PHYS1004 PHYSICS 1 (ENVIRONMENTAL & LIFE ...sydney.edu.au/.../jphys/exams/1004_exam_solutions_2008.pdfPHYS1004 Exam Solutions Semester 2, 2008 Page 1 THE UNIVERSITY OF SYDNEY PHYS1004

PHYS1004 Exam Solutions Semester 2, 2008 Page 1

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 Density of water ρ = 1.00 × 103 kg.m−3

Density of air ρ = 1.20 kg.m−3

Atmospheric pressure 1 atm = 1.01 × 105 Pa Magnitude of local gravitational field g = 9.81 m.s-2

Avogadro constant NA = 6.022 × 1023 mol-1

Permittivity of free space ε0 = 8.854 × 10−12 F.m−1

Coulomb’s law constant K = 8.99 × 109 N.m2.C-2

Permeability of free space µ0 = 4π × 10−7 T.m.A−1

Elementary charge e = 1.602 × 10−19 C

Speed of light in vacuum c = 2.998 × 108 m.s−1

Planck constant h = 6.626 × 10−34 J.s

Rest mass of an electron me = 9.110 × 10−31 kg

Rest mass of a neutron mn = 1.675 × 10−27 kg

Rest mass of a proton mp = 1.673 × 10−27 kg

Rest mass of a hydrogen atom mH = 1.674 × 10−27 kg

Boltzmann constant k = 1.381 × 10−23 J.K-1

Stefan-Boltzmann constant σ = 5.671 × 10-8 W. m -2 K-4

Atomic mass unit u = 1.661 × 10−27 kg

Energy equivalent of 1 atomic mass unit E = 931.49 MeV


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 48.00 10 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,

0

0

410

3 2

10

8.00 10 2001.55 10 Pa

(12.5 10 ) 2.1

= 1.6 10 Pa

F LY

A L

LFY

A L

! "

#=

$ =#

%= = %

%

%

(2 marks)


(b)

4

8

3 2

(8.00 10Stress _ max 1.6 10 Pa

(12.5 10 )

F

A ! "

#= = = #

#

0

2.1Strain _ max 0.011

200

L

L

!= = =

(3 marks)

(Total 5 marks)


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) 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.

(b) 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

2 /i op p r!" =

where ip and

op are the internal and external pressures respectively, ! is the surface tension,

and r is the radius.

This shows that as r decreases (breathing out) i op p! increases further, which could lead to

further air being expelled until 0r! . Similarly, if i op p! is sufficient to cause r to

increase (breathing in) then the required i op p! 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)


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

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)


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)


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

max.K 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

max

34 14 19

19

(6.626 10 )(5.0 10 ) 1.602 10 J

1.71 10 J

1.1eV

K h! "

# #

#

= #

= $ $ # $

% $

%

The minimum (threshold) frequency to produce electrons corresponds to max

0K = and is 19

14

min 34

1.602 10 J2.42 10 Hz

6.626 10 J.sh

!"

#

#

$= = % $

$.

Since the frequency 142.0 10 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)


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

3

6

2.90 10 m.K

2.90 109.73 10 m

298

m

m

T!

!

"

""

= #

#$ = = #

(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:

34 820

6

20

19

(6.63 10 )(3.0 10 )2.04 10 J

9.73 10

2.04 100.13 eV

1.6 10

m

hcE

!

""

"

"

"

# #= = = #

#

#= =

#

Correct answer is 0.1 eV.

(1 mark)

(Total 5 marks)


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 3 -31.00 10 kg.m! . 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,

1 2p p= . 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

2 2

1 1 1 2 2 2

1 1

2 2B B B B

p v g y p v g y! ! ! !+ + = + + .

But 1 2

0v v= = and 1 2y y= and so

1 2p p= (a streamline could join points 1 and 2).

(2 marks)

(b) No,

3 4p p! . Pressure

4p is atmospheric pressure since it is at the surface of the liquid,

whereas 3 atm Ap p g d!= + , where 20 14 cmd = ! . (The argument that pressures are equal at


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:

1 (0.200 0.100)atm Ap p g!= + "

and

2 (0.140 0.100)atm B

p p g!= + "

These are equal, so

3 3 3

(0.100) (0.040)

0.100(1.00 10 )(2.5) 2.5 10 kg.m

0.040

A B

B A

! !

! ! "

=

# = = $ = $

(4 marks) (d) Gauge pressure is pressure above atmospheric, so

G atmp p p g d!= " = Use this for the right hand arm since it has only a single liquid.

