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helper

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15 is going to be interesting to see what the board comes up with:

B, which will occur at the instant you apply the field. Similar to Q 8.
C If you take it to the level Who loves maths takes it.
D when you start going into the hall effect and relative mobility.
 

frenzal_dude

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HotShot said:
its ac current, plus others r not sine waves, also when the switch is closed current flows, so that means there is some current. and why would it drop after sometime?
its not ac, coz its a battery, and batteries are dc, yeh so what if current flows when the switch is closed, there is still only an instant of current flowing in the secondary coil due to the intial change in flux, but then it stops changing so current drops 2 0, therefore the answer is c.
 

haboozin

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DigitalFortress said:
DC current cannot be transformed so i'm thinking its the one with 0 current.
hey, yea your right, i thought the same thing and actually put A.

but it says "whent he switch is being closed"
and if you remember that is the first experiment which showed the motor effect, how as the switch is being closed and opened a current was induced.
 

who_loves_maths

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Originally Posted by helper
15 is going to be interesting to see what the board comes up with:
B, which will occur at the instant you apply the field. Similar to Q 8.
C If you take it to the level Who loves maths takes it.
D when you start going into the hall effect and relative mobility.
err... where did you come up with that?

first of all, it cannot be D, because that is the direct opposite of C (since they are either same side or opposite), and i have explained with evidence why it would be C and not D.

second of all, it cannot be B. the magnetic field is applied perpendicular across the direction of current - the force induced is also perpendicular to the direction of current.
so the accelerations are pendicular to each other - and you know that motion can be broken into two components at 90 degrees to each other.
hence, the magnetic force experienced on both holes and electrons do not act to "slow" or "fasten" them up, they act at 90 degrees to each other!
the electrons' and holes' procession through the conductor is thus unaffected.


BoardOS do not formulate exams in which the answers are ambiguous or the question "tricky" - they only test you on the material in the course and nothing else.
the answers to these question are not made up after the exam. they have already been decided months before the HSC.
in fact, the HSC papers all go through a rigorous system of testing and editing months before the actual HSC to try to minimise any possibilities of ambiguity or error.
so the Board won't be "tossing up" between B, C, or D. a definite answer already exists. and certainly not all three answer are correct.
 
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Look at me I can bold all my words and sound like a smart arse too!


The BoardOS do ask ambiguous questions so plz stfu, good examples are multiple choice questions, e.g 2003 Q5
Where there is "no" answer, or more than 1 answer (2002 IPT)
 

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Electrons and holes aren't going to move to the same side. The difference is holes aren't physical objects, and when "holes move" it's actually because of electrons moving in the other direction. Since electrons are the only things here that actually have a charge (discounting protons since they're fixed in position), they're the only things that are going to experience a force. When they move to one side of the conductor, it follows that holes are going to be left over on the other side.
 

helper

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who_loves_maths said:
first of all, it cannot be D, because that is the direct opposite of C (since they are either same side or opposite), and i have explained with evidence why it would be C and not D.
Go and read up on the Hall effect and relative movements.
http://www.warwick.ac.uk/~phsbm/qhe.htm

second of all, it cannot be B. the magnetic field is applied perpendicular across the direction of current - the force induced is also perpendicular to the direction of current.
Changing flux. Results in an induced EMF to oppose the change. That is going to be a back EMF. That will reduce the current by slowing down the charge flow. This will only occur for a fraction of a second but does occur when the magnetic field is first applied. Which is what the question asked about?

Which also goes on from yours. The sideways movement produces a second force, which will oppose the original force and slow down there progress along the semiconductor. This will not change the speed but will reduce the current.

This I feel is the least likely answer they were after but is a direct test of the concept of back EMF.


BoardOS do not formulate exams in which the answers are ambiguous or the question "tricky" - they only test you on the material in the course and nothing else.
the answers to these question are not made up after the exam. they have already been decided months before the HSC.
in fact, the HSC papers all go through a rigorous system of testing and editing months before the actual HSC to try to minimise any possibilities of ambiguity or error.
Which is why marking schemes change and multiple choice answers are changed after the exam because of people reading what they think is there.

You have a lot more faith in the examiners than me, since there are controversial questions every year that go on your depth of knowledge.

Eg Time dilation in 2003, Braking Effect in 2004.
Testing information removed from the syllabus. Eg Solar cell question in 2004

so the Board won't be "tossing up" between B, C, or D. a definite answer already exists. and certainly not all three answer are correct.
Yes they will have an answer they were expecting but that doesn't always mean it is correct or won't be questioned for valid reasons.
 
