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 × 106 Pa × 220 × 10−6 m2 = 2.2 × 103 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 A s     Therefore the units of C are: ___________   = kg−1 m−2 A2 s4 kg m2 A−1 s−3 4 a) 11 km; S 278 E b) Average speed = 3.0 km h−1

Average velocity = 2.2 km h−1 in the direction S 278 E

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:

Average

l/mm

w/mm

t/mm

95

62

43

96

62

43

95(.5)

62

43 Table A.1 

Volume: V = l w t

= 9.55 cm × 6.2 cm × 4.3 cm

= 255 cm3

250 g     Density = __  m = _______   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 1

Edexcel Physics for AS © Hodder Education 2009

Full Answers to Review Questions: 2 A guide to practical work

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 Table A.2 

ii) Area A = 0.297 m × 0.210 m = 0.0624 m2

5.04 g     = 80.8 g m−2 ‘gsm’ =  __________ 0.006 24 m2

iii) Percentage difference

(80.8 – 80) g m−2         × 100% = 1% =  _____________________ 80 g m−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

48 mm t =  ______      500

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.

= 0.096 mm = 0.0096 cm mass    Density of paper =  _______ 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 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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Full Answers to Review Questions: 2 A guide to practical work 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 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. 2 t  π d  b) V =  ____   4

π × (2.03 cm)2 × 0.165 cm =  _____________________        = 0.534 cm3   4

Density = _______   mass    volume

3.56 g     = 6.7 g cm–3 =  _________ 0.534 cm3

c) i) Percentage difference

(7.8 – 6.7) g cm–3        × 100% = 14% =  _____________________ 7.8 g cm–3

ii) Percentage uncertainty in 10d

× 100% = 0.5% = _______   1 mm    200 mm

Percentage uncertainty in 10t

× 100% = 3% = ________   0.5 mm     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        × 100% = 9%  ______________________ 8.15 g cm–3

Tip

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.

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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Full Answers to Review Questions: 2 A guide to practical work 2 __  2 h 4 a) If h =  _12   g t2 ⇒ t2 = ___ g   =  g  × h A graph of t2 against h therefore should be a straight line through the origin of gradient equal to __  2g  . b) h/cm

40

60

80

100

120

t/s

0.30

0.38

0.42

0.47

0.51

t2/s2

0.090

0.144

0.176

0.221

0.260 Table A.4 

Your graph should be as in Figure A.1. 0.30

0.25

0.20

t2/s2

0.15

0.10

0.05

0

0

0.20

0.40

0.60 h/m

0.80

1.00

1.20

Figure A.1 

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. 2 − 0.015 s2 _______________        = 0.206 s2 m–1 d) Gradient =  0.250 s 1.14 m − 0.00 m

2 = 0.206  __ g

⇒ g = _____   2     = 9.7 m s–2 0.206

e) Percentage difference

(9.8 – 9.7) m s–2       × 100% = 1% =  _____________ 9.8 m s–2

f) i) T he 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. 4


Full Answers to Review Questions: 3 Rectilinear motion

3 Rectilinear motion − 0) m s−1 −  u  =  (40 ____________ 1 a) a =  v_____        = 0.50 m s−2 – the answer is B. t   80 s (40 + 0) −1 +  u  b) s =  v_____  × t =  _______     × 80 s = 1600 m – the answer is B.   m s 2 2 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  δx  at an instant Instantaneous velocity =  ___ δt Dv   5 a) Acceleration is the change of velocity in unit time: a 5  ___ Dt b) i) Average acceleration ≈ 3 m s−2 ii) Gear changes are likely to affect the acceleration. 6 s =  _12 (  u + v)t v = u + at s = ut +  _12 a  t 2 v2 = u2 + 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 +  _12 a  t2 = 0 + _ 12  × 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

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Full Answers to Review Questions: 4 Forces

