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PHYS1004 Exam Solutions
Page 1
Semester 2, 2008
THE UNIVERSITY OF SYDNEY
PHYS1004
PHYSICS 1 (ENVIRONMENTAL & LIFE SCIENCES)
SOLUTIONS
NOVEMBER 2008
Time allowed: THREE Hours
MARKS FOR QUESTIONS ARE AS INDICATED
TOTAL: 90 marks
INSTRUCTIONS
•
•
•
All questions are to be answered.
Use a separate answer book for each section.
All answers should include explanations in terms of physical principles.
DATA
3
Density of water
ρ
=
1.00 × 10 kg.m−3
Density of air
ρ
=
1.20 kg.m−3
1 atm
=
1.01 × 10 Pa
g
=
9.81 m.s
=
6.022 × 10
=
8.854 × 10
=
=
8.99 × 10 N.m2.C-2
−7
−1
4π × 10 T.m.A
=
1.602 × 10
Atmospheric pressure
Magnitude of local gravitational field
Avogadro constant
N
A
Permittivity of free space
ε
Coulomb’s law constant
K
Permeability of free space
µ
Elementary charge
0
0
e
5
-2
23
-1
mol
−12
F.m−1
9
−19
C
8
Speed of light in vacuum
c
=
Planck constant
h
=
2.998 × 10 m.s−1
−34
6.626 × 10
J.s
=
9.110 × 10−31 kg
=
1.675 × 10
−27
kg
=
1.673 × 10
−27
kg
=
1.674 × 10
−27
kg
Rest mass of an electron
m
Rest mass of a neutron
m
Rest mass of a proton
m
Rest mass of a hydrogen atom
m
e
n
p
H
Boltzmann constant
k
=
Stefan-Boltzmann constant
σ
=
1.381 × 10−23 J. K -1
-8
-2 -4
5.671 × 10 W. m K
Atomic mass unit
u
=
1.661 × 10
Energy equivalent of 1 atomic mass unit
E
=
931.49 MeV
−27
kg
PHYS1004 Exam Solutions
Semester 2, 2008
Page 2
SECTION A
Question 1
A sample of animal bone is tested under tensile load (stretching). The sample may be
considered as a uniform cylinder, with diameter 25.0 mm and length (before the test) of
200.0 mm. When subjected to stretching, the sample obeys Hooke’s law up to a tensile force
of 8.00 !104 N , at which point its length is 202.1 mm. A small further increase in tensile
load breaks the bone.
(a)
Calculate Young’s modulus for this sample.
(b)
Sketch a graph of stress versus strain for this process. Label the axes and on your
graph
(i)
indicate the Hookean region,
(ii)
label the position of the breaking point.
(5 marks)
Solution
(a)
For a linear extension of a Hookean material,
F
#L
=Y
A
L0
F L0
A #L
8.00 %104
200
=
= 1.55 %1010 Pa
"3 2
! (12.5 %10 ) 2.1
$Y =
= 1.6 %1010 Pa
(2 marks)
PHYS1004 Exam Solutions
Semester 2, 2008
Page 3
(b)
Stress _ max =
F
(8.00 #104
=
= 1.6 #108 Pa
"3 2
A ! (12.5 #10 )
Strain _ max =
!L
2.1
=
= 0.011
L0
200
(3 marks)
(Total 5 marks)
PHYS1004 Exam Solutions
Semester 2, 2008
Page 4
Question 2
In the lungs, gas exchange with the blood stream is carried out at the surface of numerous
small air sacs, the alveoli. The inner surface of each air sac is wet with fluid.
(a)
(b)
Using Laplace’s Equation, explain why it would appear at first sight that the alveoli
would be liable to collapse totally when breathing out, or to expand until rupture when
breathing in.
Explain how this ‘problem’ is avoided in real lungs.
(5 marks)
Solution
(a)
Laplace’s Law for a single surface spherical drop (or air sac) is
pi " po = 2 ! / r
where pi and po are the internal and external pressures respectively, ! is the surface tension,
and r is the radius.
This shows that as r decreases (breathing out) pi ! po increases further, which could lead to
further air being expelled until r ! 0 . Similarly, if pi ! po is sufficient to cause r to
increase (breathing in) then the required pi ! po is less at larger r , and the alveoli could
expand until rupture.
(3 marks)
(b)
The above assumes surface tension ! is constant. But lung surfaces are coated with
surfactant, which results in a varying surface tension. Long molecules become aligned and
harder to separate when the film is stretched, so increasing ! as r increases. Likewise, on
relaxing the tension the molecules are more randomised, reducing the surface tension. Hence
Laplace’s law still applies, but varying ! avoids collapse or rupture of the alveoli.
