Question
(a) Define gravitational potential at a point.
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(b) TESS is a satellite of mass 360kg in a circular orbit about the Earth as shown in Fig. 1.1.
The radius of the Earth is \(6.4 × 10^6\) m and the mass of the Earth, considered to be a point mass at its centre, is \(6.0 × 10^{24}\) kg.
(i) It takes TESS 13.7 days to orbit the Earth.
Show that the radius of orbit of TESS is 2.4 × 108m.
(ii) Calculate the change in gravitational potential energy between TESS in orbit and TESS on a launch pad on the surface of the Earth.
change in gravitational potential energy = ……………………………………………… J
(iii) Use the information in (b)(i) to calculate the ratio:
\(\frac{gravitational field strength on surface of Earth}{gravitational field strength at location of TESS in orbit}\)
ratio = …………………………………………………
Answer/Explanation
Ans:
(a) work done per unit mass
work done moving mass from infinity (to the point)
(b)(i)
gravitational force provides centripetal force
mv2 / r = GMm / r2 and v = 2 πr / T OR mrω2 = GMm / r2 and ω = 2 π / T
OR r3 = GMT2 / 4 π2
r3 = 6.67 × 10-11 × 6.0 × 1024 × (13.7 × 24 × 3600)2 / 4 π2
so r = 2.4 × 108 m
b(ii)
EP = (–) GMm / r
work done = GMm / r1 – GMm / r2
= 6.67 × 10–11 × 360 × 6.0 × 1024 (1/6.4 × 106 – 1 / 2.4 × 108)
= 2.2 × 1010 J
b(iii)
g = GM / r2
\(ratio = \frac{r_{TESS}^2}{ r_{earth}^2}\)
= (2.4 × 108 / 6.4 × 106)2
= 1400
Question
A large container of volume \(85m^3\) is filled with 110kg of an ideal gas. The pressure of the gas is \(1.0 × 10^5Pa\) at temperature T.
The mass of 1.0mol of the gas is 32g.
(a) Show that the temperature T of the gas is approximately 300K.
(b) The temperature of the gas is increased to 350K at constant volume. The specific heat capacity of the gas for this change is 0.66J \(kg^{−1}K^{−1}\).
Calculate the energy supplied to the gas by heating.
energy = ……………………………………………… J
(c) Explain how movement of the gas molecules causes pressure in the container.
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(d) The temperature of a gas depends on the root-mean-square (r.m.s.) speed of its molecules.
Calculate the ratio:
\(\frac{r.m.s. speed of gas molecules at 350K}{r.m.s. speed of gas molecules at 300K}\)
ratio = …………………………………………………
Answer/Explanation
(a)
n = 110 / 0.032 or 110000 / 32 or 3440
pV = nRT
T = (1.0 × 105 × 85) / (8.31 × (110 / 0.032)) = 300 K
(b)
E = mcΔθ
= 110 × 0.66 × 50
= 3600 J
(c)
Any 3 from:
• molecule collides with wall
• momentum of molecule changes during collision (with wall)
• force on molecule so force on wall
• many forces act over surface area of container exerting a pressure
(d)
KE ∝ T
v ∝ √T
ratio = √(350 / 300)
= 1.1
Question
(a) A body undergoes simple harmonic motion.
The variation with displacement x of its velocity v is shown in Fig. 3.1.
(i) State the amplitude xo of the oscillations.
xo = …………………………………………….. m
(ii) Calculate the period T of the oscillations.
T = ……………………………………………… s
(iii) On Fig. 3.1, label with a P a point where the body has maximum potential energy.
(b) A bar magnet is suspended from the free end of a spring, as shown in Fig. 3.2.
One pole of the magnet is situated in a coil of wire. The coil is connected in series with a switch and a resistor. The switch is open.
The magnet is displaced vertically and then released. The magnet oscillates with simple harmonic motion.
(i) State Faraday’s law of electromagnetic induction.
…………………………………………………………………………………………………………………………
(ii) The switch is now closed. Explain why the oscillations of the magnet are damped.
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Answer/Explanation
a(i)
0.050 m
a(ii)
ω = vo / xo
T = 2 π / ω
0.42 = (2π × 0.050) / T
T = 0.75s
a(iii)
one point labelled P where ellipse crosses displacement axis marked
b(i)
(induced) e.m.f. proportional to rate
of change of (magnetic) flux (linkage)
b(ii)
(there is) current in the circuit
either
current causes thermal energy (dissipated) in resistor
thermal energy comes from energy of magnet
or
current causes magnetic field around coil
two fields cause an opposing force on magnet
Question
(a) (i) Explain why ultrasound used in medical diagnosis is emitted in pulses.