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leigh_cummins

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Broodchuck, do you know what you're on about?

Broodchuck said:
I admit it, I am sorry, I don't do three unit maths.



Yes, your right, batteries work on DC current but why are you talking about that? If your commenting about my answer for question 15 then there is never a mention of AC or DC.

Now, voltage and current are always proportional to each other, so if your correct, then the right answer would be the exact inverse of that graph. now, there are no other graphs that like the inverse of that graph... so i must conclude that you are wrong.



Think about it...


SINCE WHEN ARE VOLTAGE AND CURRENT PROPORTIONAL TO EACH OTHER??!! One of the simplest formulae we are given is 'P=VI'. Which means that if power is going to remain constant within a system, and one of the components, say voltage, is increased, the other must decrease to compensate!!! The same is true for an increase in current. So in fact, it is everyone else who must conclude that YOU are wrong!
 

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DigitalFortress said:
Look at me I can bold all my words and sound like a smart arse too!


The BoardOS do ask ambiguous questions so plz stfu, good examples are multiple choice questions, e.g 2003 Q5
Where there is "no" answer, or more than 1 answer (2002 IPT)
is that smart arse statement required??? :rolleyes: look if he is rong, prove him wrong, dont just go flaming people. Yes i believe wlm is rong for once :p, but u just dont flame him, just say why he is rong.

PS. to add to helpers comments, they do change the marking scheme, eg last years AC motor question, a teacher at my school is a HSC marker, and he said for that question, they basically accpeted anything.
 

who_loves_maths

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Originally Posted by Dumsum
Electrons and holes aren't going to move to the same side. The difference is holes aren't physical objects, and when "holes move" it's actually because of electrons moving in the other direction. Since electrons are the only things here that actually have a charge (discounting protons since they're fixed in position), they're the only things that are going to experience a force. When they move to one side of the conductor, it follows that holes are going to be left over on the other side.
actually, you will find that the whole point of the syllabus and Solid State Physics in general that we learn at the HSC level is to accept that "holes" are actually real physical objects.
that is why in textbooks that tell you that holes are positively charged! and not neutral.
that is also why in textbooks they tell you that holes are the major current carriers in P-type semiconductors - being a major current carrier, they cannot be "unreal", since otherwise the books would mention that electrons are major carriers in both types - which they are not.

the syllabus is trying to get us to accept that holes are real objects. and the HSC exam questions are set accordingly. if one does not understand that point, then it goes to explain why one would get such a question incorrect in an HSC exam.

also, your concept of the "hole" being imaginary just because it is created due to movement of electrons is equivalent to reasoning that entities such as quarks, or even anti-matter, which have not been "seen", do not exist and that they are imaginary.
{eg. when a virtual matter particle is created in a vacuum, its antimatter conterpart is equivalent generated - just like electron movement induces holes}

yet somehow, i feel that most ppl doing the Quanta to Quarks option get the idea from the books that these "fundamental" particles are real and do exist, even though there is no conceptual difference between them in relation to your argument of holes and electrons...


and to the others such as DigitalFortress, helper, and xrtzx: i won't continue to argue my point because 1) i have argued it already, and 2) i won't be able to change your individual opinions anyways.
so atm, we may simple agree to disagree :)
we will see what happens in due course i'm sure...

P.S. in relation to the Hall Effect helper - it is strictly not taught in the physics syllabus. so by considering that, you are digging a grave for yourself in strigent relation to HSC physics exams.
once again, i emphasise my point that the HSC is not set to trick ppl - if the question needs to take the Hall Effect into account in order to produce a correct answer, then that question would not even be in the exam!
the fact of the matter is, it is readily solvable in an unambiguous fashion using all that is included in the syllabus - and that is as i have already outlined in my previous explanations of this question...
 

Dumsum

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The syllabus mentions them as "positive charge carriers," yes, but it's open to interpretation. It also talks about holes being defined as an "absence of electrons in a nearly full band," which to me states that, under the syllabus, holes do not physically exist.
 

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haboozin said:
hey, yea your right, i thought the same thing and actually put A.

but it says "whent he switch is being closed"
and if you remember that is the first experiment which showed the motor effect, how as the switch is being closed and opened a current was induced.
maybe you should go back and read the question again and stop misquoting it, the question actually says "which of the following graphs best show the current flow in the galvanometer when the swith is closed?"

dont just add teh word "being" in to try and prove your point, all it proves is that you can't read
 

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Haboozin is interpreting it the way it has been interpreted in past HSC questions have been interpreted.
 

