 Area of trapezium
1
 sum of parallel sides   height
2
1
  2.0  10.0  10.0
2
 60.0 m

2. (a)
(i)
a)
Emr with wave length longer than visible light –
infrared/radiowave
b)
Emr with wavelength shorter than visible light – Gamma rays/
X-rays / ultraviolet rays
(ii)
Name of wave
X-ray
Gamma ray
Radio wave
(b)
(i)
Source
X-ray tube
Radioactive Nuclei
Radio / TV transmitters
Use
To take X-ray pictures
Radiotherapy
Communication
v f
v
 f 

3.0 108

3.0 1012
 1.0 1020 Hz
230
(ii)
sin iˆ (air)
sin rˆ (glass)
sin 
1.5 
sin 35
sin   1.5sin 35
 1.5  0.5736
 0.860 4
  59.4
Refractive index, n 


3. (a)
Circuit symbol
Name of components
Cell
Variable resistor
or
Filament lamp/bulb
a.c. supply
Semi-conductor diode
Fuse
or
(b)
(i)
V  IR
(ii)
Total resistance in circuit, RT 
V
I
24

2
 12 
R2  RT  R1
 12  4
 8
(iii)
Voltmeter reading, V  IR2
 28
 16 V
(iv)
With switch S closed, new total resistance, RT  4 
88
88
 44
RT  8 
231
New current, I 
V
RT
24
8
 3.0 A

4. (a)




(b)
The boiling tube is heated in a water bath until all the naphthalene is completely
melted.
The boiling tube is removed from the water bath and allowed to cool.
The temperature is recorded at fixed intervals (e.g. every minute using a stop
watch) while stirring continuously to ensure equilibrium temperature.
A graph of temperature vs time is plotted to produce the cooling curve.
(i)
Mass of melted ice  110  100
 10 g
(ii)
Heat lost by water  mc
 100  4.2   30  20 
 4 200 J
(iii)
Total heat gained by ice 
Heat required to Heat to warm melted

melt ice
ice to 20°C
 mi l f  mi  cw  
(iv)
Heat lost by water  Total heat gained by ice
4 200  mi l f  mi  cw  
4 200  10l f  10  4.2   20  0 
4 200  10l f  840
4 200  840
10
 336 J g -1
lf 
5. (a)
232



(b)
When the magnet is pushed into the coil, the change in magnetic flux
causes an induced e.m.f.
If the magnet is pushed in at a faster speed, the rate of change of magnetic
flux is increased which in turn causes an increased induced e.m.f.
If a more powerful magnet is used there will also be an increased induced
e.m.f.
(i)
The changing speed of the car will cause a changing magnetic flux and
therefore an induced e.m.f. (current).
(ii)
Increasing speed of a car will increase the rate of magnetic flux cutting
which in turn will increase the charging current to the batteries. Reducing
the car speed reduces the charging current.
(iii)
Conversion efficiency 
Power output
100%
Power input
600 100
100%
200 000
 30%

6. (a)
J.J. Thomson postulated the ‘Plum Pudding’ model. In this model the atom was
looked at as a mass of positive charge dotted with negative electrons that
neutralized the positive charge. The structure was similar to a pudding (+ve)
dotted with currants.
Ernest Rutherford’s nuclear model of the atom from the α – particle suffering
experiment conducted by his students, Geiger and Marsden. From the result of the
experiment, he concluded that the atom consists of a tiny positively charge mass
or nucleus. The rest of the atom was mainly empty space that carried a negative
charge.
(b)
(c)
(i)
Nuclides with identical mass number -
(ii)
The heaviest nuclide is
238
92
Number of neutrons in
238
92
(iii)
Isotopes –
(i)
First half life, t 1 :
40
19
P and
40
19
P and
40
18
R
Xe .
Xe  238  92
 146
39
19
Y (same atomic number, different mass number)
2
t1
t1
2
4000 
 2000 
1000
2
233
2t 1  55.0 s
2
 t1 
2
55.0
2
 27.5 s
Second half life, t 1 :
2
t1
2
1000 
 500
t 1  80.0  55.0
2
 25.0 s
(ii)
27.5  25
2
2
 26.3 s
Average t 1 
234
1. (a)
Graph of Velocity, v/ms-1 vs Time, t/s
235
y2  y1
x2  x1
57.0  11.0

50.0  10.0
 1.15 ms -2
(b)
Slope 
(c)
The slope of a velocity vs time graph is equal to the acceleration.
(d)
(i)
Total distance travelled  Area under graph
 Area under trapezium
1
 sum of parallel sides   height
2
1
  60  180   69
2
 8 280 m

Total distance travelled
Total time taken
8 280

180
 46.0 ms -1
(ii)
Average velocity of taxi 
(iii)
Momentum of taxi  Mass  Velocity
 1500  69
 103500 kg ms -1
(e)
(i)
Displacement is the distance moved in a specific direction.
(ii)
Quantity
Displacement
Acceleration
2. (a)
Scalar
Vector


(i)
Quantity
Specific heat capacity
Symbol
c
S.I. Unit
J kg-1 K-1
Specific latent heat of
vapourisation
lv
J kg-1
236
(b)
(ii)
The heat capacity of substance is the heat energy required to raise the
temperature of the substance by 1 degree Celsius or Kelvin.
(iii)
Heat capacity  Mass  Specific heat capacity
C  mc
(i)
Assuming no heat loss:
Total energy required  Energy to change ice at 0°C to water at 0°C
+Energy to heat water from 0°C to 3°C
 ml f  mc
  0.025  340 000    0.025  4 200  3
 8500  315
 8815 J
3. (a)
Energy
Time
8815

300
 29.38 J s -1 or W
(ii)
Rate of heat energy received 
(i)
In a simple cell the current is due to the flow of positive and negative ions
(i.e. cations and anions).
In the copper wire the current is due to the flow of free electrons.
(ii)
Positive terminal – Carbon rod
(iii)
Negative terminal – Zinc (casing)
(iv)
Using Q  It
Q  0.1 60
 6.0 C
(v)
Since the same charge flows through the circuit, the charge that flows
through the cell  6.0 C.
237
(b)
(i)
Waveform
Wave A
Wave B
Wave C
(ii)
Type of current
d.c.
d.c.
a.c.
For waveform C:
Period,
T  20 s
1
Frequency, f 
T
1

0.20
 5.0 Hz
4. (a)
The double slit is placed directly in front of the ray box and the white screen
placed approximately 1 m from the double slit as show in the diagram above. As
the light passes through each double slit, diffraction occurs i.e. the light spreads
out. The diffracted light from each slit interferes to produce bright and dark bands
due to constructive and destructive interference respectively.
(b)
(i)
Refractive index, n 
sin iˆ (air)
sin rˆ (prism)
sin 30
sin 20
0.500

0.342
 1.46

238
(ii)
n
1.46 
Speed in air, v1
Speed in prism, v2
3 108
v2
 v2  2.05 108 ms-1
5. (a)
v
(iii)
Using f 
(i)
Advantages of using a.c. to transmit electrical power:
 There is less transmission loss with low current a.c.
 Ability to step-up or step-down voltages using transformers

3.0 108
f 
430 10 9
 6.98 1014 Hz
(ii)
Features which enhance efficiency:
 Secondary coil is wound on top of primary to maximize flux
linkage
 Thick copper wire for coils (less resistance) to reduce energy loss
due to heating.
 Soft iron core to reduce energy loss due to magnetization and
demagnetization.
 Laminated core to reduce energy loss due to Eddy currents.
239
(b)
(i)
Using
Vs N s

Vp N p
110 000 900

11000
Np

N p  900 
11000
110 000
 90 turns
(ii)
Using
Vs I p

Vp I s
110 000 8000

11000
Is

I s  8000 
11000
110 000
 800A
(iii)
Efficiency,  

Power output
Power input
Transmission power
Vp I p
 Transmission power   Vp I p
 0.7  11000  8000
 61.6  106 W
 61.6 MW
6. (a)
(i)
Three uses of radioisotopes in medicine:
 Radio therapy - treatment of cancer
 Diagnostics - tracer studies
 Sterilization of medical instruments and bandages.
(ii)
Safety precaution when using radioactive substances:
 Avoid eating or drinking near radioactive substances.
 Use gloves, tongs, body suite, masks or robotics when handling
radioactive substances.
 Use radioactive shielding e.g. lead embedded glass.
 Keep exposure time as short as possible.
 Keep large distance away from radioactive source.
240

