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AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread]

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Original post by CD223
Is that the one with the water rising in a very thin tube?

I calculated:

The internal radius:
r=1.36×103mr = 1.36 \times 10^{-3}m

The % uncertainty in internal radius r = 2.97%

You?


Do you still have the values of R


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Original post by CD223
Is that the one with the water rising in a very thin tube?

I calculated:

The internal radius:
r=1.36×103mr = 1.36 \times 10^{-3}m

The % uncertainty in internal radius r = 2.97%

You?


Hi there,

Yeah I got exactly r=1.36×103mr = 1.36 \times 10^{-3}m for the first part :smile:

My percentage uncertainty was 3.0 but that's just because I rounded to 2 significant figures!
Reply 1602
Original post by sam_97
Hi there,

Yeah I got exactly r=1.36×103mr = 1.36 \times 10^{-3}m for the first part :smile:

My percentage uncertainty was 3.0 but that's just because I rounded to 2 significant figures!


Yeah 3.0 is the correct answer. I'll lose marks for incorrect sig figs.


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Reply 1603
Original post by Mehrdad jafari
Do you still have the values of R


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Sadly not. I don't remember the figures as I did it on 5th May.


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Original post by CD223
Yeah 3.0 is the correct answer. I'll lose marks for incorrect sig figs.


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I don't think you'll lose marks to be honest! What sort of values did you get for the gradient etc. in the first question?
The value of (l1-l2) was something like 3.37cm, based on the value of r being 1.36x10^-3
Original post by CD223
No problem! How's your revision going?



Hmmm okay, but I don't really know what to do apart from past papers (of which there is a limited amount), I've made notes already so now don't know what more to do to revise!

Does anyone here have good ideas? =D

How about you CD223?
Reply 1607
Original post by sam_97
I don't think you'll lose marks to be honest! What sort of values did you get for the gradient etc. in the first question?


Here's my full answers. I disagree with most, but they're what I basically put haha.
Gradient at V1=V2 = 0.0494 Volts per second
Ve/G = 37.7 seconds
R2/R1 = 2.16 (ratio - no unit)
V2 adds % uncertainty to R2/R1 as you draw a tangent and don't know where it exactly peaks. V1 adds uncertainty because it's a smaller value so it's nominal uncertainty is a greater portion of its value. When divided these % uncertainties are added.
Current flows from C2 to terminal Y when V2 is max - positive charge flows from positive plate through Y which discharges C2 through R2, producing the dip in the graph
Graph of V2 for student is exactly 2.5 times the y values of the V1 graph. Same shape, all values multiplied by 2.5. Y intercept = 4.0 V.
Systematic error in y would produce curve shifted up or down vertically
Plotted h against 1/r which gave gradient of 2(gamma)/gp (definitely wrong)
Units of gamma = Nm^-1
Thinner match stick would increase (steepen) gradient of y against 1/x
r = 1.36x10^-3m
% uncertainty in internal radius r = 2.97%


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Original post by sykik
It was this part!!


The answer to that is in the unit. You can also figure out what the answer is since the particles are charged. You know that you can only have charged particles (in the tube)when they lose electrons and for that to happen the electrons emitted from the positive plate (cathode) collide with gas molecules or atoms (i think it's a noble gas) ripping of electrons from the particles ( that's one possibility).
And for conduction to occur there must a flow of charged particles, whether negative or positive, but because you can't have negative ions in the tube, the conduction is due to electrons and positive ions


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Original post by CD223
Here's my full answers. I disagree with most, but they're what I basically put haha.
Gradient at V1=V2 = 0.0494 Volts per second
Ve/G = 37.7 seconds
R2/R1 = 2.16 (ratio - no unit)
V2 adds % uncertainty to R2/R1 as you draw a tangent and don't know where it exactly peaks. V1 adds uncertainty because it's a smaller value so it's nominal uncertainty is a greater portion of its value. When divided these % uncertainties are added.
Current flows from C2 to terminal Y when V2 is max - positive charge flows from positive plate through Y which discharges C2 through R2, producing the dip in the graph
Graph of V2 for student is exactly 2.5 times the y values of the V1 graph. Same shape, all values multiplied by 2.5. Y intercept = 4.0 V.
Systematic error in y would produce curve shifted up or down vertically
Plotted h against 1/r which gave gradient of 2(gamma)/gp (definitely wrong)
Units of gamma = Nm^-1
Thinner match stick would increase (steepen) gradient of y against 1/x
r = 1.36x10^-3m
% uncertainty in internal radius r = 2.97%


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My values for the gradient and Ve/G were slightly different but that's to be expected, I got 2.2 for R2/R1 so that's pretty much the same.

