Answers to end-of-chapter questions
Chapter 1
7 a volume = l × b × h
1
Mass
Length
Volume
Time
balance
metre rule
measuring
cylinder
stopclock
scales
tape measure
= 80 × 40 × 15
= 48 000 m3
b mass = volume × density
= 48 000 × 1.3
= 62 400 kg
electronic
timer
[1]
[1]
[1]
8 a Half-fill a measuring cylinder with water;
vernier calipers
record volume.
Place pebble in water, ensuring that it
is submerged.
Record new volume.
Volume of pebble equals difference in
recorded volumes.
micrometer
screw gauge
2 a density = mass
volume
b
[1]
[1]
[1]
[1]
[1]
[1]
[1]
Unit of
mass
Unit of
volume
Unit of
density
kg
m3
kg/m3
9 a V1 = 70 cm3
[1]
g
cm3
g/cm3
V2 = 95 cm3
[1]
b V = 95 − 70
= 25 cm3
[1]
[1]
mass
c density = volume
[1]
b mass of pebble
3 a vernier callipers; micrometer screw gauge
b Time, say, 40 drops and divide by 40.
4 a volume = l × b × h
= 8.4 × 8.0 × 5.5
= 369.6 cm3
mass
b density =
volume
340
=
369.6
= 0.92 g/cm3
5 a 70 – 15
= 55 cm3
b 43 – 12
= 31 s
6 mass of liquid = 203 – 147
= 56 g
mass
density =
volume
56
= 59
= 0.95 g/cm3
© Cambridge University Press 2014 IGCSE Physics
[1]
[1]
[1]
102
25
= 4.08 g/cm3
=
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
d
Sample m / g
[1]
[1]
[1]
[1]
V2 [1] / V1 [1] / V / cm3 Density /
cm3 [1] cm3 [1]
[1]
g/cm3 [1]
B
144
80
44
36 [1]
4.0 [2]
C
166
124
71
53 [1]
3.1 [2]
10 a 30.98 − 30.72
= 0.26 g
b density = mass
volume
[1]
[1]
[1]
= 0 26
200
[1]
= 0.0013
g/cm3
[1]
[1]
Answers to end-of-chapter questions: Chapter 1
1
11 a water
[1]
b volume (of water) or water level
[1]
c the stone
[1]
d volume (of water)
[1]
e subtract or calculate the difference between [1]
first volume from (or and) second volume [1]
© Cambridge University Press 2014 IGCSE Physics
Answers to end-of-chapter questions: Chapter 1
2
Answers to end-of-chapter questions
Chapter 2
7 a
1 a average speed = distance travelled
800
time taken
Distance / m
b m/s
c Graph is a horizontal straight line, showing that
speed does not change.
d distance travelled = area under graph
600
400
200
2 a graph A; speed = gradient (slope) of graph
0
0
b Graph is a straight line.
c graph B; acceleration = gradient (slope) of graph
Description
Examples
scalar
has magnitude
only
speed, distance
vector
has magnitude
and direction
velocity, acceleration,
weight
4 average speed = distance
time
400
50
= 8.0 m/s
=
5 Speed is uniform (constant) in both.
The bus travels faster during B than A.
6 distance = speed × time
= 15 × 30
= 450 m
20
30
Time / s
40
suitable scales chosen
horizontal axis and scale correct
vertical axis and scale correct
five points correctly plotted and straight
line drawn
3 a, b
Quantity
10
b Graph is straight line,
so constant speed.
[1]
[1]
graph is horizontal
[1]
[1]
[1]
[1]
[1]
[1]
[1]
b point 3,
graph is steepest
[1]
[1]
c point 2,
graph is becoming steeper
(gradient is increasing)
[1]
[1]
d point 4,
graph is becoming less steep
(gradient is decreasing)
[1]
e point 6,
distance is decreasing
[1]
[1]
[1]
9 a speed of light
[1]
b distance = speed × time
10
© Cambridge University Press 2014 IGCSE Physics
[1]
[1]
[1]
8 a point 1 or 5,
[1]
[1]
[1]
[1]
[1]
Object
Distance
travelled
Time
taken
Speed
bus
20 km
0.8 h
25 km/h
taxi
6 km
200 s [1] 30 m/s
aircraft
4950 km
snail
3 mm
[1] 5.5 h
10 s
[1]
900 km/h
0.3 mm/s [1]
Answers to end-of-chapter questions: Chapter 2
1
11 speed is constant
[1]
[1]
acceleration = 0
Speed
12 a
horizontal axis showing time
vertical axis showing speed
rising straight-line graph starting at origin
[1]
[1]
[1]
Speed
b
Time
horizontal and vertical axes showing
time and speed
horizontal straight-line graph above axis,
then decreases down to zero
change in velocity
13 acceleration =
time taken
=
b acceleration = gradient of graph
[1]
8.0
2.0
[1]
14 initial speed = 0 m/s
[1]
[1]
[1]
[1]
change in speed = acceleration × time
= 2.3 × 4.0
= 9.2 m/s
speed
acceleration
24
=
5.6
= 4.3 s
15 time =
[1]
c distance = area under speed
against time graph
= area of triangle + area of rectangle
[1]
[1]
= 12 × 30 × 27 + 20 × 27
[1]
= 405 + 540
= 945 m
[1]
[1]
17 a B, D
[2]
b A, E
[2]
c Acceleration is changing in the
other section, C.
[2]
distance
time
1425
=
75
= 19 m/s
[1]
b the direction of its motion
[1]
[1]
30
10
0
0
10
20
30
Time / s
© Cambridge University Press 2014 IGCSE Physics
40
50
[1]
[1]
b i
[1]
accelerating or increasing speed
ii steady or constant speed
[1]
iii decelerating or slowing down
[1]
c less than
20 a i
[1]
constant/steady/uniform speed or
velocity or speed or velocity = 2.5 (m/s) [1]
speed or velocity = 2.5 m/s
[1]
ii shape curving upward but not to vertical [1]
b horizontal (straight) line (parallel to time /
x-axis)
c i
20
[1]
19 a 25 km
[1]
[1]
[1]
= 0.9 m/s2
[1]
[1]
[1]
[1]
= 4.0 m/s
27
30
18 a speed =
[1]
2
Speed / m/s
[1]
[1]
[1]
[1]
=
Time
16 a
horizontal axis and scale correct
vertical axis and scale correct
six points correctly plotted
graph drawn through points
[1]
horizontal straight line at 2.5 m/s from
0 to 2 s
[1]
ii straight line rising to the right as far as
the edge of the graph area
Δv = 4 m/s or gradient clearly 2 m/s2
[1]
[1]
d horizontal straight line
at 0 m/s
Answers to end-of-chapter questions: Chapter 2
[1]
[1]
2
Answers to end-of-chapter questions
Chapter 3
11 a weight = mass × g
1 A force can make an object change direction,
= 80 × 10
= 800 N
decelerate, or accelerate.
2 resultant force
[1]
[1]
[1]
3 weight
b the same
[1]
4 a force = mass × acceleration
c less
[1]
12 a the two 5000 N forces
b, c
Quantity
Unit
Scalar or vector?
mass
kg
scalar
acceleration
m/s2
vector
force
N
vector
They are equal in size but act in
opposite directions.
[1]
[1]
[1]
c The lorry will speed up (accelerate).