(b)
(i)
Use photographic plate badges if working in radioactive
environment.
Number of half lives t 1 in going from 16 dis/min to 1 dis/min
2
t1
t1
t1
t1
2
2
2
2
16 
 8 
 4 
 2 
1
Number of half lives  4
Probable age of plant  4  5600
 22400 years
(ii)
Energy lost by the sun in 1 second, E  mc2
 2.0 109   3 108 
2
 1.8 1026 J
Power output  1.8 1026 W

1.8 1026
1000
 1.8 1023 kW
241
1. (a)
p/cm
10.0
20.0
30.0
35.0
40.0
45.0
q/cm
86.0
76.2
66.0
62.8
57.9
53.6
x/cm
40.0
30.0
20.0
15.0
10.0
5.0
y/cm
36.0
26.2
16.0
12.8
7.9
3.6
(b)
Graph of y/cm vs x/cm
242
y2  y1
x2  x1
33.5  2.5

38.0  4.0
31.0

34.0
 0.91
(c)
Slope, z 
(d)
When plasticine is 27.5 cm from pivot, R:
 x  27.5 cm
From the graph, y  24.0 cm
(e)
The principle of moments states that when a body is in equilibrium, the sum of the
clockwise moments about a point is equal to the sum of the anti-clockwise
moments about the same point.
(f)
(i)
Wp  z  Wm
 z  mg
 0.9  0.05 10
 0.46 N
(ii)
2. (a)
mp 
Wp
g
0.46

10
 0.046 kg
(i)
Name
Mass
Time
Current
Temperature
Length
Symbol
m
t
I
T
l
Base (S.I.) unit
kg
s
A
K
m
(ii)
A linear scale is one which has equal spacing between intervals.
A non-linear scale is one which has unequal spacing between intervals.
(iii)
Non-linear scale – Conical flask
243
(b)
3. (a)
Mass
Volume
Mass

l bh
15000

2  1.1 2.5
 2 727.3 kg m -3
(i)
Density 
(ii)
Pressure 
(i)
E  mc2
E  Change in energy
m  Change in mass
c  Speed of light
(ii)
For:
1) Large amount of nuclear energy produced from small
quantity of raw material
2) There are no emission of smoke or greenhouse gases.
Force
Area of base
15000 10

1.1 2
 68181.8 Nm -2
Against:
1) Accidents can be catastrophic.
2) Hazardous waste produced.
3) Weaponizing the energy threatens the existence of living things.
(b)
Conservation of mass number, A
1  235  A  90  3
 A  143
Conservation of atomic number, Z
0  92  56  Z
 Z  36
(c)
2  3  P 1
P4
11  Q
Q2
244
Hence, X is He (Helium).
4. (a)
(b)
When the switch S is closed, the current flows through the brushes and through
the coil. From Fleming’s left hand rule, the magnetic field acting on the coil
causes one side to move up and the other side to move down.The momentum of
rotation of the coil allows it to move it cross the vertical position after which the
commutator reverses the current in the coil to produce continuous rotation.
(i)
Period, T  30 ms
1
Frequency, f 
T
1

30 10 3
 33.3 Hz
(ii)
Peak current, I o  4 A
Vo  I o R
 4  45
 180 V
Peak-to-peak voltage 180 
180  360
5. (a)
(iii)
Maximum power  IoVo
 4  180
 720 W
(i)
Heat flows from Block B (higher temperature) to Block A (lower
temperature).
(ii)
T / K   / °C  273
T / K  52  273
 325 K
(b)
(i)
Total heat energy transferred to water 
40
 300 103
100
 120000 J
Using E  mc
120000  2  4.2 103 
120 000
  
8.4 103
 14.3 °C
245
Final temperature of water  27 14.3
 41.3 °C
(ii)
6. (a)
(b)
a)
The black surface makes it a good absorber of heat.
b)
In the collector reservoir, the hot water is at the top because of
convection.
c)
The glass cover of the solar collector allows short wavelength
radiation to enter and prevents the long wavelength radiation from
leaving.
d)
The copper tubing is a good conductor of heat and allows the
transfer of heat from the blank surface absorber plate to the water
in the tubing.
e)
The insulation reduces the transfer of heat by conduction from the
collector to the surroundings.
(i)
In Longitudinal waves the vibration of the parties are parallel to the
direction of wave travel.
In transverse waves the vibration of the particles are perpendicular to the
direction of wave travel.
(ii)
Using v  f 
v
 f 

3 108

700 10 9
 4.3 1014 Hz
(i)
A series of sound pulses was transmitted into the water from the ship.
The time taken, t for the reflected sound pulses to return to the ship was
measured. The depth of the oil plume was determined by multiplying the
t
speed of the pulse by .
2
(ii)
Depth of oil plume  Speed in water 
t
2
0.3
2
 217.5 m
 1450 
246
1. (a)
Graph of Potential difference (V) vs Current (I)
247
(b)
Points used (0.56, 42.0) and (0.1, 7.5).
y y
Gradient  2 1
x2  x1
42.0  7.5

0.56  0.1
34.5

0.46
 75 V/A
(c)
The gradient represents the resistance.
(d)
The potential difference is directly proportional to the current. (Graph is a straight
line through the origin.)
(e)
(f)
The readings from the voltmeter and ammeter are taken.
The rheostat is adjusted and new readings are recorded.
This is repeated at least five times. Then a graph of V vs I is plotted.
(g)
For parallel arrangement of resistors:
1
1
1
1
 

RT R1 R2 R3
1 1 1 1
  
RT 8 12 15
1 15  10  8

RT
120
33
120
11

40

248
 RT 
40
11
 3.6 
2. (a)
(i)
Physical Quantity
Area
Volume
Density
Derived S.I. unit
m2
m3
kg m-3
Physical Quantity
Mass
Time
Derived S.I. unit
kg
s
(ii)
Or
Current
Temperature
Length
(b)
A
K
m
Mass
Volume
102

150
 0.68 g cm-3 / 680 kg m-3
(i)
Density 
(ii)
Volume 
(iii)
Relative density of gasoline 
Mass
Density
325

13.6
 23.9 cm3
Density of gasoline
Density of mercury
0.68

13.6
 0.05
249
3. (a)
(b)
Weight of yacht  mg
 8300 10
 83000 N
83000
7 000
 11.9
Minimum number  12 types
Number of tyres required 

(c)
1. The resultant force in any direction is zero.
2. The sum of the moments about any point is zero.
(d)
(i)
For a body totally or partially immersed in a fluid, the weight of the fluid
displaced is equal to the upthrust.
(ii)
The entire yacht is not solid steel but constitutes other materials including
air. So its overall density is less than sea water. This means that it will
displace its own weight in sea water while it is still afloat.
(iii)
Weight of sea water displaced  Weight of yacht
 83000 N

Mass of sea water  8300 kg
Mass
Volume of sea water displaced 
Density
8300

1025
 8.1 m3
4. (a)
(b)
A narrow beam of α particles was directed towards a very thin sheet of gold foil.
The scattering was monitored using a rotating scintillation microscope which can
detect α particles.
This was enclosed in an evacuated chamber to avoid interference from air
molecules. The results showed that:
1) Most of the α particles went straight through deflected which indicates that
most of the atom was empty space.
2) Few α particles were deflected between 0 - 90º which indicates that the atom
has a small positively charged nucleus.
3) Even fewer α particles were rebounded which indicates that the nucleus has a
mass.
(i)
235
92
144
1
U  01 n 
 90
36 Kr  56 Ba  2 0 n  Energy
250
(ii)
Mass defect   235.118  1.009    89.947  143.881  2.018 
 0.281 u
E  mc2
Using
E   2811.66 10 27    3.0 108 
2
 4.189 101 J
5. (a)
(b)
p1V1 p2V2
pV

or
 constant
T1
T2
T
(i)
General Gas Law:
(ii)
According to the kinetic theory, the air molecules inside the balloon are in
continuous random motion. These molecules strike the walls of the
balloon causing a change of momentum. The rate of change of momentum
is equivalent to the force exerted by the molecules on the balloon. This
force acting on the surface of the balloon gives the pressure.
(i)
Energy used, Q  Energy required to (change ice at 0°C to water at 0°C
 Heat water 0°C to water at 100 °C + change water at
100°C to steam at 100 °C)
Q  ml f  mc  mlv
 2 000 g  2 kg 
  2  330 000    2  4 200 100    2  2 250 000 
 660 000  840 000  4500 000
 6 000 000 J
(ii)
Energy
Time
6 000 000