2 3

,5 (14.0 4.0) 10 2.45 10 PaG Bp g! "= " # = #

(Total 10 marks)

Question 8

The average flow rate of blood in a certain artery is 11.3 mL.s

! ( 6 31mL 10 m

!= ). The artery

has a radius of 2.2 mm . The viscosity of blood at body temperature is 32.1 10 Pa.s

!" and its

density is 3 31.1 10 kg.m!" .

(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


(a) Volume flow rate is

avgQ Av= where A is cross-section area, and avgv is average flow

velocity. So 6

1

3 2

1.3 100.085 m.s

(2.2 10 )avg

Qv

A !

""

"

#= = =

#.

(2 marks) (b) Evaluating the Reynolds Number for the flow

R

v dN

!

"=

we get 3 3

3

(1.1 10 )(0.085)(2 2.2 10 )

2.1 10

200 (197)

RN

!

!

" " "=

"

=

This is much less than 2000, so the flow is expected to be laminar (streamline). (3 marks)

(c) Using Poiseuille’s law for flow of a viscous liquid

4

4

3 6

3 4

8

8

(8)(2.1 10 )(0.080)(1.3 10 )

(2.2 10 )

24 Pa

R p LQQ p

L R

! "

" !

!

# #

#

$= % $ =

& &=

&

=

(3 marks) (d) Power is pQ! for viscous flow. So

6

5

(24)(1.3 10 )

3.1 10

Power

W

!

!

= "

= "

(2 marks)

(Total 10 marks)


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 " i2r .

(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, maxP , 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 ( ) 0V R

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

maxV != (Note

max0V ! ).

(2 marks)

ε 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

2 2

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

2 2

2 4

4

P i i r rr r

r r

r

! !! !

! !

!

" # " #= $ = $% & % &

' ( ' (

= $

=

(2 marks)

(Total 10 marks)

Power = 0 Power = 0


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 6 1

10 V.m! 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)

6 8(10 )(500) 5 10 VV

E V E dd

!= "! = = = #

(b)

1280 (8.85 10 )(2000 2000)

7.1 10 F500

AC

d

! ""# #

= = #= .

(c)

( )2 8 8 2 91(0.5)(7.1 10 )(5 10 ) 8.9 10 J

2C

U C V!= " = # # = # .

(d)

8 8(7.1 10 )(5 10 ) 36 CCQ C V != " " == # .


E fields goes from positive charge to negative. The upwards field means that the clouds are negatively charged. (e)

6

6

35.53.6 10

10 10

QI A

t

!= = = "! "

.

(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 2

13.6eV

nE

n

!=

(1 mark)

19

1 113.6 1.89 eV

9 4

3.02 10 J

photonE

!

" #= ! ! =$ %

& '

= (

(1 mark)


34 8

19

7

(6.63 10 )(3.00 10 )

3.02 10

6.59 10 m

659 nm

photon

photon

h c hcE h f

E!

!

"

"

"

# #= = $ = =

#

= #

(1 mark) No this is harmless as it is visible light.

(2 marks) (b) (i) Alpha decay.

(1 mark) (ii)

!+" PbPo206

82

210

84 or 210 206 4

84 82 2Po Pb 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)


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

6

7

(ln 2) / (5.0d)(24 60 60 s/d) 1.6 10 Bq

(ln 2) / (20.0d)(24 60 60 s/d) 4.0 10 Bq

nuc

bio

!

!

"

"

= # # = #

= # # = #

(1 mark) The effective decay constant is then

(1 mark) (b) Formula:

0 0nucR N!=

(1 mark for formula; 1 mark for using nuc! and not eff! )

6 11 5

0 (1.6 10 Bq)(6.2 10 ) 9.9 10 BqR!

= " " = " (1 mark)

(c) Formula: 0( ) exp( )effR t R t!= "

(1 mark) 5 6

4

(14d) (9.9 10 Bq) exp( 2.0 10 14 24 60 60)

8.9 10 Bq

R!

= " ! " " " " "

= "

(1 mark) (d) Energy absorbed:

11 13(5.0 10 decays)(0.35 MeV/decay)(1.60 10 J/MeV)

0.028 J

E!

= " "

=

(1 mark)

Dose: 4(0.028)3.7 10 Gy

75

ED

m

!= = = "

(1 mark) Effective dose: 4 4(1.5)(3.7 10 Gy) 5.6 10 SvH RBE D

! != " = " = "

(1 mark)

(Total 10 marks)

PHYS1004 PHYSICS 1 (ENVIRONMENTAL & LIFE ...sydney.edu.au/.../jphys/exams/1004_exam_solutions_2008.pdfPHYS1004 Exam Solutions Semester 2, 2008 Page 1 THE UNIVERSITY OF SYDNEY PHYS1004

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