[1]
= 20 × 5
= 100 N
6 a weight downwards, air resistance upwards
c The resultant force on it is zero, so it does not
accelerate.
force
mass
1 400 000
=
800 000
d terminal velocity
= 1.75 m/s2
14 acceleration =
b zero
7 a impulse of force = change of momentum
b F = force, t = time, m = mass, v = final velocity,
u = initial velocity
15 acceleration =
=
c Momentum is a vector quantity.
8 a kilogram (kg) or gram (g)
[1]
b newton (N)
[1]
c metre per second per second (m/s2)
[1]
direction (like an arrow).
b Mass is not a vector quantity (it is a
scalar), so it does not have direction.
10 the bigger force acting on the smaller mass,
that is, the 10 N force acting on the 5 kg mass
© Cambridge University Press 2014 IGCSE Physics
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
change in speed
time
[1]
(20 – 12)
6.4
[1]
= 1.25 m/s2
force = mass × acceleration
= 1200 × 1.25
= 1500 N
16 weight = mass × g
9 a Force is a vector quantity, that is, it has
[1]
b resultant force = 1300 – 1200
= 100 N
forwards (to the left)
13 force = mass × acceleration
5 90°
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
= 50 × 1.6
= 80 N
17 a resultant force = 680 – 600
= 80 N
upwards
b He will accelerate upwards.
Answers to end-of-chapter questions: Chapter 3
[1]
[1]
[1]
[1]
1
18 a i (engine) thrust and (air) friction
ii force shown vertically upwards,
anywhere on plane
b i speed = distance in any form
time
[1]
[1]
[1]
= 2200
2 75
[1]
= 800 (km/h)
[1]
ii idea of headwind on outward journey
or tailwind on return journey or routes
of different lengths or less friction or
less weight
© Cambridge University Press 2014 IGCSE Physics
(v − u)
or v or 8
3
t
t
= 2.7 m/s2
[1]
ii F = ma or 42 × 8/3
= 112 N
[1]
[1]
19 a i
iii distance in first 3 s = 12 m
so distance in last 11.2 s = 88 m
so final speed = 88 = 7.9 m/s
11.2
b Any two from: lower top speed, longer total
time, less steep slope at first, etc.
[1]
[1]
[1]
[1]
[2]
[1]
Answers to end-of-chapter questions: Chapter 3
2
Answers to end-of-chapter questions
Chapter 4
1 a increase
b increase
5 moment
[1]
6 no resultant force (forces balanced)
[1]
[1]
no resultant moment
2 a resultant
7 See Activity 4.3.
b zero
Make three small pinholes around the
edge of the lamina.
Suspend the lamina freely from a pin
through one hole.
Mark a vertical line below the pin
using a plumb-line.
Repeat this process for the other
two pinholes.
The centre of mass is where the
three lines intersect.
3 a, b for example
centre
of
mass
stable object
8
contact
force
A
for example
1m
0.9 m
centre
of mass
[1]
[1]
[1]
[1]
[1]
B
1.5 m
weight
of beam
centre
of
mass
unstable object
x
4 a
pivot
F
b moment = force × distance from pivot
c
Quantity
Unit
force
N
distance
m
moment of force
Nm
© Cambridge University Press 2014 IGCSE Physics
a centre of mass correctly marked,
as in diagram
[1]
b arrows and labels added correctly
[2]
c moment of weight = force × distance
= 200 N × 0.5 m
= 100 N m
moment of force F is F × 1.0 = 100 N m
so F = 100 N
[1]
[1]
[1]
[1]
[1]
d upward contact force = sum of
downward forces
= 200 N + 100 N
= 300 N
[1]
[1]
[1]
Answers to end-of-chapter questions: Chapter 4
1
9 a force and perpendicular distance (of force)
from the point
b i
downward force arrow at centre of bar
ii 0.50 m or 50 cm
[1]
© Cambridge University Press 2014 IGCSE Physics
[1]
[1]
30
= 0.60 kg
[1]
[1]
b (1.5 + 0.6) × 10 = 21 N
[1]
c i stays in position
[1]
iii moment of force = 40 × 1.2 = 48 N m
[1]
moment of weight = + 30 × 0.5 = 15 N m [1]
total clockwise moment = 48 + 15
= 63 N m
[1]
iv F × 0.2 = 63
63
= 315 N
F=
0.2
v make bar longer or move pivot/stone
to the left or move pivot to left or
increase mass of bar
10 a mass = 1 5 × 12
[1]
ii as the parrot is rotated, both distances
change in proportion
so clockwise moment = anticlockwise
moment
[1]
Answers to end-of-chapter questions: Chapter 4
2
[1]
[1]
[1]
Answers to end-of-chapter questions
Chapter 5
1 a extension = length when stretched − original
length
b graph B
2 a The extension of a spring is proportional to the
load, provided the limit of proportionality is
not exceeded.
Student must measure:
◆ length of spring when weights added
◆ unstretched length of spring
◆ repeated for at least five different weights.
8 a
Load / N
Length / cm
Extension / cm
0.0
83.0
0.0
5.0
87.0
4.0 [1]
10.0
91.0
8.0 [1]
3 a increases
15.0
95.0
12.0 [1]
b decreases
20.0
99.0
16.0 [1]
b load = stiffness × extension
c See Figure 5.7a.
[1]
[1]
[1]
c increases
b
d decreases
4 a pressure = force
1.5
Extension / cm
area
b P= F
A
c
Quantity
Unit
force
N
area
m2
pressure
Pa
2.0
1.0
0.5
0
0
5
10
Load / N
15
20
horizontal axis and scale correct
vertical axis and scale correct
five points correctly plotted and graph drawn
through points
5 a A fluid is a liquid or a gas; any substance that
can flow.
b g = 10 m/s2
c P = hρg
[1]
[1]
[1]
9 a See Figure 5.14.
6 extension = change in length
= 66 – 58
= 8.0 cm
[1]
[1]
[1]
7 See Activity 5.1.
Diagram or list indicating:
◆ spring hanging vertically from clamp
◆ weights hanging from end of spring
◆ ruler.
© Cambridge University Press 2014 IGCSE Physics
[1]
[1]
[1]
Diagram showing:
◆ vertical tube with closed upper end
◆ open end submerged in
mercury reservoir
◆ mercury in tube continuous from
reservoir up to empty (vacuum)
space near top
Answers to end-of-chapter questions: Chapter 5
[1]
[1]
[1]
1
b An increase in atmospheric pressure causes
the level of mercury in the tube to rise / the
length of the mercury column to increase. [1]
10 If you stand upright, your weight is pressing
down on a small area.
[1]
This gives a high pressure.
[1]
If you use a ladder, the pressure is less because
your weight is spread over a greater area.
[1]
11 extension for 5 N is 15 – 12 = 3.0 cm
[1]
[1]
[1]
extension for 15 N is 3 × 3 cm = 9.0 cm
length is 12 + 9 = 21 cm
12 a
Load / N
0
Length / m
3.200
3.207
Extension / mm
0
7
Extension / mm
b
10
20
30
3.215
15
40
3.222
22
50
3.230
30
3.242
42
60
70
3.255
55
80
14 pressure = height × density × g
60
= 0.760 × 13 600 × 10
= 103 400 N/m2
3.270
70
[4]
[1]
[1]
[1]
15 a extension indicated between two broken
lines
40
b i
20
0
0
20
40
Load / N
60
horizontal axis and scale correct
vertical axis and scale correct
eight points correctly plotted
graph drawn through points
c Draw up from 25 N to intersect
graph line,
from this intersection, go across
to axis, 19 mm
d the point where the graph line
ceases to be straight
40 N approximately
13 a force = pressure × area
= 100 000 × 2.0 × 1.25
= 250 000 N
b There are equal forces on both
sides of the window.