6 000
Power 
 1000 W
6. (a)
(i)
Three differences between ‘light waves’ and ‘sound waves’:
 Light waves do not require a medium whereas sound waves require
a medium for travel.
 Light waves are transverse waves whereas sound waves are
longitudinal waves.
 Light waves can be polarized whereas sound waves cannot be
polarized.
 Light waves belong to the e.m. spectrum whereas sound waves do
not.
251
(ii)
(b)
Electromagnetic waves mare transverse waves which means that they can
be polarized. They travel at a speed of 3.0 108 ms-1 in a vacuum and do
not require a medium for travel. The progressive electromagnetic waves
can transfer energy from one place to another and undergo reflection,
refraction and diffraction.
 30.3 s
30.3
Time for each echo 
50
 0.606 s
Time for 50 claps
Distance sound travelled  100  2
 200 m
Distance
Time
200

0.606
 330 ms -1
Speed 
(c)
Using v  f 
v


f
3.0 108

100 106
 3.0 m
 300 cm
252
1. (a)
Graph of Vs/V vs Vp/V
253
(b)
Points selected for gradient  (7.2, 65) and (1.6, 15)
y y
Gradient, S  2 1
x2  x1
65  15

7.2  1.6
50

5.6
 8.9
(c)
(i)
Gradient, S 
(ii)
Using
Vs
Vp
Vs N s

Using
Vp N p
750
8.9 
Np
750

Np 
8.9
 84.3

(d)
(i)
Vs I s

Vp I p
1.6
8.9 
Is
1.6
Is 
8.9
 0.18 A
Power in secondary
 100%
Power in primary
IV
 s s 100%
I pV p
Efficiency,  
0.15
 8.9 100%
1.6
 83.4%

(ii)
Ideal transformer has an efficiency of 100% i.e. no power loss.
254
(e)
For efficient function of transformer:
1) Laminated care (eliminate Eddy current)
2) Soft iron care (easy magnetization and demagnetization)
3) Thick copper wire windings (low resistance)
4) Secondary coil wound on top primary (better magnetic linkage)
2. (a)
Quantity
Diameter of wire
Volume of liquid
Temperature
Weight/Force
Time
(b)
(c)
(i)
Force acting on stone – weight due to gravity
(ii)
The velocity is increased from zero to a maximum as it hits the ground i.e.
it is accelerated.
(i)
Weight  mg
 0.06  10
 0.6 N
(ii)
a)
b)
3. (a)
Instrument
Micrometer screw gauge or Vernier
caliper
Measuring cylinder/Burette
Thermometer
Spring balance
Stop watch/clock
Resultant, R  11.8 cm  23.6 ms-1
Direction from OA  9.6
(i)
255
(ii)
Type of Thermometer
Clinical thermometer
Laboratory thermometer
Thermocouple
(b)
(i)
Operating Temperature
Range/°C
20 to 110
250 to 800
35 to 43
Pressure at 20 m below surface  Atmospheric pressure + Pressure due to
Depth
 PA   gh
 1.01105  1025 10  20 
 3.06 105 Pa
(ii)
Using

p1V1

T1
V2

V1
p2V2
T2
p1T2
p2T1
3.06 105   25  273

1.01105  10  273
 3.19
4. (a)
(i)
The normal is an imaginary line draw perpendicular to the reflecting
surface at the point of incidence.
(ii)
Angle of incidence is the angle between the incident ray and the normal at
the point of incidence.
(iii)
Angle of reflection is the angle between the reflected ray and the normal at
the point of incidence.
(b)
Features of the image produced by plane mirror:
 Image is literally inverted
 Image is erect
 Image is same size as object
 Image is virtual
 Image distance is the same as object distance
(c)
To reverse the laterally inverted word when viewed on the rear-view mirror. This
will allow the correct wording ‘AMBULANCE’ to be seen from the reflection of
the mirror.
256
(d)
(i)
Angle of incidence, iˆ  90  30
 60
sin iˆ
sin rˆ
sin 60
1.5 
sin rˆ
sin 60
 sin rˆ 
1.5
0.866

1.5
 0.577
ˆr  35

Using n 
(ii)
5. (a)
(i)
ˆ  60 
Angle of refraction on QR boundary  CBD
The semi-conductor diode is defective if:
 It conducts in both forward and reverse bias orientations
 It does not conduct in both forward and reverse bias orientations.
(ii)
(iii)
Not-gate
Input
0
1
(b)
(i)
Output
1
0
Equivalent resistance (series), RT  R1  R2  R3
 2  6  12
 20 
257
(ii)
Equivalent resistance (parallel),
1
1
1
1
 

RT R1 R2 R3
1 1 1 1
  
RT 2 6 12
6 2 1
 
12 12 12
9

12
3

4
4
RT  
3


(c)
6. (a)
The decision for series circuit was not wise because:
In series circuit, if one bulb blows all the others would not light
The p.d. across each bulb is reduced in a series circuit.
Alternative sources of Energy in the Caribbean:
 Solar – Water heating, generate electricity
 Wind – Generate electricity
 Geothermal – Generate electricity
 Hydro – Generate electricity
Importance:
 Less reliance on depleting fossil fuel
 Less emission of CO2 and other pollutants
 Renewable sources of energy
 Cheaper production of energy
(b)
(i)
When the ball is kicked, it is given kinetic energy. As it moves towards the
goalkeeper the kinetic energy is converted into potential energy as it rises.
At the highest point the kinetic energy is minimum and the potential
energy is maximum. The potential energy decreases to zero and kinetic
energy to a maximum on striking the ground.
(ii)
Difference in energy  P.E.  K.E.
1
 mgh  mv 2
2
(iii)
Momentum of ball  Mass  Velocity
 0.43  7
 3.01 kg ms-1
258
1. (a)
Graph of Length/m vs Load/N
259
(b)
Points considered for gradient: (9.6, 0.50), (2.0, 0.29)
y y
Gradient  2 1
x2  x1
0.50  0.29

9.6  2.0
0.21

7.6
 0.028 m N -1
(c)
The gradient can be used to determine the spring constant, k, where k 
(d)
(i)
The original length of this spring is the length when F  0 N .
Form graph when F  0 , Length  0.24 m
(ii)
Using F  kx (Hooke's Law)
(where x is extension)
0.7 10  1 x

x  0.20 m
1
.
Gradient
 New length  0.24  0.2
 0.44 m
(e)
(iii)
The graph would change from straight line to curve.
(i)
The region of proportionality exist in the straight line section of the graph.
(ii)
Quantity
Load
Extension
2. (a)
(i)
Scalar
Vector


The specific heat capacity of a substance is the heat energy required to
change the temperature of 1 kg of the substance by 1 degree Kelvin or
Celsius.
(ii)
Quantity
Heat capacity
Specific latent heat of fusion
(iii)
General Gas Law:
Symbol
C
lf
S.I. Unit
J K -1
J kg
p1V1 p2V2

T1
T2
260
where p1  intial pressure
V1  initial volume
T1  initial temperature
(b)
(i)
p2  final pressure
V2  final volume
T2  final temperature
Energy, Q1 , required to heat water to 100°C:
Q1  mc
 8  4 200  100  33
 2 251200 J
 22.5 105 J
(ii)
Energy, Q2 , required to change water at 100°C into steam at 100°C:
Q2  mlv
 8  2300 000
 18 400 000 J
(iii)
Total energy, QT , required to heat 8 kg of water at 33º C to steam at
100ºC:
QT  Q1  Q2
 2 251200  18 400 000
 20 651200 J
 20.7 MJ
3. (a)
(i)
Snell’s law states that for light rays passing from one transparent medium
to another, the ratio of the sine of the angle of incidence to the sine of the
sin iˆ
angle of refraction i.e.
is a constant called the refractive index, n.
sin rˆ
(ii)
261
(b)
4. (a)
(b)
(i)
The ray of light will enter the prism undeviated since it enters at 90 º. The
ray is then incident on AC at an angle of 45º. Since this angle is greater
that the critical angle for glass (42º), total internal reflection occurs at AC.
This reflected ray strikes BC at an angle of incidence of 45º, again causing
total internal reflection.
The reflected ray from BC is incident on AB at 90º and therefore passes
through undeviated.
(ii)
After emerging, the ray had turned through 180º.
(i)
When two or more bodies collide, the total momentum of the bodies
before collision is equal to the total momentum of the bodies after
collision provided no external forces are acting on the bodies.
(ii)
For a launching rocket, the momentum of the rocket in the upward
direction is equal to the momentum of the extruded burnt fuel in a
downward direction.
(i)
Initial momentum of track  mv
 1250  25
 31250 kg ms -1
(ii)
Initial momentum of car  625  30
 18750 kg ms-1
(iii)
By the principle of conservation of momentum:
Total momentum before collision  Total momentum after collision
31250  18750  1250  625   v
12500  1875v

v  6.7 ms-1 in direction due North
5. (a)
262




(b)
The circuit is set up as shown in diagram with the ammeter in series with
the test resistor R and the voltmeter in parallel to R.
The variable resistor Q is adjusted and the values of I and V are recorded
from the ammeter and voltmeter respectively.
This is repeated to obtain at least 5 pairs of values for I and V.
A graph V vs I is plotted and the gradient will give the resistance of the
metallic conductor R.
R2 R3
 R4
R2  R3
3 3
RT  3 
3
33
 3  1.5  3
Total resistance, RT  R1 
(i)