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My answers with explanations

Hi guys!

Firstly i would like to say, AHAHAHAHA LOL this is the funniest forum thread i have ever read, some of you are hilarious, so direct, 'if you picked this or this you're an idiot'.....ehehe.

Anyway, I tried to read the whole thing as thoroughly as I could. And I would like to add my 2c.

Being my first post, before i start i would just like to add, physics is my favourite subject and probably my best, i certainly HATE English and thus the board of studies, but I never have to do it again, until it bites me back in the butt when i get my results back. I was so enthusiastic about physics in year 11, but then got glandular fever during the transition of yr11 to yr12. It screwed me up bigtime, major fatigue. In every class, most of the time i just rested on the desk usually with a headache, i tried to take things in as much as i could. When it came down to studying, well i did very small amounts, it was so hard. Being an insomniac doesn't help either. I need a UAI of 75 to get into electrical engineering at UTS. Me being too tired I didnt do that misadventure thing or whatever to get more marks because of damn glandular, and I thought I could get at least 75 without it. I just have to wait and see...


Anyway, what I'll do is for every question explain as simply as possible why I think the answer is whatever it may be and also try to explain why i dont think the other 3 are right. Well for most of the Q's anyway, some i had no idea, like all of the history questions, yeah couldnt have been worse doing quanta to quarks. In blue ill put the answer/s i thought were right, I didnt necessarily choose those in the exam, if i put the wrong one ill say so next to the one.


Q1. A

a) The velocity changes since its a composition of both X and Y and since Y changes, the overall velocity changes.

b) The acceleration is always g = 9.8m/s/s, since its parabolic, it must be.

c) Only the Y component of velocity is zero, X is always constant and does exist unless it was thrown directly upward, the picture shows it had X component.

d) As per b, the acceleration doesnt change, g = 9.8m/s/s


Q2. D

d) This one is a little obvious, Atmosphere creates drag and slows it down so it can no longer sustain an orbit and falls to earth. This is similar to Q4.


Q3. B

b) All a, c and d are in the formula for escape velocity, mass of probe is not


Q4. Either C or D

Lots of controversy.
a)There is no way it will move away from the earth if it is slowed down. It would have to fire boosters toward the earth to do that. forward boosters fire perpendicular to the earth so just affects the speed of the craft, (which will affect the altitude due to downward acceleration).

b) It would definitely make a difference to the orbit as the new slower velocity cant maintain the same orbit. (see orbital velocity formula)

c) I picked this in the test because i liked the idea of the spaceship crashing into the earth, well partially, my argument for c is that because if the ship slows down it cant maintain the same orbit so decays toward the earth, the question is does it take up a lower orbit or does it crash to the earth? According to the orbital velocity formula a slower ship has a greater orbital radius, and since its now closer to the earth, ie lesser orbital radius, then it cant be in orbit and hence falls to the earth.

d) (Probably the better answer) Since the ship is falling it does speed up (slingshot effect) and so might be able to maintain a stable orbit, however I am not sure whether it will gain enough speed to maintain the orbit (since it already lost half), plus drag of the atmosphere will slow it down more. So it would keep losing speed and thus altitude and then fall to the earth slowly. (I hope its not this for my sake, lol, it crashes into the earth damnit!)


Q5. C

a)b)d) It has been proved previously to be wrong.

c) Only logical answer, just throw a ball as hard as you can straight ahead then time it to hit the ground, then throw it as high as you can, it will take much longer to hit the ground.


Q6. B

a) Length WILL affect the current due to the formula posted previously.

b) Since it states in the question that it is 'of very low resistance' you can assume that it is negligible thus thickness of the wire wont make a difference (or the LEAST, which is what the questions asks).
(Where on earth did someone pull out the explanation that thicker wire made it heavier and so they rotated it more slowly? that made me crack up ehehe, its a good observation but isnt relevant for this question, and plus that would mean that its still the speed that affects the current, not the thickness, but due to the thickness, you can assume the experiment was controlled)

c) Speed is a change in mag field so more of it will make more current.

d) The wire is assumably using the earth's magnetic field, and since they go from top of the world to bottom, it will matter which way they are holding the wire.