4 Forces 1 a) The resultant force acting on the system = 7000 N − (1000 + 1000) N = 5000 N. 5000 N    F _______ Using Newton’s second law: a =  __ = 1.25 m s−2 – the answer m   =  4000 kg  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. × 9.8 m s−2 × 2.4 m mgh 1000 kg work done ____ 3 Power =  _________        =  ______________________        = 2940 W ≈   =       t time 8.0 s 3000 W – the answer is C. 4 Useful power output = 75% of 120 W = 90 W

90 W     Pv   =  _______ = 45 N – the answer is A. P = F × v ⇒ F = __ 2.0 m s−1

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. Edexcel Physics for AS © Hodder Education 2009

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Full Answers to Review Questions: 6 Fluids b) 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 Fave, the gain in EPE will be Fave∆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 _ 12 m   v 2. b) KE of the golf ball = _ 12 (  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:

(length of card)      v =  _____________ t 

Repeat for a range of values of h.

v2       ∆KE    =  ____ ∆GPE = mgh and ∆KE = __  1  mv 2 ⇒ ______ 2 ∆GPE 2gh

Plot a graph of v2 against h. The percentage of GPE converted to KE is found by 2g  _______      × 100%. gradient

useful work output 10 Efficiency =  ________________        × 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

3430 J work done      = 690 W = ______   Power =  _________ time   5.0 s

c) The student will do more work moving limbs, contracting muscles, pumping blood, etc. 12 a) Using v2 = u2 + 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 ii) P = Fv = (60 N) × (5.0 m s−1) = 300 W

6 Fluids 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. 50 × 10−3 kg   mass           = 1.5 kg m–3 =  _______________ 2 a) ρ = _______ 4 _ volume  3 π   (20 × 10–2 m)3 7


Full Answers to Review Questions: 7 Solid materials b) 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 m3 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.

( 

F   = (6 π) η  __   v −1   . Units of η are: N(m2 s−1)−1, i.e. N s m−2 6  __ (  mr    ) _____ N m s

)

7 a) v ≈ 1 mm s−1 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. 8.0 N = 64 N m−1 F   =  _______ 2 b) k =  __ x 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  ∆l   ii) Strain (ε) = __ l

σ   iii) Young modulus E =  __ ε 50 N  b) σ =  ______________     = 1.8 × 108 Pa π (0.3 × 10−3 m)2

× 10−3 m __________ ε =  2.5      = 1.25 × 10−3 2.00 m

× 108 Pa __________ E =  1.8    = 1.4 × 1011 Pa 1.25 × 10−3

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Full Answers to Review Questions: 8 Nature of waves 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 λ is lower than that of light – the answer is C. 8 −1 × 10 m s 4 Using c = f λ; f = __  c   =  3___________      = 2.5 × 109 Hz = 2.5 GHz – the answer 0.12 m λ 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 _ 14  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 1  ;  so v = f λ v = __  λ   but f =  __ T 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 × 108 m s−1.

10 a) i) 0.75 m s−1

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Full Answers to Review Questions: 9 Transmission and reflection of waves 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 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. sin 40 7 a) 1.55 = _____   sin r  

⇒ r = 24.58

b) 1.55 sin 24.5 = μ sin 28

⇒ μ = 1.37

 1.52   8 sin θ = ____ 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. Edexcel Physics for AS © Hodder Education 2009

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Full Answers to Review Questions: 11 Charge and current 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 × 108 m s−1 v  ≈  3___________ f =  __      ≈ 2.5 × 109 Hz 0.12 m λ 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

λ   in the fundamental mode; hence λ = b) ii) For an open-ended pipe l =  __ 2 2l. v  =  __ v    f =  __ λ 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: 11


Full Answers to Review Questions: 11 Charge and current

115.2 C      = 7.2 × 1020 Number of electrons per hour =  ___________ 1.6 × 10–19 C

The answer is B.

b) The time taken for an electron to cross the tube is given by

–2 m distance 21 × 10   time =  _______      = 5.0 × 10−9 s    =  ____________ 7 speed 4.2 × 10 m s–1

Substituting into Q = It gives

Q = 32 × 10−3 A × 5.0 × 10−9 s = 1.6 × 10−10 C

Dividing by the charge on an electron:

× 10–10 C ___________ Number of electrons in beam =  1.6     = 1.0 × 109 1.6 × 10–19 C

The answer is A.