 7.5 
V
RT
12

7.5
 1.6 A
(ii)
Using I 
(iii)
Using V  IR
 V  I   Combined resistance of R2  R3 
 1.6  1.5
 2.4 V
6. (a)



The background count is measured without any radioactive source in
place.
With the β source in place the count rate is obtained for increasing number
of Aluminium sheets until it drops to the background count.
The β – source is replaced by the γ source, the process is repeated and the
results compared.
263
(b)
(i)
1
0
n  23290Th 
 23390Th
Th 
 01 e  233
91 Pa
233
90
233
91
(ii)
Pa 
 01 e  233
92 U
Mass defect  L.H.S.  R.H.S.
132.91525  97.91033  
  233.03964  1.00867   

 3  1.00867

 234.04831  233.85159
 0.19672 u
Energy released, E  mc2
 0.19672 1.66 10 27   3.0 108 
2
 2.94 1011 J
264
1. (a)
Graph of Image size, I vs Object size, O
265
(b)
Points selected for gradient: (2.8, 1.40), (0.8, 4.0)
y y
Gradient, G  2 1
x2  x1
14.0  4.0

2.8  0.8
10.0

2.0
 5.0
(c)
The gradient G represents magnification.
(d)
The focal length in the distance between the principal focus and the optical centre
of the lens.
(e)
Focal length is associated with lens.
(f)
(i)
For a plane mirror:
Image size  Object size  10.0 cm
(ii)
Magnification 
2. (a)
Image size
Image distance
or
Object size
Object distance
10.0

10.0
1
(i)
266
(ii)
1)
2)
3)
(b)
Force
Gravitational force (weight)
Drag force (air resistance)
Upthrust
Centripetal, magnetic,
nuclear, tension, etc.
Situation
A javelin falling in the air
An object immersed in a liquid
Object in circular motion, between
magnets or current carrying
conductors, nucleus of atoms,
stretched springs, etc.
(i)
(ii)
By the principle of moments:
Clockwise moments  Anti-clockwise moments
W  0.5   500  1   400  0.5 


0.5W  700
700
W
0.5
 1400 N
3. (a)
(iii)
Taking moments about X:
500  y  1400  0.5
700
 y
500
 1.4 m (on the same side of the fulcrum)
(i)
A laboratory thermometer – volume of a liquid
(ii)
A thermocouple – E.m.f.
(b)
Use of thermometer
To measure body temperature
To measure temperature lower than
40C
Rapidly changing temperature
Design feature
Small temperature range, constriction
in the bore
Alcohol thermometer
Junction of small mass
267
(c)
(i)
Charles’ law states that the volume of a fixed mass of gas is directly
proportional to the absolute temperature provided the pressure remains
constant.
(ii)
From Charles’ Law:
V1 V2

T 1 T2
V2
40


 273  30   273  70 
40  343

V2 
303
 45.28 cm3
Increase in volume  45.28  40
 5.28 cm3
5.28
100%
40
 13.2%
Percentage increase 
4. (a)
(b)
(i)
Properties of electromagnetic waves:
 They are all transverse waves
 They travel at the same speed in a vacuum i.e. 3 x 108 ms-1
 Can travel in a vacuum
 Can be polarized
(ii)
Other types of electromagnetic waves:
 Gamma rays - lower
 X-rays
- lower
 UV
- lower
 IR
- higher
 Radio
- higher
(i)
a)
Using f 

c

3 108
f1 
2 10 7
 1.5 1015 Hz
3 108
f2 
6.5 10 5
 4.6 1012 Hz
268
b)
Decrease in frequency, f  1.5 1015  4.6 1012
 1.495 1015 Hz
(ii)
E1  k 1.495 1015  J
(iii)
New, frequency f 
…Equation 
3 108
6 10 7
 0.5 1015 Hz
f  0.5 1015  4.6 1012
 4.954 1014
E  k  4.954 1014  J
…Equation 
Equation   Equation :
14
E k  4.954 10 

E1 k 1.495 1015 
 0.33
 E  0.33E1
5. (a)




The circuit is set up as shown in diagram with the ammeter in series with
the filament lamp, L and the voltmeter in parallel to L.
The variable resistor Q is adjusted and the value of I and V are recorded
from the ammeter and voltmeter respectively.
This is repeated to obtain at least 5 pairs of wide range values of I and V.
A graph of I vs V is plotted.
269
(b)
(i)
Total resistance, RT  2 
3 5
35
 3.785 kΩ
Using V  IR
 V  1103  3.875 103
 3.875 V
(ii)
6. (a)
 3 
Current in 5 k   
1 mA
 35
 0.375 mA
(iii)
If the 2 kΩ resistor ,burns out the circuit is now open and no current will
flow.
(i)
226
86
(ii)
Number of neutrons in
43
Ra 
 222
84 Rn  2 He  
226
86
Ra  A  Z
 226  86
 140
(b)
An atom is normally neutral because it has equal number of protons and
electrons i.e. equal number of oppositely charged particles. The atoms are stable
because the nuclear forces holding the protons and neutrons together in the
nucleus is greater than the repulsive electrostatic forces. Larger atoms require a
greater neutron : proton ratio for stability.
(c)
Isotopes are atoms of the same element with the same atomic number but
different mass number i.e. same number of protons but different number of
neutrons.
270
(d)
For element X:
(i)
Mass number  5  6
 11
Atomic number  5
Charge on nucleus  Positive (protons)
(ii)
Isotope of X  125 X
271
1. (a)
Graph of Temperature,  /°C vs Time, t/min
(b)
Melting point of substance  70.0°C
(c)
(i)
The substance is changing between liquid and solid state between B and
C.
272
(d)
(ii)
As the substance is changing state, latent heat of fusion is removed from it
resulting in no temperature change.
(i)
At C, the substance is in solid phase.
(ii)
Between C and D the substance is cooling.
Heat loss, Q  Heat loss from A to B + Heat loss from B to C + Heat loss
from C to D
Q  mc  ml f  mc
(e)
  0.015 1763  90.0  70.0   0.015  215000 
  0.015 1760   70.0  57.5
 528.9  3225  330.56
 4 084.46 J
 4.08 kJ
(f)
Physical quantity
Heat capacity
Symbol
C
S.I. Unit
J K-1
lv
J kg-1
Specific latent heat of
vapourisation
2. (a)
(i)
Equivalent derived unit for Joule:
 kg m2 s-2
 Nm
(ii)
Application of solar energy:
 Solar water heater
 Watches
 Calculators
 Satellites
(iii)
Advantage of using solar energy in the Caribbean:
 plenty of sunlight in the Caribbean
 renewable source
(iv)
Alternative energy
Geothermal
Wind
Hydro-electricity
Source
Hot rocks deep in the earth
Wind, especially along coastline
Flowing rivers, waterfalls
273
(b)
(i)
Gravitational potential energy, E p  mgh

E p  0.44  9.8 12
 51.74 J
(ii)
Assuming no energy loss:
Kinetic energy, Ek  Initial potential energy, E p
1
Ek  mv 2
2
 51.74
2  51.74
v
m

2  51.74
0.44
 15.3 ms -1

(iii)
Momentum, p  mv
 0.44 15.3
 6.7 kg ms -1
3. (a)
(b)
(i)
p1V1 p2V2

T1
T2
(ii)
p1  initial pressure
T1  initial temperature
V1  initial volume
(i)
Using
p2  final pressure
T2  final temperature
V2  final volume
p1 p2

T1 T2
p2
2  105

 273  23  273  34 

p
2  105
 2
296
307
2  105  307
p2 
296
 2.07 105 Nm -2
274
(ii)
An increase in pressure causes an increase in the kinetic energy of the air
molecules in the tyre. This causes the air molecules to strike the inner
walls of the tyre with greater frequency and momentum, thus increasing
the pressure.
(iii)
Using