(2 marks)
(Total 5 marks)
PHYS1004 Exam Solutions
Page 5
Semester 2, 2008
Question 3
In a cyclotron, a positively charged H+ ion (a proton) moves in a spiral ‘orbit’ at right angles
to an applied magnetic field as its energy increases. For most of the time on each orbit the ion
moves at constant speed on a circular path with the centripetal acceleration produced by
magnetic forces.
(a)
Derive an expression for the period of the orbit in terms of the ion mass (m), charge
(q), and magnetic field strength (B), and show that it does not depend on the speed of
the ion.
(b)
Why is it important to the operation of the cyclotron that the period does not depend
on the speed of the ion?
(5 marks)
Solution
(a)
The basic principle is that the centripetal force is supplied by the magnetic force (the question
says this) - so
FB = Fcentripetal
q vB =
mv 2
r
(1 mark)
hence,
r=
mv
qB
Period is then the circumference of the ‘orbit’ divided by the velocity
T=
2! r 2! m
=
v
qB
(Correct procedure – 1 mark)
PHYS1004 Exam Solutions
Page 6
Semester 2, 2008
which independent of v, as required.
(Correct result and conclusion – 1 mark)
(b)
A constant period means that the brief acceleration phase can occur at regularly spaced
intervals, irrespective of the particle velocity. The E field causing the acceleration can
oscillate at a fixed frequency.
(2 marks)
(Total 5 marks)
Question 4
As a honey bee flies through the air it collides with minute dust particles and acquires an
electric charge. The charge on the bee attracts pollen grains when it lands on a flower.
(a)
Explain why the bee becomes charged.
(b)
Explain with the aid of a diagram why an uncharged conducting pollen grain will be
attracted to a positively charged bee.
(5 marks)
Solution
(a)
Electrons are added or removed from the bee by friction with the dust such that a net charge
results.
(2 marks allocated as 1 mark for friction; 1 mark for electrons)
(b)
A diagram showing negative and positive charges separated on the pollen grain. The negative
charge is closer to the bee.
The reason the neutral pollen grain is attracted to the bee is because the attractive force
between the negative charges is slightly greater than the repulsive force of the positive
charges because the negative charge is slightly closer.
(3 marks: 1 for diagram showing separation of charge, 1 for showing negative charge
closer to bee, 1 mark for showing greater attractive force from negative charge)
(Total 5 marks)
PHYS1004 Exam Solutions
Page 7
Semester 2, 2008
Question 5
Light with frequency ν is incident on a metal with a work function φ = 1.0 eV. Electrons are
observed to be ejected from the metal, up to a maximum kinetic energy Kmax described by
K max = h! # " .
(a)
What is the maximum kinetic energy of photo-electrons ejected when the light has
frequency ν = 5.0 × 1014 Hz?
(b)
What is the maximum kinetic energy of ejected electrons when the metal is
illuminated with light with frequency ν = 2.0 × 1014 Hz?
(c)
If the intensity of the incident light is doubled, does the maximum kinetic energy of
ejected electrons change? Does the number of ejected electrons change? Explain each
of your answers.
(d)
Briefly explain how these results are incompatible with classical physics.
(5 marks)
Solution
(a)
The maximum kinetic energy is
K max = h ! # "
= (6.626 $10#34 )(5.0 $1014 ) # 1.602 $10#19 J
% 1.71$10#19 J
% 1.1 eV
The minimum (threshold) frequency to produce electrons corresponds to K max = 0 and is
"min =
! 1.602 $10#19 J
=
% 2.42 $1014 Hz .
h 6.626 $10#34 J.s
Since the frequency ! = 2.0 "1014 Hz is less than this no electrons will be produced.
(2 marks)
(b)
When the intensity of light is doubled the maximum KE of electrons does not change.
However, the number of electrons ejected per unit time will double.
(1 mark)
(c)
In the classical picture the light is a wave which delivers energy continuously to electrons in
the metal. The energy in the wave is proportional to the intensity.
In the quantum picture the light consists of photons which give up energy discretely to
electrons. The energy of a photon is E = h ! , and the number of photons per unit time is
proportional to the intensity. The result in part (a) is inconsistent with the classical picture
because in the classical picture the energy in the wave is independent of the frequency.
Hence, irrespective of ! , the wave should be able to supply sufficient energy (in some time)
to lead to electron ejection. The result in part (b) is inconsistent with the classical picture
because in the classical picture when the intensity is increased the amount of energy supplied
per unit time is increased, so the maximum KE of ejected electrons would increase.
