Suggested Solutions (VCAA Physics Unit 4 2008)

[Mao]

...stuff chemistry... I can't be bothered studying:

VCE Physics 2008 Unit 4 Exam
Suggested Solutions

AoS 1 - Electric Power

Question 1 2 marks
the key word here was "through the coil".... I wonder what they want... 5 lines towards the right, or 5 rings [because last year there was something in the ass report on this]. Go with your own interpretations.

Question 2 2 marks
AB

Question 3 2 marks


Question 4 3 marks


Question 5 2 marks
B, the absolute sine graph

Question 6 2 marks
D, the sine graph [AC]

Question 7 2 marks
2.83 V

Question 8 2 marks
the one up one down graph, C

Question 9 3 marks


Question 10 4 marks
Q to P
The net change of flux is out of the page [from into the page --> nil]
Hence by Lenz's law, the induced current would have a flux opposing this, i.e. into the page.
Clockwise flow, Q to P in the square coil.

Question 11 3 marks
48W

Question 12 3 marks
9V

Question 13 3 marks
4.8A

Question 14 2 marks
400 turns

Question 15 3 marks
0.167 A_RMS <--- VCAA didn't put RMS in... lol


AoS 2 - Light and Matter

Question 1 2 marks
Thermal excitation.
Electrons are heated to varying energies, and their thermal motion emits a range of energies as EM radiation, which translates to a continuous range of frequencies, including radiation in the visible spectrum.

Question 2 2 marks
a Mercury Vapour lamp works by photon emissions in electron shells, and its spectrum is discrete [mostly black with lines where discrete energy levels exist within the atom]
in an incandescent light globe, a broad range of EM is emitted, and its spectrum is continuous [no black bands]

Question 3 2 marks
4.5 cm

Question 4 2 marks
Path difference = 3 cm, an exact multiple of the wavelength.
Constructive interference --> antinode --> max intensity

Question 5 2 marks

decreasing d increases w (fringe spacing), hence WXYZ are more spaced-out.

Question 6 3 marks
Graph.
According to my teacher who was an assessor once upon a time, the section below the x axis CANNOT be solid. (because it is an extrapolation, and experimental data can't really be obtained... or can it?)

Question 7 3 marks


Question 8 2 marks
2.5 eV
[negative would not be accepted... i don't think]

Question 9 2 marks
0.0364 nm [now WHY do they want this answer in nm?! escapes me....]

Question 10 2 marks
The spectral line, indicated with the arrow on Figure 3 (), is in the visible region of the spectrum.
What is the energy, in eV, of a photon of this wavelength?
3.03 eV

Question 11 2 marks
n=4 --> n=2

Question 12 2 marks
97 nm

Detailed Study - Sound
please forgive me for not learning the other detailed studies... I am a very lazy person

also - WHY THE HELL WAS THERE SO MANY Bs?!
AND, if you ask me how many marks each question are worth in this, prepared to be epically trout-slap'd.

question 1
B

Question 2
B

Question 3
B

Question 4
C

Question 5
B

Question 6
B

Question 7
C

Question 8
C

Question 9
This is a question under discussion by people who think that "2nd harmonic" does not exist in a close-ended pipe, but in my experience, I have come across two ways of dealing with this, one is where the harmonic refers to how many fundamental frequencies are fitted into it, the other one is just the nth "possible" harmonic.
Another interesting thing was placing signal generator right in front of the pipe, thus making it almost a "closed at both end" pipe... It's possible though, just that it'll be very difficult to hear. But for VCAA's sake, we should just assume that the signal generator magically make the open end a node.
C

Question 10
C

Question 11
A, now WHY OH WHY does anyone want a 8.1 cm tube?! [but the funny thing is, for the wavelength of 0.325 to be the first overtone of a pipe closed at one end, the pipe needs to be about a metre long.... so i'm just going to suggest LEARN2FLUTE]

Question 12
B

Question 13
Contending between C and D.
I chose D, which said "multiple paths" thus causing distortion and sound loss due to interference.
The thing about C that i didn't like is it said "different frequencies will reflect differently". As far as I know, that is true for diffraction [has got to do with wavelengths], but not reflection [which almost always behaves nicely with the angle of reflection equating to the angle of incident]

that makes it, 1 A, 1 D, a lot of Bs and Cs having an orgy.

GG, hope you've done well.

[chid]

Thanks Mao!

Electric power Q15: 0.17A?
Light and matter: was it 5x10^(-15) eVs

[champorado]

For Q15 in electric power, wasn't it P=40W, V=240V?

And how many marks was Q6-8 worth in light and matter?

Thanks.

[Blizz]

thanks man

Wow i actually did ok for multiple
am hoping Mao is wrong for Qu 13 coz i went for C haha (with no real logic, just a random guess b/w B and C really)

Lmao i did shit for the year 11 physics part.. for shame.

[JackOfSpades]

good work. you don't have the blank exam do you?

[Umesh]

q15 electric power .167A?

transformer ideal, 40w power, 240v input, thereofre .167A current?

[BiG DaN]

Thanks Mao!

Electric power Q15: 0.17A?
Light and matter: was it 5x10^(-15) eVs

i agree with both of urs there

[JackOfSpades]

saw your post on the other thread... ignore my question.

[Mao]

Thanks Mao!

Electric power Q15: 0.17A?
Light and matter: was it 5x10^(-15) eVs

yes and yes.

q15 electric power .167A?

transformer ideal, 40w power, 240v input, thereofre .167A current?

yes... I am being a rebel and not wearing glasses (YAY for deteriorating health)... as a result I can't read my own hand-writing hurriedly scribbled on the formula sheet. ahhh =]

champorado - I will update with mark allocations.
ADDED. DONE.

[Umesh]

lol

[vacuous]

Thanks Mao.

Question 12 in AOS 2 I did differently.

wavelength = hc/E = 4.14e-15 * 3e8 / .7 = 1774 nm

How did you get 97 nm?

[Blizz]

Thanks Mao.

Question 12 in AOS 2 I did differently.

wavelength = hc/E = 4.14e-15 * 3e8 / .7 = 1774 nm

How did you get 97 nm?

n=4 to ground zero gives you a smaller wavelength

[jamestaR`]

I thought the last one was C, becuase it included the destructive interference resulting in cancellation of sound and therefore apparent distortion. The bit about "different freqs reflect differently" is just saying that theres a lot of freqs hitting the wall and bouncing back to superimpose...

but i could be wrong

[Fidgey]

Thanks Mao.

Question 12 in AOS 2 I did differently.

wavelength = hc/E = 4.14e-15 * 3e8 / .7 = 1774 nm

How did you get 97 nm?

i got 1774... i thought it said with the smallest energy... not wavelength

[charm]

Thanks Mao.

Question 12 in AOS 2 I did differently.

wavelength = hc/E = 4.14e-15 * 3e8 / .7 = 1774 nm

How did you get 97 nm?

i got 1774... i thought it said with the smallest energy... not wavelength

oh...i think it said "what is the shortest possible wavelength" so i got 97 nm using the highest energy.