4. (a)
(b)
V1 T1

V2 T2
V2 273  34

V1 273  23
307

296
 1.04
Laws of reflection:
 The incident ray, refracted ray and normal at the point of incidence are all
on the same plane.
 For light rays passing from one transparent medium to another, the ratio of
the sine of the angle of incidence to the sine of the angle of refraction
sin iˆ
is a constant called refractive index.
sin rˆ
(i)
Assuming Nemo is along the straight line with BA:
ˆ
Angle C  NBA
 90  42
 48
(ii)
Using n 
1
sin cˆ
1
n
sin 48
1

0.7431
for critical angle
 1.35
(iii)
For Nemo to see Bruce it means that is there is total internal reflection i.e.
angle of incidence on the water-air boundary is greater than the critical
angle c. Therefore the horizontal distance from Bruce’s eye to B is 5m.
So that the distance from Bruce’s eye from Nemo’s eye is 10m.
275
5. (a)
(b)
Ways of conserving existing energy sources:
 Replace incandescent bulbs with CFL bulbs to conserve electrical energy.
 Switch off lights and appliances when not in use to conserve electrical
energy.
 Use car-pooling and more walking to save fuel or chemical energy.
 Use energy efficient appliance to conserve electrical energy.
(i)
Number of hours in two weeks  2  7  24
 336 hours
Power of bulbs in kWh  0.06  336
 20.16 kWh
6. (a)
(ii)
Electrical charges for two weeks  20.16  $0.26
 $5.24
(iii)
Efficiency 
(i)
‘Half-life’ is the time taken for half the number of radioactive atoms to
disintegrate.
Useful energy output
 100%
Energy input
20.16  15.5

100%
20.16
4.66

100%
20.16
 23%
t1
(ii)
N t 12 N t 12 N t 12 N
N 
 
 
 

2
4
8
16
 5t 1  20 days
2
2
t1 
2
20
5
 4 days
(b)
Uses of radioactive isotopes:
 Tracers e.g. detect leaking gas lines
 Dating i.e. estimating age of artifacts
 Nuclear energy
 Radio therapy
276
Precautions when handling radioisotopes:
 Uses thick gloves or remote control mechanical arms.
 Store radioactive materials in thick lead containers.
 Use protective clothing that must not be removed from lab.
 Use proper labelling.
(c)
Using E  mc2
E

m  2
c
6.7 1010

2
 3 108 
 7.4  10 7 kg
 New mass  1  7.4 10 7
 0.9999926 kg
277
Corection: This is June 2015 Paper 2
1. (a)
Graph of Activity (A) vs Time (t)
278
(b)
t  1.5s
t  9.5s  t 1  8.0 s
At 36s-1 ,
At 18s-1 ,
2
t  7.0 s
t  14.0s  t 1  7.0 s
-1
At 24s ,
At 12s-1 ,
2
t  10.5s
t  18.0s  t 1  7.5s
-1
At 16s ,
At 8s-1 ,
2
 t 1  average  
2
8.0  7.0  7.5
3
 7.5 s
(c)
After 25 days, activity  4.5 s-1
(d)
(i)
Radioactive emissions:
 Alpha particles
 Beta particles
 Gamma Rays
(ii)
Most dangerous radioactive emission is alpha particles.
(i)
C  mc
(ii)
C  Heat capacity
c  Specific heat capacity
2. (a)
(iii)
Specific heat capacity, c
The quantity of heat energy
required to change the temperature
of 1kg of a substance by 1 degree
C or K.
Unit: J kg-1 K-1
E
c H
m
Each specific substance has a
constant value
Heat capacity, C
The quantity of heat energy
required to change the temperature
of the total mass of a substance by
1 degree C or K.
Unit: J K-1
E
C H

Varies according to the mass of the
substance
279
(b)
(c)
(i)
Using E  mc
 2  4 200  100  37 
 529 200 J
(ii)
Using E  mlv
 2  2.3 106
 4.6 106 J
3. (a)
(iii)
Total heat energy  529200  4.6 106
 5129200 J
(i)
‘Electrical resistance’ is the opposition to the flow of an electrical current.
V
It is calculated from R  .
I
(ii)
Meter
Ammeter
How connected in a
circuit (series or
parallel)
Series
Resistance
(high or
low)
Low
Voltmeter
Parallel
High
Reason for size of
resistance
So as not to affect the
current in the circuit
So as to draw as little
current as possible
from the circuit
280
(b)
(i)
2 6
26
12

8
 1.5 
RT 
I  A1  
V
RT
12
1.5
8 A

(ii)
4. (a)
(b)
Since power supply is connected directly across the 2  resistor, then
V
Current I  A2  
R
12

2
6 A
Newton’s three laws of motion:
 A body will continue in a state of rest or of uniform motion in a straight
line unless compelled by an external force to act differently.
 The rate of change of momentum is directly proportional to the applied
force and takes place in the direction in which the force acts.
 If an object A exerts a force on object B then object B will exert an equal
but opposite force on object A i.e. to every action there is an equal and
opposite reaction.
(i)
Initial momentum  mv
 70  26
 1820 kg ms-1 (in a direction towards the wall)
(ii)
Change in momentum, p
Time
mv  mu
F
t
m v  u 

t
70  0  26 

0.1
 18 200 N (in a direction against the dummy)
Average force, F 
281
(iii)


5. (a)
(b)
p
t
p
t
F
1820

45000
t  0.040 s
Using F 
(i)
The graph of volume versus temperature in degrees Celsius will produce
A straight line intercepting the volume axis (i.e. not through the origin).
When this graph is extrapolated to the temperature axis, it cuts the axis at
273°C which represents the absolute zero on the Kelvin scale.
(ii)
T / K   / C  273
(i)
Using pV
1 1  p2V2
5  50  1V2

V2  5  50
 250 ml
(ii)
Using
p1 p2

T1 T2
p2
5

 273  25  273  60 
p
5
 2
298 333
333  5
p2 
298
 5.6 atm
282
6. (a)
(b)
(i)
(ii)
Magnification, m 
(i)
Using



Image distance, v
height of image
or
Object distance, u
height of object
1 1 1
 
f u v
1
1 1


10 20 v
1 1 1
 
v 10 20
2 1

20
1

20
v  20 cm
Image is on opposite side of lens.
(ii)

(iii)
v
u
20
m
20
1
Using m 
The image formed is real.
283
Correction: This is January 2015 P2
1. (a)
Length of
Pendulum
Ɩ (m)
0.20
0.30
0.40
0.50
0.60
0.70
Time for 20
Oscillations,
t (s)
18.00
21.91
25.40
28.28
31.10
33.80
Time for 1
Oscillation
(period) T (s)
0.90
1.10
1.27
1.41
1.56
1.69
Period
squared
T2 (s2)
0.81
1.21
1.61
1.99
2.43
2.86
Graph of Period squared ( T 2 ) vs Length (L)
284
(b)
Points selected for gradient: (0.10, 0.40), (0.62, 2.50)
y y
Gradient  2 1
x2  x1
2.50  0.40
0.62  0.10
2.1

0.52
 4.0 s 2 m -1

(c)
1
2
g  4 2  T
l
1
gradient
1
 4  3.142 
4.0
-2
 9.86 ms
 4 2 
(d)
2. (a)
(i)
Forms of energy
Mechanical
Chemical
Thermal
(ii)
Example
A moving car/object
A car battery
A pot of boiling water
Unit of energy : Joule (J)
285
(b)
(iii)
A Joule is the work done by a force of one Newton when its point of
application moves one metre in the direction of action of the force.
(i)
Assuming no friction:
Amount of energy  Work done
 Force  Distance
 F s
 mgs
 60 10 150
 90 000 J
(ii)
Power 
Work done
Time taken
90 000
25
 3600 W
 3.6 kW

(iii)
3. (a)
More power would be needed since energy would be lost due to friction.
(i)
Temperature-Fixed point
Upper fixed point
Lower fixed point
(ii)
Value from Dr. T’s Bag
100°C
0°C
The lower fixed point is the temperature of pure melting ice at normal
atmospheric pressure.
(iii)
Type of thermometer
Liquid-in-glass thermometer
Platinum resistance thermometer
OR
Constant volume gas thermometer
Thermocouple
(b)
Physical quality
Volume of a liquid
Resistance of platinum
Pressure of gas
E.m.f. between the junctions
Initial temperature, T1  27  273  300 K
Initial pressure,
p1  220  100  320 kN m-2
Final pressure,
p2  250  100  250 kN m-2
286
p1 p2