(2 marks)
(Total 5 marks)
PHYS1004 Exam Solutions
Page 8
Semester 2, 2008
Question 6
An object at room temperature (298 K) is emitting electromagnetic radiation as a blackbody.
(a)
What is the peak wavelength of the blackbody spectrum at this temperature?
(b)
What type of electromagnetic radiation does this correspond to? (Choose from radio,
infrared, visible, ultraviolet, X-ray, gamma ray.)
(c)
Given that this radiation is emitted by thermally vibrating atoms, which of the
following would you expect the thermal energy of the atoms to be closer to: 0.1 eV,
1eV, 10eV, or 100eV.
(5 marks)
Solution
(a)
Using Wien’s displacement law
(1 mark)
"3
!m T = 2.90 #10 m.K
$ !m =
2.90 #10"3
= 9.73 #10"6 m
298
(1 mark)
(b)
Infrared
(1 mark)
(c)
The thermal energy of the atoms should be comparable to the energy of a photon at this peak
wavelength.
(1 mark)
Calculate the energy of a photon at this wavelength:
h c (6.63 #10"34 )(3.0 #108 )
E=
=
= 2.04 #10"20 J
!m
9.73 #10"6
=
2.04 #10"20
= 0.13 eV
1.6 #10"19
Correct answer is 0.1 eV.
(1 mark)
(Total 5 marks)
PHYS1004 Exam Solutions
Page 9
Semester 2, 2008
Question 7
The figure shows a U-tube containing two liquids (A and B) which do not mix together and
are in a stable condition (i.e. not moving).
(a)
Is the pressure at point 1 the same as that at point 2? Explain why or why not. Both
points are 10.0 cm above the reference level.
(b)
Is the pressure at point 3 the same as at point 4? Explain why or why not. Both points
are 14.0 cm above the reference level.
(c)
The density of liquid A is 1.00 !103 kg.m -3 . Calculate the density of liquid B.
(d)
Calculate the gauge pressure at point 5, which is just inside the base of the tube at
4.0 cm above the reference level.
(10 marks)
Solution
(a)
Yes, p1 = p2 . They are connected through a uniform fluid in the lower part of the tube. If
there were a pressure difference it would cause fluid to flow. The same can be seen from
Bernoulli’s equation
1
1
p1 + ! B v12 + ! B g y1 = p2 + ! B v2 2 + ! B g y2 .
2
2
But v1 = v2 = 0 and y1 = y2 and so p1 = p2 (a streamline could join points 1 and 2).
(2 marks)
(b)
No, p3 ! p4 . Pressure p4 is atmospheric pressure since it is at the surface of the liquid,
whereas p3 = patm + ! A g d , where d = 20 ! 14 cm . (The argument that pressures are equal at
PHYS1004 Exam Solutions
Page 10
Semester 2, 2008
the same horizontal level does not apply directly because the liquids A and B have unequal
densities).
(c)
Evaluate pressures for the two arms, at point(s) known to have equal pressures. This could be
points 1 and 2 or point 5. Using points 1 & 2:
p1 = patm + ! A g (0.200 " 0.100)
and
p2 = patm + ! B g (0.140 " 0.100)
These are equal, so
! A (0.100) = ! B (0.040)
0.100
# !B = ! A
= (1.00 $103 )(2.5) = 2.5 $103 kg.m "3
0.040
(4 marks)
(d)
Gauge pressure is pressure above atmospheric, so
pG = p " patm = ! g d
Use this for the right hand arm since it has only a single liquid.
pG ,5 = ! B g (14.0 " 4.0) #10"2 = 2.45 #103 Pa
(Total 10 marks)
Question 8
The average flow rate of blood in a certain artery is 1.3 mL.s !1 ( 1 mL = 10!6 m3 ). The artery
has a radius of 2.2 mm . The viscosity of blood at body temperature is 2.1"10!3 Pa.s and its
density is 1.1"103 kg.m !3 .
(a)
Calculate the average flow velocity of blood in the artery.
(b)
Is the flow in the artery likely to be laminar or turbulent? Justify your answer.
(c)
What is the pressure difference over a length of 80 mm along the artery?
(d)
Calculate the power needed to maintain this flow.
(10 marks)
Solution
PHYS1004 Exam Solutions
Page 11
Semester 2, 2008
(a)
Volume flow rate is Q = Avavg where A is cross-section area, and vavg is average flow
velocity. So
Q
1.3 #10"6
vavg =
=
= 0.085 m.s "1 .
