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THE UNIVERSITY OF SYDNEY
PHYS1003 – PHYSICS 1 (TECHNOLOGICAL)
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
Density of water ρ = 1.00 × 10
3
kg.m
− 3
Density of air ρ = 1.20 kg.m
− 3
Atmospheric pressure 1 atm = 1.01 × 10
5
Pa
Magnitude of local gravitational field g = 9.81 m.s
Avogadro constant N
A
= 6.022 × 10
23
mol
Permittivity of free space ε
0
= 8.854 × 10
− 12
F.m
− 1
Permeability of free space μ
0
= 4 π × 10
− 7
T.m.A
− 1
Elementary charge e = 1.602 × 10
− 19
C
Speed of light in vacuum c = 2.998 × 10
8
m.s
− 1
Planck constant h = 6.626 × 10
− 34
J.s
Rest mass of an electron m
e
= 9.110 × 10
− 31 kg
Rest mass of a neutron m
n
= 1.675 × 10
− 27
kg
Rest mass of a proton m
p
= 1.673 × 10
− 27
kg
Rest mass of a hydrogen atom m
H
= 1.674 × 10
− 27
kg
Boltzmann constant k = 1.381 × 10
− 23 J.K
Atomic mass unit u = 1.661 × 10
− 27
kg
Rydberg constant R = 1.097 × 10
7
m
SECTION A
Question 1
Water is flowing in a horizontal pipe due to a pressure difference between the ends of the pipe
created by a pump (see diagram). A student measures flow rate, Q , from the pipe as a
function of the pressure difference,! p , between the ends of the pipe. She plots her results
and finds that there is a change in the character of the flow over the range of pressure used in
her experiment.
(a) Draw a sketch showing typical results for flow rate (vertical axis) against the pressure
difference (horizontal axis). On your sketch show the regions of different type of fluid
flow behaviour.
(b) Describe the characteristics of the different types of fluid flow that were found in her
experiment.
(c) The experiment is repeated for the same range of pressure difference but using oil
rather than water. The oil used had a much larger viscosity than water but is similar in
density. Explain the effect of using oil on
(i) the flow rate,
(ii) the transition between the different types of fluid flow behaviour.
(5 marks)
Solution
(a)
(2 marks for diagram showing regimes and transition in correct direction)
Question 2
St Elmo’s fire is a bluish flickering glow sometimes seen at tips of the masts of sailing ships
when storm clouds are overhead and especially when the masts are wet. The tips of the masts
appear to be ‘on fire’ but don’t burn. Saint Erasmus (also known as Saint Elmo) became the
patron saint of sailors who regarded the blue glow at their mastheads as a sign of Saint
Erasmus’ protection and thus a good omen. Unfortunately for the sailors, the appearance of
St Elmo’s fire actually indicates that lightning is more likely to strike the ship.
(a) What is the glow around the tip of the mast?
(b) Why is the glow most likely to be seen around the tips of the masts?
(c) Why is the mast the part of the ship most likely to be hit by lightning?
(5 marks)
Solution
This is an example of a “glow discharge” or “coronal discharge”. The origin of St Elmo’s fire
is closely related to the physics behind lightning rods.
(a)
Where the field exceeds ~3 x 10
6
V.m
, the air can spontaneously break down as electrons are
ejected from their atoms, in turn ejecting further electrons.
(Explain breakdown in air – 1 mark)
These electrons later re-combine with their atoms, which emit light at characteristic
wavelengths. In the nitrogen/oxygen mixture of air this colour is characteristically blue/violet
- as observed in St Elmo’s fire.
(Explain why light is emitted – 1 mark)
(b)
During a thunderstorm, the electric field in the atmosphere is ~
6
V.m
clear weather value of ~100 V.m
- 1 . This field attracts charge to high places – closer to the
opposite charges in the clouds. In particular, charge accumulates in pointy conductive objects
- trees, church steeples, lightning rods – or the masts (or yards) of sailing ships.
(Charge accumulation at points – 1 mark)
Since the charge density is highest at these places, the electric field is also high there (since E
~ σ/ε o
, where σ is the charge density).