T1 T2
320 350

300 T2
350
T2 
 300
320
Using


4. (a)
 328 K
T2  382  273
 55°C
(i)
The laws of reflection :


(ii)
The angle of incident, iˆ is equal to the angle of reflection, r̂ .
The incident ray, the reflected ray and the normal at the point of
incidence are all on the same plane.
The raindrops on the windscreen act like prisms which cause dispersion of
light to produce the ‘glare’.
287
(b)
(i)
Using refractive index, nw 
sin bˆ (air)
sin aˆ (water)
sin 45
sin 32
0.7071

0.5229
 1.33
nw 
(ii)
Using n 
sin bˆ
sin aˆ
 sin bˆ  n sin aˆ
 1.36  sin 32
 1.36  0.5299
 0.720 7
bˆ  46

5. (a)
(b)
(iii)
The pencil ‘bends’ more in ethanol than in water.
Ethanol has a higher refractive index which implies a decrease in angle of
refraction i.e. more ‘bending’.
(i)
V  IR , where R  resistance of conductor
(ii)
For resistors in parallel:
1
1
RT  
R1 R2
(i)
Using V  IR
V

R
I
3.0

0.30
 10.0 
288
(ii)
Total resistance, RT 
R1 R2
R1  R2
10 100
10  100
1000

110
 9.1

V
R
3.0

9.1
 0.33 A
Using I 
(iii)
If the rheostat is reduced too much the overall resistance in the circuit will
decrease causing the current to increase. If this reaches above the current
rating of the bulb it can cause it to blow as well as the heating up of the
connecting wires.
(iv)
For small currents the resistance is constant, hence the straight line. As the
current increases, heat is produced causing the resistance to increase hence
the curved line in the graph.
289
6. (a)
(i)



(ii)
(b)
(c)
The alpha particles are deflected in the same direction as the
electric field i.e. away from positive and towards negative.
The beta particles are deflected opposite to the direction of the
electric field with greater deflection since they are lighter than
alpha particles.
Gamma rays carry no charge and are therefore undeflected by the
electric field.
For a uniform magnetic field into the page:
 Alpha particles would be deflected towards the left (Fleming’s left
hand rule) in a circular arc.
 Beta particles would undergo greater deflection towards the left.
 Gamma rays would not be deflected.
210
82
0
Pb 
 210
83 Bi  1 e (beta)
210
83
0
Bi 
 210
84 Po  1 e
210
84
Po 
 82 Pb  24 He (alpha)
206
Mass defect, m  0.214 1.66 10 27
 3.3432 10 28 kg
Energy released, E  mc2
 3.3432 10 28   3 108 
2
 3.0 1011 J
290
1. (a)
Graph of Velocity, V vs Time, t
291
(b)
(i)
Acceleration during AB  Slope of AB
Vertical displacement

Horizontal displacement
30

20
 1.5 ms -2
(ii)
Velocity after 70 seconds  15 ms-1
(iii)
Total distance travelled  Area under graph (trapezium)
1
  sum of parallel sides   height
2
1
  40  80   30
2
1
 120  30
2
 1800 m
(c)
Over the period BC the car is travelling with a constant velocity of 30 ms-1. There
is no acceleration.
(d)
Velocity is defined as the rate of change of displacement or the rate of change of
distance in a particular direction.
2. (a)
Physical quantity
(b)
Unit
(i)
Kinetic energy is the energy of a body due to its motion.
(ii)
Potential energy is the energy of a body due to its state or position.
(iii)
E p  mg h
292
(c)
(d)
(iv)
At the top of the waterfall the water has potential energy and as it fall
down the waterfall the potential energy is converted into kinetic energy.
(i)
A - potential energy
(ii)
B – kinetic energy
(iii)
C – potential energy
(iv)
At, D, the type of energy increasing is potential energy.
Reasons: The pendulum bob is gaining height as it approaches A
1 2
mv
2
1
  0.4  52
2
5 J
Ek 
3. (a)
Type of thermometer
Design feature
Mercury-in-glass
Narrow bore
laboratory
thermometer
Clinical thermometer
Constriction in bore
Thermocouple
Small junctions with
a low heat capacity
Reason for design feature
To detect small changes in
temperature i.e. sensitive
Retaining a measured
temperature
Ability to measure rapidly
changing temperatures
(b)
The upper fixed point is the temperature of steam at standard atmospheric
pressure and is 100 ºC.
The lower fixed point is the temperature of pure melting ice and is 0º C.
(c)
(i)
Pressure (Pa)
1.1105
1.2 105
1.3 105
(ii)
Temperature(°C)
35
63
91
Temperature(K)
308
336
364
According to the Pressure Law p  T (absolute temperature) provided V
p
is constant   constant, k
T
Testing data:
293
p1 1.1105

 357
T1
308
p2 1.2 105

 357
T2
336
p3 1.3 105

 357
T3
364
 Pressure Law is supported by the set of data.
4. (a)
(b)
(c)
(i)
The incident ray, reflected ray and normal at the point of incidence are all
on the same plane.
(ii)
Properties of the image formed by a plane mirror:
 Same size as object
 Same distance from mirror as object
 Virtual
 Laterally inverted
 Upright
(i)
Image distance  Object distance
 85
 13 m
(ii)
Since light travels in straight lines the truck driver can only view to car
from the side mirror as the truck is opaque. Car drives on the other hand
can use the rear view mirror that can view vehicles through the back
windscreen.
(i)
Using n 
(ii)
An increase in refractive index implies a greater angle of refraction and
therefore increased lateral displacement.
sin iˆ
sin rˆ
sin iˆ
 sin rˆ 
n
sin 30

1.3
0.5

1.3
 0.384 6

rˆ  22.6
294
5. (a)
(i)


(ii)
A real image is formed when the object distance is greater than the focal
length.
A virtual image is formed when the object distance is less than the focal
length.
Real images are inverted and virtual images are upright.
(iii)
(b)
Image size
Object size
3.6

2.4
 1.5
(i)
Magnification, m 
(ii)
Using m 

(iii)
Using


v
u
v  mu
 15  2.0
 30 cm
1 1 1
 
f u v
1 1
1


f 20 30
3 2

60
5

60
1

12
f  12 cm
295
6. (a)
(b)
The circuit consists of a 3 V battery power supply which is connected in series
with a switch, a rheostat, a fixed resistor and an ammeter. A voltmeter is
connected in parallel to the fixed resistor.
(i)
R1 R2
R1  R2
1 2
R
1 2
2
 
3
Resistance across BC: R 

(ii)
 1 2 
Total resistance in circuit, RT  
3
 1 2 
2

RT   3
3
2
3 
3
 3.67 
V
R
12
I
3.67
 3.27 A
Using I 

(iii)
Using P  I 2 R
 3.27 2  3
 32.1 W
296
1. (a)
Graph of Activity (A) vs Time (t)
297
(b)
(i)
For activity
A0  80, t0  0 hours
For half the activity, A1  40, ti  1.6 hours
t 1  1.6 hours
 Half life,
2
(ii)
When
A2  50, t2  1.0 hour
For half the activity, A3  25, t3  2.6 hours
t 1  1.6 hours
 Half life,
2
When
A4  20, t4  3.0 hours
For half the activity, A5  10, t5  4.5 hours
t 1  1.5 hours
 Half life,
2
Average, t 1 
2
1.6  1.6  1.5
3
 1.6 hours
(c)
For A  10 diss-1 , t  4.5 hours
(d)
The line is not perfectly smooth because of the random nature of decay.
(e)
Atomic number is the number of protons in the atom of the element.
Mass number is the number of protons and neutrons in the atom of the element.
(f)
The number ‘123’ represents a specific isotope of Iodine having a mass number
of 123.
2. (a)
Physical quantity
(b)
(i)
S.I. unit
If a body A , exerts a force on body B, then body B will exert an equal and
opposite force on body A. i.e to every action there is an equal and opposite
reaction.
298
(ii)
As the plane engine expels the air with a force towards the tail end of the
plane, an equal and opposite force is exerted on the plane that propels it
forward. The weight of the plane is offset by the lift force caused by the
wing which keeps it flying horizontally.
(c)
(i)
Linear momentum is the product of mass and velocity of a moving body.
(ii)
By principle of conservation of momentum:
Total momentum before collision  Total momentum after collision
8 10    2  5  8  2  v
80  10  10v
70
v
10
 7 ms-1 due East