2 a) Rearranging Q = It,

Q _________ 8 C  = 16 000 A = 16 kA I =  __ –3 s  t   =  0.5 × 10 

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):

8 C     = 5 × 1019 ions Number of ions in each strike =  ___________ 1.6 × 10–19 C

3 a) In the equation I = nAvq

I = current in the conductor

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

Tip The words in italics are important to use; candidates often omit these and therefore lose marks.

b) Rearranging I = nAvq and remembering to put the current in amperes and the dimensions in metres: –3

10 × 10 A I    =  ______________________________________________ v =  ____             nAq 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 106 times less than in the semiconductor), the number of charge carriers per unit volume in the copper must be about 106 more than in the semiconductor – that is in the order of 1029 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:

12 C   Number of electrons =  ___________    = 7.5 × 1019 electrons 1.6 × 10–19 C

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 Tip metals, the charge carriers will be ‘free’ electrons in both cases, so q 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. 12 tungsten much greater. As both materials are metals, the number of Edexcel Physics for AS © Hodder Education 2009

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)

J = N m (work = force × distance)

N = kg m s−2 (from F = ma)

W = kg m s−2 × m × s−1 = kg m2 s−3

The answer is D.

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:

b) Working from first principles, potential difference = work done per charge passing: –2 ×m J Nm kg m s V =  __   =  ____     =  __________    = kg m2 s−3 A−1 C As As

From P = VI we have kg m2 s–3 V = __  P   = ________        = kg m2 s−3 A−1 I A as before. Clearly this is a much quicker way, but it does depend on you getting the answer to part a) correct!

The answer is B.

2 Quantity or quantities Which quantity is the product of two other quantities?

Power = p.d. × current (from P = VI)

Which quantity is one of the quantities divided by one of the other quantities?

Current = power 4 p.d. (from P = VI)

(3 possible answers) Which quantity, when divided by time, gives another quantity in the table? (2 possible answers)

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,

1500 W   P  =  _______ = 6.5 A I =  __ 230 V 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

5.5(272) J work done       = 0.72 W    =  _________ b) Power =  _________ 7.7 s time taken c) P = VI = 3.2 V × 0.75 A = 2.4 W useful power taken out _______ 0.718 W       × 100% ≈ 30%  =      d) Efficiency =  ___________________ 2.40 W total power put in 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 13


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. 5 a)

A

copper wire

V

 Figure A.2

b) Rate of doing work = power = VI = 0.53 V × 1.8 A = 0.95(4) W c) Rearranging I = nAvq: 1.8 A I    =  ______________________________________ v =  ____           = 2.4 mm s−1 nAq 8.0 × 1028 m–3 × 0.059 × 10–6 m2 × 1.6 × 1019 C d) Using P = Fv (see page 48):

–1

0.954 J s P   =  _____________      F =  __ –1  = 400 N (to 2 significant figures) v 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:

1500 W P   =  _______ I =  __      = 125 A V 12 V

c) For each headlight:

P  =  60 W _____  = 5 A I =  __ 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

864 000 J fraction used =  _________     = 0.32 ≈  _13   2 678 400 J

f) Using P = Fv (see page 48), power to drive at 90 km per hour up a gradient of 10% is given by: × 103 m _________ P = 0.1 × mg × v = 0.1 × 1740 kg × 9.8 N kg−1 ×  90    60 × 60 s = 43 kW (to 2 significant figures)

Maximum power is stated as 180 kW, so fraction of maximum power

43 kW =  _______     = 0.24 ≈  _14   180 kW

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:

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.