© Cambridge University Press 2014 IGCSE Physics
80
[1]
[1]
[1]
[1]
[1]
[1]
[1]
four points correctly plotted
straight line through points and origin
[2]
[1]
ii proportional
[1]
iii 1 newton(s)
2 extension = 25 − 26 mm
length = 75 − 76 mm
[1]
[1]
[1]
16 a wall A has bigger area
so lower pressure (on soil)
b i two from
◆ depth/height of air/atmosphere
◆ density of air/atmosphere
◆ acceleration due to gravity or weight
of air
ii 1 the same as
2 greater than or four times
[1]
[1]
[2]
[1]
[2]
[1]
[1]
[1]
[1]
[1]
[1]
Answers to end-of-chapter questions: Chapter 5
2
Answers to end-of-chapter questions
Chapter 6
6 energy supplied = 100 J
1
Name
Description
kinetic energy
energy of a moving object
internal energy
energy stored in a hot object
chemical energy
energy stored in a fuel
light energy
energy that we can see
sound energy
energy that we can hear
strain (elastic)
energy
energy stored in a squashed
spring
electrical energy
energy released = 93 + 7 = 100 J
Energy before is equal to energy after,
so energy is conserved.
[1]
[1]
[1]
7 a gravitational potential energy
→ kinetic energy
b kinetic energy
→ gravitational potential energy
[2]
[2]
energy carried by an electric
current
c Some energy is lost as heat due
to friction and/or air resistance,
so the final g.p.e. cannot equal
the original g.p.e.
[1]
nuclear energy
energy stored in the nucleus of
an atom
d She needs to supply energy,
by jumping up as she starts off.
[1]
[1]
heat thermal
energy
energy escaping from a hot
object
8 a, b
2 a heat energy
b efficiency
c conservation
3 a k.e. = 1 mv 2 (m = mass, v = speed)
2
b g.p.e. = mgh (m = mass, g = acceleration due to
gravity, h = height)
4 a waste energy = energy input − useful energy
output
b efficiency =
[1]
Low-energy
bulb
Filament
bulbs
cost of one bulb
400 p
50 p
number of bulbs
required for 10 000
hours
1
10
cost of electricity for
1 hour
0.2 p
1.0 p
total cost of electricity
for 10 000 hours
2 000 p [1]
10 000 p [1]
total cost of bulbs and
electricity
2 400 p [1]
10 500 p [1]
useful energy
gy output
p
× 100%
energy input
5 a chemical energy → light + heat
[2]
b electrical energy → kinetic energy
[2]
c kinetic energy → electrical energy
[2]
d kinetic energy → thermal (heat) energy
[2]
© Cambridge University Press 2014 IGCSE Physics
c money saved = 10 500 – 2400
= 8100 p
[1]
[1]
d initial cost is high
difficult to dispose of
[1]
[1]
Answers to end-of-chapter questions: Chapter 6
1
9 a thermal (heat) energy, electrical energy
[2]
b thermal (heat) energy
[1]
c Yes, because 90% of the energy is used,
and only 10% is wasted.
[1]
[1]
10 a k.e. of moving air → electrical energy
electrical energy → k.e. of car
b Process is less than 100% efficient,
so car will not gain speed.
11 a weight = mass × g
= 180 × 1.6
= 288 N
[1]
[1]
[1]
[1]
[1]
[1]
[1]
b change in g.p.e. = weight × change
in height
= 288 × 100
= 28 800 J
[1]
[1]
[1]
c g.p.e. increases
[1]
© Cambridge University Press 2014 IGCSE Physics
12 work
potential / gravitational / p.e. / g.p.e. / position
kinetic / k.e. / movement
constant / the same / uniform
joule(s) or J
13 a mgh = 0.5 × 10 × 1.1
= 5.5 J
b i 1.5 (J)
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
ii energy used to deform ball/ground
or strain energy stored in (deformed)
ball/ground or
heat generated in deformed ball/ground [1]
c 9 + 5.5 = 14.5 J
[1]
k.e. = 1 mv 2
2
v = 7.6 m/s
[1]
Answers to end-of-chapter questions: Chapter 6
[1]
2
Answers to end-of-chapter questions
Chapter 7
1 a resource
6 a i
fission
ii uranium (or plutonium)
b Sun
b i fusion
ii hydrogen
iii helium
c renewable
d fossil fuels; non-renewable
e wind, electricity
b nuclear fission
using sunlight as their energy source.
b Sunlight causes evaporation,
producing clouds;
rain falls, and finally enters rivers,
whose water is trapped behind a dam.
[1]
[1]
[1]
7 Renewable: two from hydroelectricity, solar,
2 a nuclear fusion
3 a Trees and plants grow
[1]
[1]
[1]
[1]
tidal, wind
[2]
Non-renewable: two from coal, oil, nuclear
[2]
(At least two correct in each column for 4 marks;
deduct 1 mark for any in incorrect column.)
8 a oil
nuclear fission
[1]
[1]
[1]
4 a Sunlight is always available in space, and
b i
gas lamp
[1]
[1]
[1]
ii electric motor or loudspeaker
[1]
iii microphone
[1]
not much power is needed on a spacecraft. [1]
But in cities, there are large numbers of
people in a small area,
[1]
so there is not enough roof space for all the
solar cells that would be needed to generate
enough power.
[1]
b (for example) In a desert for
roadside phones,
because there is then no need to connect
the phone to the mains electricity supply.
c A rechargeable battery can store
the energy produced by solar cells,
and can therefore supply electricity
when the sun is not shining.
[1]
[1]
[1]
[1]
5 a g.p.e.
[1]
b k.e.
[1]
c g.p.e. → k.e. → electrical energy
[2]
© Cambridge University Press 2014 IGCSE Physics
Answers to end-of-chapter questions: Chapter 7
1
Answers to end-of-chapter questions
Chapter 8
8 a work done = force × distance moved
1 a more
= 250 × 12.0
= 3000 J
b more
2 a energy
b gain in g.p.e. = weight × increase in height
= 700 × 2.5
= 1750 J
b work
3 work done = energy transferred
9 a weight = mass × g
4 a work done = force × distance moved (in the
= 45 × 10
= 450 N
direction of the force)
b
Quantity
Unit
ΔW
joule, J
F
newton, N
x
metre, m
b gain in g.p.e. = weight × increase in height
= 450 × 0.20 × 36
= 3240 J
work done
time taken
3240
=
4.2
c power =
c ΔE = energy transferred
5 a power = work done
= 770 W
= 0.77 kW
time taken
b
Quantity
Unit
P
watt, W
ΔW
joule, J
t
second, s
6 Ahmed
He lifts them to a greater height.
7 a Millie: speed =
Lily: speed =
25
= 0.50 m/s
50
100
= 0.40 m/s
250
10 a work done = force × distance moved
[1]
[1]
[1]
[1]
b Millie
[1]
Because they are identical, the one with the
greater speed has the greater power.