3. (a)
EH :
Heat energy
Unit: Joule (J)
c:
 :
Specific heat capacity
Change in temperature
Unit: J kg-1 K-1
Unit: K or °C
(b)
The symbol ‘Ɩ’ represent the specific latent heat.
(c)
(i)
Heat lost by water:
Using E  mc
 100  4.2   30  20 
 100  4.2  10
 4 200 J
(ii)
Energy required to change melted ice at 0º C to water at 20º C:
Using E  mc
 10  4.2  20
 840 J
299
(iii)
Heat lost by water  Heat gained by ice
4 200  840  mi l f
4 200  840  10l f
l f  336 J g -1
4. (a)
(b)
Electromagnetic waves and use: (any three)
 Gamma rays radiotherapy (cancer treatment)
 X-rays
radiography (imaging)
 UV
medical treatment of skin disorder
 Visible
analytical equipment to detect elements
 IR
radiator heating
 Radio
satellite communication
Using v  f 
v


f
For rattlesnakes:
3 108

3.5 1014
 8.6 10 7 m
For honeybee:
3 108

11015
 3 10 7 m
Distance
Time
750

2.3
 326 ms -1
Speed 
5. (a)
(i)
The electric current in a metal is due to the free electrons as the only
charge carrier. The direction of flow of these electrons is opposite to the
conventional current direction. In electrolyte the electric current is due to
the flow of negative and positive ions. The conventional current direction
is the same as the direction of flow of positive ions and opposite to the
flow of negative ions.
(ii)
The current flow in semiconductor is similar to that of electrolyte since the
semiconductor has both free electrons and positive holes a charge carriers.
300
(b)
Using Q  It
Q

t
I
For 4320 C battery:
4320
t1 
0.6
 7 200 s
(i)
For 9000 C battery:
9 000
t2 
0.6
 15000 s
Time difference, t  t2  t1
 15000  7 200
 7800 s
Charging time for old model  7 200 s
Q
Current rating,
I
t
9 000

7 200
 1.25 A
(ii)
6. (a)
(b)
The candle wax in the test tube was first melted to liquid state by placing it in the
hot water bath using a test tube holder. The tube was then placed on a rack and a
thermometer was inserted. While in liquid state it was stirred for even distribution
of heat. Starting from t = 0 s, the temperature was measured at regular time
intervals. A graph of temperature vs time was then plotted.
(i)
(ii)
As the water drops from the top of a waterfall, its potential energy is
converted to kinetic energy. This energy is transferred to the rotating
blades of the turbine which is converted to electrical energy by the
generator.



The presence of waterfall in Dominica ensures a renewable source of
energy.
Hydroelectricity produces no waste or greenhouse gases and therefore
does not affect global warming.
The long term cost of hydroelectricity production is cheaper than energy
from fossil fuel which makes it a viable alternative.
301
1. (a)
Graph of Electrical energy, E vs Temperature rise, T
302
(b)
Points selected for slope: (2.0, 2.4), (13.4, 16.0)
y y
Slope, S  2 1
x2  x1
16.0  2.4

13.4  2.0
13.6

11.4
 1.2 kJ K -1
(c)
The slope represents heat capacity, C.
(d)
Specific heat capacity, c 
(e)
(i)
The procedure will minimize the error since at 10º C below room
temperature heat is transferred from the environment to the liquid. But at
10º C above room temperature, heat is transferred from the liquid to the
environment, resulting in negligible net heat transfer.
(ii)
The liquid should be stirred while heating to ensure equilibrium measured
temperature.
C
m
1.2 1000

250
 4.8 J g -1 K -1
Energy, E
Time, t
E
Time, t 
P
18000

40
 450 s
Using power, P 
(f)

 7.5 minutes
2. (a)
(b)
1) Carbon rod (+ve electrode)
2) Electrolyte (paste of Ammonium Chloride)


A primary cell cannot be recharged whereas a secondary cell can be recharged.
A primary cell has high internal resistance whereas a secondary cell has low
internal resistance.
303

(c)
3. (a)
The chemical reaction in a primary cell is irreversible whereas in a secondary cell
the chemical reaction is reversible.
(i)
Charge added, Q  It
 1 4  60  60
 14 400 C
(ii)
Energy added, E  QV
 14 400 12
 172800 J
 172.8 kJ
(iii)
The charging voltage for a solar module can be 12, 24 or 48 VDC in order
to handle the current (A) from the solar module.
(i)
The principle of moments states that when a body is in equilibrium, the
sum of the clockwise moments about any point (pivot) is equal to the sum
of the anticlockwise moments about the same point.
(ii)
(b)
(i)




A force applied to a spanner to rotate a nut.
A force applied to swing-open a door.
A force applied to push a swing.
A driver turning a steering wheel.
By the principle of moments:
Clockwise moments (Kyle)  Anti-clockwise moments (Keion)
500 1  300  x
500

x
300
 1.67 m
(ii)
304
At equilibrium:
Total upward force  Total downward forces
R  500  300
 800 N
(iii)
4. (a)
(i)


Moment of reaction force about the pivot  0 Nm
A swimming pool seemed shallow when viewed from above
A straight stick seemed to bend when dipped in water.
(ii)
(b)
(iii)
The speed of light will decrease in travelling from air to water since water
is a denser transparent medium than air.
(i)
The prism causes dispersion of white light.
(ii)
305
(c)
(i)
Angle of
incidence, iˆ
30°
50°
60°
(ii)
Using n 
 sin rˆ 

5. (a)
(b)
Angle of
refraction, r̂
sin iˆ
sin rˆ
20°
31°
35°
0.50
0.77
0.87
0.34
0.52
0.57
sin iˆ
sin rˆ
sin iˆ
n
sin 70

1.52
0.94

1.52
 0.62
ˆr  38
(i)
NOT gate
(ii)
AND gate
(iii)
NOR gate
(i)
NOT gate
Input
0
1
(ii)
(i)
Output
1
0
NOR gate
Input
0
0
1
1
(c)
sin iˆ
sin rˆ
1.47
1.48
1.53
0
1
0
1
Output
1
0
0
0
A 1
B0
C 1
306
(d)
6. (a)
(ii)
A 1
B 1
C 1
(i)
The use of technology has improved transportation in land, air, sea and
space travel.
(ii)
Advances in electronic technology have improved radio, television,
internet and satellite communication.
(iii)
Improved operating efficiency in equipment and household appliances.
(i)
A – Thin sheet of gold foil
B – Rotating scintillation microscope
C – Alpha source (in lead container)
(b)
(c)
(ii)
Gold foil was used because gold is very malleable and therefore very thin
sheets of gold foil can be produced.
(iii)
1) Most of the atom was empty space
2) The atom has a positively charged center that contains most of its mass.
(i)
Number of neutrons  28 14
 14
(ii)
29
14
(i)
The decay process represents alpha decay.
(ii)
p  222  4
Si or
30
14
Si or
31
14
Si
 218
q  76  2
 74
0
Y 
 218
75 Z  1 e
(iii)
218
74
(iv)
The particle’s mass is reduced in an alpha decay and almost unchanged in
a beta decay.
307
1. (a)
Graph of Induced E.m.f./V vs Time/ms
(b)
(i)
Induced e.m.f. after 12.5 ms  1.50 V
(ii)
Another time when induced e.m.f. was 0V  27.75 ms
308
(c)
(i)
(ii)
A changing magnetic flux due to the approaching magnet causes an
induced e.m.f. in the coil which in turn produces an induced current.
(iii)
The sensitive galvanometer can detect and measure the current flowing in
both directions.
(iv)
(d)
(v)
To increase the induced e.m.f:
 Move the magnet faster
 Use a stronger magnet
 Use a coil with more turns
(vi)
With the magnet stationary in the coil there would be no change in
magnetic flux and therefore no induced current.
(i)
309
(ii)
1
T
1
f 
0.02
 50 Hz
Using f 
(iii)
2. (a)
(b)
(i)
A transverse wave is one in which the displacement of the particles is at
right angles to the direction of travel of the wave. A longitudinal wave is
one in which the displacement of the particles is parallel to the direction of
travel of the wave.
(ii)
Transverse wave: surface water waves, electromagnetic waves, waves in
strings.
Longitudinal wave: Sound waves, Pressure waves, Seismic P-waves.
(iii)
v f
(iv)
v  10  250
 2500 ms-1
(i)
(ii)
Amplitude, A  0.5 103 m
(iii)
Period, T  4.0 103 s
310
(iv)
Frequency, f 
1
T
1
4.0 10 3
 250 Hz