(8.0 – 0.0) V       = 32 Ω R =  ____________ (0.25 – 0.0) A

in    values ________________ ii) Percentage difference =  difference     × 100% stated value (33 – 32) Ω   × 100% = 3% =  __________    33 Ω 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:

12.0 V     = 32 Ω R = __  V   =  _______ I 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:

1.0 V  V   =  ______    = 10 Ω R =  __ I 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. 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 I2R. 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. 2.0 V  V   =  ______    = 0.177 A = 177 mA b) I =  __ R 11.3 Ω c)

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.

I/A

0.200 0.177

0.100

0

1.0

2.0 V/V  Figure A.3

d)

12 V voltage sensor

nichrome ribbon

V

current sensor A

data logger

computer

Figure A.4 

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. Edexcel Physics for AS © Hodder Education 2009

16

Full Answers to Review Questions: 14 Resistance and resistivity e) The resistance of the whole ribbon when the toaster is operating at 240 V, 1000 W is given by

(240 V)2 V2  = ________ V2  ⇒ R =  ___      = 57.6 Ω P =  ___ P 1000 W R

If a length of 1.00 m has a resistance of 11.3 Ω, then the total length of the ribbon will be:

57.6 Ω   × 1.00 m = 5.10 m length of ribbon =  ______ 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.

4 a)

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.

V  : The required value of resistance is given by R =  __ I

0.83 V    = 33.2 Ω R =  __________ 25 × 10–3 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:

0.7 V  = 17.5 Ω ≈ 18 Ω    R =  __________ 40 × 10–3 A

17


Full Answers to Review Questions: 14 Resistance and resistivity b) 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:

0.6 V    V   =  __________  = 50 Ω R =  __ I 12 × 10–3 A

Note The scale of the graph is such that these answers are only approximate.

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 m2

Resistance is:

ρl ______________________ × 10–5 Ω m × 5 × 10−3 m          = 0.015 Ω R =  __   =  3.0 A 10 × 10–6 m2

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 m2

Resistance is:

ρl ______________________ × 10–5 Ω m × 1 × 10−3 m          = 6 × 10−4 Ω R =  __   =  3.0 A 50 × 10–6 m2

4 a) i) Rearranging P = VI,

3000 W   = 12.5 A I = __  P  =  _______ V 240 V

So the kettle will operate safely with a 13 A fuse.

V  , ii) From R =  __ I 240 V    = 19.2 Ω ≈ 19 Ω R =  ______  12.5 A V  , b) i) At 110 V, from I =  __ R 110 V ______      = 5.7 A I =   19.2 Ω

Tip Note that you can check this by using your answer to part i):

2

V   : ii) The power dissipated is given by P =  ___ R (110 V)2 ________    = 630 W P =   19.2 Ω

P = VI = 110 V × 5.7 A = 630 W

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:

Tip

ρ = ___  RA      l

where R is the resistance, l is the length and A is area of cross-section perpendicular to the length. ρl b) i) Rearranging R = __     : A –8 Ω m ρ × 10 1.72 R ______________       = 0.070 Ω m−1      =    __   = __ l A 0.2453 × 10–6 m2 2 πd  ii) Rearranging A =  ___  : 4 _____________ 4 × 0.1110 mm2 4A   π  ⇒ d =     _____________      = 0.3759 mm d2 = ___ π  



It is easiest to define resistivity by means of the appropriate equation, but remember that all the quantities in the equation must be defined.

18


Full Answers to Review Questions: 14 Resistance and resistivity  RA    : iii) Rearranging ρ = ___ l

ρ = A × __  R   = 0.1110 × 10−6 m2 × 4.41 Ω m−1 = 4.90 × 10−7 Ω m l 2 πd  iv) From A =  ___  : 4

π × (0.2743 mm)2 A =  _______________        = 0.0591 mm2 4 v) Rearranging ρ = ___  RA    : l

ρ = A × __  R   = 0.0591 × 10−6 m2 × 18.3 Ω m−1 = 1.08 × 10−6 Ω m l 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.

