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As Physics Unit 3B 11th May 2012

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Original post by PhantomPhreek
For the question where you had to find the average of the wavelength, did you people use all the values that were given?. One of them were way too off from the rest.

328 was the answer.


Yeah, i used all values, bc the difference between the values were more or less the same, right?
Reply 101
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Sellord
Original post by aliveyuen
should be r=-gradient???


Yes indeed, as the equation is V = -rI + E.
But of course, negative resistance wouldn't make sense.
Original post by solittletime
Yeah, i used all values, bc the difference between the values were more or less the same, right?


I got 328 too
Original post by Lackadaisical
What do you guys think the grade boundaries will be like for an A?


Probably above 27 or so
Original post by solittletime
Yeah, i used all values, bc the difference between the values were more or less the same, right?


nah' it wasn't more or less the same, it was a difference of 8, the other values all had a difference of about 2 or 3 from each other. If it's +- 3 or 4 of a difference then it is more or less the same.
(edited 11 years ago)
Reply 105
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Ankurr1
actually what was the answer for uncertainty and velocity ?? will anyone get marks for working or will get zero if the answer is wrong ? can the uncertainty be given as % since it was written in the q that with a suitable uncertainty ?
Reply 106
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Ankurr1
what was the answer for uncertainty and velocity?
Original post by Ankurr1
what was the answer for uncertainty and velocity?

uncertainty was 2mm, the velocity was 328ms-1 if I remember right.
Guys I have a confusion.
Although the mark was 3 but I elaborated it.
The question was how to find the emf and internal resistance using the data collected.
I wrote, plot a graph of Voltage against Current.....the intercept on y-axis (voltage) is the emf. On the other hand, use the equation power=voltage*current
and resistance=(voltage /current).
plot a graph of power delivered to the resistor verses resistance.The resistance at which maximum power is delivered is the internal resistance.

since the question asked for graphical method I answered this way.
do you guys think its okay?
Original post by PhantomPhreek
uncertainty was 2mm, the velocity was 328ms-1 if I remember right.


I remember the answer to wavelength is 320mm(+/-)8mm.
the speed is 328 m/s.
Original post by The_New_Guy
Guys I have a confusion.
Although the mark was 3 but I elaborated it.
The question was how to find the emf and internal resistance using the data collected.
I wrote, plot a graph of Voltage against Current.....the intercept on y-axis (voltage) is the emf. On the other hand, use the equation power=voltage*current
and resistance=(voltage /current).
plot a graph of power delivered to the resistor verses resistance.The resistance at which maximum power is delivered is the internal resistance.

since the question asked for graphical method I answered this way.
do you guys think its okay?

Can't say for sure if this is correct or not, but personally I think it's wrong but that's just me. Here's what I did, plot a graph of voltage against current, the y-intercept is the EMF and the NEGATIVE of the gradient is the internal resistance.
(edited 11 years ago)
Original post by The_New_Guy
I remember the answer to wavelength is 320mm(+/-)8mm.
the speed is 328 m/s.

8mm because you took the value that was way off from the rest of the values (all values except one were off from each other by just 2-3, except the one value which was off by 8 from all other values which I didn't take in my calculation of the mean as I considered it an anomaly and I get the uncertainty 2mm), I'm not sure if this is correct or incorrect but I've got a feeling both would get the marks.
Another reason why I left it out was because it was the first reading, the first reading has a higher chance of being an anomaly, which I thought was supported by the fact that it was 8mm apart from the rest of the values.
(edited 11 years ago)
Original post by PhantomPhreek
Can't say for sure if this is correct or not, but personally I think it's wrong but that's just me. Here's what I did, plot a graph of voltage against current, the y-intercept is the EMF and the NEGATIVE of the gradient is the internal resistance.


Initially,I thought the same way,but then I realised that the gradient would give the value of internal resistance NEGATIVE.The 'r' is a scaler quantity.
Anyway,if mine is wrong I guess I will get a mark to find the EMF using the intercept on y-axis.
Whatever mistakes we do,lets hope we gain marks that are above the grade boundary for an A.
Original post by The_New_Guy
Initially,I thought the same way,but then I realised that the gradient would give the value of internal resistance NEGATIVE.The 'r' is a scaler quantity.
Anyway,if mine is wrong I guess I will get a mark to find the EMF using the intercept on y-axis.
Whatever mistakes we do,lets hope we gain marks that are above the grade boundary for an A.

Yeah, you would get a NEGATIVE gradient, that's why you say "the internal resistance is the NEGATIVE of the gradient", the two negatives become +. Suppose the gradient was -5, the negative of the gradient would then be (+)5.
Reply 114
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Sellord
Guys I dont understand why are you giving importance to the sign value of the gradient. Effectively, it is the negative of the gradient, but it doesnt matter, because you obviously wouldn't write -2ohms as the internal resistance right?

If you look at January 2012 for example, in the MCQ Nº5, "Which of the following would give the magnitude of the internal resistance of the cell?"
Answer is "gradient of the graph"; problem solved.
Reply 115
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aymaanx
8
agreed ^
Original post by Sellord
Guys I dont understand why are you giving importance to the sign value of the gradient. Effectively, it is the negative of the gradient, but it doesnt matter, because you obviously wouldn't write -2ohms as the internal resistance right?

If you look at January 2012 for example, in the MCQ Nº5, "Which of the following would give the magnitude of the internal resistance of the cell?"
Answer is "gradient of the graph"; problem solved.


O SNAP! SHI* just got real! :P
but yeah i dont see the point of emphasizing the point NEGATIVE OF the gradient. but hey my opinion might be biased. :biggrin:
Original post by Sellord
Guys I dont understand why are you giving importance to the sign value of the gradient. Effectively, it is the negative of the gradient, but it doesnt matter, because you obviously wouldn't write -2ohms as the internal resistance right?

If you look at January 2012 for example, in the MCQ Nº5, "Which of the following would give the magnitude of the internal resistance of the cell?"
Answer is "gradient of the graph"; problem solved.

Hey calm down, look, I'm not trying to say anybody's wrong here I'm just saying what I did, and I sure as hell know that it is correct.
Reply 118
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Sellord
I never tried to seem offensive xD, just wanted to clear the point.
Reply 119
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Vocker
anyone think i might get any marks (like at least 2 marks) for the wave length question? coz i messed up and didnt find the difference. so my speed was 540 lol, i know thats stupid to make. but anyone think i might get any marks for trying to find the average/uncertainty or using the equation V=F(lamda)??

i hope so, coz that exam was very nice, but that mistake just keeps getting on my nerves!! lool :tongue:

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