# Edexcel physics unit 6B (15 May 2014)Watch

4 years ago
#121
(Original post by Cosmologist)
It's 8am in the UK. My answers for the paper:

1. a) Why micrometer?
-The precision (0.001mm) of the micrometer allows to obtain values of the diameter to 3 d.p.
- The scale division is 0.001mm
- It gives small %U
(1 mark)

b) What technique would you use to make you results as accurate as possible?
- Measure the diameter of the wire at multiple orientations and then average the result
- Check for zero error on the micrometer
(1 mark)

c) Find the value for the resistivity.
- I really forgot the answer, but it was smth*10^-7
(3 marks)

d) Percentage uncertainty in resistivity.
- I found it 3.5% (calculated value was 3.45%, so I rounded it up to 3.5%, but I think 3.4% is ok)
(2 marks)

2. a)How would you use a metre rule to make your readings of h as reliable as possible?
- Use set square on the floor to make sure the metre rule is vertically placed
- Align your eyes with a scale to avoid parallax, whilst measuring the height h
- Mark the centre of the ball to measure the height precisely to the centre of the ball
- Repeat taking measurements of h to obtain many values (reduce random error) and then average the result
(3 marks)

b) Describe a technique to measure the period of oscillations T as accurate as possible?
- Record a time for 10-15 complete oscillations (using stopwatch 0.01s) and divide by the number of oscillations to obtain a value for the period of oscillations T. This method would reduce %U.
- Use timing marker in the equilibrium position to see full oscillations (easier to count oscillations)
- Repeat recording the time for 15 oscillations and average to minimise random error (use lap-timer for successive recording of time for 15 oscillations)
(3 marks)

c) Find the gradient of the graph?
- It was (-4pi^2/g)
(1 mark)

d) Show that the ratio intercept/ gradient is equal to H.
- It's weird, since it turns out to be (-H), but I took the modulus of the gradient and obtained H.
(2 marks)

3.a) Find the value of k.
I filled out the table, as following:
k /10^-18Vm^2
1.26
1.14
and then averaged to get 1.20*10^-18 to 3 s.f.
(3 marks)

b) The percentage uncertainty in k
- I came up with 2 ways to do it, but finally realised that %U in k is best found just taking the half-range divided by mean (as we had only 1 mark for that), so (1.26-1.14)/ 1.20 * 100% = 10%
(1 mark)

c) Find the h from h^2=2*m*e*k.
- I computed it to be 5.91*10^-34 to 3 s.f.
(2 marks)

d) Hence, write down the %U in h.
- It's just %U in k halved, so 5%
(1 mark)

e) Comment on the validity of your experimental value of h.
%D = (6.63-5.91)/6.63 *100%=10.8% ~11%
Since %D (between theoretical value for h and our 5.91*10^-34) is significantly greater than %U in experimental value, then h is likely to be invalid.
(2 marks)

4. a)Thermistor is a semiconductor, so as the temperature increases, more electrons get released and the resistance reduces.
(2 marks)

b) Plan the experiment to investigate how the resistance of the thermistor varies with temperature. (i) Describe the apparatus to be used, you may draw a diagram, if you wish. (ii) How would you vary the temperature in the range 0â„ƒ and 100â„ƒ? (iii) Some words about tech precautions to improve the accuracy of results.
- Well, I wont draw it here, but there should be the water bath, Bunsen burner, thermistor and ohmmeter/ multimeter connected by wire, thermometer (both thermometer and the circuit are fixed to the wall, or smth)
- Put ice in the icy water to approach 0â„ƒ and boil the water to reach 100â„ƒ.
- Remove the heat source, while taking the readings.
- Place thermometer and thermistor close together and away from walls and bottom of the container.
- Allow some time for the apparatus to come to thermal equilibrium
- Make sure the thermistor is fully immersed in the water.
(5 marks)

c) Prove that lnR against Éµ would produce the straight graph.
- Just compare to y= mx+c and state that (-a) is the gradient and lnR0 is the intercept
(1 mark)