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: 2 π × (0.25 × 10−3 m)2 πD  A =  ____    =  _________________        = 4.91 × 10−8 m2 4 4 4.0 V  = 44.4 Ω R =  __________    90 × 10−3 A × 4.91 × 10–8 m2 ___________________  RA       = 1.1 × 10−6 Ω m ρ = ___    =  44.4 Ω   2.00 m l 7 a)

Tip Note the importance of including through the origin, without which the answer would be of little value.

I/A 0.50 0.42

0.25

0

6

12 V/V  Figure A.6

b) Rearranging P = V2/R:

(12 V)2 V2  = _______      = 29 Ω   R =  ___ 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:

1.0 V     = 50 Ω R = __  V   =  __________ I 20 × 10−3 A

ii) From the graph, when V = 3.0 V, I = 100 mA:

3.0 V   V   =  ___________    = 30 Ω R =  __ I 100 × 10−3 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 19


Full Answers to Review Questions: 15 Electric circuits 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 = R1 + R2 + R3 = 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   =  ___ 1  =  ____ + 2  = ____ 1    =  1_____ 1   + ___ 1    +  ____  __  1   ⇒  __   3    ⇒ R = 10 Ω R R1 R2 R 30 Ω 15 Ω 30 Ω 30 Ω Y: Combining the two parallel resistors: 1 1 1    =  ____ 1  + 1    +  ____ 1  =  ___ 2    ⇒ R = 7.5 Ω    ___   ⇒  __  =  ____  __ R R1 R2 R 15 Ω 15 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: 3    ⇒ R = 5 Ω 1   +  ___ 1   ⇒  __ 1  =  ____ 1    +  ____ 1    =  ____ 1  =  ___ 1   +  ___ 1    +  ____  __ R R1 R2 R3 R 15 Ω 15 Ω 15 Ω 15 Ω a) We can now see that Z = W/9 – the answer is A. b) W = 2Y – the answer is D.

Tip

c) X = 2 × Z – the answer is D. 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  =  ___ 1   ⇒ __ 1    =  _____ 1   +  ___ 1    +  _____ 2    ⇒ R = 4.7 Ω  __  1  =  _____ R R1 R2 R 9.4 Ω 9.4 Ω 9.4 Ω

It is worth remembering that if two resistors of the same value are connected in parallel, their combined resistance is half the individual resistance.

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 9 V ______     = 18 A I =   0.50 Ω This is a large current and will generate a power of P = I2 R = (18 A)2 × 0.50 Ω = 162 W inside the battery, which will make the battery hot.

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.

20


Full Answers to Review Questions: 15 Electric circuits 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. V   the maximum current will be: b) From I =  __ R 3

5 × 10 V   = 1.0 × 10−4 A = 0.10 mA   I =  _________ 50 × 106 Ω

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 = I2 R, the power dissipated in the girl would be:

P = (1.0 × 10−4 A)2 × 10 × 103 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: 1.44 V R = __  V   =  ______   = 7.20 Ω I 0.20 A and 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 Table A.6 

P/W 1.1 1.0 0.9 0.8 0.7 maximum power when R = 0.60 Ω

0.6 0.5

0

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

R/Ω  Figure A.7

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

Full Answers to Review Questions: 15 Electric circuits

5.90 V  V   =  ________    = 39.3 kΩ R =  __ I 0.15 mA

and

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 Table A.7 

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  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: (6.50 – 4.25) V       = 3.7 kΩ r =  _______________ (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.

22


Full Answers to Review Questions: 15 Electric circuits 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

 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: + 2  = ____ 1   = ___  1   ⇒ __  1   = ____  __    = 13.2 Ω   1    =  3_____   5    ⇒ R = ____  66 Ω  1   + ___   1    + ____ R R1 R2 R 22 Ω 33 Ω 66 Ω 66 Ω 5

The total circuit resistance is therefore 47 Ω + 13.2 Ω = 60.2 Ω.

