Solved physics Paper 5??

[Soldier313]

Finally! Anyone?

okay so the eqn is
(hc/lambda) = B + eV
If V is one the y-axis and 1/lambda on the x-axis
write it into the form y = mx + c
where y is V and x = 1/lambda
So:
eV = (hc/lambda) - B
Divide both sides by e to make V the subject
V = [ ( hc/lambda) - B ] /e
so V = [ (hc/e) x 1/lambda ] - B/e
therefore gradient is hc/e
y - intercept is -B/e

 

[Soldier313]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_53.pdf
some1 plz help me with the diagram for question 1...
ps: i'l be quite grateful if some1 postd the procedure as well!

That's a tough one
Help anyone, am stuck on the same qn too....

 

[raamish]

hey guys in n09/51 what is your gradient and error in 2c. mine is coming -3.78+-0.24. Not too confident abt the error too. And in drawing graphs do all points have to be equally on both sides of lines cuz there are some graphs in which i cant do that. Is that wrong?

 

[ash369]

okay so the eqn is
(hc/lambda) = B + eV
If V is one the y-axis and 1/lambda on the x-axis
write it into the form y = mx + c
where y is V and x = 1/lambda
So:
eV = (hc/lambda) - B
Divide both sides by e to make V the subject
V = [ ( hc/lambda) - B ] /e
so V = [ (hc/e) x 1/lambda ] - B/e
therefore gradient is hc/e
y - intercept is -B/e

JazakAllahu Khairan! Pretty much all of my question was going wrong because of this one part! May Allah SWT reward you infinitely in this life and the hereafter! (Amen!)

 

[Soldier313]

JazakAllahu Khairan! Pretty much all of my question was going wrong because of this one part! May Allah SWT reward you infinitely in this life and the hereafter! (Amen!)

Wa iyyaki sis. Aameen. Thank you for the beautiful du'a I pray the same for you

 

[blueberryyums]

Would love your input on my diagram for may/june 2008. I am no artist, and no it is not a snake eating the glass sheet, it is my micrometer screw guage.. I've completed the paper and marked it and it seems right. If my diagram is approved, I'll scan my answer and put it up here

 

[ash369]

If L=54.5+-0.5 (10^-2)
then 1/L=1.83 but what about the uncertainty? :-/

 

[Omar99]

hey guys in n09/51 what is your gradient and error in 2c. mine is coming -3.78+-0.24. Not too confident abt the error too. And in drawing graphs do all points have to be equally on both sides of lines cuz there are some graphs in which i cant do that. Is that wrong?

how did u get this error, please explain all though i got almost the same gradient as yours but the errors dont match

 

[Soldier313]

If L=54.5+-0.5 (10^-2)
then 1/L=1.83 but what about the uncertainty? :-/

let p = 1/L
so (error in p)/p = (error in L)/L ........... 1 is a constant so no error
so (error in p) = [ (0.5x10^-2)/(54.5 x 10^-2) ] x 1.83
error in p is 0.02
hence error in 1/L = 0.02

 

[raamish]

how did u get this error, please explain all though i got almost the same gradient as yours but the errors dont match

hmm this error may be wrong of mine. u can ask other people on the thread they may explain it to u better to confirm your answer.

 

[raamish]

guys in nov11/51 this reason is given:
reasoned method
to keep Hall probe in constant orientation (e.g. use of set square, fix to
rule, optical bench or equivalent).
i want to know that whenever we want something to be in orientation we use the set square or are there any uses of set square. And by being in constant orientation what do they mean? When should we write this that we want to keep an object in constant orientation?

 

[Praveena]

thnk u

 

[ash369]

let p = 1/L
so (error in p)/p = (error in L)/L ........... 1 is a constant so no error
so (error in p) = [ (0.5x10^-2)/(54.5 x 10^-2) ] x 1.83
error in p is 0.02
hence error in 1/L = 0.02

JazakAllahu Khairan again! Seems you're going to get flooded by my prayers in no time InshaAllah!

 

[Praveena]

May June 2008 p5 Question no 2
How to solve 10% of R part?
Help Help

 

[Soldier313]

JazakAllahu Khairan again! Seems you're going to get flooded by my prayers in no time InshaAllah!

wa iyyaki Awwh i'd love that ♥

 

[Soldier313]

May June 2008 p5 Question no 2
How to solve 10% of R part?
Help Help

10% of R0 = 0.01 R0
R = R0 x exp (-pnx)
0.01Ro = Ro x exp (-11300)(gradient)(x)
0.01 = exp (-11300)(gradient)(x)
ln(0.1) = (-11300)(gradient)(x)
make x the subject.


For the error in x:
ln(0.1) = (-11300)(gradient)(x)
ln(0.1) (-11300) are constants so,

the (error in x) / x = (error in n) / n

I'm not quite sure about the error part, it'd be nice if someone could clarify.

 

[biba]

That's a tough one
Help anyone, am stuck on the same qn too....

i know.. mark scheme isnt helping either, i thnk its written in gibberish

 

[SararaIH]

I have a few questions regarding paper 5 Physics:
1. Are the error bars in the graph all of the same length?
2. Can Error bars be really large or are they meant to be really small?
3. When calculating the gradient ,do we draw the triangle on the graph between the two points that LIE on the line of best fit?
4. Should the reading and absolute uncertainty have the same no of significant figures

Also can someone tell me how to calculate the uncertainity in l^2 question 2 in
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s07_qp_5.pdf

I can answer your 1st and 4th question.
The error bars aren't always the same length.
For example, if there are values in the table such at 4.5 +/- 0.2 , 5.6 +/- 0.3 ,
Then definitely, the error bar of the the first value (4.5 +/- 0.2) will be smaller than (5.6 +/- 0.3)

The reading should have the same significant figures as the raw data or ONE MORE significant figure than the raw data.
The absolute uncertainty should also be the same significant figures as the raw data that is given. Most of the times, the uncertainty that is given in the raw data is of one s.f. so the absolute uncertainty should also be of one s.f.

And I'm sorry but I can't answer your 2nd and 3rd question because I'm not sure about those myself

------

And this is how you calculate the uncertainty in l^2:
I'll show you with the first value. And then you can probably do the rest
The first value on the table is 6 +/- o.4

First find the square of l.
l^2
= (6)^2
= 36

And then you find the fractional uncertainty, which is,
(Uncertainty/original value)
so,
(0.4/6) x 2 (It is being multiplied by 2 because they are asking you for a squared value of L )
= 0.13333...

And then you multiply the fractional uncertainty with the square of L that you have found previously.
(0.133333...) x 36
= 4.8

so the final answer will be : 36 +/- 4.8

This process is used whenever they ask you for a square value.

I hope you have understood!
Don't hesitate to ask me if you have any other questions!
Best of luck!

 

[Silent Hunter]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_52.pdf

anyone did the graph portion ov this one? and the errors one too obviously ?

can anybody show the graph plotted or the errors ?

SararaIH
biba
Soldier313

or anyone?

 

[SararaIH]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_52.pdf

anyone did the graph portion ov this one? and the errors one too obviously ?

can anybody show the graph plotted or the errors ?

SararaIH
biba
Soldier313

or anyone?

I have just started solving this! After I'm done, I'll surely post it for you... If I am able to draw everything correctly lol