Full Answers to Review Questions - Hodder Plus Home

Full Answers to Review Questions
1 Quantities and units
1 a) Metre and second are base units, and speed is a derived quantity; the only
base quantity is therefore length – the answer is A.
b) Force is a derived quantity; metre and second are both base units; the
newton, N, is the only derived unit – the answer is C.
c) Acceleration and force both have implied direction and so are vector
quantities; metre is a unit; speed has magnitude only, and so is a scalar
quantity – the answer is D.
d) Distance, mass and time have magnitude but not direction and are all
scalar quantities; velocity is the only vector quantity – the answer is D.
2 F = PA = 10 × 10 6 Pa × 220 × 10 −6 m 2 = 2.2 × 10 3 N = 2.2 kN
3 From Table 1.2 (page 3):
• The units of Q are: A s
• Those of V are: kg m2 A−1 s−3 Therefore the units of C are: ___________ A s
kg m2 A−1 s−3 = kg−1 m−2 A2 s4 4 a) 11 km; S 278 E
b) Average speed = 3.0 km h −1
Average velocity = 2.2kmh −1 in the direction S 278E
5 Horizontal component = 120 cos 30 = 104 m s −1
Vertical component = 120 sin 30 = 60 m s −1
2 A guide to practical work
1 a) For example:
l/mm w/mm t/mm
95 62 43
96 62 43
Average 95(.5) 62 43
Volume: V = l w t
= 9.55 cm × 6.2 cm × 4.3 cm
= 255 cm 3
Density = m __ = _______ 250 g
V 255 cm3 = 0.98 g cm −3 (980 kg m −3 )
b) Percentage uncertainty in l
= 1 mm
________
× 100% = 1.0%
95.5 mm
Percentage uncertainty in w
= 1 mm
______
× 100% = 1.6%
62 mm
Percentage uncertainty in t
= 1 mm
______
× 100% = 2.3%
43 mm
Table A.1 �
Edexcel Physics for AS © Hodder Education 2009
1

The overall percentage uncertainty is therefore in the order of 5%
(probably more when also taking into account any manufacturing
tolerance and the mass of the wrapper). This means that the density of
the butter could lie between about 0.95 g cm −3 and 1.03 g cm −3 .
Although the experimental value of 0.98 g cm −3 suggests that butter
will float, the experiment may not be sensitive enough to confirm this
beyond all doubt. (You might like to check what happens with a small
piece of butter in a cup of water!)
2 a) i) For example, mass of packet of paper
M = 2.52 kg
ii) Mass of single sheet
2520 g
m = ______ = 5.04 g
500
b) i) For example:
l/mm 297 297 Average: 297
w/mm 210 210 Average: 210
ii) Area A = 0.297 m × 0.210 m = 0.0624 m 2
5.04 g
‘gsm’ = __________ = 80.8 g m−2
2 0.006 24 m
iii) Percentage difference
= (80.8 – 80) g m−2
_____________________
80 g m−2 × 100% = 1%
Full Answers to Review Questions: 2 A guide to practical work
Table A.2 �
This is acceptable experimental error, particularly when taking into
account the mass of the packet and the lack of sensitivity of the
kitchen scales.
c) i) For example, measured thickness of packet/mm
= 48, 47, 48, 49 ⇒ average = 48 mm
Thickness of single sheet
t =
48 mm
______
500
= 0.096 mm = 0.0096 cm
Density of paper = mass _______
volume
=
5.04 g
_________________________
29.7 cm × 21.0 cm × 0.0096 cm
= 0.84 g cm –3 (840 kg m –3 )
ii) The thickness of a single sheet of paper could be checked as follows:
• First check the micrometer screw gauge or digital callipers for zero
error.
• Fold the paper four times to get 16 thicknesses.
• Compress to remove any air.
• Measure 16 t in four different places.
• Take the average and hence find t.
Tip
Note that your final answer can
be quoted only to the number of
significant figures of the least precise
of your measurements. In this case,
the answer can only be stated to two
significant figures as t has only been
measured to 2 s.f.
Tip
Note the experimental techniques
given here – multiple readings (16 t)
taken in different places – and the
use of bullet points. This is a good
strategy as it helps you set out your
answer in a clear and logical manner.
Examiners love bullet points!
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2

3 a) For example:
10d/mm 10d/mm mean d/mm
203 203 20.3
10t/mm 10t/mm mean t/mm
16.5 16.5 1.65
Full Answers to Review Questions: 2 A guide to practical work
Table A.3 �
Note that d and t have been found by measuring the length of ten coins
in a row and the height of ten coins, respectively.
b) V = π d2 t
____
4
= π × (2.03 cm)2 _____________________ × 0.165 cm
= 0.534 cm3
4
Density = mass
_______
volume
= 3.56 g
_________
0.534 cm 3 = 6.7 g cm –3
c) i) Percentage difference
= (7.8 – 6.7) g cm–3
_____________________
7.8 g cm –3 × 100% = 14%
ii) Percentage uncertainty in 10d
= 1 mm
_______
× 100% = 0.5%
200 mm
Percentage uncertainty in 10t
= 0.5 mm
________
× 100% = 3%
16.5 mm
The fact that the percentage difference between the experimental
value for the density of the coins and the given value for the
density of mild steel differs by 14%, which is much more than the
experimental uncertainty, suggests that the coins are not made of
mild steel. However, the true value for the average thickness is
considerably less than that measured. When 10 coins are stacked on
top of each other, the thickness measured is actually the thickness
of the rim of the coin and not its average thickness. A better average
value could be obtained by using a micrometer and measuring the
thickness in several places.
d) The percentage difference between the density of brass (8.5 g cm –3 ) and
that of mild steel (7.8 g cm –3 ) is:
(8.5 – 7.8) g cm–3
______________________
8.15 g cm –3
× 100% = 9%
This is more than the estimated experimental uncertainties, so it should
be possible to distinguish between the two types of coin. In particular,
the difference in 10 thicknesses would be (16.5 – 15.2) mm = 1.3 mm,
which can easily be detected with a rule. If only one coin were available,
the difference in thickness would be 0.13 mm, which could be detected
easily with a micrometer or digital callipers.
Tip
Note that the discussion and
conclusions in parts c) and d)
have been argued on the basis of
quantitative evidence. You must
remember to do this wherever
possible.
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3

4 a) If h = 1 _
2 g t2 ⇒ t2 =
2 h
___
g
= __ 2
g × h
A graph of t2 against h therefore should be a straight line through the
origin of gradient equal to 2 __
g
.
b)
h/cm 40 60 80 100 120
t/s 0.30 0.38 0.42 0.47 0.51
t 2 /s 2 0.090 0.144 0.176 0.221 0.260
Your graph should be as in Figure A.1.
t 2 /s 2
0.30
0.25
0.20
0.15
0.10
0.05
0
0
Figure A.1 �
0.20 0.40 0.60
h/m
0.80 1.00 1.20
Full Answers to Review Questions: 2 A guide to practical work
Table A.4 �
c) The graph is a straight line but does not pass through the origin. This
suggests a small systematic error. There has possibly been a systematic
error in measuring the distance, h (perhaps caused by measuring to the
wrong place each time) or, more likely, there has been a systematic error
in t due to time delays in releasing the sphere or opening the trap door.
d) Gradient = 0.250 s2 − 0.015 s2 _______________
2 –1
= 0.206 s m
1.14 m − 0.00 m
2
__
g = 0.206
⇒ g = 2 _____ = 9.7 m s–2
0.206
e) Percentage difference
= (9.8 – 9.7) m s–2
_____________
9.8 m s
–2 × 100% = 1%
f) i) The values of h have been recorded only to the nearest centimetre
when they would have, presumably, been set to a precision of 1 mm.
The values should therefore have been recorded as 40.0 cm, 60.0 cm,
etc. to reflect this.
ii) Repeat timings should definitely have been recorded, with at least
three values for each height. More readings, with larger values of h if
possible, would also improve the results.
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4