[1]
© Cambridge University Press 2014 IGCSE Physics
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
= 780 × 100
= 78 000 J
[1]
[1]
[1]
b work done = force × distance moved
= 240 × 100
= 24 000 J
[1]
[1]
[1]
c k.e. = 12 × 750 × 122
[1]
= 54 000 J
d work done by engine = work done against
friction + k.e.
78 000 = 24 000 + 54 000
so energy is conserved
Answers to end-of-chapter questions: Chapter 8
[1]
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1
11 a i gravitational potential energy (g.p.e.)
ii force/mass/weight of (basket of) rocks
and height of cliff
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b chemical energy
[1]
c time taken
to raise basket up cliff
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12 a M = V × D = 103 × 10−3
= 1.0 kg
b mgh = 1 × 10 × 0.8
= 8.0 J (or 8.0 N m)
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c P = E = 8 × 90
t
60
[1]
= 12 W (or 12 J/s or 12 N m/s)
d P = ρgh
8000 Pa (or 8000 N/m2)
© Cambridge University Press 2014 IGCSE Physics
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Answers to end-of-chapter questions: Chapter 8
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2
Answers to end-of-chapter questions
Chapter 9
1 See Figure 9.2.
11 a Molecules of ethanol leave the surface
of the liquid
so that its mass decreases.
2 See Figure 9.5.
3 a energy
b The more energetic molecules of ethanol
are more likely to leave the liquid,
so the average energy of the molecules
remaining decreases.
Hence its temperature decreases.
b temperature
4 a evaporation
b faster-moving or more energetic; decrease or
fall/drop
5 a smoke particles
b molecules of the air
b The pressure will decrease.
[1]
b In the liquid, forces between the
particles hold them together.
If it is to become a gas, energy must be
supplied to overcome these forces and
separate the particles.
c quickly
7 a pressure × volume = constant
[1]
9 a solid
b The particles are well separated and
can move about within the volume
of their container,
colliding with its walls and with
each other.
10 a particles of smoke
b The smoke particles are moving
because the particles of the air are
continually colliding with them,
changing their speed and direction
of motion.
© Cambridge University Press 2014 IGCSE Physics
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15
a
Solid
Liquid
[1]
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120 000 × 20 = 160 000 × V2
V2 = 15 m3
b
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c liquid
[1]
p1V1 = p2V2
14
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b quickly
b solid
[1]
12 a The pressure will increase.
13 a evaporation (or vaporisation)
6 a slowly
b pV = constant
p1V1 = p2V2
p∝ 1
V
8 a gas
[1]
[1]
c
Gas
Shape
Molecules
fixed shape
vibrate about a
fixed position
[2]
shape fills the
container from
the bottom
move around,
close together
[1]
completely fills
the container
move around,
far apart
[1]
[1]
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Answers to end-of-chapter questions: Chapter 9
1
b i increases
16 a i bombardment/collisions with air
molecules/particles
[1]
ii any two from
lighter / very small / smaller than smoke
particles / too small to be seen
fast-moving / high kinetic energy
random movement / movement in all
directions
[2]
© Cambridge University Press 2014 IGCSE Physics
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ii air molecules/particles/atoms
bombard/hit walls
molecules faster / higher energy when
temperature raised
(not vibrate faster)
greater force (per unit area) or more
collisions per second
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Answers to end-of-chapter questions: Chapter 9
2
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Answers to end-of-chapter questions
Chapter 10
1 a Liquid in bulb absorbs energy; gets hotter;
7 a Mercury expands as its temperature
expands; pushes up tube.
b melting point of pure ice (0 °C); boiling
point of pure water (100 °C)
2 a A has greater range (120 °C, from −10 °C to
+110 °C).
(B’s range is only 60 °C, from −10 °C
to +50 °C.)
b B is more sensitive. Each degree is a wider
interval on the scale, so smaller changes
can be measured.
3 solids, liquids, gases
4 more, greater
or less, smaller
5 a specific heat capacity – the energy required
per kilogram and per degree celsius to raise
the temperature of a substance.
b specific latent heat of fusion – the energy
per kilogram required to cause a substance
to change state from solid to liquid at its
melting point.
c specific latent heat of vaporisation – the
energy per kilogram required to cause a
substance to change state from liquid to
gas at its boiling point.
6 a energy = mass × specific heat capacity
× change in temperature
energy in J, mass in kg, s.h.c. in J/(kg °C),
change in temperature in °C
b energy = mass × specific latent heat
energy in J, mass in kg, specific latent
heat in J/kg
© Cambridge University Press 2014 IGCSE Physics
increases.
[1]
b
Definition
lower
fixed point
melting point
of pure ice
upper
fixed point
boiling point of
pure water
[1]
Value
0 °C [1]
100 °C
c (for example) the resistance of a
resistor or thermistor
8 a internal energy
b the steel block
It takes more energy to raise the
temperature of the steel block by a
certain amount than that of the
copper block.
9 a the thermocouple thermometer
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b 100 °C
This is a fixed point on the Celsius scale.
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c the liquid-in-glass thermometer
It can measure to 0.5 °C (or better);
the other measures to the nearest 1 °C.
[1]
d The properties of the two materials
used in the thermometers do not
vary linearly with temperature.
The voltage of the thermocouple
does not increase at a steady rate as
the temperature goes up.
10 a the final temperature of the block
the mass of the block
[1]
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b If poorly insulated, some energy
will be lost.
[1]
c too high (because the heater will have
to supply more energy to make the
temperature rise by 1 °C)
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Answers to end-of-chapter questions: Chapter 10
1
11 a 0 and 100 (°C)
[1]
b i expands
[1]
ii moves along the tube/up/to the right
stops at/near 100 mark
c arrow slightly to left of −10 mark
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b i Q = mcθ
= 100.6 − 12.0 = 88.6
= 0.80 × 3900 × 88.6
= 276 432 J
ii Q = Wt so t = 276 432
620
= 446 s
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12 a i electrical method – 3 marks for all 6 points
(deduct 1 mark for each point omitted)
lagged container + lid
liquid (allow water)
heater in liquid
heater connected to electrical supply
voltmeter and ammeter appropriately
connected
thermometer
[3]
or mixtures method – 3 marks for all 6 points
(deduct 1 mark for each point omitted)
lagged container
liquid
hot solid/hot liquid
means of heating hot solid/liquid (seen or
stated)
means of weighing hot solid/liquid/use of
known mass (seen or stated)
thermometer
[3]
ii electrical method – 3 marks for all 5 points
(deduct 1 mark for each point omitted)
initial and final temperatures of liquid or
temperature rise
voltmeter reading
ammeter reading
heating time
mass of liquid
[3]
or mixtures method – 3 marks for all 5 points
(deduct 1 mark for each point omitted)
initial and final temperatures of liquid or
temperature rise
initial and final temperatures of added solid/
liquid or temperature drop
mass of added solid/liquid
mass of liquid
s.h.c. of added solid/liquid
[3]
© Cambridge University Press 2014 IGCSE Physics
Answers to end-of-chapter questions: Chapter 10
2
Answers to end-of-chapter questions
Chapter 11
1 a temperature; higher; lower
7 a As the air is heated, it expands.
Its density decreases.
It is lighter than the surrounding air,
so it floats upwards.
b metal; non-metal
2
convection
Warm fluid moves, carrying energy
with it.
radiation
Energy travels as infrared waves.
conduction
Energy travels through a material
without the material moving.
b The surrounding air is cooler and
so less dense.