Q7. A

Again, lots of controversy.
a)OK there are actually TWO right hand rules relating to induced forces. ONE of them is for MOTOR effect and the other is for GENERATORS. You have the mag field and the direction of rotation, and you need to find the current direction.

- Ok the motor effect is where your fingers are the magnetic field. Your thumb points in the direction of the conventional current, and your palm points in the direction of force on the wire. This question is NOT that one, resulting in D.

- The generator one is where fingers are mag field, your thumb points in the direction that the wire is moving, and the palm points in the direction of current.

Take the diagram for a, in your head rotate the coil 90 degrees clockwise and picture your fingers pointing downward with the field, then since its clockwise point your thumb to the right, your palm points into the page so thats the direction of current. Thus answer is A.

b,c) Visual trick but not very effective. Only possible answers are a and d.

d) This would be true if this was a MOTOR, and you knew the current and mag field but not the direction of rotation.


Q8. C

a) Yes its a DC current (batteries are DC), but its the CHANGE in mag field that induces current. And closing the switch causes current to go from ZERO to MAX, therefore there is a change and an induced current.

b) Once the current flows and stops changing, there is no induced current so cant be continuous.

c) Since the change is only as soon as you turn it on and stops shortly after you leave it on the induced current stops as illustrated.

d) I am a complete moron, in the exam i chose this because I thought the question said the switch was closed and opened again straight away, just exam stress i guess. Well yeah the switch definitely stayed on so cannot be this answer.


Q9. B

I am definite, I saw it with my own eyes! (it was inverted as in the copper was a big tube and we put a super magnet down it, it went slowly, then a tube with a slit cut through the whole thing, the magnet went slowly but quicker than the solid tube, then we got a plastic pipe and mag dropped right through, this is basically the same experiment)

Ring P is plastic so no eddy currents slowing it down, falls at 9.8m/s/s

Ring Q is copper with completed loop, allows lots of eddy's to slow it down.

Ring R is copper with slit, eddy CAN STILL EXIST, just less of them and so it slows but not as much. Definitely slower than ring Plastic


Q10. Not Sure, I picked C, here's why:

I was stuck in the exam, but then thought which way does mag field go around a wire, its right hand cork screw. So put my hand horizontal and it goes in the Y plane, so i thought putting sheets perpendicular will lessen them. And so therefore B would not. However I have reason to believe that the answer is B, after seeing a picture posted on a link in a previous post.

a) der

d) I have never seen this in my life so if it worked i probably would have seen it. I would have thought eddys spin in in the direction of the circles.


Q11. B

This is one of the pracs you were supposed to do. Its just a matter of remembering what this particular one proves. Straight lines is the Maltese cross one.


Q12. - Beats the crap out of me.

Like I said before, Im bad with history, probably because remembering names is boring so i didnt bother, even after studying it. I can't even remember what I put in the exam. hopefully guessed right. Never liked this black body stuff anyway.


Q13. Again, dunno. But i picked C, for some unknown reason. Phosphorus caught my eye, i might have worked it out in the exam with a good reason for picking C but cant be bothered right now, its over.


Q14. B

Pretty straight forward formula substitution work. Using E=hf


Q15. D seemed the only logical one to me.

I don't know how this thing works, but if holes are opposite charge to electrons it makes sense they would be forced in opposite directions.


HMMMM thats it.......pretty interesting. Ill sum it up:

1. A
2. D
3. B
4. C -> prolly D
5. C
6. B
7. A
8. C
9. B
10. Maybe C
11. B
12. -
13. Some say A or C
14. B
15. Most say D


I am Pretty confident with the answers I gave a single answer to.

I conclude Q4 the most interesting.

So yeah any problems with my explanations please reply.

Wow nearly 2am, better get to bed.

CYA!
 

Abtari

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sasso said:
Q6. B

a) Length WILL affect the current due to the formula posted previously.

b) Since it states in the question that it is 'of very low resistance' you can assume that it is negligible thus thickness of the wire wont make a difference (or the LEAST, which is what the questions asks).
(Where on earth did someone pull out the explanation that thicker wire made it heavier and so they rotated it more slowly? that made me crack up ehehe, its a good observation but isnt relevant for this question, and plus that would mean that its still the speed that affects the current, not the thickness, but due to the thickness, you can assume the experiment was controlled)

c) Speed is a change in mag field so more of it will make more current.

d) The wire is assumably using the earth's magnetic field, and since they go from top of the world to bottom, it will matter which way they are holding the wire.
NO. if u are under the assumption that the resistance is negligible, and henceforth if u think that the thickness won't affect current much, than by the same token, the length of the rotating wire wouldn't affect current much either.. after all, it is negligible resistance...rite?.. dont think so

"which of the following is least likely to affect the amount of current produced?"
amount of current produced is detected by the attached galvanometer..and this reading depends primarily on the resistance offered by the wires...