d) Plot this weird graph.
- This data table for the graph is quite unreasonable as ln0.906<0 and we're provided with positive sector of the axis. I multiplied values in R-column by 10 to get 9.06, and changed R/kâ„¦ to R/10^2â„¦, then plotted a graph with a good enough range and scale. I am sure, many guys just sketched it as it was given with negative sign, but I struggled with that for too long. Doubtful task...
(4 marks)

e) Use your graph to calculate the value for a.
Gradient= -a=-0.03884...So a=3.88*10^-2 (~3.9*10^-2 or 3.8*10^-2)
(3 marks)

had similar answers but isn't the uncertainty for a micrometer +/- 0.01 mm ? And for 4b do u think "remove parallax error" while taking the temperature readings for better accuracy be awarded a mark ? For the graph I calculated lnR after changing kR to R
0
4 years ago
#122
Oh and I didn't write a unit for lnR, thought it's no longer required once u multiply with ln, so I'll lose a mark there :/
0
4 years ago
#123
And I think for 4a the key points were "it has a negative temperature coefficient" and " more charge carriers are released at higher temperature and as resistance and current are inversely proportional, the resistance decreases"
0
4 years ago
#124
(Original post by jtbteddy)
Yuppp, all my answers are same as yours, except for 2a I think Id lose a mark because I only gabe two points

I had same problem with 2d but I looked at it this way:

Theres a negative correlation between both variables, hence the negative sogn, so it doesnt mean H is negative.. So in my final answer I said that H= -intercept/gradient so that the negative signs would cancel out and give me the value of H

3b) I used half the range, leading to 5% and then 2.5% for the uncertainty of h.. But its the same thing

4a) i forgot to mention its a semiconductor

4b) I pretty much invented it, and somehow it coincides with what you described. I dont think I mentioned all the points though, so Id lose a mark or two.

4d) i drew the graph of (lnR) just bcoz it gave me larger values, my values ranged from 6.5 to 8.5 i think, so i could spread out my graph

Rest of it.. Did the same as you

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Great! Absolutely agree with your explanation of why H is positive in 4d)
Yeah, I might be penalised for overthinking in graph plotting, but I was just not really satisfied with a negative point there (ln0.906)...Have never seen negative data before
All in all, I hope we scored above 50 out of 60 in ums
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4 years ago
#125
(Original post by BNaeem)
had similar answers but isn't the uncertainty for a micrometer +/- 0.01 mm ? And for 4b do u think "remove parallax error" while taking the temperature readings for better accuracy be awarded a mark ? For the graph I calculated lnR after changing kR to R
It seems to vary: some give precision 0.005 (close to your value), others (mentioned in past papers) produce 0.001mm uncertainty.
And micro- is 10^-6, so I thought logically the scale division should be 10^-6m=10^-3mm)

Yes, it's never useless to point out, I would only say that if you wrote concisely (without being more specific how to avoid this parallax error:saying that, for instance, that your eyes should be aligned with the scale or simply drawing eye next to the reading on the thermometer), they might not award you a mark. Your point is valid enough here
0
4 years ago
#126
(Original post by BNaeem)
And I think for 4a the key points were "it has a negative temperature coefficient" and " more charge carriers are released at higher temperature and as resistance and current are inversely proportional, the resistance decreases"
Charge carriers here are electrons, they get released with a rise in temperature. You were more precise in this question (as I forgot to say about the inverse proportionality and current itself), I might have lost one mark here...
0
4 years ago
#127
(Original post by Cosmologist)
Great! Absolutely agree with your explanation of why H is positive in 4d)
Yeah, I might be penalised for overthinking in graph plotting, but I was just not really satisfied with a negative point there (ln0.906)...Have never seen negative data before
All in all, I hope we scored above 50 out of 60 in ums
Yeah same here, which is why i ln'd the R value, instead of the kR value.. So i was free of negatives too. I didnt wqnt toplot a negative bcoz the poitive axes were already marked on the graph pper

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0
4 years ago
#128
For that temperature experiment precautions I wrote room temperature and something else I forgot just now. And for the graph question I started my y axis graph wih 0 because my last InR was -0.20
0
4 years ago
#129
guys in uncertainty for k question, why is it wrong to take the two values of k, subtract them and divide by 2 (range/ 2).. i did that and then calculated percentage uncertainty, i got 2.5 %
wrong? :/