A ! (2.2 #10"3 ) 2
(2 marks)
(b)
Evaluating the Reynolds Number for the flow
NR =
! vd
"
we get
(1.1"103 )(0.085)(2 " 2.2 "10!3 )
2.1"10!3
= 200 (197)
This is much less than 2000, so the flow is expected to be laminar (streamline).
NR =
(3 marks)
(c)
Using Poiseuille’s law for flow of a viscous liquid
! R 4 $p
8" L Q
Q=
% $p =
8" L
! R4
(8)(2.1&10#3 )(0.080)(1.3 &10#6 )
! (2.2 &10#3 ) 4
= 24 Pa
=
(3 marks)
(d)
Power is !p Q for viscous flow. So
Power = (24)(1.3 "10!6 )
= 3.1"10!5 W
(2 marks)
(Total 10 marks)
PHYS1004 Exam Solutions
Page 12
Semester 2, 2008
Question 9
A battery can be thought of as an ideal source of emf " with an internal resistance r. It puts a
potential difference V across an external load R, and delivers a current i.
!
battery
i
ε
r
R
V
(a)
Write down an expression for the current i through the external load in terms of these
quantities.
(b)
What value of the external load will produce the maximum possible current ( imax )
through the load? What is the current through the load and the potential difference
across the load in this situation? (Don’t actually try this!)
(c)
What value of the external load will produce the maximum possible potential
difference ( Vmax ) across the load? What is the current through the load and the
potential difference across the load in this situation?
(d)
In each of the cases (b) and (c), no power is dissipated in the external load. Show that
in the general case, the power dissipated in the external load is given by P = ! i " i 2 r .
(e)
Draw a sketch showing how the power dissipated in an external load depends on
current. Indicate the two values of current at which the power is zero.
(f)
What is the the maximum power, Pmax , that can be dissipated in an external load and at
what value of current does this occur?
(10 marks)
Solution
(a)
Current is determined by the total resistance of the circuit, so i = ! (r + R ) .
(1 mark)
(b)
Current is maximum when R is minimised – i.e. R = 0
Then imax = ! r and V = ! R = 0
r
( )
(2 marks)
(c)
Maximum potential difference Vmax across the external load is " , but only when the potential
difference across r is zero - i.e. when i = 0 because R ! " . Then i = 0 and Vmax = ! (Note
Vmax ! 0 ).
!
(2 marks)
!
PHYS1004 Exam Solutions
Page 13
Semester 2, 2008
(d)
P = V i = (! " i r )i = ! i " i 2 r
(1 mark)
(e)
Power = 0
Power = 0
(shape of curve 1 mark), (labelled axes and power=0 points 1 mark)
(f)
imax
!
. This can be obtained by the differential, from
=
2
2r
the properties of a parabola in the graph, or a sensible guess (mark as correct). The power is
then
Maximum power occurs at a current
2
" ! # " ! #
P = ! i $ i r = ! % & $% & r
' 2r ( ' 2r (
!2 !2
=
$
2r 4r
!
=
4r
2
(2 marks)
(Total 10 marks)
PHYS1004 Exam Solutions
Page 14
Semester 2, 2008
Question 10
The base of a large storm cloud is 500 m above the surface of the Earth and has an area of
2 km × 2 km. The base of the cloud and the Earth can be considered to be a parallel plate
capacitor. Just before a lightning strike to ground, the electric field between the cloud and the
ground is measured to be 106 V.m !1 vertically upwards.
(a)
Calculate the potential difference between the surface of the Earth and the base of the
cloud.
(b)
Calculate the capacitance of this cloud-Earth combination.
(c)
Calculate the electrical energy stored in this cloud-Earth capacitor.
(d)
Calculate the amount of charge on the base of the cloud and determine its sign.
(e)
A lightning flash from this cloud lasts for 10 µs and transfers all the charge from the
base of the cloud to the surface of the Earth. Calculate the average current that flows
during the lightning discharge.
(10 marks)
Solution
(a)
E=
!V
" !V = E d = (106 )(500) = 5 #108 V
d
(b)
C=
! 0 A (8.85 #10"12 )(2000 # 2000)
=
= 7.1# 10"8 F .
d
500
(c)
UC =
1
2
C ("V ) = (0.5)(7.1#10!8 )(5 #108 ) 2 = 8.9 #109 J .
2
(d)
Q = C #VC = (7.1"10!8 )(5 "108 ) = 36 C .
