Physics practical tips

[arlery]

AsSalamoAlaikum Wr Wb!

for vernier calipers...liked the animation found it here: http://www.animatedscience.co.uk/00-introduction-to-gce-physics

arlery you may add this to your first page.....

View attachment 11602​

Wa alaikum Assalam!
Changes done, and a few more tips added.

 

[user]

Aoa wr wb!
arlery
can u explain me the last point? i'm confused with that...
somewhere earlier in ur tips u mentioned that if it's less than the uncertainty u calculated

plz elaborate how to prove that the suggested relationship is correct
jazakAllah

 

[arlery]

Aoa wr wb!
arlery
can u explain me the last point? i'm confused with that...
somewhere earlier in ur tips u mentioned that if it's less than the uncertainty u calculated

plz elaborate how to prove that the suggested relationship is correct
jazakAllah

Well we find the difference between both values of k. For eg for a value of 1.02 and 0.50, the difference is obviously larger than 10% therefore the relation is not proved, that is, they're not proportional. However, if the difference between both values of k is very minor then the relationship is proved correct. For that, you'll have to find the difference between both values taking the original k value in the denominator and then multiply it by hundred to find the percentage difference. If this falls within 10% then the relation is proved correct.

Erm I believe you're taking about this point:
* If two values, for say, a density are available, calculate the % difference between them. If a value is given by the examiner, then use this as the "correct" value, and calculate the % difference the following way: where x is your measured value, and c is the examiner's value. If you have obtained two values, then the expression changes: where x1 and x2 are your measured values, and is the median of the two, (not necessarily the mean!).
* Compare the % difference with your % uncertainty. Any relationship suggested, such as that the two densities should be equal, can be considered correct if your % difference is less than your % uncertainty.

^I've seen this method in many mark schemes.

Here's what an examiner report says:

A minority of candidates compared their values of k using a percentage difference and linked
this to a judgement of whether or not their results supported the given relationship, by
comparing this percentage difference with an experimental percentage error either taken
from (e) or estimated themselves. Candidates are encouraged to work out the percentage
difference between the two k values and then make a judgement whether this is above or
below what is expected for this particular experiment. Candidates are encouraged to state
what they think is a sensible limit for the percentage uncertainty for this particular
experiment.

I hope that cleared up the confusion.

Wa iyakum!

 

[Wanzi21]

my god I changed the all the given units to SI units.. and I didnt realised that A.ohm is actually V. but I wrote A.ohm.. was what I did acceptable??? :/

 

[hmlahori]

arlery all that was really helpful.........thanks.

 

[user]

arlery
jazakAllah again....
ermm..now i m confused abt...if we are finding the difference in the 2 values 'we' measured...then what shall we take in the denominator?

and see the question of nov 2011 paper 31

9702_w11_qp_31.pdf

In b(ii) my answer is 0.179%

my T1 = 1.053 and L1 = 28.0 ...k1 comes as 0.0396

T2 = 0.0593 and L2 = 10.0 k2 = 0.0352

now if i do.... (k1 - k2)/k1 x 100 = 11.1 %

so what does that show?

am i supposed to calculate the total uncerainty...like including the uncertainty in T and then compare the two percentages?

 

[arlery]

arlery
jazakAllah again....
ermm..now i m confused abt...if we are finding the difference in the 2 values 'we' measured...then what shall we take in the denominator?

and see the question of nov 2011 paper 31

9702_w11_qp_31.pdf

In b(ii) my answer is 0.179%

my T1 = 1.053 and L1 = 28.0 ...k1 comes as 0.0396

T2 = 0.0593 and L2 = 10.0 k2 = 0.0352

now if i do.... (k1 - k2)/k1 x 100 = 11.1 %

so what does that show?

am i supposed to calculate the total uncerainty...like including the uncertainty in T and then compare the two percentages?

It shows that the relation is correct because there's a very minor difference between both values.
How many significany figures have you allotted in the previous part?
Because both your values of T and L need to have the same amount of s.f.
If you're taking that as 4 then your percentage difference (11%) proves that the relation is correct. However, if you take the s.f. as 3 then your k values and %age difference will be quite different [I think its a pretty large %age difference] hence the relation will be proved wrong. It all depends on your previous answers. According to the ms there's no proper answer. Marks will be allotted according to your reasoning.

 

[arlery]

arlery
jazakAllah again....
ermm..now i m confused abt...if we are finding the difference in the 2 values 'we' measured...then what shall we take in the denominator?

If you take k1-k2 in the numerator then take k1 in the denominator. I asked my teacher and she said its better to subtract the smaller value from the larger value. Although, if your k2 value is larger than it should be k2-k1 in the numerator and k2 in the denominator.