It sinks and replaces the warm air rising
above the flame.
with cooler neighbours and share energy; these
vibrate more, pass energy on to their neighbours;
and so on.
Electrons collide with particles in hotter region,
gain energy; move randomly to cooler region,
collide with particles there, give them energy.
4 expands; greater; less; lighter; rises; more;
gravity; convection
energy, so vibrate more.
They collide with neighbours,
sharing energy with them.
Energy is thus transferred from
the hot end to the cold end.
Good absorber
Good emitter
Good reflector
matt
matt
shiny
black
black
white
6 a Air is a good insulator, so less heat
c The glass wool prevents the movement
of air in the gap, so it is difficult for a
convection current to be set up,
which would transfer energy from
the inner wall to the outer wall.
© Cambridge University Press 2014 IGCSE Physics
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b The temperature of the cold end of
the rod would rise more rapidly,
because metals are better conductors
than plastics.
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c electrons
[1]
9 a walls made of glass – poor conductor
5
b Infrared (heat) radiation from below is
reflected back into the house,
so that less escapes from the house.
[1]
8 a Particles at the hot end have greater
3 Particles in hotter region vibrate more; collide
is lost by conduction.
Cold air from the window cannot
flow into the room, so convection
current losses are reduced.
[1]
[1]
vacuum between walls – no
conduction or convection
silvering – reflects away
infrared radiation
lid – prevents convection losses
(but see part b)
b A liquid that is colder than its
surroundings does not heat
the air above it,
so no convection current rises
above it. Hence a lid is not essential.
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10 a i conduction
[1]
ii convection
[1]
b heat lost at same rate as heat supplied
[1]
c i boiling
[1]
ii steam
[1]
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Answers to end-of-chapter questions: Chapter 11
1
11 a i conduction
ii atoms/free electrons at hot end vibrate
more/have more energy
share energy with others by collisions
b copper is a better conductor or iron is a
worse conductor
c iron conducts heat slowly
so gas above gauze is hot enough to burn
copper conducts heat rapidly
so gas above gauze is not hot enough
to burn
© Cambridge University Press 2014 IGCSE Physics
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Answers to end-of-chapter questions: Chapter 11
2
Answers to end-of-chapter questions
Chapter 12
1 a source
7 a trace A
The amplitude of trace A is the greatest.
b vibrations
b trace C
The frequency of trace C is the greatest
(because more waves are contained in
the same time interval).
c echo
d frequency; second
e hertz
8 You need a source of sound,
f gases; vacuum
and two detectors in line with the sound.
You need to measure the distance between
the two detectors,
and the time interval between the sound
reaching them.
distance
Then use speed =
time
to calculate the speed of sound.
2 a greater frequency
b greater amplitude
3 a
B A
b
D
9 a solid
C
4 a shaded from 20 Hz to 20 kHz
b region beyond 20 kHz
5
rarefaction
where particles of the medium are
spread out
compression
where particles of the medium are
squashed together
6 a the air inside the instrument
[1]
b the strings of the instrument
[1]
c The vibrations of the instrument cause
the air near the instrument to vibrate.
Compressions and rarefactions
are formed,
and these propagate through the air
to the listener’s ear.
© Cambridge University Press 2014 IGCSE Physics
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b (for example) Place ear against table,
tap table at a distance
and hear the sound through the wood.
[1]
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c distance travelled = 2 times length
of rod
= 800 m
[1]
distance
time
[1]
800 m
0.16 s
[1]
= 5000 m/s
[1]
speed =
=
10 a i
reflection or wave bounces back
from large object/sea bed
[1]
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ii distance = speed × time
= 1500 × 0.80
= 1200 (m)
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iii 1200/2 = 600 (m)
[1]
b graph should show
uniformly sloping line
with positive gradient
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Answers to end-of-chapter questions: Chapter 12
1
11 a any large surface, e.g. wall/cliff/mountain
b i when hears bang/sees flash
ii when hears echo
c i reading = 2.25 s
speed = distance
time
[1]
[1]
[1]
[1]
[1]
= 720
2.25
[1]
= 320 (m/s)
[1]
ii one from
inaccurate distance from firework
reaction time
wind
© Cambridge University Press 2014 IGCSE Physics
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Answers to end-of-chapter questions: Chapter 12
2
Answers to end-of-chapter questions
Chapter 13
1 a See Figure 13.5.
9 a Ray diagram correctly drawn showing
b angle of incidence = angle of reflection
i=r
2 a virtual
that the ray passes through both surfaces
undeflected, that is, the ray remains
straight.
[2]
ray bends towards normal
then away again
so that it ends up parallel to original path
[1]
[1]
[1]
b
b the same size as
c object
d left–right inverted
3 See Figure 13.9a.
4 a n=
speed of light in a vacuum
speed of light in the material
b n = sin i
sin r
c Parallel rays remain parallel.
n = refractive index, i = angle of incidence,
r = angle of refraction
10 a
5 See Figure 13.12a.
30° 30°
A
[1]
50° 50°
B
6 See Figure 13.16a.
7 a converging
b closer than
c virtual; magnified
8
normal
mirror
angle of
incidence i
angle of
reflection r
incident ray
reflected ray
incident and reflected rays
correctly drawn
normal correctly drawn
angle of incidence correctly marked
angle of reflection correctly marked
© Cambridge University Press 2014 IGCSE Physics
In block A, reflected ray at equal angle
and refracted ray bent away from normal.
In block B, reflected ray only,
at equal angle.
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b When the angle of incidence is greater
than the critical angle,
there is only an internally reflected ray;
all of the ray is totally internally reflected.
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Answers to end-of-chapter questions: Chapter 13
1
11 a
13 a i image behind mirror
image
image same distance from mirror,
along line perpendicular to mirror
ii reflected ray reaching eye
direction of reflected ray coming
from image
mirror
object
Two rays in different directions
from a single point on the lamp
reflect off the mirror correctly
and are extrapolated back behind
the mirror,
so that the image is at the point
where they cross.
[1]
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b Each ray is reflected
so that angle of incidence equals
angle of reflection.
[1]
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c Light appears to come from a point
behind the mirror
but no light actually travels behind
the mirror.
12 a
[1]
[1]
ray 2
I
F
ray 1
F
O
[1]
[1]
[1]
c both rays straight on at first surface
30° prism ray refracted down in air at
second surface
45° prism ray reflected down in glass at
second surface
90° reflection
straight on at third surface
[1]
[1]
[1]
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14 a i any two of these three rays from top
of object:
parallel to axis to lens and on through
focal point
undeviated through centre of lens
as if from focal point to lens and then
parallel to axis
traced back to locate image
[2]
[1]
ii any two of:
virtual / upright / magnified / further
from lens / dimmer
[1]
ii magnifying glass
[1]
[1]
[1]
[1]
[1]
[1]
b The image is real
because it is formed where rays
of light meet.
[1]
c The image is diminished
because it is shorter than the object.
[1]
[1]
d The image is inverted because it is
below the axis.
[1]
© Cambridge University Press 2014 IGCSE Physics
[1]
b HIS, because S is not its own mirror image
b i 3.4–3.6 cm
Ray 1 continues straight through the
centre of the lens,
ray 2 bends at the lens
and passes through the principal
focus F,
so that the image is at the point
where they cross.