(Field highest at points – 1 mark)
(c)
The electric field is highest at the top of the masts, making them the most likely places for
‘risers’ to move upwards to join the downward path of charge from the cloud.
(1 mark)
(Total 5 marks)
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
F
B
= F
centripetal
q vB =
mv
2
r
(1 mark)
hence,
r =
mv
q B
Period is then the circumference of the ‘orbit’ divided by the velocity
T =
2! r
v
2! m
qB
(Correct procedure – 1 mark)
(b)
Sinusoidal graph as below - also accept a decaying sine curve which would result if the coil
has some resistance.
max
T 2 L C
Q
i
L C
(Graph 1 mark; correct labels on graph 1 mark)
(c)
The circuit oscillates at the resonant frequency. Energy stored in the capacitor’s electric field
when fully charged is stored in the inductor’s magnetic field when the capacitor is discharged
and current is maximum.
(1 mark)
(Total 5 marks)
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 K max
described by
max
K = h! #".
(a) What is the maximum kinetic energy of photo-electrons ejected when the light has
frequency ν = 5.0 × 10
14
Hz?
(b) What is the maximum kinetic energy of ejected electrons when the metal is
illuminated with light with frequency ν = 2.0 × 10
14
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
max
34 14 19
19
( 6. 626 10 )( 5. 0 10 ) 1. 602 10 J
1. 71 10 J
- 1 eV
K h! "
The minimum (threshold) frequency to produce electrons corresponds to max
K = 0 and is
19
14
min (^34)
1. 602 10 J
- 42 10 Hz
h 6. 626 10 J.s
Since the frequency
14
! = 2. 0 " 10 Hzis 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)
Question 7
A Venturi meter (used to measure flow rate in pipes) is installed in a horizontal water pipe, as
shown in the diagram. The pipe has circular cross-section, with diameter D 1
in the first
segment and D 2
in the second segment, with D 2
< D
1
. The density of water is ρ. The volume
flow rate of the water in the pipe is R (measured in m
3
.s
).
(a) Express the speed of flow v 1
in the first section of pipe in terms of R and 1
D.
(b) What is the speed of flow v 2
in the second section of pipe?
(c) Using the result in (a) find an expression for the pressure difference between point 1
and point 2.
(d) What is the difference h in the water level in the two tubes expressed in terms of R ,
1
D , and 2
D?
(e) Suppose the volume flow rate through the pipe is
3 3 1
- 0 10 m .s
!!
" and the pipe has
diameter 0.1 0 m at point 1. Is the flow through the pipe turbulent? Note that the
density of water is 1000 kg.m
and the viscosity of water is
3
- 0 10 Pa.s
!
".
(f) What change could you make to the system so that the flow is exactly on the boundary
between turbulent and laminar flow?
(10 marks)
Solution
(a)
By the equation of continuity, the volume flow rate is constant, so
1 1 2 2
A v = A v = R
where
2
D
A!
is the cross-sectional area of the pipe and v is the speed of the flow. Hence
(^1 )
1 1
R 4 R
v
A! D
( 2 marks)
(b)
Similarly,
(^2 )
2 2
R 4 R
v
A! D
(1 mark)
(c)
Apply Bernoulli’s equation between points 1 and 2 to find the pressures 1
P and 2
P in the large
and small segments. They are on the same streamline, so Bernoulli’s equation applies; they
are at the same height so 1 2
y = y = 0. Hence from Bernoulli’s equation,
2 2
1 1 2 2
P +! v = P +! v
so
( )
2 2
1 2 2 1
" P = P # P =! v # v
or
4
(^2 )
(^1 )
2
D
v
D
(2 marks)
(d)
By Pascal’s law, the pressure P at the base of each tube is equal to the pressure exerted by the
column of water. The tops are open to the air, so the pressure at the surface is atmospheric
pressure 0
P.