The circuit current will be given by:

6.02 V   = 0.10 A I = __  V   =  ______ 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:

V   =  ______ 1.32V   I =  __ = 0.060 A = 60 mA 22 Ω R

and the current in the 33 Ω resistor as

1.32V   V   =  ______ = 0.040 A = 40 mA I =  __ 33 Ω R

ii) The power generated is given by P = I2 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:

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Full Answers to Review Questions: 15 Electric circuits

7.5 V  V   =  _____    = 0.136 mA I =  __ 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:

V   =  _____ 3.5 V  I =  __    = 0.159 mA R 22 kΩ

The combined resistance of the parallel arrangement of the voltmeter and the 33 kΩ resistor will therefore be:

4.0 V    V   =  _________ = 25.1 kΩ R =  __ I 0.159 mA

For this parallel arrangement:

1   ⇒  ___ 1   =  __ 1  –  ___ 1     1     1    +  ___ 1    =  _______ 1   =  ___ –  ______  __ R R33 RV RV R R33 25.1 kΩ 33 kΩ 1   = 0.0398 kΩ−1 − 0.0303 kΩ−1 = 0.0095 kΩ−1  ___ RV

RV = 106 kΩ

c) If the voltmeter is rated as 10 V/100 μA, its resistance should be:

10 V   R = __  V   =  ___________    = 1.0 × 105 Ω = 100 kΩ I 100 × 10−6 A

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 Vout will fall. b) i) Reading from the graph, at 0 8C the thermistor has a resistance of 18.0 kΩ.

When Vout 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:

4.0 V     = 0.22 mA I =  _______ 18.0 kΩ

The value of R is then given by:

Vout ________ 5.0 V     =   = 22.5 kΩ    R =  ____ 0.22 mA I

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Full Answers to Review Questions: 16 Nature of light

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:

power ___________ 3 W     =      = 0.11 W m−2 I = ______   4π × (1.5 m)2 4πr2

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:

−1 × 108 m s _____________  = 4.55 × 1014 s−1       f = __  c   =  3.00 −9 λ 660 × 10 m

Then:

E = hf = 6.63 × 10−34 J s × 4.55 × 1014 s−1 = 3.01 × 10−19 J

Converting to electron-volts using 1 eV = 1.6 × 10−19 J: 3.01 × 10−19 J E =  _______________        ≈ 2 eV 1.6 × 10−19 J eV−1 The answer is B.

2 a) Rearranging φ = hf0 and remembering to convert eV into J:

φ 2.28 eV × 1.6 × 10−19 J eV−1        =  _______________________        f0 = __ h 6.63 × 10−34 J s

= 5.5 × 1014 Hz

The answer is C.

b) We will need to use E = hf, so we must find the frequency corresponding to a wavelength of 447 nm. Rearranging c = fλ:

−1 × 108 m s _____________  = 6.71 × 1014 s−1       f = __  c   =  3.00 −9 λ 447 × 10 m

Then:

E = hf = 6.63 × 10−34 J s × 6.71 × 1014 s−1 = 4.45 × 10−19 J

Converting to electron-volts using 1 eV = 1.6 × 10−19 J: 4.45 × 10−19 J E =  ______________       = 2.78 eV 1.6 × 10−19 J eV−1 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:

(1.51 – 0.54) × 1.6 × 10–19 J  = 2.34 × 1014 s−1        f =  _______________________ 6.63 × 10–34 J s

Rearranging c = fλ:

25



× 108 m s−1 _____________ λ =  3.00        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:

7.68 × 10–20 J    efficiency =  ___________    × 100% = 19% 4 × 10–19 J

d) Rearranging E = hf:

4.0 × 10–19 J       = 6.03 × 1014 s–1 f = __  E   =  ____________ h 6.63 × 10−34 J s

Rearranging c = fλ:

8 m s–1 λ = _ c  = ____________     = 4.97 × 10–7 m = 500 nm (to 2 significant figures)  3.0 × 10    f 6.03 × 1014 s–1

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.