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(ii) Explain the principles of the detection of ultrasound waves used in medical diagnosis.
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(b) The specific acoustic impedances Z of some media are given in Table 4.1.
Table 4.1
| media | Z/ kgm−2 s−1 |
air gel soft tissue | 4.3 × 102 1.5 × 106 1.6 × 106 |
(i) The specific acoustic impedances of two media are Z1 and Z2. The intensity reflection coefficient α for the boundary of these two media is given by:
\(\alpha = \frac{\left ( Z_{1}-Z_{2} \right )^{2}}{\left ( Z_{1}+Z_{2} \right )^{2}}\)
Calculate, to three significant figures, the fraction of the ultrasound intensity that is reflected at a boundary between air and soft tissue.
α = …………………………………………………
(ii) Use your value in (b)(i) to explain why gel is applied to the surface of the skin during an ultrasound scan.
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Answer/Explanation
a(i)
Any 2 from:
• allows the reflected signal to be distinguished from the emitted signal
• detection occurs in the time between emitted pulses
• (reflection of ultrasound) detected by same probe / transducer / crystal
• cannot emit and detect at same time (hence pulses)
a(ii)
piezo-electric crystal
ultrasound makes crystal vibrate / resonate
vibration produces (alternating) e.m.f. / p.d. across crystal
b(i)
= (1.6 × 106 – 4.3 × 102)2 / (1.6 × 106 + 4.3 × 102)2
= 0.999
b(ii)
without the gel most of the ultrasound is reflected
Z values more similar / α reduces
so less (ultrasound) is reflected / more (ultrasound) is transmitted
Question
(a) State two advantages of the transmission of data in digital form, rather than analogue form.
1. …………………………………………………………………………………………………………………………….
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2. …………………………………………………………………………………………………………………………….
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(b) Optic fibres are used for the transmission of data.
(i) A signal in an optic fibre is carried by an electromagnetic wave of frequency 1.36 × 1014Hz. The speed of the wave in the fibre is 2.07 × 108ms−1.
For this electromagnetic wave, determine the ratio:
\(\frac{wavelength in free space}{wavelength in fibre}\)
ratio = …………………………………………………
(ii) The attenuation per unit length of the signal in the fibre is 0.40 dB km−1. The input power is 1.5mW and the output power is 0.060 mW.
Calculate the length of the fibre.
length = …………………………………………… km
Answer/Explanation
(a)
Any 2 from:
• noise can be filtered out / noise can be removed / signal can be regenerated
• can carry more information per unit time / greater rate of transmission of data
• can have extra bits of data to check for errors
• can be encrypted
b(i)
v ∝ λ
ratio = vair / vfibre = 3.00 × 108 / 2.07 × 108
= 1.45
b(ii)
attenuation = 10 log (P2/P1)
0.40 × L = 10 log (1.5 / 0.06)
0.40 × L = 13.979
L = 35 km
Question
Two positively charged identical metal spheres A and B have their centres separated by a distance of 24cm, as shown in Fig. 6.1.
The variation with distance x from the centre of A of the electric field strength E due to the two spheres, along the line joining their centres, is represented in Fig. 6.2.
(a) State the radius of the two spheres.
radius = …………………………………………… cm
(b) The charge on sphere A is \(3.6 × 10^{−9}\)C. Determine the charge \(Q_B\) on sphere B.
Assume that spheres A and B can be treated as point charges at their centres.
Explain your working.
\(Q_B\) = …………………………………………….. C
(c) (i) Sphere B is removed.
Use information from (b) to determine the electric potential on the surface of sphere A.
electric potential = ……………………………………………… V
(ii) Calculate the capacitance of sphere A.
capacitance = ……………………………………………… F
Answer/Explanation
(a)
2.0 cm
(b)
At 16 (cm) from A the electric fields are equal or EA = EB
E = Q / 4 π εor2
QA / (4 πεorA2) =QB / (4πεorB2)
3.6 × 10-9 / 0.162 = QB / 0.082
QB = 9.0 × 10–10 C
(c)(i)
V =Q / 4πεorA
V = 3.6 × 10–9 / (4 × π × 8.85 × 10–12 × 0.020)
V = 1600 V
(c)(ii)
C = Q / V
= 3.6 × 10–9 / 1600
= 2.3 × 10–12 F
Question
(a) On Fig. 7.1, sketch the temperature characteristic of a negative temperature coefficient (n.t.c.) thermistor. Label the axes with quantity and unit.