3 Rectilinear motion
1 a) a =
b) s =
v − u
_____
t
v + u
_____
2
(40 − 0) m s−1
= ____________
80 s = 0.50 m s−2 – the answer is B.
(40 + 0)
× t = _______
2 m s−1 × 80 s = 1600 m – the answer is B.
2 The gradient of a displacement–time graph represents the velocity – the
answer is D.
3 The area under a velocity–time graph represents the displacement – the
answer is B.
total displacement
4 Average velocity = ________________
time
Instantaneous velocity = δx ___ at an instant
δt
5 a) Acceleration is the change of velocity in unit time: a 5 Dv ___
Dt
b) i) Average acceleration ≈ 3 m s−2 ii) Gear changes are likely to affect the acceleration.
6 s = 1 _
2 (u + v)t
v = u + at
s = ut + 1 _
2 at 2
v 2 = u 2 + 2as
7 a) v = 12 m s −1
b) s = 160 m
8 b) For a height of about 2 m, the time is about 0.6 s. The % uncertainty
in h is (±0.001 m)/(2.000 m) × 100% = 0.05%; for t, the value is
(0.01 s)/(0.60 s) × 100% = 1.7%. Therefore, time has the greater effect
on the uncertainty in g.
9 a) s = u t + 1 _
2 at2 = 0 + 1 _
2 × 9.8 m s –2 × (2.2 s) 2 = 24 m
b) v = u + at = 0 + 9.8 m s −2 × 2.2 s = 22 m s −1
10 a) 7.3 m
b) 1.2 s
c) 7.6 m s −1 (downward)
11 a) Vertical component = 10 m s −1
Horizontal component = 17 m s −1
b) t = 2.0 s
c) x = 36 m
12 a) A = stationary; B = uniform velocity; C = uniform velocity in the
opposite direction to B; D = increasing velocity (acceleration)
b) A = constant velocity; B = uniform acceleration; C = uniform
deceleration; D = increasing acceleration.
13 a) 20.2 m s −2
b) 1800 m
c) 10 m s −1
Full Answers to Review Questions: 3 Rectilinear motion
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4 Forces
1 a) The resultant force acting on the system = 7000 N − (1000 + 1000) N =
5000 N.
Using Newton’s second law: a = F __
m = _______ 5000 N
4000 kg = 1.25 m s−2 – the answer
is A.
b) Resultant force on the trailer = (T − 1000) N = 2500 kg × 1.00 m s −2
T = 2500 N + 1000 N = 3500 N – the answer is B.
2 a) The weight is a downward gravitational force of the Earth on the man.
By Newton’s third law there must be an equal upward gravitational force
of the man on the Earth – the answer is C.
b) Only two forces act on the man; his weight pulling him down, and the
upward push of the table on his feet. For the man to be in equilibrium
these must be equal and opposite – the answer is B.
3 Examples of distant forces are: gravitational, electrostatic and magnetic
(electromagnetic) and nuclear.
Examples of contact forces are: friction, air resistance and normal reaction
forces.
4 Examples of gravitational forces are: sun and planets, satellites, weight of
objects on Earth.
Examples of electromagnetic forces are: motors, electrons orbiting the
nucleus, forces on static electrical charges.
Examples of nuclear forces are: strong forces binding protons and neutrons,
weak forces involved in beta decay.
5 A body is in equilibrium if the vector sum of all the forces acting on it is zero.
6 b) Weight
c) i) F = (4.0 kN) cos 15 = 3.86 kN ≈ 3.9 kN
ii) R + (4.0 kN) sin 15 = 800 kg × 9.8 m s −2 ⇒ R = 6.81 kN ≈ 7 kN
5 Work, energy and power
1 Work done by each force = F s cos θ = 100 N × 100 m × cos 60 = 5000 J
Total work done by both forces = 10 000 J – the answer is C.
2 At the lowest point the jumper is stationary, and the kinetic energy must
be zero.
The elastic cord will have elastic strain energy stored within it, but some
of the initial gravitational potential energy will have been converted
to internal energy as the molecules are displaced within the cord – the
answer is B.
3 Power = _________ work done mgh
time
= ____
t = 1000 kg × 9.8 m s−2 ______________________ × 2.4 m
= 2940 W ≈
8.0 s
3000 W – the answer is C.
4 Useful power output = 75% of 120 W = 90 W
P = F × v ⇒ F = P __
v = _______ 90 W
2.0 m s−1 = 45 N – the answer is A.
5 Work is the product of the force and the distance moved in the direction
of the applied force. Work is measured in joules (J).
6 a) W = F∆x = 60 N × 100 m = 6000 J
b) W = F∆x cos θ = 80 N × 50 m cos 30 = 3500 J (to 2 significant figures)
7 a) Potential energy is the ability of a body to do work by virtue of its
position or state.
Full Answers to Review Questions: 4 Forces
Edexcel Physics for AS © Hodder Education 2009
6