What you put for the gradient is correct I think! Although I did it slightly differently, I said plot (h + r/3) against 1/gρr and the gradient will be so γ = gradient/2
Reply 1610
Original post by Somniare
Hmmm okay, but I don't really know what to do apart from past papers (of which there is a limited amount), I've made notes already so now don't know what more to do to revise!

Does anyone here have good ideas? =D

How about you CD223?


Ah I just use past papers. I've done all bar 2013 papers now.

Have you got any predictions for the six marker?


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Reply 1611
Original post by sam_97
My values for the gradient and Ve/G were slightly different but that's to be expected, I got 2.2 for R2/R1 so that's pretty much the same.

What you put for the gradient is correct I think! Although I did it slightly differently, I said plot (h + r/3) against 1/gρr and the gradient will be so γ = gradient/2


I've no clue what the correct answer to find γ\gamma was. It didn't seem correct what I put as plotting h against 1/r left you with a y intercept of:
[br]r3[br][br]\dfrac{-r}{3}[br]

Which includes an r term.


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Reply 1612
Original post by Mehrdad jafari
The answer to that is in the unit. You can also figure out what the answer is since the particles are charged. You know that you can only have charged particles (in the tube)when they lose electrons and for that to happen the electrons emitted from the positive plate (cathode) collide with gas molecules or atoms (i think it's a noble gas) ripping of electrons from the particles ( that's one possibility).
And for conduction to occur there must a flow of charged particles, whether negative or positive, but because you can't have negative ions in the tube, the conduction is due to electrons and positive ions


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My book Just says what can we deduce from Gas discharge tube and why it glows.... it doesn't even state what happens
Reply 1613
Original post by sam_97
My values for the gradient and Ve/G were slightly different but that's to be expected, I got 2.2 for R2/R1 so that's pretty much the same.

What you put for the gradient is correct I think! Although I did it slightly differently, I said plot (h + r/3) against 1/gρr and the gradient will be so γ = gradient/2


I got Nm^-3 hmm do u know what the equation was ?
Reply 1614
Original post by sykik
I got Nm^-3 hmm do u know what the equation was ?


The equation was something like:

[br]γ=2ygrρxt[br][br]\gamma = \dfrac{2ygr \rho x}{t}[br]

In units this gives:

[br]γ=m×Nkg1×m×kgm3×mm[br][br]\gamma = \dfrac{m \times Nkg^{-1} \times m \times kgm^{-3} \times m}{m}[br]

Which simplifies to:

[br]γ=Nm[br][br]\gamma = \dfrac{N}{m}[br]

[br]γ=Nm1[br][br]\Rightarrow \gamma = Nm^{-1}[br]


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Reply 1615
I donno but i did it right after the exam aswell and i got the same answer
Original post by sykik
My book Just says what can we deduce from Gas discharge tube and why it glows.... it doesn't even state what happens


Yeah you're right. The book only explains why light is emitted, it doesn't say anything about conduction. That's a big problem, especially in the exam


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Reply 1617
Original post by sykik
I donno but i did it right after the exam aswell and i got the same answer


Most of my friends got Nm^-1. There were three m terms on the top which cancelled with kgm^-3 and one m term left on the bottom.

The kg terms on top (Nkg^-1 and kgm^-3) also cancelled out on top.


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Reply 1618
Original post by Mehrdad jafari
Yeah you're right. The book only explains why light is emitted, it doesn't say anything about conduction. That's a big problem, especially in the exam


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So what do you suggest... How to tackle these questions... should i get a new book or something?
We discussing EMPA answers? I know most of my question one was wrong, although I also put the current flows from the capacitor to Y I think. Definitely didn't put zero which apparently is the answer. I thought it meant after when it started to discharge again, so I described the charge flowing and current flowing in the opposite direction to electrons.

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