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Want to compare multiple-choice answers? (1 Viewer)

timkaldor

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For q15 you use E=V/d then c=fwavelength to find frequency of 1.6 then follow the graph up to 2.5 then see that its aluminium. hence A
 

Lions_Fist

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shinji said:
yay another person who got 8. a .. lol

i made n educated gues at 9

how did u work out 15?

also i disagree with #7 .. i think it 's c. lol
15 was just working out off the frequency from the wavelength, then applying it to the graph.
 

Mumma

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timkaldor said:
for no. 9 i thought of Lenz's law and i didnt think any current would be produced just eddy currents in the disc
The disc is connected to a circuit...
 

Irskin

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1. C - g varies according to Gm/r2 and G is the universal gravitation constant

2. B - when the centripetal force of the string is removed, the mass will move horizontally perpendicular to its centre of motion.

3. D - M-M experiment shows the constancy of speed of light which provides evidence for Einstein's theory of relativity

4. B - The only acceleration on a projectile is gravity (g) which remains constant

5. D - Using Kepler's ratios

6. A - The wire will rotate around the magnet in a clockwise direction (i think).

7. C - Using F = BIlsin@ where l is route 1.40 squared

8. A - Im not sure whether flux is maximum or minimum at 0 (possibly B)

9. B - Using right hand palm rule where force is in opposite direction of motion of disc. Therfore current runs from Y to X.

10. A - Not sure about this, something about changing magnetic fields

11. A - Braggs used X-ray crystallography/diffraction

12. C - F = qvBsin@ and F = 0 when velocity is parallel

13. D - Metal lattice does not vibrate as frequently so resuistance is decreased

14. C - Aliminum; E = 5000 Vm-1. So we need E = 10000. This occurs for both perspex and copper, but only copper is suitable

15. D - Im not sure on this one either. Using the wavelength we find that the freqeuncy corresponds to maximum KE for aluminium but work function has to be factored in to find maximum KE. ie. KE = hf - work function of metal



Let me know what you think of these answers. Im hoping for 12-13/15
 

shinji

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lol, i found out the frequency to be 1.6x10^15

but didn't know where to go after that. haha
 

outlawjo3

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I think i got the answer to question 8 --- B

For proof go to the 2002 HSC....QS 10

Very similar and the answer is B

Keep in mind that induced emf is proportional to the rate of change of magnetic flux. Therefore the question is essentially the same

Look where the coil is perpendular to the field

From that...the answer is B
 

shinji

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@ Irskin, i agree with all ur answers besides 10 and 15. lol
=D

woot! ^^
 

philthy

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I had

8. B - Because torque is cos@, so starts at 0
9. D - Because induction = AC
10. B - Because i thought it was a DC circuit, being switched on and off hence a current was produced
 

Mumma

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8.A - because maximum FLUX goes through the loop when its perpendicular to field

9.B - Not too sure, but no way is it D? The conductor is always cutting flux at different parts, but its always at the same rate, so it had to be DC...

10.A Shouldn't the voltage be OPPOSITE? (ie. negative) Because of Lenz Law? ALSO, the powersource seems to be a battery, provides constant current.

1 A
2 B
3 D
4 B
5 D
6 A
7 B
8 A
9 B
10 A
11 A
12 C
13 D
14 C
15 A
 

matty fwd

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8 B ... its talking about the plane of the loop, not an individual side of the loop..it came up in a past paper
 

matty fwd

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ianc said:
Wasn't 7 C because you had do pythag to get the length of the wire, not just use the 0.4m?
i did this...but my answer didn't match C....anyone else have this issue?
 

Bricktop

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I'm not sure if I'm right or not, but I think a lot of you are thinking of TORQUE instead of FLUX for question 8. If the coil is in the position shown at t=0, that's the position of maximum magnetic flux; cos curve.

I put A. I hope that's right anyway because I'm not doing too well in multiple choice from what I can tell haha.
 

shinji

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ianc said:
Wasn't 7 C because you had do pythag to get the length of the wire, not just use the 0.4m?
yep. tht's what i did. =]
 

timkaldor

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8 is A I read the question wrong.

9 is A because only eddy currents are produced (not so confidant about this)

10 is A because a battery is a DC power source so there will only be a change in flux when it is switched on and off. So there is a current in X the whole time but only a current in Y at the beginning and the end cause that is when there is a change in flux
 
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XcarvengerX

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This is what I got with a small analysis. Around 6 mistakes I think.
1. C (Good)
2. B (Tangential velocity VS centripetal force)
3. D (The others are quite incorrect)
4. B (Acceleration does not change in horizontal motion)
5. B (I know it is D. I use T=2piR/v which is incorrect. You should use Kepler's Third Law)
6. A (North meet with north, so the wire will just rotate around)
7. B (May be C if you use Pythagoras to calculate the diagonal length. Can also be D. Not sure)
8. B (I know it is A. Because it is FLUX not EMF)
9. B (Can be D if somehow it is an AC, but still think it is DC. Or A if no current is produced. C is for path of electron (opposite direction from current) so pretty sure C is incorrect)
10. B (I thought because the coil is coiled with some space, so there is 0 volt... How wrong I was. Yes, I know it is either A or C. And I think now that it is A because the secondary in transformer has to be AC)
11. A (Surprisingly easy question)
12. C (Not D because when perpendicular, it means 0 degree and sin 0 is 0 thus affect the magnetic force on particle. When parallel, it means 90 degrees and sin 90 is 1 and anything times 1 is the same, thus NOT affect the magnetic force on particle)
13. D (It decreases and B is incorrect)
14. D (C? A? Don't know... B is incorrect though as persplex does not conduct electricity)
15. D (Stupid of me. It is A for sure. Use c = f times lambda, so to find f, just divide the bloody thing and you get around 1.6 times 10 to the power of 15 which according to graph at 2.5 eV belongs to Aluminium. If you want to know what I did, I times something with Planck's constant and charge of electron)

That's it. Feel free to object any of my comments.
 
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