) The gravitational potential energy (GPE) of a body is the ability of
the body to do work by virtue of its position in a gravitational field.
An object of mass m raised through a height ∆h will gain GPE of
mg∆h.
However, the elastic potential energy (EPE) of an object is the energy
stored in the object when it is stretched or compressed. For a cord
extended ∆x by an average force F ave , the gain in EPE will be F ave ∆x.
8 a) Kinetic energy (KE) is the energy of a body by virtue of its motion.
The KE of a body of mass m moving with a velocity v is 1 _
2 mv 2 .
b) KE of the golf ball = 1 _
2 (0.05 kg)(20 m s−1 ) 2 = 10 J
9 After drawing the diagram, measure the vertical height, h, of the trolley
above the base of the ramp. Record the time, t, for the interrupter card to
cut through the light beam, and find the velocity:
v =
(length of card)
_____________
t
Repeat for a range of values of h.
∆GPE = mgh and ∆KE = 1 __ mv
2 2 ⇒ ∆KE ______
∆GPE
v2
= ____
2gh
Plot a graph of v2 against h. The percentage of GPE converted to KE is
found by
_______ 2g
× 100%.
gradient
10 Efficiency =
useful ________________ work output
× 100%
work input
11 a) Power is the rate of doing work. It is measured in watts (W).
b) Work done = force × distance = (70 × 9.8) N × (20 × 0.25) m = 3430 J
Power =
work done
_________
time
3430 J
= ______ = 690 W
5.0 s
c) The student will do more work moving limbs, contracting muscles,
pumping blood, etc.
12 a) Using v 2 = u 2 + 2as where v = 0; u = 5.0 m s −1 ; s = 20 m
0 = (5.0 m s −1 ) 2 + 2a(20 m) ⇒ a = −0.625 m s −2 ≈ −0.6 m s −2
b) i) F = ma = (100 kg) × (0.6 m s −2 ) = 60 N
6 Fluids
ii) P = Fv = (60 N) × (5.0 m s −1 ) = 300 W
1 a) At the instant of release, the viscous force is zero; the ball will begin to
fall at the rate of the acceleration due to gravity, 9.8 m s −2 – the answer
is D.
b) At point 3 the sphere is travelling at constant velocity; the acceleration
is zero – the answer is A.
c) Between points 1 and 2 the sphere is still accelerating downwards; there
must be a resultant downward force, so the weight is greater than the
sum of the two upward forces, U and F – the answer is D.
d) At point 3 the sphere is moving at the terminal velocity; the resultant
force is zero, so the weight must equal the sum of the two upward forces
– the answer is C.
2 a) ρ = mass _______
volume = 50 × 10−3 _______________ kg
4 _
= 1.5 kg m–3
3
3 π (20 × 10 –2 m)
Full Answers to Review Questions: 6 Fluids
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) The air is compressed so its density inside the balloon is greater than
that at normal atmospheric pressure.
3 0.76 m of mercury
4 a) 5.1 × 10 −2 m 3
b) The above volume is less than the volume displaced by the less dense
water in the pool.
5 a) The temperature and the place of origin will both affect the viscosity
of the oil. The rate of flow is inversely proportional to the viscosity –
stickier fluids move more slowly through the pipeline, so the rate of flow
is greater at higher temperatures.
Increasing the diameter of the pipe will greatly increase the rate at
which the oil flows through it (at the same pressure).
b) If the flow rate is too fast, turbulence occurs and much more energy is
needed to transport the oil.
6 F __ = (6 π) η
N ( r __
m ) ( v _____
7 a) v ≈ 1 mm s −1
m s −1 ) . Units of η are: N(m2 s −1 ) −1 , i.e. N s m −2
b) The larger raindrops have a much bigger terminal velocity.
7 Solid materials
1 a) The toughest material has the largest area beneath the curve – the
answer is C.
b) The strongest material has the greatest breaking stress – the answer is B.
c) A polymer stretches easily as the chain molecules untangle, and then
stiffens (the gradient rises) when they are aligned – the answer is D.
d) A brittle material has little or no plastic extension before it breaks – the
answer is B.
2 b) k = F __
x = _______ 8.0 N
= 64 N m−1
0.125 m
3 a) i) C
ii) O – A
iii) C – D
b) Calculate the area under the line.
c) i) The wire returns to its original length.
ii) The wire will be permanently extended.
d) The wire regains its original stiffness – it will follow an identical loading
curve.
4 a) i) Stress (σ) = F __
A
ii) Strain (ε) = ∆l __
l
iii) Young modulus E = σ __
ε
b) σ =
50 N
______________
π (0.3 × 10 −3 m) 2 = 1.8 × 108 Pa
ε = 2.5 × 10−3 __________ m
= 1.25 × 10−3
2.00 m
E = 1.8 × 108 Pa
__________
1.25 × 10 −3 = 1.4 × 1011 Pa
Full Answers to Review Questions: 7 Solid materials
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5 Ductile materials can be drawn into wires; malleable metals can be readily
hammered into thin sheets. Copper can be drawn into wires for electrical
work, and gold can be hammered into very thin leaves for decoration.
6 a) Hysteresis occurs when the unloading curve ‘lags’ behind the loading
curve.
b) The blue (upper) line is the loading curve.
c) The area enclosed by the loop represents the energy per unit volume
(energy density) transferred to internal energy of the rubber during each
cycle.
d) The band is initially slightly stiff until the weak cross-links between the
tangled chains are broken. The chains are then uncoiled, giving a large
increase in strain for a little extra stress until the molecules become
aligned. The band now becomes stiff as the strong covalent bonds
between the atoms are stretched. On releasing the stress, the chains
recoil until the initial amorphous state is regained.
8 Nature of waves
1 Waves are in antiphase when they are half a cycle out of phase. One cycle
represents a phase difference of 2π radians; the phases of oscillations in
antiphase therefore differ by π radians – the answer is B.
2 Sound waves are longitudinal waves that transfer energy from vibrating
sources. The speed of sound in air is about 340 m s −1 at 20 8C, but it varies
with air temperature – the answer is A.
3 X-rays are produced by electron bombardment, travel at the speed of light
and can produce images on a photographic plate. X-ray photons are much
more energetic than those of visible light, and as E = hc __ , their wavelength
is lower than that of light – the answer is C. λ
4 Using c = f λ; f = c __ =
λ 3 × 108 m s−1 ___________
0.12 m = 2.5 × 109 Hz = 2.5 GHz – the answer
is C.
5 Amplitude is the maximum displacement from the mean position. Its SI
unit is metre.
Frequency is the number of complete oscillations per second. Its SI unit is
hertz.
Period is the time taken for one oscillation. Its SI unit is second.
6 The particles in longitudinal waves oscillate along the line of the
direction of propagation of the wave, e.g. sound waves. The particles in a
transverse wave oscillate at right angles to the direction of propagation of
the wave, e.g. waves on a stretched wire.
7 a) T = 40 ms; f = 1 __ = 25 Hz
T
b) Amplitude of the upper trace is 4.0 cm; the lower trace 2.0 cm.
c) The oscillations are 1 _
4 of a cycle out of phase; a phase difference of π __
2
radians.
8 Wavelength is the distance between two adjacent points that are in
phase.
v = distance/time; for one oscillation distance = 1 wavelength, time =
1 period
v = λ __ __
but f =
T 1
; so v = f λ
T
9 Mechanical waves require particles to oscillate and so must have a
medium for transmission. Electromagnetic waves are associated variations
of electric and magnetic fields and they travel through vacuum with a
speed of 3 × 10 8 m s −1 .
10 a) i) 0.75 m s −1
Full Answers to Review Questions: 8 Nature of waves
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ii) 6.0 × 10 −8 m
iii) 11 GHz
b) The waves in (ii) are in the ultraviolet (UV) region.
9 Transmission and reflection of waves
Full Answers to Review Questions: 9 Transmission and reflection of waves
1 Blue light has a shorter wavelength and higher frequency than red light.
Blue light is slowed down more than red light (it has a higher refractive
index) and so it is deviated more towards the normal when it enters the
glass – the answer is D.
2 Passing light through a narrow slit may produce diffraction but the
amplitude of oscillations within photons is so small that vibrations in all
planes will be transmitted. Reflection, scattering and Polaroid films can
all create polarised light – the answer is A.
3 Ultrasound is non-ionising radiation; it is much less likely to kill or
mutate cells than X-rays – the answer is C.
4 The observed frequency of a moving source changes according to the
Doppler effect. The observed frequency becomes lower when the source
moves away from the observer, with the change in frequency depending
on the speed of the source. The firework is accelerating away from us; so
the observed frequency, and hence the pitch of the sound, will continue
to fall as the speed increases – the answer is C.
5 Sound travels more slowly than light. It takes the sound 0.3 s to travel
100 m so the speed is 300 m s −1 (estimate to 1 significant figure).
6 b) Light travels faster in water than in glass so in your diagram you need
to show that the wavelength will increase. The direction of the wave
moves away from the normal.
7 a) 1.55 =
sin 40
_____
sin r
⇒ r = 24.58
b) 1.55 sin 24.5 = μ sin 28
⇒ μ = 1.37
8 sin θ = 1.52
____
1.60
⇒ θ = 728
9 Sound waves cannot be polarised because they are longitudinal. No
particles vibrate perpendicular to the direction of the wave.
10 Reflected glare is completely or partially polarised in the plane at right
angles to the plane of transmission of the lens.
11 0.44 g ml −1
12 1200 m
13 Advantage: better resolution; disadvantage: shorter range.
14 As the space probe moves away from the Earth, the received signal will
be at a lower frequency than the transmitted one (Doppler effect). Any
changes in speed relative to the receiver will require it to be tuned to the
appropriate frequency.
10 Superposition of waves
1 Although a constant phase relationship is needed for coherence, it is not
essential that the sources are always in phase; it is essential that they have
the same frequency – the answer is B.
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10

2 To form a minimum on an interference pattern the waves must be in
antiphase at that point. The path difference must therefore be half a
wavelength – the answer is B.
3 A guitar string vibrating in the fundamental mode has a node at each
end with a single antinode in between. The length is equal to half a
wavelength, so wavelength = 1.24 m – the answer is D.
4 The second harmonic has a frequency twice that of the fundamental – the
answer is D.
5 The engine noise is picked up by a microphone, electronically processed
and delivered to the pilot’s headphones exactly out of phase with the noise
so that destructive superposition occurs. Other signals to the headphones
are not processed.
6 At the edges of the bumps light reflected from the top interferes
destructively with reflections from the ‘well’. The path difference between
the reflections at the receiver must be half a wavelength, and so the height
of the hump will be one quarter of a wavelength:
λ __ = 160 nm
4
λ = 640 nm
7 The distance between adjacent antinodes in a standing wave is half a
wavelength:
λ __ ≈ 6 cm
2
λ ≈ 12 cm
f = v __ ≈
λ 3 × 108 m s−1 ___________
0.12 m ≈ 2.5 × 109 Hz
8 a) i) Narrower slit gives more diffraction and so the central maximum
should be wider on your sketch.
ii) and iii) Red light has a longer wavelength than blue so the central
maximum will be wider for the red light sketch.
b) The peak of the central maximum of the blue pattern falls within the
first minimum and so the two images can be resolved. The central
maximum of the red light lies outside the first minimum and so only one
image is detected.
9 a) i) See Figure 10.19, page 113.
ii) Fundamental: λ = 2l
First overtone: λ = l
b) ii) For an open-ended pipe l = λ __ in the fundamental mode; hence λ =
2
2l.
f = v __ =
λ v __
2l
iii) Gradient = 170 m s−1 = v __
2
v = 340 m s−1 c) The speed of sound in air increases as the temperature rises. As the
wavelength is fixed the frequency is proportional to the speed, and will
therefore rise and fall with the temperature.
11 Charge and current
1 a) Substituting into Q = It, and remembering to put the current in amperes
and the time in seconds:
Q = 32 × 10 –3 A × 60 × 60 s = 115.2 C
This now has to be divided by the charge on each electron:
Full Answers to Review Questions: 11 Charge and current
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11