(b) An n.t.c. thermistor and a resistor are connected as shown in Fig. 7.2.
The temperature of the thermistor is increased.
State and explain the change, if any, to the reading on the voltmeter.
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(c) The variation with the fractional change in length Δx /x of the fractional change in resistance ΔR/R for a strain gauge is shown in Fig. 7.3.
The unstrained resistance of the gauge is 120Ω. Calculate the new resistance of the gauge when it is extended to a strain of 0.020.
resistance = …………………………………………….. Ω
Answer/Explanation
(a)
axes labelled with resistance and temperature
concave curve not touching temperature axis
line with negative gradient throughout
(b)
resistance of thermistor decreases
total circuit resistance decreases so voltmeter reading increases
or
current increases so voltmeter reading increases
or
greater proportion of resistance in fixed resistor so voltmeter reading increases
or
p.d. across thermistor decreases so voltmeter reading increases
(c)
(0.020 strain means) ΔR / R = 0.090
ΔR = 0.090 × 120 = 10.8 Ω
resistance = 120 + 10.8 = 130 Ω
Question
(a) Explain what is meant by a magnetic field.
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(b) The apparatus shown in Fig. 8.1 is used in an experiment to find the magnetic flux density B between the poles of a horseshoe magnet. Assume the magnetic field is uniform between the poles of the magnet and zero elsewhere.
The rigid metal rod of length 300mm is fixed in position perpendicular to the direction of the magnetic field. The poles of the magnet are both 45mm long. There is a current in the rod that causes a force on the rod. The balance is used to determine the magnitude of the force.
The variation with current I of the force F on the rod is shown in Fig. 8.2.
Calculate the magnetic flux density B.
B = ……………………………………………… T
(c) In a different experiment, electrons are accelerated through a potential difference and then enter a region of magnetic field. The magnetic field is into the plane of the paper and is perpendicular to the direction of travel of the electrons, as illustrated in Fig. 8.3.
(i) Explain why the electrons follow a circular path when inside the region of the magnetic
field.
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(ii) State the measurements needed in order to determine the charge to mass ratio, e/me, of an electron.
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Answer/Explanation
(a)
a region where a magnet / magnetic material / moving charge / current carrying conductor experiences a force
(b)
B = F / I L
e.g. = 9 × 10–3 / (5.0 × 0.045)
= 0.040 T
(c)(i)
force is (always) perpendicular to the velocity / direction of motion
magnetic force provides the centripetal force
or force perpendicular to motion causes circular motion
magnitude of force (due to the magnetic field) is constant
or
no work done by force
or
the force does not change the speed
(c)(ii)
Applying the list rule, any 2 from:
accelerating p.d.
radius of path / radius of semicircle
magnetic flux density
Question
(a) The output of a power supply is represented by:
V = 9.0 sin 20t
where V is the potential difference in volts and t is the time in seconds.
Determine, for the output of the supply:
(i) the root-mean-square (r.m.s.) voltage, Vr.m.s.
Vr.m.s = ……………………………………………….V
(ii) the period T.
T = ……………………………………………… s
(b) The variations with time t of the output potential difference V from two different power supplies are shown in Fig. 9.1 and Fig. 9.2.
The graphs are drawn to the same scale.
State and explain whether the same power would be dissipated in a 1.0Ω resistor connected to each power supply.
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(c) (i) The power supply in (a) is connected to a transformer. The input power to the transformer is 80W.
The secondary coil is connected to a resistor. The r.m.s. voltage across the resistor is 120V. The r.m.s. current in the secondary coil is 0.64A.
Calculate the efficiency of the transformer.
efficiency = …………………………………………………
(ii) State one reason why the transformer is not 100% efficient.
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Answer/Explanation
(a)(i)
9.0 / √2 = 6.4 V
(a)(ii)
ω = 20
ω = 2π / T
T = 2π / 20
T = 0.31 s
(b)
the r.m.s. voltages are different, so no
(c)(i)
P = Vr.m.s. × Ir.m.s
= 120 × 0.64 = 76.8 W
efficiency = (76.8 / 80) × 100
= 0.96 or 96 %
(c)(ii)
Any one from:
• heat losses due to resistance of windings / coils
• heat losses in magnetising and demagnetising core / hysteresis losses in core
• heat losses due to eddy currents in (iron) core
• loss of flux linkage
Question
(a) By reference to the photoelectric effect, explain what is meant by work function energy.
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(b) In an experiment, electromagnetic radiation of frequency f is incident on a metal surface.