Resistance = resistivity. area/length is the formula...

as can be seen, length of the wire and the thickness directly have impact on the resistance of the wire, and changing these variables can significantly affect the amount of current produced...
the speed with which the wire is rotating also affects the amount of current, as it means the rate at which the wire cuts its own magnetic field changes.. hence it induces current, and affects the amount of current produced.

D, 'whether the wire is oriented north-south or east-west' seems the most plausible as, although the earth's magnetic field is present, it is not significantly large enough to create any sizeable change in the amount of current produced.. the magnetic force exerted by the earth is one of the weakest forces..
 
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sasso

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Abtari said:
NO. if u are under the assumption that the resistance is negligible, and henceforth if u think that the thickness won't affect current much, than by the same token, the length of the rotating wire wouldn't affect current much either.. after all, it is negligible resistance...rite?.. dont think so

"which of the following is least likely to affect the amount of current produced?"
amount of current produced is detected by the attached galvanometer..and this reading depends primarily on the resistance offered by the wires...

Resistance = resistivity. area/length is the formula...

as can be seen, length of the wire and the thickness directly have impact on the resistance of the wire, and changing these variables can significantly affect the amount of current produced...
the speed with which the wire is rotating also affects the amount of current, as it means the rate at which the wire cuts its own magnetic field changes.. hence it induces current, and affects the amount of current produced.

D, 'whether the wire is oriented north-south or east-west' seems the most plausible as, although the earth's magnetic field is present, it is not significantly large enough to create any sizeable change in the amount of current produced.. the magnetic force exerted by the earth is one of the weakest forces..

The question specifically says that it has very low resisitance, hinting that changing the thickness wont have much of an effect. yes, changing the length will increase resisitance but its already 'of low resisitance' so it wont affect the resistance significantly, as you said. But i am not talking about how the length affects resisitance, but how it affects the specific component that when increased, more wire passes through the earths mag field and so more current is produced.
Just look at all the formulas, L=Length of wire, F/L=kII/d, F=BILsinx, t=BIAcosx (A being Length times width). All these say that length will always affect the result. So length affecting the current, and thickness much less so.


for d) Its not saying, 'the strength of the field' but the direction the wire is moving relative to the field. If the earths mag field is going north to south, then you want the wire going east to west so it would actually move in the perpendicular component, just think of a motor/generator, the field goes across and the armature rotates like it does, look at Q7 to get the picture. The coil is rotating in, lets call it the Z plane, so that when the coils are at the top and bottom then it produces the most current or has the most torque. The magnetic field is going down the Y plane, so rotating the thing about the Y plane wont give hardly any perpendicular component, if any, depending on how much it bows.
Therefore the orientation of the wire is very important. And since the length of the wire creates more component, it is important, and speed is obviously important. So that leaves thickness to be the only one to have the LEAST effect, since the question specifically states that is has low resisitance, hence making it thicker wont make much difference (the current produced will be tiny relative to the original thickness of the wire, since its a glavanometer, they measure tiny current dont they, otherwise it would have an ammeter)

So answer is B
 
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fizzwizz

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sasso said:
Just look at all the formulas, L=Length of wire, F/L=kII/d, F=BILsinx, t=BIAcosx (A being Length times width). All these say that length will always affect the result. So length affecting the current, and thickness much less so.

You have stated three formulae, involving length, which have nothing to do with the physical situation DECRIBED IN QUESTION. Why not use Einsteins length contraction formula into the mix and completely confuse yourself You should at least talk about using a formula in its correct context.

Anyway the answer is D
 

Abtari

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sasso said:
Just look at all the formulas, L=Length of wire, F/L=kII/d, F=BILsinx, t=BIAcosx
let's not..

sasso said:
So answer is B
no it isnt.. the answer should be D

sasso said:
(the current produced will be tiny relative to the original thickness of the wire, since its a glavanometer, they measure tiny current dont they, otherwise it would have an ammeter)
ur reading too much into the question..
 
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