Number of electrons per hour =
The answer is B.
___________ 115.2 C
1.6 × 10 –19 = 7.2 × 1020
C
b) The time taken for an electron to cross the tube is given by
time = distance _______
speed = 21 × 10–2 ____________ m
Substituting into Q = It gives
4.2 × 10 7 m s –1 = 5.0 × 10−9 s
Q = 32 × 10 −3 A × 5.0 × 10 −9 s = 1.6 × 10 −10 C
Dividing by the charge on an electron:
Number of electrons in beam = 1.6 × 10–10 ___________ C
1.6 × 10 –19 = 1.0 × 109
C
The answer is A.
2 a) Rearranging Q = It,
I = Q
__
t =
_________ 8 C
0.5 × 10 –3 = 16 000 A = 16 kA
s
b) To find the number of ions in each strike, divide the total charge (8 C)
by the charge on each ion (1.6 × 10 −19 C):
Number of ions in each strike =
3 a) In the equation I = nAvq
I = current in the conductor
8 C
___________
1.6 × 10 –19 C = 5 × 1019 ions
n = number of charge carriers per unit volume of the conductor
A = area of cross-section perpendicular to direction of current
v = drift speed of charge carriers
q = charge on each charge carrier
b) Rearranging I = nAvq and remembering to put the current in amperes
and the dimensions in metres:
v = I ____
nAq =
10 × 10 –3 ______________________________________________
A
7.0 × 1022 m –3 × 6 × 10 –3 m × 0.5 × 10 –3 m × 1.6 × 10 –19 C
= 0.298 m s −1 ≈ 0.3 m s −1
c) From the rearranged equation v = I/nAq we can see that the drift speed,
v, is inversely proportional to n, the number of charge carriers per unit
volume. It follows that if the drift speed in copper is about 10 −7 m s −1
(i.e. about 10 6 times less than in the semiconductor), the number of
charge carriers per unit volume in the copper must be about 10 6 more
than in the semiconductor – that is in the order of 10 29 m −3 , which is,
indeed, the case.
4 a) i) Substituting into Q = It, and remembering to put the current in
amperes and the time in seconds:
Q = 25 × 10 –3 A × 8.0 × 60 s = 12 C
ii) Now divide this charge by the charge on an electron:
Number of electrons =
12 C
___________
1.6 × 10 –19 C = 7.5 × 1019 electrons
Full Answers to Review Questions: 11 Charge and current
Tip
The words in italics are important to
use; candidates often omit these and
therefore lose marks.
b) As the filament is in series with the leads, the current I must be the
same in each. As the copper leads and the tungsten filament are both
metals, the charge carriers will be ‘free’ electrons in both cases, so q Tip
will be the same for both. The tungsten filament will be much thinner Note that the possible effect of each
than the copper leads and so the area of cross-section A will be much quantity in the equation has been
less for the tungsten. This would tend to make the drift speed in the considered in turn.
tungsten much greater. As both materials are metals, the number of
Edexcel Physics for AS © Hodder Education 2009
12

Full Answers to Review Questions: 12 Potential difference, electromotive force and power
charge carriers per unit volume, n, is unlikely to be more than an order
of magnitude different and so will have a relatively small effect. The
drift speed in the tungsten filament is therefore likely to be greater than
in the copper leads.
12 Potential difference, electromotive force and
power
1 a) W = J s −1 (power = rate of doing work)
2
J = N m (work = force × distance)
N = kg m s −2 (from F = ma)
W = kg m s −2 × m × s −1 = kg m 2 s −3
The answer is D.
b) Working from first principles, potential difference = work done per
charge passing:
V = J __ =
C Nm ____
As = kg m s–2 __________ × m
As = kg m2 s−3 A−1 The answer is B.
Which quantity is the product of two
other quantities?
Which quantity is one of the quantities
divided by one of the other quantities?
(3 possible answers)
Which quantity, when divided by time,
gives another quantity in the table?
(2 possible answers)
Quantity or quantities
Power = p.d. × current (from P = VI)
Current = power 4 p.d. (from P = VI)
P.d. = power 4 current (from P = VI)
P.d. = energy 4 charge (from V = W/Q)
Current = charge 4 time (from I = Q/t)
Power = energy 4 time (from P = W/t)
Table A.5 �
3 a) Rearranging P = VI,
I = P __ = _______ 1500 W
= 6.5 A
V 230 V
b) Substituting into Q = It, and remembering to put the time in seconds:
Q = 6.5 A × 20 × 60 s = 7.8 × 103 C (to 2 significant figures)
c) Rearranging the definition that power = work done/time taken:
Thermal energy = power × time of operation
= 1500 J s21 × 20 × 60 s = 1.8 MJ
4 a) Work done = force × distance moved
= mg × h = 0.600 kg × 9.8 N kg−1 × 0.94 m = 5.5 J
b) Power =
work done
_________
time taken
5.5(272) J
= _________ = 0.72 W
7.7 s
c) P = VI = 3.2 V × 0.75 A = 2.4 W
d) Efficiency =
useful power taken out
___________________
total power put in
0.718 W
= _______
× 100% ≈ 30%
2.40 W
e) The rest of the energy will be converted into heat by the work done
by the motor against friction as it rotates and by the masses against air
Tip
It is always a good idea to check
answers as far as possible, if time
permits. In this case we could use our
answer to part a) as follows:
From P = VI we have
V = P __
I = kg m2 s –3
________
A = kg m2 s−3 A−1 as before.
Clearly this is a much quicker way,
but it does depend on you getting the
answer to part a) correct!
Edexcel Physics for AS © Hodder Education 2009
13

5 a)
Full Answers to Review Questions: 12 Potential difference, electromotive force and power
resistance. Also heat will be generated in the coil of the motor and in
the connecting wires due to the current in them having to overcome
their electrical resistance.
copper wire
V
A
� Figure A.2
b) Rate of doing work = power = VI = 0.53 V × 1.8 A = 0.95(4) W
c) Rearranging I = nAvq:
v = I ____
nAq = ______________________________________
1.8 A
8.0 × 1028 m –3 × 0.059 × 10 –6 m2 × 1.6 × 1019 C
d) Using P = Fv (see page 48):
F = P
__
v =
_____________ 0.954 J s–1
–1 = 400 N (to 2 significant figures)
2.4 × 10 –3 m s
6 a) Power delivered by alternator = VI = 14 V × 70 A = 980 W
b) Current taken by starter motor is given by rearranging P = VI:
I = P __ = _______ 1500 W
= 125 A
V 12 V
c) For each headlight:
I = P __ =
60 _____ W
= 5 A
V 12 V
so next fuse up, i.e. 8 A, would be suitable.
d) Energy = VIt = 12 V × 1 A × 62 × 60 × 60 s ≈ 2.7 MJ
e) If the four sidelights are left on for 12 hours:
energy used = 4 × 5 W × 12 × 60 × 60 s = 864 000 J
battery capacity, from part d), is 2 678 400 J
fraction used =
864 000 J
_________
2 678 400 J
1
= 0.32 ≈ _
3
= 2.4 mm s−1
f) Using P = Fv (see page 48), power to drive at 90 km per hour up a
gradient of 10% is given by:
P = 0.1 × mg × v = 0.1 × 1740 kg × 9.8 N kg−1 × 90 × 103 _________ m
60 × 60 s
= 43 kW (to 2 significant figures)
Maximum power is stated as 180 kW, so fraction of maximum power
= 43 kW
_______
180 kW
1
= 0.24 ≈ _
4
7 a) Energy saved over 8000 hours = (60 − 18) W × 8000 h = 42 × 8 kWh =
336 kWh
cost of energy saved will be = 336 × 15 p = £50.40
Allowing for £2 cost of lamp, total cost saved will be £48.40, which
comfortably exceeds the manufacturer’s claim.
b) Apart from monetary cost, the saving in energy is important in
conserving the world’s energy resources as well as reducing the carbon
footprint (less electrical energy consumed means less carbon dioxide is
Edexcel Physics for AS © Hodder Education 2009
14

Full Answers to Review Questions: 13 Current–potential difference relationships
used in its production). Against this, there may be a greater carbon cost
in manufacturing a low-energy lamp compared with a filament lamp,
although it will almost certainly last longer. From a safety aspect, the
low-energy lamp will probably not get as hot.
13 Current–potential difference relationships
1 a) A typical semiconductor diode will not start to conduct until a potential
difference of about 0.5 V is applied – the answer is C.
b) The filament will have resistance even when there is no potential
difference applied – its resistance is a property of the wire from which it
is constructed. This resistance will increase when a potential difference
is applied as the current heats the filament – the answer is D.
2 a) Graph A, for the carbon resistor, is a straight line through the origin.
This indicates that the current is proportional to the potential difference
so that the resistor obeys Ohm’s law and has a constant resistance. As
graph B, for the filament lamp, is a curve, it shows that the lamp does
not obey Ohm’s law. As the curve gets less steep, it indicates that the
resistance of the lamp increases with increased potential difference due
to the heating effect of the current in the tungsten filament wire.
b) i) As the resistor is ohmic, R = V __ = constant = inverse gradient. Using
I
a large triangle:
R =
____________
(8.0 – 0.0) V
= 32 Ω
(0.25 – 0.0) A
ii) Percentage difference =
= (33 – 32) Ω
__________
33 Ω
× 100% = 3%
difference ________________ in values
× 100%
stated value
which is within the stated 5% tolerance.
or
The nominal value is 33 Ω ± 5% = (33 ± 1.65) Ω, so the measured
value of 32 Ω is within the stated tolerance.
c) i) Reading off from graph B, when the potential difference across the
lamp is 12 V, the current in it is 0.375 A, so:
P = VI = 12 V × 0.375 A = 4.5 W
ii) Percentage difference between this value and the stated value of 5 W
= (5.0 – 4.5) W
___________
× 100% = 10%
5.0 W
d) i) From graph B, when the potential difference across the lamp is
12.0 V, the current in it is 0.375 A, so:
R = V
__
I
= _______ 12.0 V
= 32 Ω
0.375 A
ii) From graph B, when the potential difference across the lamp is 1.0 V,
the current in it is 0.10 A, so:
R = V
__
I
= ______ 1.0 V
= 10 Ω
0.10 A
e) i) When the resistor and lamp are connected in series, the 12 V
potential difference is shared between them, so the potential
difference across the lamp will be less than 12 V. As the lamp has the
same value of resistance as the resistor at 12 V (32 Ω), its resistance
at a lower potential difference will be less than that of the resistor.
This means the lamp will take less than half of the 12 V and so will
only glow dimly.
Tip
A common mistake that candidates
make is to assume that the resistance
of a filament lamp is zero when the
potential difference across it is zero.
This is not the case.
Edexcel Physics for AS © Hodder Education 2009
15