[1]
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Answers to end-of-chapter questions: Chapter 13
2
Answers to end-of-chapter questions
Chapter 14
1 energy; matter
7
2
transverse
describes a wave that varies from side to
side, at right angles to the direction of
travel
longitudinal
describes a wave that varies back and
forth along the direction of travel
3
Symbol
Quantity
Unit
v
speed
m/s
f
frequency
Hz
λ
wavelength
m
waves correctly reflected at barrier
separation remains as before
8 a decreases
4 a bounces off
[1]
b stays the same
[1]
c decreases
[1]
9 a speed = frequency × wavelength, v = f λ
b speed
5 reflection, refraction, diffraction (in any order)
6 a 4.0 cm
[1]
b 3.0 cm
[1]
c one wave = 4 cm so 10 cm = 2.5 waves
so 2.5 waves pass in 1 s
frequency = 2.5 Hz
[1]
[1]
[1]
[1]
[1]
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b v = f λ = 6 × 1014 × 3.75 × 10−7
= 2.25 × 108 m/s
[1]
[1]
waves are curved in space beyond barrier
separation remains as before
[1]
[1]
10
d
2
y / cm
1
0
1
2
3
4
5
–1
6
7
8
9
x / cm
–2
correct value of amplitude
correct value of wavelength
© Cambridge University Press 2014 IGCSE Physics
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Answers to end-of-chapter questions: Chapter 14
1
11 a i amplitude
ii wavelength
b i string moves air
backwards and forwards or up and
down or produces compressions and
rarefactions
[1]
[1]
[1]
[1]
ii gets quieter/softer/less loud
[1]
12 a i R in correct position, by eye
[1]
ii three reflected waves correctly
meeting mirror
three reflected waves equidistant
and centred on R
b first ray plus reflection correct
second ray plus reflection correct
reflected rays projected back, to meet
behind mirror or labelled I and in
correct position
© Cambridge University Press 2014 IGCSE Physics
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Answers to end-of-chapter questions: Chapter 14
2
Answers to end-of-chapter questions
Chapter 15
1 a red, orange, yellow, green, blue, indigo, violet
8 White light is dispersed because it is a
b red = lowest frequency, longest wavelength
violet = highest frequency, shortest wavelength
2 a radio waves, microwaves, infrared, visible
light, ultraviolet, X-rays, gamma rays
b radio waves = lowest frequency, longest
wavelength
gamma rays = highest frequency, shortest
wavelength
3 300 000 000 m/s or 3.0 × 108 m/s
4 monochromatic; frequency (or wavelength)
5 a
prism
ray broadens on entering prism,
and again on leaving
red and violet ends of spectrum
correctly indicated
ii dispersion
[1]
b i A (red, at top)
ii C (yellow, third from top)
[1]
[1]
c any two from:
gamma, cosmic, X-rays, UV, IR,
microwaves, radio, TV
[2]
b violet
c Different colours travel at different
speeds in glass.
The slowest colour is the most
strongly dispersed.
b film or photographic film or electronic
detector or charge-coupled device (CCD)
[1]
[1]
[1]
[1]
[1]
[1]
short
red orange
yellow
green
blue
violet indigo
[1]
[1]
9 a i refraction
10 a electromagnetic
spectrum
white light
mixture of different colours/wavelengths,
which travel at different speeds in glass.
Laser light is a single wavelength
(monochromatic) and so cannot
be dispersed.
[1]
[1]
[1]
c absorbed/stopped by bone (not deflected/
reflected)
[1]
less absorption by flesh or penetrates/passes
through flesh
[1]
d any one of:
photographic film badges, behind screen
when operating X-ray machine, protective
clothing, minimise exposure
[1]
[1]
6 a false
b true
c true
7 300 000 000 m/s = 3.0 × 108 m/s
© Cambridge University Press 2014 IGCSE Physics
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Answers to end-of-chapter questions: Chapter 15
1
Answers to end-of-chapter questions
Chapter 16
b
1 a i repel
N
ii attract
b See Figure 16.3.
S
S
N
N
S
2 a i soft
ii hard
b i for example: steel
ii for example: soft iron
S
3 stroking with one pole of a permanent magnet
magnets in a square
arranged N–S–N–S–N–S–N–S
with attractive forces shown
place in electromagnet connected to d.c. supply
4 hammer it
place in electromagnet connected to a.c. supply
6 a one of the following:
2
1
attract
N
N
S
attract
3
S
N
S
[1]
9 a A soft magnetic material is easy
b one of the following:
N and S poles at opposite ends
field lines have same pattern
S
[1]
[1]
b (for example) in a scrapyard crane or an
electromagnetic door bolt
electromagnet can be switched on and off
strength can be varied by changing current
poles can be reversed by reversing current
N
[1]
more turns of wire or turns of wire
closer together
add an iron core
b See Figure 16.7b.
repel
[1]
[1]
8 a bigger current
5 a See Figure 16.7a.
7 a
N
to magnetise
and to demagnetise.
A hard magnetic material is difficult to
magnetise and demagnetise.
[1]
[1]
b A hard material
because it retains its magnetisation well.
[1]
[1]
c A soft material
because its magnetisation can
change easily.
[1]
[1]
[1]
10
repel
4
each correct pair of attractive or
repulsive forces
© Cambridge University Press 2014 IGCSE Physics
N
N
S
S
[4]
Answers to end-of-chapter questions: Chapter 16
1
a left-hand end of solenoid N
right-hand end of solenoid S
b lines of force out of N poles and
into S poles
lines close together at poles, farther
apart elsewhere
similar pattern for both magnet
and solenoid
repulsion indicated by distortion
of pattern
11 a (S) N S N
b i
Switch
closed
[1]
[1]
ii attractive force
[1]
iii with soft iron core
[1]
iv can be switched on and off
(or can be stronger)
[1]
12 a can be switched off
can vary the strength
[1]
b i
[1]
[1]
1000 turns
[1]
ii iron
[1]
iii 3.0 A
[2]
Switch open
Soft
iron
magnetised
loses its
magnetism
Steel
magnetised
keeps its
magnetism
© Cambridge University Press 2014 IGCSE Physics
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Answers to end-of-chapter questions: Chapter 16
2
Answers to end-of-chapter questions
Chapter 17
1 a rubbed, friction, opposite
Because the electrons have negative
charge, this gives the balloon a
negative charge.
b repel, attract
2 a electrons
b negative
b The balloon repels electrons in
the paper,
so that there is a positive charge on
the area of paper closest to the balloon.
This positive charge and the negative
charge of the balloon attract each other.
c positively
3 charged, attract, attract, attract, induction
4 a
electric force
8 a i iron or ferromagnetic
+
unmagnetised (before being brought
near magnet)
(not non-magnetic)
b The field is in same direction as the force on a
positive charge.
5
Symbol for
unit
Quantity
Unit
force
newton
N
electric charge
coulomb
C
6 a positive
[1]
[1]
c Suspend one so that it can turn freely.
Bring the other close to one end and
observe repulsion.
[1]
causes electrons to be transferred
from atoms in the wool to atoms in
the balloon.
© Cambridge University Press 2014 IGCSE Physics
b attracts (at first)
repels after touching or angle of thread
increases as XY decreases
9 a rub/rubbing
b They are equal.
