Solved physics Paper 5??

[meobunny]

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?

 

[Nibz]

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?

Calculating the gradient of a worst acceptable line is similar to calculating the gradient of best fit line.

Just draw a triangle whose hypotenuse is the worst-acceptable line (make sure this hypotenuse covers more than half of the line), and find the gradient the way you would for a normal line.

Good Luck.

 

[meobunny]

Yes oh okayy. Thanks do u have a link from where i can find complete solved p5 please?

 

[Nibz]

Yes oh okayy. Thanks do u have a link from where i can find complete solved p5 please?

You will find some solved questions in the first few pages of this thread.

 

[SararaIH]

Awww okay thank you very much

No problem

 

[SararaIH]

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

Thank you so much!

Can you please show me the diagram for question #1 of this same paper please?

 

[radhika bhandari]

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?

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.
y component is difference of (last y value+error)-(first y value-error)
if by joining bottom of last error bar and top of first error bar then viceversa

 

[Ruman Tiwari]

For Physics paper 5 as i think following things may help to score a good marks for planning.
1) Read the question well and try to rearrange the given equation in the form of y=mx or y=mx+c. Logarithm is used depending upon question.
2) Answer the question in the following way.
i) First identify and write down the independent variable, dependent variable and control variable. Independent variable i.e. x(the one that you get from your rearranged equation) (that you are going to change during the experiment) and Dependent variable i.e. y(the one that you get from your rearranged equation) (that depends on independent variable). Write usually more than 1 control variable.(this will depend on the experiment)
ii) A proper diagram that describes your experiment. You must include meter rule, voltmeter, ammeter, ohmmeter, rehostat and other devices you use. Don't just usually copy the diagram if it is given in question. A complete diagram can help you to score good marks even if you can't express your writing properly.
iii) Then in analyzing data there comes plotting of graph with six sets of data. If possible give the outline of graph and table. If you have to test the relationship then state the nature of graph and if we are to find value of some constant then state the gradient in terms of the constant.
iv) Stating safety precautions depends on the question. Using goggles, gloves, not touching the wire, keeping distance, using sand tray are some safety tips.
v) Finally in Additional details there must be at least 4 points stated clearly that will make your experiment more reliable. Some common points can be repeat and average of data.
3) Slight knowledge of some devices and their work might be helpful.

Take your time to do your paper. Hope you all will be helped slightly by me. If any mistakes are there you may help me too. Best of luck for june 4. I am also giving my paper.

 

[SararaIH]

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

 

[magnesium]

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.

 

[Ruman Tiwari]

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.

Either you join the lower error bar of bottom to upper error bar of top or upper error bar of bottom to lower error bar of top..hope u understood me.

 

[magnesium]

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) ??????

 

[sagar65265]

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) ??????

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!

 

[Ruman Tiwari]

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)^2
percentage uncertainity will be [2 del(gradient)/gradient] *100% + [del(T)/T] * 100%

 

[Ruman Tiwari]

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!

tx for e ii)

 

[magnesium]

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!

but my gradient is 3.5 x 10^-4 and uncertainity is 0.2 .......percentage uncertainity is tooooo large...what is wrong ?

 

[blueberryyums]

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

 

[sagar65265]

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?

 

[magnesium]

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?

sorry my gradient is 146 ...... i was confusing it with value of miu ...thank u!

 

[Omar99]

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

Jazzak Allah khair, for your wonderfull help of question 2.,
but i want to ask u that in some cases u r doing best fit -worstfit. and some time worst fit -best fit, can u explain why.