Full Answers to Review Questions: 13 Current–potential difference relationships
A similar argument in terms of current can be made. As the
resistance of the lamp is about the same as that of the resistor, the
circuit resistance is approximately doubled when they are connected
in series. This means the current in the lamp is only about half the
normal operating current and so the lamp will only glow dimly.
ii) The latter argument is better for considering the power, which
is given by I 2 R. In series, each will take the same current. As the
resistance of the lamp at a potential difference of less than 12 V is
less than that of the resistor (which remains at 32 Ω), the power
developed in the resistor is greater than that in the lamp.
3 a) i) Ohm’s law states that for a metallic conductor at a constant temperature
the current in the conductor is proportional to the potential
difference across the conductor.
ii) The resistance of any conductor is defined as the potential difference
across the conductor divided by the current in the conductor.
b) I = V
__
c)
d)
I/A
0.200
0.177
0.100
12 V
R =
______ 2.0 V
= 0.177 A = 177 mA
11.3 Ω
0 1.0 2.0 V/V � Figure A.3
Figure A.4 �
nichrome
ribbon
current
sensor
A
voltage
sensor
V
data
logger
computer
The current sensor and voltage sensor are connected to the data logger
(analogue to digital converter), the output of which is fed into a
computer. The potential divider is used to vary the potential difference
from zero to 2.0 V (note that a series variable resistor cannot do this –
see page 133) and the current and potential difference are recorded at
set intervals. This data is stored in the data logger and can then be used
to plot a graph via the computer.
The main advantage of using a data logger is that a large amount of data
can be collected and processed in a relatively short time, thus giving
better average values.
The only real disadvantage is the complexity of the set-up compared
with just using digital meters.
Tip
The conditions in italics must be
included in order to give a complete
definition and get full marks.
Tip
If you state the formula R = V __ , you
I
must define each term in the formula.
Edexcel Physics for AS © Hodder Education 2009
16

4 a)
e) The resistance of the whole ribbon when the toaster is operating at
240 V, 1000 W is given by
P = V2 ___
⇒ R =
R V2 ___ = ________ (240 V)2
= 57.6 Ω
P 1000 W
If a length of 1.00 m has a resistance of 11.3 Ω, then the total length of
the ribbon will be:
length of ribbon =
______ 57.6 Ω
× 1.00 m = 5.10 m
11.3 Ω
This assumes that the resistivity of the nichrome is the same when it is
at the operating temperature of the toaster as it is at 2.0 V. In practice,
the resistivity will increase with temperature and so the length of ribbon
will be less than the calculated value. An estimate of ‘about 5 m’ would
not be unreasonable.
I/mA
100
50
0 0.5 1.0 V/V � Figure A.5
b) Reading from the graph, when the current in the diode is 25 mA, its
resistance is 30 Ω.
The potential difference across the diode is then given by V = IR:
V = 25 × 10 −3 A × 30 Ω = 0.75 V
The potential difference across the resistor will therefore be
(1.58 − 0.75) V = 0.83 V.
The required value of resistance is given by R = V
__
I :
R =
__________ 0.83 V
25 × 10 –3 = 33.2 Ω
A
The most suitable value would therefore be the 33 Ω resistor.
14 Resistance and resistivity
1 a) The number of charge carriers per unit volume will remain virtually
constant with temperature (it would actually reduce very slightly as
the copper expanded, but this would be almost negligible over the
temperature range of 0 8C–100 8C) – the answer is B.
b) The resistance would increase linearly with temperature from a finite
value at 0 8C (not zero) – the answer is C.
c) The resistance of a thermistor decreases with temperature due to the
significant increase in the number of charge carriers – the answer is D.
2 a) From Figure 14.1b on page 140, the potential difference is about 0.7 V
for a current of 40 mA. Using R = V/I gives:
R =
__________ 0.7 V
40 × 10 –3 = 17.5 Ω ≈ 18 Ω
A
Full Answers to Review Questions: 14 Resistance and resistivity
Edexcel Physics for AS © Hodder Education 2009
17

) Draw a straight line through the origin and the point (0.8 V, 16 mA).
This represents a resistance of 50 Ω. This cuts the curve at about 0.6 V,
so the component will have a resistance of 50 Ω when the potential
difference across it is about 0.6 V.
As a check, the current at 0.6 V is about 12 mA, giving:
R = V
__
I =
__________ 0.6 V
12 × 10 –3 = 50 Ω
A
3 a) Between the longer edges of the chip:
l = 5 mm = 5 × 10 −3 m
A = 10 mm × 1 mm = 10 × 10 −3 m × 1 × 10 −3 m = 10 × 10 −6 m 2
Resistance is:
R = ρl __ =
A 3.0 × 10–5 Ω m × 5 × 10−3 ______________________ m
2 = 0.015 Ω
10 × 10 –6 m
b) Between the faces of the chip:
l = 1 mm = 1 × 10 −3 m
A = 10 mm × 5 mm = 10 × 10 −3 m × 5 × 10 −3 m = 50 × 10 −6 m 2
Resistance is:
R = ρl __ =
A 3.0 × 10–5 Ω m × 1 × 10−3 ______________________ m
4 a) i) Rearranging P = VI,
I = P __ = _______ 3000 W
= 12.5 A
V 240 V
50 × 10 –6 m 2 = 6 × 10−4 Ω
So the kettle will operate safely with a 13 A fuse.
ii) From R = V
__
I ,
R =
______ 240 V
= 19.2 Ω ≈ 19 Ω
12.5 A
b) i) At 110 V, from I = V
__
I =
______ 110 V
= 5.7 A
19.2 Ω
R ,
ii) The power dissipated is given by P = V2 ___
P =
________ (110 V)2
= 630 W
19.2 Ω
c) The assumption is reasonable. As a kettle is used to boil water, the
element will heat up the water until both reach a temperature of 100 8C
whether the voltage is 240 V or 110 V – it will just take longer at
110 V. As the temperature of the element is the same in both cases, its
resistance will be the same.
5 a) Resistivity, symbol ρ, is defined by the equation:
ρ = RA ___
l
where R is the resistance, l is the length and A is area of cross-section
perpendicular to the length.
b) i) Rearranging R = ρl
__
A :
R
__
l
= ρ
__
Full Answers to Review Questions: 14 Resistance and resistivity
R :
=
A 1.72 × 10–8 ______________ Ω m
2 = 0.070 Ω m−1
0.2453 × 10 –6 m
ii) Rearranging A = πd2 ___
4 :
d2 = 4A ___
π ⇒ d =
�
_____________
4 _____________
× 0.1110 mm2
π
= 0.3759 mm
Note
The scale of the graph is such that
these answers are only approximate.
Tip
Note that you can check this by using
your answer to part i):
P = VI = 110 V × 5.7 A = 630 W
Tip
It is easiest to define resistivity by
means of the appropriate equation,
but remember that all the quantities in
the equation must be defined.
Edexcel Physics for AS © Hodder Education 2009
18