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: −1 × 108 m s _____________ f = __  c   =  3.00  = 5.64 × 1014 s−1       −9 λ 532 × 10 m Then:

E = hf = 6.63 × 10−34 J s × 5.64 × 1014 s−1 = 3.74 × 10−19 J

Converting to electron-volts using 1 eV = 1.6 × 10−19 J: 3.74 × 10−19 J E =  ______________       = 2.34 eV ≈ 2.3 eV 1.6 × 10−19 J eV−1 c) Rearranging hf = φ + _ 12  mv 2max gives:

 _12  mv2max = hf − φ = (2.34 2 1.90) eV = 0.44 eV

= 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:

(  )

2.34 λ = ____        × 532 nm = 655 nm 1.90 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.

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.

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. b) i) We will need to use E = hf, so we must find the frequency 26 corresponding to a wavelength of 630 nm. Rearranging c = fλ: Edexcel Physics for AS © Hodder Education 2009


−1 × 108 m s _____________  = 4.76 × 1014 s−1       f = __  c   =  3.00 −9 λ 630 × 10 m Then:

E = hf = 6.63 × 10−34 J s × 4.76 × 1014 s−1

= 3.16 × 10−19 J ≈ 3 × 10−19 J

ii) As 1 eV = 1.6 × 10−19 J: 3.16 × 10−19 J E =  ______________       = 2.0 eV 1.6 × 10−19 J eV−1 c) i) We then need use the relationship that intensity I at a distance r is given by: power ____________    =   18 mW     = 15.9 mW m−2 ≈ 16 mW m−2 I = ______   4πr2 4π × (0.30 m)2 ii) The area of the pupil is given by: 2 π(5 × 10−3 m)2 A =____  πD    =  ____________        = 1.96 × 10−5 m2 4 4

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 m2

= 3.12 × 10−7 J s−1

From part b) i), the energy of each photon is 3.16 × 10−19 J, so: 3.12 × 10–7 J s–1 number of photons per second =  _____________       = 1.0 × 1012 s−1 3.16 × 10–19 J

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 =  _________________    × 100% power consumption × 100% = 15% = _______   18 mW     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) Vs/V 1.2 1.0 0.8 0.6 0.4 0.2 +

f0 = 4.3 1014 Hz 0

0

4.0

5.0

6.0

7.0

8.0 f/1014 Hz

 Figure A.10

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Full Answers to Review Questions: 16 Nature of light d) The term eVs is equal to the maximum kinetic energy ½mv2max 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 f0 is the frequency when the stopping potential Vs = 0. From the graph, f0 = 4.3 × 1014 Hz. ii) The work function φ is given by:

φ = hf0 = 6.63 × 10−34 J s × 4.3 × 1014 s−1 = 2.9 × 10−19 J

As 1 eV = 1.6 × 10−19 J:

2.9 × 10−19 J φ =  ______________       = 1.8 eV 1.6 × 10−19 J eV–1

f) The energy to release a photoelectron comes from one photon. Below the threshold frequency f0 no electrons are emitted because the energy of each photon, hf0, is not sufficient to provide an electron near the surface with enough energy to escape. 8 a) Using hf = E2 − E1 and remembering to convert the energy from eV to J:

3

−19

–1

(69.6 – 1.8) × 10 eV × 1.6 × 10 J eV  = 1.64 × 1019 s−1          f =  _________________________________ 6.63 × 10–34 J s

Rearranging c = fλ: × 108 m s−1 _____________ λ =  3.00       = 1.83 × 10−11 m ≈ 0.02 nm 1.64 × 1019 s−1 b) This is in the X-ray region of the electromagnetic spectrum.

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.

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 = E2 − E1. 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. Edexcel Physics for AS © Hodder Education 2009

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Full Answers to Review Questions: 16 Nature of light c) The energy of the photon in the absorption diagram is:

hf = E2 − E1

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.

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