1 0 1
P = P +! g y
and
2 0 2
P = P +! g y
Hence
2 2
1 2 2 1
1 2
P P P v v
h y y
! g! g g
From the results for parts (a) and (b) this can be written as
2
2 4 2
2 1
8 R 1 1
h
! g D D
or
2 4
1 1
4
2
v D
g D
(3 marks)
(e)
If
3 3 - 1
R 10 m .s
!
= then from part (a)
3
(^1 2 )
1
- 127 m.s
R
v
! D!
"
Question 8
A conducting sphere of radius R 1
has a charge of +Q. It is surrounded by a larger conducting
concentric spherical shell of inner radius R 2
and outer radius R 3
. The shell has charge – 2 Q.
(a) With the aid of a diagram describe how charge is arranged on the sphere and on the
spherical shell.
(b) Find expressions for the magnitude of the electric field as a function of the distance r
from the centre and give its direction:
(i) inside the sphere;
(ii) between the sphere and shell;
(iii) inside the shell;
(iv) outside the shell.
Explain your conclusions.
(c) Plot the magnitude of electric field as a function of r. Label the axes on your graph and
mark in R 1
, R
2
and R 3
(10 marks)
Solution
(a)
Electric field inside a conductor must be zero
(½ mark)
A charge of +Q is located on the outer surface of the conducting sphere.
(½ mark)
A charge of - Q is located on the inside surface of the conducting shell.
(½ mark)
A charge of - Q is located on the outer surface of the conducting shell.
(½ mark)
(b)
Gauss’s Law (^) E. dA =
q enclosed
0
"
(i) E = 0 because enclosed charge is zero; electrostatic E always zero inside a conductor.
(1 mark)
(ii) E =
0
Q
r
2
outward from the positive charges from Gauss’ Law since there is a net
positive charge enclosed; field is radial because of symmetry.
(2 marks)
(iii) E = 0 because net enclosed charge is zero; electrostatic E always zero inside a
conductor.
(1 mark)
(iv) E =
0
Q
r
2
inward to the negative charges from Gauss’ Law since there is a net
negative charge enclosed; field is radial because of symmetry.
(2 marks)
Students must give explanations to justify their answers to get full marks, e.g. in terms of
Gauss’ Law, symmetry and charge distribution.
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.
(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 ( (^) i max
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 max
( V ) 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, max
P , 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 i max
!
r
and
V R 0
r
(2 marks)
(c)
Maximum potential difference
V
max
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 max
V =! (Note
max
V! 0 ).
(2 marks)
R
r
battery
V
i
(d)
( )
2
P = V i =! " i r i =! i " i r
(1 mark)
(e)
(shape of curve 1 mark), (labelled axes and power=0 points 1 mark)
(f)
Maximum power occurs at a current
max
i
r
=. This can be obtained by the differential, from
the properties of a parabola in the graph, or a sensible guess (mark as correct). The power is
then
2
2
2 2
P i i r r
r r
r r
r
(2 marks)
(Total 10 marks)
Power = 0 Power = 0
(b)
A similar argument but with ( )
2
2
r
i r i
R
= inside the cylinder
(1 mark)
yields ( )
0
2
2
u i r
B r R
! R
(1 mark)
(c)
Use ( )
0
2
2
u i r
B r R
! R
< = and u =
B
2
2 μ o
So at any point within the cylinder the energy density is
2 2
2 4
8
o
i r
du
R
μ
Consider a thin cylindrical shell of radius r as in the above diagram. The volume of this
cylindrical shell is
dV = ( 2! r ) dr L
Then the energy within this volume element is
2 2
2 4
2
3
4
o
o
dU du dV
i r
r L dr
R
i L
r dr
R
μ
μ
Integrate over the radius of the cylinder
2 3 4 2 3 4
(^2 )
4
0
2
0
0
o
o
r R
o
r
o
r R
r
r R
r
U dU
i L
r dr
R
i L
r dr
R
i L r
R
i L
μ! μ! μ! μ!
=
=
=
=
=
=
.
.
.
As required, this result is independent of R.
(procedure 3 marks)
(result 1 mark)
(Total 10 marks)