iii) Rearranging ρ = RA
___
l :
ρ = A × R
__
l = 0.1110 × 10−6 m 2 × 4.41 Ω m −1 = 4.90 × 10 −7 Ω m
iv) From A = πd2 ___
4 :
π × (0.2743 mm)2
A = _______________ = 0.0591 mm2
4
v) Rearranging ρ = RA ___
l :
ρ = A × R
__
l = 0.0591 × 10−6 m 2 × 18.3 Ω m −1 = 1.08 × 10 −6 Ω m
6 a) As the graph is a straight line through the origin, it shows that the current
is proportional to the potential difference, which is Ohm’s law.
7 a)
b) Resistivity is given by ρ = RA/l. We are given l (= 2.00 m), A can be
found from the diameter (D = 0.25 mm) and R can be found from the
inverse gradient of the graph:
A = πD2 ____
R =
4 = π × (0.25 × 10−3 m) 2
_________________
__________ 4.0 V
90 × 10−3 = 44.4 Ω
A
ρ = RA ___
l = 44.4 Ω × 4.91 × 10–8 m2 ___________________
I/A
0.50
0.42
0.25
4 = 4.91 × 10−8 m 2
2.00 m = 1.1 × 10−6 Ω m
0 6 12 V/V � Figure A.6
b) Rearranging P = V 2 /R:
R = V2 ___ = _______ (12 V)2
= 29 Ω
P 5 W
c) When the lamp is ‘off’ it is at a much lower temperature than when it
is ‘on’ and glowing white hot. At a lower temperature, the lattice ions
vibrate much less and therefore less impede the flow of electron charge
carriers. The drift velocity of the electrons is therefore much greater.
This means that in the equation I = nAvq, v is much greater whilst n,
A and q remain the same. The current is therefore much greater, which
means the resistance is much less.
8 a) i) From the graph, when V = 1.0 V, I = 20 mA:
R = V
__
I =
__________ 1.0 V
20 × 10−3 = 50 Ω
A
ii) From the graph, when V = 3.0 V, I = 100 mA:
R = V
__
I =
___________ 3.0 V
100 × 10−3 = 30 Ω
A
b) i) P = VI = 1.0 V × 20 × 10 −3 A = 20 mW
ii) P = VI = 3.0 V × 100 × 10 −3 A = 300 mW
Full Answers to Review Questions: 14 Resistance and resistivity
Tip
Note the importance of including
through the origin, without which the
answer would be of little value.
Edexcel Physics for AS © Hodder Education 2009
19

c) i) The above answers suggest that as energy is being converted at a
much greater rate at 3.0 V, then the temperature of the thermistor
will be greater at 3.0 V than at 1.0 V.
ii) A thermistor is a semiconductor. In semiconductors the number
of charge carriers increases exponentially with temperature. This
means that in the equation I = nAvq, n increases significantly
with temperature whilst the quantities A and q remain the same.
Although v is slightly reduced due to increased lattice vibrations,
this is nowhere near as significant as the increase in n. The overall
effect is that the current increases, and therefore the resistance gets
less, as the temperature rises.
15 Electric circuits
1 Start by working out the resistance of each of the combinations:
W: In series R = R 1 + R 2 + R 3 = 15 Ω + 15 Ω + 15 Ω = 45 Ω
X: Start by adding the two series resistors: 15 Ω + 15 Ω = 30 Ω
Then combine this with the parallel resistor:
1 __ =
R 1 ___ +
R1 1 ___ ⇒
R2 1 __ =
R 1 ____ + ____ 1
=
1 _____ + 2
= ____ 3
⇒ R = 10 Ω
30 Ω 15 Ω 30 Ω 30 Ω
Y: Combining the two parallel resistors:
1 __ =
R 1 ___ +
R1 1 ___
R2 ⇒ 1
__
=
R 1 ____
15 Ω
+ ____ 1
15 Ω
= ____ 2
⇒ R = 7.5 Ω
15 Ω
Then adding the 15 Ω in series gives 15 Ω + 7.5 Ω = 22.5 Ω
Z: Combining the three parallel resistors:
1 __ =
R 1 ___
R1 + 1 ___
R2 + 1 ___
R3 ⇒ 1
__
=
R 1 ____
15 Ω
+ ____ 1
15 Ω
+ ____ 1
15 Ω
a) We can now see that Z = W/9 – the answer is A.
b) W = 2Y – the answer is D.
c) X = 2 × Z – the answer is D.
= ____ 3
⇒ R = 5 Ω
15 Ω
2 a) Each ‘arm’ has two 4.7 Ω resistors in series, giving 9.4 Ω in total for each
‘arm’. These ‘arms’ are in parallel, so:
1 __ =
R 1 ___ +
R1 1 ___ ⇒
R2 1 __ =
R 1 _____ + _____ 1
= _____ 2
⇒ R = 4.7 Ω
9.4 Ω 9.4 Ω 9.4 Ω
b) This network of four resistors has a resistance equal to that of each of
the single resistors. It might be preferable to use this network rather
than a single resistor as the current will split, with half going through
each ‘arm’. This means that the current in each resistor is only half
what it would have been for a single resistor. As the power depends on
the square of the current (P = I2 R), the power in each resistor will be
a quarter of what it would be in a single resistor. This means that the
resistors will not heat as much.
Note that the total power developed in the network is the same as for a
single resistor – it is shared equally by the four resistors.
3 If the battery is short-circuited, the only resistance will be the internal
resistance of the battery. Using I = V __ :
R
I = ______ 9 V
= 18 A
0.50 Ω
This is a large current and will generate a power of P = I 2 R = (18 A) 2 ×
0.50 Ω = 162 W inside the battery, which will make the battery hot.
Full Answers to Review Questions: 15 Electric circuits
Tip
It is worth remembering that if two
resistors of the same value are
connected in parallel, their combined
resistance is half the individual
resistance.
Tip
You should always back up your
argument with quantitative evidence
as far as possible, in this example by
showing that the power developed in
the battery would be 162 W.
Edexcel Physics for AS © Hodder Education 2009
20

4 a) A 50 MΩ resistor in series with the output will keep the current very
small. As the severity of an electric shock depends on the current
passing through your body to earth, the resistor acts as a safety device.
b) From I = V
__
R the maximum current will be:
I = 5 × 103 _________ V
50 × 106 Ω = 1.0 × 10−4 A = 0.10 mA
c) The total resistance between the terminal of the supply and the ground
will be 50 MΩ + 10 kΩ. As 10 kΩ is only 0.01 MΩ, the girl’s resistance
has very little effect and so the current in the girl would be virtually the
same as in part b).
d) Using P = I 2 R, the power dissipated in the girl would be:
P = (1.0 × 10 −4 A) 2 × 10 × 10 3 Ω = 1.0 × 10 −4 W = 0.10 mW
This is considerably less than the 14 mW from the car battery, and so
shows the effectiveness of the 50 MΩ resistor as a safety device.
Warning: high voltage supplies, even when safety protected, are still very
dangerous and must be treated with the utmost caution.
5 The table is completed by adding values for the resistance and power, for
example V = 1.44 V and I = 0.20 A gives:
R = V
__
I
and
1.44 V
= ______ = 7.20 Ω
0.20 A
P = VI = 1.44 V × 0.20 A = 0.29 W
I/A 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60
V/V 1.44 1.32 1.20 1.09 0.95 0.84 0.73 0.59
R/Ω 7.20 3.30 2.00 1.36 0.95 0.70 0.52 0.37
P/W 0.29 0.53 0.72 0.87 0.95 1.01 1.02 0.94
P/W
1.1
1.0
0.9
0.8
0.7
0.6
0.5
maximum power
when R = 0.60 Ω
0
0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 R/Ω
Full Answers to Review Questions: 15 Electric circuits
� Figure A.7
Table A.6 �
Note that in order to achieve a sensible scale, the first data point has been
omitted.
From the graph it can be seen that the power has a maximum value when
the load R = 0.60 Ω, which is equal to the internal resistance of the cell.
6 a) The table is completed by adding values for the resistance and power, for
example V = 5.90 V and I = 0.15 mA gives:
Edexcel Physics for AS © Hodder Education 2009
21

R = V
__
I
and
= ________ 5.90 V
= 39.3 kΩ
0.15 mA
P = VI = 5.90 V × 0.15 mA = 0.88 mW
I/mA 0.10 0.15 0.20 0.30 0.40 0.50 0.60 0.65
V/V 6.07 5.90 5.72 5.35 4.97 4.20 3.00 2.01
R/kΩ 60.7 39.3 28.6 17.8 12.4 8.4 5.0 3.1
P/mW 0.61 0.88 1.14 1.60 1.99 2.10 1.80 1.31
b) i)
V/V
6.0
5.0
4.0
3.0
2.0
1.0
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 I/mA
Full Answers to Review Questions: 15 Electric circuits
Table A.7 �
� Figure A.8
For small currents the graph, as shown in Figure A.8, is linear,
indicating a constant internal resistance. For larger currents the
graph clearly curves downwards, showing that the internal resistance
increases as the current gets larger.
ii) The e.m.f. of the cell will be the voltage when the current is zero, i.e.
the intercept on the voltage axis. From the graph this is 6.5 V.
The internal resistance for low current values is given by the
numerical value of the gradient of the linear part of the graph.
Extending the linear part to give a large triangle:
r =
_______________
(6.50 – 4.25) V
= 3.7 kΩ
(0.60 – 0.00) mA
Note
Your answers may differ slightly from
those given. Drawing a graph and
extracting data from it is not an exact
science as it is always subject to
human error.
Edexcel Physics for AS © Hodder Education 2009
22