(c)
3. (a)
(b)
(c)
Property of electromagnetic waves:
 Transverse waves
 Travel at a speed of 3 x 108 ms-1 in a vacuum
 Carry no charge
 Do not need a vacuum to travel
(i)
Nuclear fission is the splitting of large unstable nucleus into smaller, more
stable nuclei with the release of energy.
(ii)
Two advantages of utilizing nuclear energy:
 No greenhouse gases or smoke produced
 Small quantities of raw material produce large quantities of energy
(i)
Precautions taken by workers in a nuclear power plant:
 Use protective clothing and shielding
 Handle radioactive material with robotic arms
 Use film badge to indicate exposure to radiation
(ii)
Disadvantages of using nuclear reactor to generate energy:
 Nuclear waste is difficult and expensive to process.
 Nuclear accidents can be devastating
 Nuclear plants have limited life and therefore has to be dismantled
at a cost.
 Nuclear power plants generate external dependence since not many
countries have Uranium.
(i)
P  92  36
 56
(ii)
‘c’ represents the speed of light.
(iii)
Total mass of elements  Total mass of product side of equation
  232.560  152.620  3 1.670   10 27
 390.190 10 27 kg
(iv)
Mass defect, m   398.350  1.670   10 27  390.190  10 27
 9.8310 27 kg
311
Energy released, E  mc2
 9.83 10 27   3 108 
2
 8.85 1010 J
4. (a)
(b)
(v)
This energy can be used to boil water to make steam which drives turbine
to generate electricity.
(i)
For carriage moving horizontally at a constant speed in a straight line:
Newton’s first law of motion applies: Everybody continues in a state of
rest or of uniform motion in a straight line unless compelled by an external
force to act differently.
(ii)
Carriage is in free fall:
Newton’s second law of motion applies: The rate of change of momentum
is proportional to the applied force and takes place in the direction in
which the force acts.
(i)
Using
v  u  at
64.8 1000 m
64.8 kmh -1
60  60 s
64800

3600
 18.0 ms-1


(ii)
5. (a)
(i)
18.0  0  10t
a  g 
t  1.8 s
1
Using s  ut  at 2
2
1

s  gt 2
u  0
2
1

s  10 1.82
2
 16.2 m
A car radiator: needs to lose heat efficiently
 Painted black since a black surface is a better radiator of heat than
a white surface.
 Has a large surface area with the use of ‘fins’ to ensure greater
emission.
312
(b)
(ii)
The roof of a Caribbean home: needs to reflect radiation
 Usually a shiny surface since it reflects radiation better than a dull
surface.
 It is painted white since this colour is a poor absorber of heat
radiation.
(i)
Assuming negligible heat loss:
Heat supplied by immersion heater  Latent heat required to evaporate
Water
Pt  mlv


(ii)
6. (a)
150  5  60   0.28  0.26   lv
45000  0.20  lv
45000
lv 
0.02
 2250000 J kg-1
If the coil of the immersion heater is not completely submerged it will
take a larger time to evaporate the same mass of water. If the same power
rating value (150 W) is used in calculation, the value of the specific latent
heat of vaporization will increase.
(i)
The principal focus of a converging lens is a point on the principal axis
which all rays initially parallel to the principal axis will converge on after
refraction by the lens.
313
(ii)
The principal focus of a diverging lens is a point on the principal axis
which all rays initially parallel to the principal axis appear to diverge from
after refraction by the lens.
(b)
(i)
Using


1 1 1
 
f v u
1 1 1
 
v f u
1
1


12.0 18.0
3 2

36
1

36
v  36.0 cm
v
u
36.0

18.0
2
(ii)
Magnification, m 
(iii)
Height of image  m  7.2
 2  7.2
 14.4 cm
(iv)
The image formed is real.
314
1. (a)
(i)
Angle of reflection, r̂
10.0
30.0
50.0
70.0
90.0
(ii)
Angle of incidence, iˆ
6.5
19.0
30.0
38.0
41.0
sin rˆ
0.174
0.500
0.766
0.940
1.000
sin iˆ
0.113
0.326
0.500
0.616
0.656
Graph of sin rˆ vs sin iˆ
315
(iii)
Points selected for gradient
(0.60, 0.91), (0.10, 0.15)
y y
Gradient  2 1
x2  x1
0.91  0.15

0.60  0.10
 1.52
(b)
The physical property represented by gradient is refractive index.
(c)
The critical angle is the angle of incidence that produces an angle of refraction of
90º for light travelling from one medium to an optically less dense medium.
(d)
From Table 1, critical angle for glass  41.0º
(e)
Using


n(cladding) sin 90

n(core)
sin c
1
1.03 
sin c
1
sin c 
1.03
 0.9709
c  76.1
2. (a)
Solid
Liquid
Gas
(b)
(i)
Shape
Volume
Movement of
molecules
Definite/Fixed
Takes shape of
container
No fixed shape
Definite/Fixed
Definite/Fixed
Vibrate
Move amongst
one another
Move freely
Full space
Intramolecular
forces
Strong
Very weak
Negligible
weak
Time of heating   5  60   45
 345 s
(ii)
Mass of water evaporated  375  360
 15 g
 0.015 kg
316
(c)
3. (a)
(b)
Assuming no heat loss:
Energy supplied by electric heater  heat energy required to evaporate water
Pt  mlv
Pt
lv 
m
100  345

0.015
 2.3106 J kg-1
Energy is the stored ability to do work.
S.I. unit of energy  Joule (J)
(i)
Forms of energy
Nuclear Energy
Electromagnetic Radiation Energy
Kinetic Energy
(ii)
(c)
(i)
Example
Radioactive decay
Radio waves, X-rays
Objects in motion
Chemical energy stored in the battery is converted to light energy (and
some heat energy).
1
Maximum kinetic energy, Ek  mv 2
2
1
  0.5 1.82
2
 0.81 J
(ii)
Assuming no energy is lost in moving from Y to X
Maximum gravitational potential energy  Maximum kinetic energy
 0.81 J
(iii)
Maximum E p  mgh

h
Ep
mg
0.81

0.5  9.8
 0.165 m
 16.5 cm
317
4. (a)
Arguments for nuclear fission reactors:
1) There is a reduction in environmental pollution as a result of little or no
carbon dioxide emission.
2) Produce high amounts of nuclear fission energy using relatively small
quantities of raw materials.
Arguments against nuclear fission reactors:
1) Accidents can be devastating, resulting in radiation exposure.
2) They create harmful nuclear waste.
(b)
Advantages of nuclear fusion over nuclear fission:
1) Nuclear fusion produces much more energy than nuclear fission.
2) Nuclear fusion does not produce the level of toxic radioactive waste, like
nuclear fission.
(c)
2
1
(d)
H  21 H 
 31 H  11 H
Mass of 21 H  31 H  2.01410178  3.01604927
 5.03015105 u
Mass of 42 He  01 n  4.00260325  1.00866492
 5.01126817 u
Mass defect   5.03015105  5.01126817  1.66 10 27 kg
 3.135 10 29 kg
Energy released  mc2
 3.13455808 10 29   3.0 108 
2
 2.82 1012 J
5. (a)
Newton supported the particle theory of light based on the evidence that light
travels in straight lines and can travel through a vacuum. It is also supported by
the phenomena of photo electric effect.
Huygens supported the wave nature of light based on the evidence that light can
undergo reflection, refraction, diffraction and interference.
(b)
(i)
Objects placed at a distance between f and 2f would produce a real
magnified image. The phone should therefore be placed at a distance
greater than 15.0 cm but less than 30.0 cm.
318
(ii)
Using


1 1 1
 
f u v
1 1 1
 
v f u
1 1
 
15 20
1

60
v  60 cm
v
u
60

20
3
(iii)
Magnification of image, m 
(iv)
Dimension of image of phone screen  11.0  6.0  3
 33.0 cm 18.0 cm
6. (a)
(v)
Image formed in (b)(ii) is inverted.
(vi)
The image can be made larger by moving the phone closer to the lens or
moving the screen further from the lens.
(i)
Magnetic field due to a current carrying conductor
319
(b)
(ii)
Resulting magnetic field when a current carrying conductor is placed
between the poles of the magnet.
(i)
The part labeled X is the split ring or commutator.
(ii)
When the switch is closed the current flows through the carbon brushes
and commutator into the coil. The current flows from D to C and B to A.
From Fleming’s Left Hand Rule, a downward force is exerted on DC
while an upward force is exerted on AB. This is due to the magnetic fields
produced by the coil and the magnet. The momentum of the coil allows it
to cross the vertical position. The commutator reverses the direction of the
current in the loop as the contact changes from one brush to the other. At
this point AB then moves down while DC moves up resulting in
continuous rotation in one direction.
320