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How do we cqlculate gradient for worst fit,line. I never get my answers right. Ifmits on x axis ( error bar) how to take y axis value.? Its confusing?
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How do we cqlculate gradient for worst fit,line. I never get my answers right. Ifmits on x axis ( error bar) how to take y axis value.? Its confusing?
Yes oh okayy. Thanks do u have a link from where i can find complete solved p5 please?
Awww okay thank you very much
Thank you so much!Alhamdullilah!! I am able to solve the question u mentioned. I checked it from mark scheme also and found everything correct. Hope it vl help u View attachment 28292View attachment 28293View attachment 28294View attachment 28295View attachment 28296
if u are making straight lineby joining top of last error bar and bottom of first error bar then then calculating gradient by ycomponent upon x-component.How do we cqlculate gradient for worst fit,line. I never get my answers right. Ifmits on x axis ( error bar) how to take y axis value.? Its confusing?
The worst line should pass through the first and last tip of the error bars ?
Is it compulsory ?
What about the best fit line ?
worst fit line Should pass from left of top error bar to right
of bottom error bar or right of top error bar to
left of bottom error bar.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_52.pdf
Q2 d ii) how to find percentage uncertainity????plz help!!!!!
e ii) ??????
I think for d ii)-- equate (1/2) * sqrt(T/u) with gradient and find expression in terms of u i.e. u=T/ (2*gradient)^2http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_52.pdf
Q2 d ii) how to find percentage uncertainity????plz help!!!!!
e ii) ??????
tx for e ii)For question 2 (d) (ii), the basic idea is that when two quantities are multiplied or divided together, the percentage uncertainties in their values are added together and used to find the absolute uncertainty in the product. So, over here, the calculation required for μ is:
μ = Tension/(4 * Gradient^2)
So, there are two quantities on the RHS that have uncertainties in them, the Tension in the rope and the Gradient.
Since the Tension is 30 ± 3 N, the percentage uncertainty in the tension is
(3/30) * 100 = 10% = 30 ± 10%
Using the value of the gradient you have obtained and the uncertainty in it, you can calculate the percentage uncertainty in the gradient:
(Uncertainty in Gradient/Gradient) * 100
Since you're squaring the gradient, the percentage uncertainty in the gradient is going to double (Because Gradient ^ 2 is the same as Gradient * Gradient and according to the multiplication rules, you add the percentage uncertainty, resulting in twice the percentage uncertainty here).
Adding the two values together, the answer is
10% + 2 * (Percentage Uncertainty in Gradient).
For e (ii), you need to find r and the percentage uncertainty in its value. To calculate r, you can transpose the equation to give
r = sqrt(μ / ρ π)
and substitute the values into the equation. This time, however, there is no percentage uncertainty in any value other than in μ; ρ and π are all given exactly, so you can assume there are no percentage errors in them. Therefore, the only arising error will be from μ.
But since you're rooting the value of μ, you halve the percentage uncertainty in it. So the answer is
0.5 * (percentage uncertainty in μ) (The marking scheme says "(d)(ii) / 2", which is the same thing)
Hope this helped!
Good Luck for all your exams!
For question 2 (d) (ii), the basic idea is that when two quantities are multiplied or divided together, the percentage uncertainties in their values are added together and used to find the absolute uncertainty in the product. So, over here, the calculation required for μ is:
μ = Tension/(4 * Gradient^2)
So, there are two quantities on the RHS that have uncertainties in them, the Tension in the rope and the Gradient.
Since the Tension is 30 ± 3 N, the percentage uncertainty in the tension is
(3/30) * 100 = 10% = 30 ± 10%
Using the value of the gradient you have obtained and the uncertainty in it, you can calculate the percentage uncertainty in the gradient:
(Uncertainty in Gradient/Gradient) * 100
Since you're squaring the gradient, the percentage uncertainty in the gradient is going to double (Because Gradient ^ 2 is the same as Gradient * Gradient and according to the multiplication rules, you add the percentage uncertainty, resulting in twice the percentage uncertainty here).
Adding the two values together, the answer is
10% + 2 * (Percentage Uncertainty in Gradient).
For e (ii), you need to find r and the percentage uncertainty in its value. To calculate r, you can transpose the equation to give
r = sqrt(μ / ρ π)
and substitute the values into the equation. This time, however, there is no percentage uncertainty in any value other than in μ; ρ and π are all given exactly, so you can assume there are no percentage errors in them. Therefore, the only arising error will be from μ.
But since you're rooting the value of μ, you halve the percentage uncertainty in it. So the answer is
0.5 * (percentage uncertainty in μ) (The marking scheme says "(d)(ii) / 2", which is the same thing)
Hope this helped!
Good Luck for all your exams!
Can anybody show me the diagram for #1 of this paper please? :
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_52.pdf
but my gradient is 3.5 x 10^-4 and uncertainity is 0.2 .......percentage uncertainity is tooooo large...what is wrong ?
One question; have you included the powers of 10 in your uncertainty calculation?
Because if the uncertainty in your gradient is 0.2, your worst fit gradient would have to be 0.20035, which is over 500 times your best fit gradient!
Could you please check if you have missed out any powers of ten in your gradient calculation?
Jazzak Allah khair, for your wonderfull help of question 2.,Alhamdullilah!! I am able to solve the question u mentioned. I checked it from mark scheme also and found everything correct. Hope it vl help u View attachment 28292View attachment 28293View attachment 28294View attachment 28295View attachment 28296
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