c) i)
2.2
P/mW
2.0
1.8
1.6
1.4
1.2
1.0
0.8
0.6
0
0
10 20 30 40 50 60 R/kΩ
Full Answers to Review Questions: 15 Electric circuits
� Figure A.9
ii) From the graph, the maximum power is approximately 2.1 mW.
7 a) i) Combining the 22 Ω and 33 Ω parallel resistors:
1 __ =
R 1 ___
R1 + 1 ___
R2 ⇒ 1
__
R
= ____ 1
+ ____ 1
=
3 _____ + 2
= ____ 5
22 Ω 33 Ω 66 Ω 66 Ω
⇒ R = ____ 66 Ω
= 13.2 Ω
5
The total circuit resistance is therefore 47 Ω + 13.2 Ω = 60.2 Ω.
The circuit current will be given by:
I = V __ = ______ 6.02 V
= 0.10 A
R 60.2 Ω
The current through the 47 Ω resistor is therefore 100 mA. When
this current comes to the parallel network, it will split in the inverse
ratio of the resistances, i.e. 22/55 (= 40 mA) through the 33 Ω
resistor and 33/55 (= 60 mA) through the 22 Ω resistor.
Alternatively, the potential difference across the parallel network is:
V = IR = 0.10 A × 13.2 Ω = 1.32 V
giving the current in the 22 Ω resistor as:
I = V __ =
R 1.32V ______
22 Ω
= 0.060 A = 60 mA
and the current in the 33 Ω resistor as
I = V __ =
R 1.32V ______
33 Ω
= 0.040 A = 40 mA
ii) The power generated is given by P = I 2 R in each case:
47 Ω: (0.10 A) 2 × 47 Ω = 0.47 W
22 Ω: (0.06 A) 2 × 22 Ω = 0.079 W
33 Ω: (0.04 A) 2 × 33 Ω = 0.053 W
b) The 47 Ω resistor takes 470 mW, which is just on the limit of the power
rating of 500 mW. The other two resistors are comfortably within the
rating. Thus the resistors would be suitable, although a 1 W rating for
the 47 Ω resistor might be more prudent.
8 a) Before the voltmeter is connected, the total circuit resistance is
R = 22 kΩ + 33 kΩ = 55 kΩ
This gives a circuit current of:
Edexcel Physics for AS © Hodder Education 2009
23

I = V __ = _____ 7.5 V
= 0.136 mA
R 55 kΩ
The potential difference across the 33 kΩ resistor is therefore:
V = IR = 0.136 mA × 33 kΩ = 4.5 V
Alternatively, if you are good at ratios you might spot that the 7.5 V will
split up in the ratio of the two resistors:
V across 33 kΩ = 33/55 × 7.5 V = 4.5 V
b) i) When the switch is closed, the voltmeter is connected in parallel
with the 33 kΩ resistor. The resistance of this parallel combination
will be less than 33 kΩ, so the voltage dropped across the
combination will be less than the previous 4.5 V.
ii) If the voltmeter reads 4.0 V, the potential difference across the 22 kΩ
resistor must be (7.5 − 4.0) V = 3.5 V.
The current in the 22 kΩ resistor, which is also the circuit current,
will be:
I = V __ = _____ 3.5 V
= 0.159 mA
R 22 kΩ
The combined resistance of the parallel arrangement of the
voltmeter and the 33 kΩ resistor will therefore be:
R = V
__
I
= _________ 4.0 V
= 25.1 kΩ
0.159 mA
For this parallel arrangement:
1 __ =
R 1 ___ +
R33 1 ___ ⇒
RV 1 ___ =
RV 1 __ –
R 1 ___ = _______ 1
– ______ 1
R33 25.1 kΩ 33 kΩ
1 ___ = 0.0398 kΩ
RV −1 − 0.0303 kΩ−1 = 0.0095 kΩ−1 R V = 106 kΩ
c) If the voltmeter is rated as 10 V/100 μA, its resistance should be:
R = V
__
I =
10 V
___________
100 × 10 −6 A = 1.0 × 105 Ω = 100 kΩ
This differs by 6% from the experimental value. As each resistor has a
tolerance of 5%, the experimental value is within the overall tolerance
and so is compatible with the stated rating of the voltmeter.
9 a) As the temperature of the thermistor falls, its resistance increases as
there will be less charge carriers per unit volume. If the resistance of the
thermistor increases, the proportion of the supply voltage dropped across
it will also increase and so the proportion of the supply voltage across
the resistor R will decrease and V out will fall.
b) i) Reading from the graph, at 0 8C the thermistor has a resistance of
18.0 kΩ.
When V out is 5.0 V, the voltage across the thermistor and the
potentiometer will be (9.0 − 5.0) V = 4.0 V. Assuming that the
potentiometer is set at zero, the voltage across the thermistor will be
4.0 V.
The current in the thermistor (and therefore the circuit current as it
is a series circuit) will be given by:
I =
_______ 4.0 V
= 0.22 mA
18.0 kΩ
The value of R is then given by:
R = V ____ out
I
= ________ 5.0 V
= 22.5 kΩ
0.22 mA
Full Answers to Review Questions: 15 Electric circuits
Edexcel Physics for AS © Hodder Education 2009
24

The best value to use would therefore be the 22 kΩ resistor.
ii) The purpose of the potentiometer is to fine tune the circuit. It is
adjusted so that the lamp is switched on at exactly 0 8C.
16 Nature of light
1 a) First of all, 5% of 60 W = 3 W. We then need to use the relationship
that intensity I at a distance r is given by:
I = power ______
4πr
3 W
4π × (1.5 m) 2 = 0.11 W m−2
= ___________
2
The answer is A.
b) We will need to use E = hf, so we must find the frequency
corresponding to a wavelength of 660 nm. Rearranging c = fλ:
f = c
__
Then:
=
λ 3.00 × 108 m s−1 _____________
660 × 10 −9 m = 4.55 × 1014 s −1
E = hf = 6.63 × 10 −34 J s × 4.55 × 10 14 s −1 = 3.01 × 10 −19 J
Converting to electron-volts using 1 eV = 1.6 × 10 −19 J:
E = 3.01 × 10−19 _______________ J
−1 ≈ 2 eV
1.6 × 10 −19 J eV
The answer is B.
2 a) Rearranging φ = hf 0 and remembering to convert eV into J:
f0 = φ __ =
h 2.28 eV × 1.6 × 10−19 J eV−1 _______________________
= 5.5 × 10 14 Hz
The answer is C.
6.63 × 10 −34 J s
b) We will need to use E = hf, so we must find the frequency
corresponding to a wavelength of 447 nm. Rearranging c = fλ:
f = c
__
Then:
=
λ 3.00 × 108 m s−1 _____________
447 × 10 −9 m = 6.71 × 1014 s −1
E = hf = 6.63 × 10 −34 J s × 6.71 × 10 14 s −1 = 4.45 × 10 −19 J
Converting to electron-volts using 1 eV = 1.6 × 10 −19 J:
E = 4.45 × 10−19 ______________ J
−1 = 2.78 eV
1.6 × 10 −19 J eV
The answer is C.
c) In part b) we found that the energy of a photon of blue light is
2.78 eV. As the work function is 2.28 eV, the reverse potential
difference that would have to be applied to just prevent photoemission
would be:
V = (2.78 − 2.28) V = 0.50 V
The answer is A.
3 a) Using hf = E2 − E1 and remembering to convert the energy from eV to
J:
f = (1.51 – 0.54) × 1.6 × 10–19 _______________________ J
6.63 × 10 –34 J s = 2.34 × 1014 s−1 Rearranging c = fλ:
Full Answers to Review Questions: 16 Nature of light
Edexcel Physics for AS © Hodder Education 2009
25

