Edexcel physics unit 6B (15 May 2014) Watch

jtbteddy
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#101
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#101
(Original post by FC.Exams)
k=1.2 X 10^(-18) .....and h=5.91 X 10^(-34)
(Original post by FC.Exams)
0.0381......hey, what was the uncertainty percentage in question 1?
(Original post by KrishHook)
On question 3 (I think?) you had to find H by Intercept/Gradient, but it came out as -H, what's up with that?
Ditto

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BNaeem
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#102
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@crazysashgirl, I had .04 as well and @krishhook how was it -H man
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KrishHook
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#103
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(Original post by BNaeem)
@crazysashgirl, I had .04 as well and @krishhook how was it -H man
The gradient was -(4pi2/2) wasn't it?
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KrishHook
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(Original post by KrishHook)
The gradient was -(4pi2/2) wasn't it?
*-(4pi2​/g) not letting me edit for some reason..
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arshia
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can someone upload the unofficial mark scheme pleaseeeeeeeeeeee?
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kcapsoccer
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guys, in the graph will they cut marks for y-axes lnR being in kΩ? i did not convert it into Ω so i started from -0.20 i think.. also what did you say in the thermistor decreasing resistance, increasing temperature question?


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Wollveren
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#107
(Original post by FC.Exams)
k=1.2 X 10^(-18) .....and h=5.91 X 10^(-34)
Most of my friends got that value for h, but I got 6.36 x 10 ^-34... My value although more accurate is not necessarily right. Do you think it could be one of the possible values to get in the mark scheme?
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BNaeem
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(Original post by kcapsoccer)
guys, in the graph will they cut marks for y-axes lnR being in kΩ? i did not convert it into Ω so i started from -0.20 i think.. also what did you say in the thermistor decreasing resistance, increasing temperature question?


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Nash man I doubt it would make a difference. I didn't write the units because I though lnwhatever doesn't have any units and I think that'll cost me a mark :
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jtbteddy
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(Original post by BNaeem)
Nash man I doubt it would make a difference. I didn't write the units because I though lnwhatever doesn't have any units and I think that'll cost me a mark :
I think they will because on the graph paper provided, there was a bold line around the edgds which marks the axes... You cant ammend that and add negative axis

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ASNspc
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(Original post by FC.Exams)
k=1.2 X 10^(-18) .....and h=5.91 X 10^(-34)
Hi guys I got 1,18 not 1,2 is it correct???
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Cemao
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(Original post by ASNspc)
Hi guys I got 1,18 not 1,2 is it correct???

(Original post by ASNspc)
Hi guys I got 1,18 not 1,2 is it correct???
I got k=1200!!! God! I'll lost all the marks in this question?!
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NilFBosh
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#112
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What did you guys get for % uncertainty of k?


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JAYANI ANN
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#113
guys deres dis new thing its called DONT TALK ABOUT THE EXAM AFTER ITS OVER ...WHATS DONE IS DONE
the best thing u can do rite now is to focus on the next exam goddamnit
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saifulahmed
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(Original post by NilFBosh)
What did you guys get for % uncertainty of k?


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4.9... or something like that...
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crazzysashgirl
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(Original post by BNaeem)
@crazysashgirl, I had .04 as well and @krishhook how was it -H man
i had 0.04 actually..
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crazzysashgirl
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(Original post by KrishHook)
*-(4pi2​/g) not letting me edit for some reason..
omg,, i nw i rmmbrd i got d same.. i just removed d minus in d final answer :3
i knw,, but i didnt hv time to check it again.. it was intercept thng ryt?
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crazzysashgirl
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(Original post by saifulahmed)
4.9... or something like that...
oops..
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ASNspc
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Guys is 45.1% correct for percentage uncertainty of k???
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Cosmologist
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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 0℃ and 100℃? (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 0℃ and boil the water to reach 100℃.
- 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 ɵ 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/kΩ to R/10^2Ω, 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)


Please, do share your answers)
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jtbteddy
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#120
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(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 0℃ and 100℃? (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 0℃ and boil the water to reach 100℃.
- 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 ɵ 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/kΩ to R/10^2Ω, 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)


Please, do share your answers)
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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