7 a The force of friction
ii magnet
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
with dry cloth
[1]
[1]
b i negative (−)
[1]
ii opposite charges attract
[1]
c horizontal arrow to L, starting or ending
on sphere
[1]
d swings to R / moves away / is repelled
[1]
[1]
[1]
Answers to end-of-chapter questions: Chapter 17
1
Answers to end-of-chapter questions
Chapter 18
1 a charge
7
cell
b positive, negative
2 a ammeter, series
current
switch
b voltmeter, parallel
3 a
lamp
A
V
b R=V
I
a series circuit correctly drawn
correct symbols with labels
[1]
[3]
b at least two arrows around circuit
from positive of cell
[1]
[1]
c voltmeter
[1]
d volt (V)
[1]
8
4
Unit
+
–
Symbol
for unit
Potential difference
volt
V
Current
ampere
A
Resistance
ohm
Ω
A
R
5 a positive, charge
b negative, positive
V
6
Equation
In words
In units
Q = It
charge = current × time
coulomb = ampere × second
(C = A s)
R=V
I
resistance =
P = IV
power = current × p.d.
watt = ampere × volt
(W = A V)
E = IVt
energy = current × p.d. × time
joule = ampere × volt × second
(J = A V s)
© Cambridge University Press 2014 IGCSE Physics
p.d.
current
ohm = volt/ampere
(Ω = V/A)
Answers to end-of-chapter questions: Chapter 18
1
a correct symbols for resistor,
ammeter and power supply
connected in series
with voltmeter in parallel with resistor
[3]
[1]
[1]
b current
[1]
c potential difference (p.d.)
V
d R=
I
6.5
=
1.25
= 5.2 Ω
[1]
[1]
[1]
[1]
9 a light
[1]
b heat
[1]
c power = 36 W
[1]
d energy = power × time
= 36 × 60
= 2160 J
P
e I=
V
36
=
12
[1]
[1]
[1]
= 3A
[1]
[1]
[1]
Q
10 a I =
t
30
=
20
= 1.5 A
[1]
b E = Pt
battery/cell, ammeter, coil in series
voltmeter in parallel with coil
standard symbols used for battery/cell,
voltmeter and ammeter
ii R = V
I
iii any two of:
length (of wire)
diameter/cross-section/area (of wire)
resistivity/type of material
temperature
b R=60
15
= 4.0 Ω
resistance of AB = 1.0 Ω
resistance per metre = 0.50 Ω/m
12 a increases as current increases
at an increasing rate
b i
25 Ω
[1]
[1]
[1]
[1]
[2]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
ii V = I R = 0.070 × 25
= 1.8 V
iii P = I V (or P = I 2 R etc.)
= 0.12 W
c i 1.8 V (same as answer to b ii)
[1]
[1]
[1]
[1]
[1]
[1]
ii 1 = 1 + 1
R R1 R2
[1]
[1]
R = 12.5 Ω
[1]
[1]
= IVt
[1]
= 1.5 × 10 × 20
[1]
= 300 J
[1]
11 a i
A
coil of
wire
V
© Cambridge University Press 2014 IGCSE Physics
Answers to end-of-chapter questions: Chapter 18
2
Answers to end-of-chapter questions
Chapter 19
1
Name of
device
b i
Circuit symbol
Description
lightdependent
resistor
(LDR)
resistance
decreases when
light falls on it
thermistor
resistance
changes when
temperature
changes
relay
See Figure 19.15c.
ii See Figure 19.15a.
iii See Figure 19.15b.
iv See Figure 19.18a.
v
an
electromagnetic
switch
See Figure 19.18b.
6 a melting, burning, fumes
b wire melts, breaks circuit
c circuit breaker
d Fuse will not break for normal current, but will
break when current rises above this value.
7
2 a current
b sum
3 a voltage (or p.d.)
b shared
each symbol correctly drawn
8 a
[4]
6V
c more (greater)
4 a series
b parallel
c series
d parallel
5 a i
series circuit
correct symbols for resistor, switch
and power supply
[1]
[3]
ii
b 10 + 40
= 50 Ω
[1]
[1]
iii
c 0.12 A
[1]
d 0.12 A
[1]
iv
v
© Cambridge University Press 2014 IGCSE Physics
9 a capacitor
[1]
b light-dependent resistor
[1]
c relay
[1]
Answers to end-of-chapter questions: Chapter 19
1
10 a wires overheat (risk of fire)
[1]
b fuse, trip switch
[2]
c Use thicker wires,
which have lower resistance,
so there is less heating.
[1]
[1]
[1]
11 a in parallel
[1]
b 6.0 V
[1]
across each resistor
[1]
14 a
[1]
1 1 1
d = +
R 2 3
3 2 5
= + =
6 6 6
[1]
[1]
R=
6
= 1.2 Ω
5
[1]
I=
V 6
=
R 1.2
[1]
= 5.0 A
[1]
A
B
electrons
diode
a diode correctly labelled
[1]
b cell correct way round
[1]
c arrow in correct direction
[1]
13 a AND gate
Output is ON only if both inputs are ON.
Input 1
d OFF (0)
[1]
because the resistance is lower.
b
[1]
[1]
[1]
c The 2 Ω resistor,
12
c AND gate
NOT gate
Input 2
[1]
[1]
[1]
[1]
[1]
c L2 has blown or is missing
[1]
d i
[1]
blows
0
0
0
[1]
1
0
1
[1]
0
1
1
[1]
1
1
0
[1]
[1]
[1]
[1]
[1]
ii nothing / does not light / off
(not turns off )
[1]
iii nothing / does not light / off
(not turns off )
[1]
15 a i 4.0 V
[1]
ii 12 V
[1]
b i 6Ω
ii 1 = 1 + 1
R 3 6
R = 2Ω
c I=V
R
= 6.0 A
d i stays the same
ii decreases
Output
© Cambridge University Press 2014 IGCSE Physics
circuit diagram with two lamps in parallel
switch alongside power supply
correct symbols for lamps and switch used
b R=V
I
= 12
16
= 7.5
ohm(s) or Ω
Answers to end-of-chapter questions: Chapter 19
[1]
[1]
[1]
[1]
[1]
[1]
[1]
2
Answers to end-of-chapter questions
Chapter 20
1 relays, motors, electric bells (any order)
9 a i
magnetised
ii attracted or magnetised
iii close
2 current, magnetic, turning, rotate
3 a force (motion)
b any two from:
armature becomes permanently magnetised
would not release from core
contacts always closed
[2]
b magnetic field
c current
4 a charged, field, force
10 a i
b electrons, cathode ray (or television)
5 Someone presses the bell push.
[1]
A current flows through the electromagnet.
[1]
The electromagnet attracts the iron armature. [1]
The hammer strikes the gong.
[1]
At the same time, the circuit is broken
at point A.
[1]
The springy metal pulls the hammer back.
[1]
The circuit is completed again at A.
[1]
6 a The wire will swing the other way.
[1]
b The wire will swing the other way.
[1]
7 a downwards
b to the right,
by Fleming’s left-hand rule
current clockwise when viewed from top [1]
ii anticlockwise or down on left and/or up
on right
[1]
b i
faster or greater turning effect
[1]
ii faster or greater turning effect
[1]
iii faster or greater turning effect
[1]
[1]
[1]
[1]
8 a downwards
[1]
b upwards
[1]
c The forces are unbalanced,
and so provide a turning effect.
[1]
[1]
d The force is zero,
because the current does not cut
across the magnetic field (it is parallel
to the field).