λ = 3.00 × 108 m s−1 _____________
2.34 × 1014 s−1 = 1.28 × 10−6 m ≈ 1.3 μm
The answer is C.
b) The red end of the visible is about 700 nm (= 0.7 μm), so 1.3 μm will
be just beyond the red, i.e. in the infrared.
The answer is A.
4 a) A photon is a small, discrete amount, or quantum, of energy associated
with electromagnetic radiation.
b) The energy to move an electron through a potential difference of
0.48 V is given by:
E = QV = 1.6 × 10 −19 C × 0.48 V = 7.68 × 10 −20 J ≈ 8 × 10 −20 J
c) The efficiency of energy conversion is therefore:
efficiency = 7.68 × 10–20 ___________ J
4 × 10 –19 × 100% = 19%
J
d) Rearranging E = hf:
f = E __ =
h 4.0 × 10–19 ____________ J
6.63 × 10−34 J s = 6.03 × 1014 s –1
Rearranging c = fλ:
λ = c _
f = 3.0 × 108 m s –1
____________
6.03 × 1014 s –1 = 4.97 × 10–7 m = 500 nm (to 2 significant figures)
5 a) The work function of a surface is the amount of energy that is needed
to just remove an electron from the surface.
b) We will need to use E = hf, so we must find the frequency
corresponding to a wavelength of 532 nm. Rearranging c = fλ:
f = c __ =
λ 3.00 × 108 m s−1 _____________
532 × 10−9 m = 5.64 × 1014 s−1 Then:
E = hf = 6.63 × 10 −34 J s × 5.64 × 10 14 s −1 = 3.74 × 10 −19 J
Converting to electron-volts using 1 eV = 1.6 × 10−19 J:
E = 3.74 × 10−19 ______________ J
1.6 × 10−19 −1 = 2.34 eV ≈ 2.3 eV
J eV _
2 mv2 max gives:
c) Rearranging hf = φ + 1
1 _
2 mv2 = hf − φ = (2.34 2 1.90) eV = 0.44 eV
max
= 0.44 eV × 1.6 × 10 −19 J eV −1 = 7.0 × 10 −20 J
d) The frequency of the red light will be less than that of the green light,
so a quantum of the red light may not have sufficient energy (hf) to
overcome the work function and release photoelectrons.
A quick way of calculating the wavelength corresponding to the work
function is to say that 2.34 eV corresponds to 532 nm (from part b)),
therefore 1.90 eV (the work function) will correspond to:
λ = ( 2.34 ____
1.90 ) × 532 nm = 655 nm
This is in the red region of the spectrum and so if the laser has a
wavelength greater than this, it will not cause any photoelectrons to
be emitted.
6 a) Green light has the shortest wavelength and therefore the highest
frequency. The green LED will therefore emit photons of the highest
energy as the photon energy is given by E = hf.
Full Answers to Review Questions: 16 Nature of light
b) i) We will need to use E = hf, so we must find the frequency
corresponding to a wavelength of 630 nm. Rearranging c = fλ:
Edexcel Physics for AS © Hodder Education 2009
Tip
If you use the ‘show that’ value of
8 × 10 −20 J and get 20% you will still
get full marks, but it is better to use
the actual calculated value.
Tip
As the question asks you to suggest
the reason, you would not be
expected to go into as much detail
as this – the first paragraph would be
sufficient.
26

f = c
__
Then:
=
λ 3.00 × 108 m s−1 _____________
630 × 10 −9 m = 4.76 × 1014 s −1
E = hf = 6.63 × 10 −34 J s × 4.76 × 10 14 s −1
= 3.16 × 10 −19 J ≈ 3 × 10 −19 J
ii) As 1 eV = 1.6 × 10 −19 J:
E = 3.16 × 10−19 ______________ J
−1 = 2.0 eV
1.6 × 10 −19 J eV
c) i) We then need use the relationship that intensity I at a distance r is
given by:
I = power ______
4πr
2 =
18 mW
____________
4π × (0.30 m) 2 = 15.9 mW m−2 ≈ 16 mW m −2
ii) The area of the pupil is given by:
A = πD2 ____
4 = π(5 × 10−3 m) 2
____________
4 = 1.96 × 10−5 m2 The energy of the photons entering the eye per second is given by:
energy per second = intensity × area
= 15.9 mW m −2 × 1.96 × 10 −5 m 2
= 3.12 × 10 −7 J s −1
From part b) i), the energy of each photon is 3.16 × 10 −19 J, so:
number of photons per second = 3.12 × 10–7 J s –1
_____________
3.16 × 10 –19 J = 1.0 × 1012 s −1
iii) As you move further away, the number of photons entering your
eye per second will get less and so the intensity will get less.
visible light emitted
d) Efficiency = _________________
×
power consumption
100%
= 18 mW _______
× 100% = 15%
120 mW
e) Even at 15% efficiency, LEDs are far more efficient than filament
lamps, which typically have an efficiency of less than 5%. In
addition, the coloured glass necessary for filament lamps absorbs a
certain amount of the light, and LEDs tend to last much longer than
filaments. It is therefore much cheaper and far more energy-efficient
to use LEDs for traffic lights.
7 a) Diffraction and interference suggest that light can behave as a wave.
b) Monochromatic literally means ‘one colour’, in other words light all
having the same frequency or wavelength.
c)
V s /V
1.2
1.0
0.8
0.6
0.4
0.2
0
0 4.0 5.0 6.0 7.0 8.0
+
f0 = 4.3 10 14 Hz
f/10 14 Hz
Full Answers to Review Questions: 16 Nature of light
� Figure A.10
Edexcel Physics for AS © Hodder Education 2009
27

d) The term eVs is equal to the maximum kinetic energy ½mv2 max of the
photoelectrons. It is a measure of the work that has to be done to just
stop the most energetic photoelectrons.
e) i) The threshold frequency f 0 is the frequency when the stopping
potential V s = 0. From the graph, f 0 = 4.3 × 10 14 Hz.
ii) The work function φ is given by:
φ = hf 0 = 6.63 × 10 −34 J s × 4.3 × 10 14 s −1 = 2.9 × 10 −19 J
As 1 eV = 1.6 × 10−19 J:
φ = 2.9 × 10−19 ______________ J
1.6 × 10−19 J eV
–1 = 1.8 eV
f) The energy to release a photoelectron comes from one photon. Below
the threshold frequency f 0 no electrons are emitted because the energy
of each photon, hf 0 , is not sufficient to provide an electron near the
surface with enough energy to escape.
8 a) Using hf = E 2 − E 1 and remembering to convert the energy from eV
to J:
f = (69.6 – 1.8) × 103 eV × 1.6 × 10−19 J eV –1
_________________________________
6.63 × 10 –34 J s = 1.64 × 1019 s−1 Rearranging c = fλ:
λ = 3.00 × 108 m s−1 _____________
1.64 × 1019 s−1 = 1.83 × 10−11 m ≈ 0.02 nm
b) This is in the X-ray region of the electromagnetic spectrum.
9 a) i) Excited means that the electrons have been given energy to raise
them to higher energy levels than their normal lowest energy level
stable state.
ii) The electrons in the excited mercury atoms are in an unstable
state. In order to achieve stability, they emit energy in the form of
quanta of electromagnetic radiation, thus returning to lower, more
stable, energy levels.
iii) According to quantum theory, the electrons can only exist in
certain allowed discrete energy levels. The frequency of the
emitted radiation corresponds exactly to the energy released when
an electron drops from one of these allowed energy levels to a
lower energy level, given by hf = E 2 − E 1 . Therefore only certain
wavelengths of radiation are emitted.
b) i) The energy levels for the phosphor electrons are different from
those of mercury and so the energy transitions will give rise to
different wavelengths from the mercury.
ii) Using Q = It and remembering to convert 1½ hours to seconds:
Q = 200 × 10 −3 A × 1.5 × 60 × 60 s = 1080 C
iii) Fluorescent tubes are more efficient, insomuch as they give out
more light than a tungsten filament light for the same amount of
electrical energy supplied. They therefore cost less to run and are
more environmentally friendly. They are, however, more costly to
manufacture, both in terms of money and carbon footprint.
10 a) Consider a simple model of an atom consisting of a nucleus
surrounded by electrons in some form of ‘orbits’. An electron in a
particular ‘orbit’ will have a certain amount of energy associated with
it, made up of its kinetic energy of rotation and its potential energy
due to the electric field of the nucleus. This is called the ‘energy level’
of the electron.
b) A photon is a small, discrete amount, or quantum, of energy associated
with electromagnetic radiation.
Full Answers to Review Questions: 16 Nature of light
Note
Although γ-rays could have the same
wavelength, γ-rays come from the
nucleus and not from transitions of
electron energy levels as we have
here. An answer of γ-radiation would
therefore be wrong.
Edexcel Physics for AS © Hodder Education 2009
28

c) The energy of the photon in the absorption diagram is:
hf = E 2 − E 1
d) When a photon is absorbed by an electron, the increase in the energy
level of the electron is exactly equal to the energy of the photon.
When the electron returns to its lower energy level the photon
emitted has an amount of energy exactly equal to the difference
between the energy levels. Therefore the laser light emitted by the
stimulated emission process must have the same wavelength as the
photon in the spontaneous emission diagram.
e) In general, ‘coherent’ means that there is a constant phase relationship
between two waves. In the case of a laser, it means that the emitted
photons are all of the same frequency and in phase.
Full Answers to Review Questions: 16 Nature of light
Edexcel Physics for AS © Hodder Education 2009
29