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

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

I calculated:

The % uncertainty in internal radius r = 2.97%

You?
Do you still have the values of R

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

I calculated:

The % uncertainty in internal radius r = 2.97%

You?
Hi there,

Yeah I got exactly for the first part

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

Yeah I got exactly for the first part

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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4. (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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5. (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?
6. The value of (l1-l2) was something like 3.37cm, based on the value of r being 1.36x10^-3
7. (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

8. (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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9. (Original post by sykik)
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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10. (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 2γ so γ = gradient/2
11. (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

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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12. (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 2γ so γ = gradient/2
I've no clue what the correct answer to find was. It didn't seem correct what I put as plotting h against 1/r left you with a y intercept of:

Which includes an r term.

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13. (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
14. (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 2γ so γ = gradient/2
I got Nm^-3 hmm do u know what the equation was ?
15. (Original post by sykik)
I got Nm^-3 hmm do u know what the equation was ?
The equation was something like:

In units this gives:

Which simplifies to:

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16. I donno but i did it right after the exam aswell and i got the same answer
17. (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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18. (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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19. (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?
20. 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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