The results in Fig. 10.1 show the variation with frequency f of the maximum kinetic energy EMAX of electrons emitted from the surface.
(i) Determine the work function energy in J of the metal used in the experiment.
work function energy = ……………………………………………… J
(ii) The work function energy in eV for some metals is given in Table 10.1.
Table 10.1
| metal | work function/eV |
| tungsten magnesium potassium | 4.49 3.68 2.26 |
Determine the metal used in the experiment. Show your working.
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(c) The intensity of the electromagnetic radiation for one particular frequency in (b) is increased.
State and explain the change, if any, in:
(i) the maximum kinetic energy of the emitted electrons
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(ii) the rate of emission of photoelectrons.
………………………………………………………………………………………………………………………….
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Answer/Explanation
(a)
energy of a photon required to remove an electron
either: energy to remove electron from a surface
or: minimum energy to remove electron
or: energy to remove electron with zero kinetic energy
(b)(i)
Correct read off from graph of f as 5.45 × 1014 Hz when \(E_{MAX} = 0\)
5.45 × 1014 × 6.63 × 10–34
= 3.6 × 10–19 J
(b)(ii)
3.6 × 10–19 / 1.6 × 10–19 = 2.3 eV so potassium
(c)(i)
each photon has same energy so no change
(c)(ii)
more photons (per unit time) so (rate of emission) increases
Question
Electrons are accelerated through a potential difference of 100kV. They are then incident on a metal target, they decelerate, and X-ray photons are emitted.
(a) Calculate the maximum possible frequency of the emitted X-ray photons.
frequency = ……………………………………………. Hz
(b) Explain why an aluminium filter may be placed in the X-ray beam when producing an X-ray
image of a patient.
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(c) The linear attenuation (absorption) coefficients μ for X-rays in bone, blood and muscle are given in Table 11.1.
Table 11.1
| μ/cm−1 | |
| bone | 3.0 |
| blood | 0.23 |
| muscle | 0.22 |
(i) A beam of these X-rays is incident on a person.
Calculate the percentage of the intensity of the X-ray beam that has been absorbed after passing through 0.80cm of blood.
percentage of intensity absorbed = …………………………………………….. %
(ii) In an X-ray image, white regions show greater absorption of X-rays than dark regions.
State and explain the difference between the X-ray image of bone compared to that of muscle.
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Answer/Explanation
(a)
eV = hf
f = 1.60 × 10–19 × 100 000 / 6.63 × 10–34
\(= 2.41 × 10^{19}\) Hz
(b)
(aluminium filter) absorbs (most) low energy X-rays
Any 2 from
• X-ray beam contains many wavelengths
• so low energy X-rays are not absorbed in the body
• low energy X-rays can can cause harm but do not contribute to the image
(c)(i)
I / Io = e– μx
e–0.23× 0.80 = 0.83
17% absorbed
(c)(ii)
bone is seen as lighter / muscle is seen as darker
either bone has a higher µ value so absorbs more
or muscle has a lower µ value so transmits more
Question
(a) Explain what is meant by the binding energy of a nucleus.
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(b) The following nuclear reaction takes place:
\(_{92}^{235}\textrm{U} + _{0}^{1}\textrm{n}\rightarrow _{55}^{144}\textrm{Cs} + _{x}^{90}\textrm{Rb} + y_{0}^{1}\textrm{n}\)
(i) Determine the values of x and y.
x = ………………………………………………………
y = ………………………………………………………
(ii) State the name of this type of nuclear reaction.
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(iii) Compare the binding energy per nucleon of uranium-235 with the binding energy per nucleon of caesium-144.
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(c) Yttrium-90 decays into zirconium-90, a stable isotope.
A sample initially consists of pure yttrium-90.
Calculate the time, in days, when the ratio of the number of yttrium-90 nuclei to the number of zirconium-90 nuclei would be 2.0.
The half-life of yttrium-90 is 2.7 days.
time = ………………………………………… days
Answer/Explanation
(a)
(minimum) energy required to separate the nucleons
to infinity
(b)(i)
X=37
Y=2
(b)(ii)
fission
(b)(iii)
binding energy per nucleon smaller for U than for Cs
(c)
Current ratio 2 Y to 1 Zr, so initially 3 Y
2 = 3 e–λt
λ = 0.693 / 2.7
ln(2 / 3) = – (ln 2 / 2.7)t
t = 1.6 days
or
(½)n = 2 / 3
n = 0.585
time = 0.585 × 2.7
= 1.6 days