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4 years ago
#130
(Original post by kcapsoccer)
guys in uncertainty for k question, why is it wrong to take the two values of k, subtract them and divide by 2 (range/ 2).. i did that and then calculated percentage uncertainty, i got 2.5 %
wrong? :/

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It's correct to use half-range, but in that case you should have gotten 5% for %U in k and 2.5% for %U in h.
k1= 1.26*10^-18
k2= 1.14*10^-18
=> k= 1.20*10^-18
so, %U in k=(1.26-1.14)/(2*1.20) * 100% = 5%

I always use half-range, but here I forgot to divide by 2 and got 10%, but it's also correct, I was really lucky
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4 years ago
#131
(Original post by Cosmologist)
It's correct to use half-range, but in that case you should have gotten 5% for %U in k and 2.5% for %U in h.
k1= 1.26*10^-18
k2= 1.14*10^-18
=> k= 1.20*10^-18
so, %U in k=(1.26-1.14)/(2*1.20) * 100% = 5%

I always use half-range, but here I forgot to divide by 2 and got 10%, but it's also correct, I was really lucky
I agree. Thats what i did- half the range

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4 years ago
#132
(Original post by Cosmologist)
It's correct to use half-range, but in that case you should have gotten 5% for %U in k and 2.5% for %U in h.
k1= 1.26*10^-18
k2= 1.14*10^-18
=> k= 1.20*10^-18
so, %U in k=(1.26-1.14)/(2*1.20) * 100% = 5%

I always use half-range, but here I forgot to divide by 2 and got 10%, but it's also correct, I was really lucky
I did the same thing. Lucky

For finding alpha and stating it units are important isn't it? Is it a must to give it in 3 SF?

For the graph I didn't start from the origin. I started from - 0.1 or something. Previous examiners reports had graphs like this, didn't they?

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4 years ago
#133
(Original post by RoyalBlue7)
I did the same thing. Lucky

For finding alpha and stating it units are important isn't it? Is it a must to give it in 3 SF?

For the graph I didn't start from the origin. I started from - 0.1 or something. Previous examiners reports had graphs like this, didn't they?

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If the question was stated "find a, not just a gradient"+it was 3-mark task, then probably, units are required, though in some past papers they ignore the absence of units. Wish you good luck in that too
did you get 0.0388 (0.038/0.039)?
0
4 years ago
#134
(Original post by Cosmologist)
If the question was stated "find a, not just a gradient"+it was 3-mark task, then probably, units are required, though in some past papers they ignore the absence of units. Wish you good luck in that too
did you get 0.0388 (0.038/0.039)?
Yes I got 0.0388. It should be to 3 SF shouldn't it?

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4 years ago
#135
(Original post by RoyalBlue7)
Yes I got 0.0388. It should be to 3 SF shouldn't it?

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Perfect! Well, I always write like that 0.0388~0.039 (0.038)
It works for examiners since this way you show your calculated value (with better precision) and then approximate it further
But for this particular task, it's clear that the answer should be to 3 s.f. as in the plotting data.
0
4 years ago
#136
(Original post by Cosmologist)
Perfect! Well, I always write like that 0.0388~0.039 (0.038)
It works for examiners since this way you show your calculated value (with better precision) and then approximate it further
But for this particular task, it's clear that the answer should be to 3 s.f. as in the plotting data.
Good for me!

What are you expecting for this paper? By your answers I assume 60?

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4 years ago
#137
(Original post by RoyalBlue7)
Good for me!

What are you expecting for this paper? By your answers I assume 60?

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I took unit 6 in June 2013 and got 46/60, but I need A* in physics, so I really hope for the best
Did you take other exams in physics?
0
4 years ago
#138
(Original post by Cosmologist)
I took unit 6 in June 2013 and got 46/60, but I need A* in physics, so I really hope for the best
Did you take other exams in physics?
Yeah I took.

Why didn't you do it in Jan?

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4 years ago
#139
(Original post by RoyalBlue7)
Yeah I took.

Why didn't you do it in Jan?

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I did, but got sick and could barely finish the paper, so it dropped to 36/60
1
4 years ago
#140
(Original post by Cosmologist)
I did, but got sick and could barely finish the paper, so it dropped to 36/60
That's so unfortunate

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