[1]
© Cambridge University Press 2014 IGCSE Physics
[1]
[1]
[1]
[1]
Answers to end-of-chapter questions: Chapter 20
1
Answers to end-of-chapter questions
Chapter 21
1 conductor, magnetic, induced, circuit, current
2 (answers from the top) d.c.; a.c.; a.c.; d.c.; a.c.; d.c.; a.c.
c Use the primary coil as the secondary
and the secondary as the primary.
11 a i
X: coil
ii Y: slip rings
iii Z: brushes
3 coil, rotate/turn, magnetic, e.m.f., current
4 a movement
b field
[1]
[1]
[1]
b A.c. flows back and forth,
changing direction.
D.c. flows in one direction only.
c current
5 high, smaller, less
[1]
[1]
+
6 a primary, core, secondary
d.c.
Current
b step-up, e.m.f./voltage
c step-down, e.m.f./voltage
0
Time
7 Vp = p.d. across primary coil
a.c.
Vs = p.d. across secondary coil
Np = number of turns on primary coil
Ns = number of turns on secondary coil
–
correct (labelled) diagram
8 Ip = current in primary coil
12 a more turns
Vp = p.d. across primary coil
Is = current in secondary coil
Vs = p.d. across secondary coil
bigger area
b stronger magnetic field
turn the coil faster
9 a The magnetic field around the wire is
changing (it is cutting across field lines).
[1]
b It will change sign / direction (from positive
to negative, or the other way round).
[1]
c She should move the wire more quickly.
[1]
d No,
because it is not cutting across the field
lines / the magnetic field is not changing.
[1]
13 a
Vp Np
=
Vs
Ns
Ns =
Ip =
Ns
Np
3 × 200
=
= 60 V
10
Vs =Vp ×
© Cambridge University Press 2014 IGCSE Physics
[1]
14 a i
[1]
[1]
[1]
[1]
[1]
[1]
0.40 × 12
230
= 0.021 A
[1]
[1]
5000 × 12
230
b Ip × Vp = Is × Vs
[1]
[1]
[1]
= 261
10 a So that less energy is lost during
transmission.
Vp Np
b
=
Vs
Ns
[1]
[1]
[1]
[1]
deflection to one side
then goes back to zero again
[1]
[1]
ii same as i but opposite direction
[1]
b larger
[1]
[1]
Answers to end-of-chapter questions: Chapter 21
1
c smaller
d nothing (or small oscillations about zero
position or blurred light spot)
15 a i
no deflection on voltmeter
ii deflection on voltmeter
© Cambridge University Press 2014 IGCSE Physics
[1]
[1]
[1]
iii deflection on voltmeter
(in same direction as in ii)
b use a stronger magnet
move coil or magnet faster
add more turns to coil
[1]
[1]
[1]
[1]
[1]
Answers to end-of-chapter questions: Chapter 21
2
Answers to end-of-chapter questions
Chapter 22
1
9 a i alpha, positive
electron
gold nucleus, positive
electron, negative
nucleus
proton
neutron
4
2
2
[3]
b Electrons (the ‘plums’) are distributed
through
a sphere of positive charge (the ‘pudding’).
[1]
[1]
What it tells us
X
chemical symbol
name of element
Z
proton number
number of protons
in nucleus
number of nucleons
in nucleus
3 proton number + neutron number = nucleon
number
4 a different numbers
alpha particle smaller than gold
nucleus, with positive charges marked
track of alpha particle correctly shown
d Most of the gold atom is empty space / the
nucleus makes up a small
fraction of the volume of the atom,
so the chance of a head-on collision
between an alpha particle and a gold
nucleus is very small.
10 a i
b the same number
c different numbers
5 alpha, deflected, thin, mass, positive, centre
6 a 6 protons
[1]
b 6 neutrons
[1]
c 6 electrons
[1]
7 a 79 + 118
gold nucleus
alpha
Name
nucleon number
ii electron, alpha particle, gold nucleus
c
He
Symbol
A
[1]
[1]
[1]
ii 3
[1]
iii 4
[1]
iv 3 + 4 = 7
[1]
[1]
Particle
Charge
Mass
m
[1]
[1]
electron
−1
neutron
0
[1]
2000m
[1]
197
79
[2]
proton
+1
[1]
2000m
[1]
8 a 19
[1]
b 39
[1]
c
Au
40
19
K
© Cambridge University Press 2014 IGCSE Physics
[2]
[1]
[1]
= 197
b
[1]
3
b 73 Li
11 a
[1]
[1]
b i 92
ii 146
iii 92
[1]
[1]
[1]
Answers to end-of-chapter questions: Chapter 22
1
Answers to end-of-chapter questions
Chapter 23
1
Radiation
2
Symbol
Type of particle or
electromagnetic radiation
Mass
alpha
α
2 protons + 2 neutrons (He nucleus)
beta
β
electron
gamma
γ
electromagnetic radiation
Charge
1
+2
small
−1
0
0
radiation in the environment
background
detectors of ionising radiation
Geiger counter, photographic film
three types of ionising radiation from radioactive
substances
alpha, beta, gamma
3 a it has negative charge
b charged; Fleming’s left-hand rule
c it is uncharged
4
Radiation
Penetration
Absorption
Absorbed by
alpha
least penetrating
most easily absorbed
thin paper, a few cm of air
beta
in between
in between
thin metal foil
gamma
most penetrating
least easily absorbed
thick lead or concrete
5 a (average) time, half, decay
b See Figure 23.10a.
6
Use
… because …
Finding the age of an object
radioactive substances decay at a known rate.
Seeing through solid objects
radiation can penetrate matter.
Sterilising medical equipment
radiation can destroy living cells.
Tracing the movement of hazardous
substances
small amounts of radiation can be detected.
© Cambridge University Press 2014 IGCSE Physics
Answers to end-of-chapter questions: Chapter 23
1
7 β is more penetrating than α.
[1]
[1]
Detect using Geiger counter.
Place thin paper over sources − α does
not pass through.
Place thin aluminium foil over
sources − neither passes through.
[1]
[1]
8 a 15
[1]
[1]
counts per minute
b 65 − 15
= 50 counts per minute
c
[1]
[1]
Count rate /
counts per minute
65
40
11 a i background
or any of the following:
contaminated surfaces
other radioactive material nearby
radiation from rocks/soil
cosmic rays/radiation from space
radon gas from ground
ii count rate = 136
4
= 34 counts/min
b i alpha or α
ii 876
4
− 34
= 185 counts/min
12 a i proton
15
0 half- 2
4
6
Time / hours
life
8
correct graph drawn
After one half-life, the measured
count rate will be down to 25 + 15 = 40
Reading across from 40 on the graph,
and then down, half-life = 1.3 h
approximately.
9 a The formation of an ion
by the removal of one or more
electrons from an atom.
b X-rays
10 a (for example) A patient with cancer is
exposed to γ-radiation.
This damages the cancerous cells,
which then die.
b (for example) During the manufacture
of cardboard, β-radiation is passed
through the card.
If the card is too thick (too thin),
the amount of radiation detected will be
too low (too high).
The machinery is automatically
adjusted to give the correct thickness.
© Cambridge University Press 2014 IGCSE Physics
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
ii proton and neutron
0
[1]
[1]
b number of protons = 47
number of neutrons = 107 − 47 = 60
[1]
[1]
c i 8 h ± 0.25 h
[1]
ii Choose two points on the graph; for
each, halve the value and add 8 h to
the time.
[2]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
Answers to end-of-chapter questions: Chapter 23
2