PHYS1004 Exam Solutions
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Semester 2, 2008
E fields goes from positive charge to negative. The upwards field means that the clouds are
negatively charged.
(e)
I=
!Q
35.5
=
= 3.6 "106 A .
6
!t 10 "10
(Total 10 marks)
Question 11
(a)
The electron in a hydrogen atom is initially at the energy level n = 3. It emits a
photon and drops down to the energy level n = 2.
(i)
Calculate the energy of the emitted photon in joules (J).
(ii)
Calculate the wavelength of the emitted photon.
(iii)
Is this radiation hazardous to biological samples? Justify your answer.
(b)
A polonium-210 atom (symbol Po, atomic number Z = 84, mass m = 209.982 848 u)
decays to a lead-206 atom (symbol Pb, atomic number Z = 82, mass m = 205.974 440 u).
(i)
What is the name for this decay process? Write out the reaction.
(ii)
How much energy is released in this decay process? Express your answer in Joules.
You may find the following values useful:
mass of electron
0.000 55 u
mass of proton
1.007 28 u
mass of neutron
1.008 66 u
mass of helium atom
4.002 603 u
(c)
How does this energy compare to the energy of the photon emission from the
hydrogen atom described in (a)? Is the radiation from the decay process hazardous to
biological samples?
(10 marks)
Solution
(a)
Use the formula En =
!13.6
eV
n2
(1 mark)
"1 1#
E photon = ! 13.6 $ ! % = 1.89 eV
&9 4'
= 3.02 (10!19 J
(1 mark)
PHYS1004 Exam Solutions
E photon = h f =
Semester 2, 2008
Page 16
hc
hc
(6.63 #10"34 )(3.00 #108 )
$! =
=
!
E photon
3.02 #10"19
= 6.59 #10"7 m
659 nm
(1 mark)
No this is harmless as it is visible light.
(2 marks)
(b)
(i)
Alpha decay.
(1 mark)
(ii)
210
206
84 Po" 82 Pb + !
or
210
206
4
84 Po ! 82 Pb + 2 He
(1 mark)
(iii)
Use helium mass in place of alpha particle, to account for the two electrons
Q = 209.982 848 u - 205.974 440 u - 4.002 603 u = 0.005 805 u
(1 mark)
Convert to Joules.
Q = (0.005 805 u) (931.49 MeV/u) (1.6 x 10-13 J/MeV) = 8.65 x 10-13 J
(1 mark)
(c)
This energy is over a million times that of the photon emitted from the hydrogen atom. It is
hazardous to biological samples.
(1 mark)
(Total 10 marks)
PHYS1004 Exam Solutions
Page 17
Semester 2, 2008
Question 12
A 75 kg person swallows 6.2 x 1011 nuclei of a beta emitter with a half-life of 5.0 days and a
biological half-life of 20.0 days.
(a)
What is the effective decay constant λeff of the beta emitter inside this person? Express
your answer in becquerels (Bq).
(b)
What is the initial activity R0 in the body?
(c)
What is the activity R in the body after 14 days?
(d)
The average beta particle has an energy of 0.35 MeV and a relative biological
effectiveness (RBE) of 1.5. If 5.0 x 1011 decays occurred inside the body, what is the
equivalent dose? Express your answer in sieverts (Sv).
(10 marks)
Solution
(a)
The nuclear and biological decay constants are
!nuc = (ln 2) / (5.0 d)(24 # 60 # 60 s/d) = 1.6 #10"6 Bq
!bio = (ln 2) / (20.0 d)(24 # 60 # 60 s/d) = 4.0 #10"7 Bq
(1 mark)
The effective decay constant is then
(1 mark)
(b)
Formula: R0 = !nuc N 0
(1 mark for formula; 1 mark for using !nuc and not !eff )
R0 = (1.6 "10!6 Bq)(6.2 "1011 ) = 9.9 "105 Bq
(1 mark)
(c)
Formula: R(t ) = R0 exp("!eff t )
(1 mark)
5
!6
R(14 d) = (9.9 "10 Bq) exp(!2.0 "10 "14 " 24 " 60 " 60)
= 8.9 "104 Bq
(1 mark)
(d)
Energy absorbed:
E = (5.0 "1011 decays)(0.35 MeV/decay)(1.60 "10!13 J/MeV)
= 0.028 J
(1 mark)
Dose: D =
E (0.028)
=
= 3.7 "10!4 Gy
m
75
(1 mark)
!4
!4
Effective dose: H = RBE " D = (1.5)(3.7 "10 Gy) = 5.6 "10 Sv
(1 mark)
(Total 10 marks)