 

[user]

It shows that the relation is correct because there's a very minor difference between both values.
How many significany figures have you allotted in the previous part?
Because both your values of T and L need to have the same amount of s.f.
If you're taking that as 4 then your percentage difference (11%) proves that the relation is correct. However, if you take the s.f. as 3 then your k values and %age difference will be quite different [I think its a pretty large %age difference] hence the relation will be proved wrong. It all depends on your previous answers. According to the ms there's no proper answer. Marks will be allotted according to your reasoning.

i think it should be 3 sf right? seeing the sf in the reading L

but if it's 11 % ...how do u say it's proved? sorry...but i'm bit confused abt the conclusion part..

 

[arlery]

i think it should be 3 sf right? seeing the sf in the reading L

but if it's 11 % ...how do u say it's proved? sorry...but i'm bit confused abt the conclusion part..

That's alright np.
Well you can say that %age difference of the two values is 11% which is very small hence the results support the suggested relationship. But make sure that you show the working too alright?

 

[user]

That's alright np.
Well you can say that %age difference of the two values is 11% which is very small hence the results support the suggested relationship. But make sure that you show the working too alright?

but isnt it like we have to compare it with the percentage uncertainty? that's what that cie book says...

 

[arlery]

but isnt it like we have to compare it with the percentage uncertainty? that's what that cie book says...

Which CIE book? :S
No its not necessary but if you really want to you can talk about a preset rate for eg. say that the percentage difference is under 15% and the %age difference you found is 11% hence the relation is proved correct.

 

[user]

Which CIE book? :S
No its not necessary but if you really want to you can talk about a preset rate for eg. say that the percentage difference is under 15% and the %age difference you found is 11% hence the relation is proved correct.

CIE AS and A level Physics coursebook..

 

[user]

It shows that the relation is correct because there's a very minor difference between both values.
How many significany figures have you allotted in the previous part?
Because both your values of T and L need to have the same amount of s.f.
If you're taking that as 4 then your percentage difference (11%) proves that the relation is correct. However, if you take the s.f. as 3 then your k values and %age difference will be quite different [I think its a pretty large %age difference] hence the relation will be proved wrong. It all depends on your previous answers. According to the ms there's no proper answer. Marks will be allotted according to your reasoning.

This is what the er says:

Some candidates correctly compared their values of k using a percentage difference, and then linked this to a judgement of whether or not their results supported the given relationship by comparing the percentage difference in k with some sort of experimental criterion. The criterion can come from the percentage uncertainty in (b)(ii) or with an estimated or calculated experimental uncertainty. Some candidates worked out the percentage difference correctly but omitted to compare it with any criterion. A few candidates did not carry out any calculations.

 

[user]

^this suggests we SHOULD have some link or something...I guess..!

 

[user]

^this suggests we SHOULD have some link or something...I guess..!

 

[smzimran]

arlery
jazakAllah again....
ermm..now i m confused abt...if we are finding the difference in the 2 values 'we' measured...then what shall we take in the denominator?

and see the question of nov 2011 paper 31

9702_w11_qp_31.pdf

In b(ii) my answer is 0.179%

my T1 = 1.053 and L1 = 28.0 ...k1 comes as 0.0396

T2 = 0.0593 and L2 = 10.0 k2 = 0.0352

now if i do.... (k1 - k2)/k1 x 100 = 11.1 %

so what does that show?

am i supposed to calculate the total uncerainty...like including the uncertainty in T and then compare the two percentages?

AoA,
In h(iii) you have to testify or nullify the relationship
if the two values of k are same or almost same, the relationship is testified, otherwise it is nullified

The method you are using i.e. (k1 - k2)/k1 x 100
^ The more closer its answer is to zero, the more chances that the relation is correct!

However, i would advise not to use this, just show that the two values of k are same! (Just the 1 mark!!!)

We do compare the uncertainties but in cases when marks alloted are more, not just one!

 

[arlery]

This is what the er says:

Some candidates correctly compared their values of k using a percentage difference, and then linked this to a judgement of whether or not their results supported the given relationship by comparing the percentage difference in k with some sort of experimental criterion. The criterion can come from the percentage uncertainty in (b)(ii) or with an estimated or calculated experimental uncertainty. Some candidates worked out the percentage difference correctly but omitted to compare it with any criterion. A few candidates did not carry out any calculations.

That's an alternative method yes, and its in the course book too. If you want to use that, then find the %age uncertainty, from b(ii) and then compare it. If your k %age difference is greater than that value then the relation is not proved.
You could do that OR use that 10% method I told you about, which is an estimated difference not a calculated one and 11% is just 1% bigger than the value so the relation is proved correct.

 

[arlery]

AoA,
In h(iii) you have to testify or nullify the relationship
if the two values of k are same or almost same, the relationship is testified, otherwise it is nullified

The method you are using i.e. (k1 - k2)/k1 x 100
^ The more closer its answer is to zero, the more chances that the relation is correct!

However, i would advise not to use this, just show that the two values of k are same! (Just the 1 mark!!!)

We do compare the uncertainties but in cases when marks alloted are more, not just one!

You're sure we don't have to find the %age difference? :S
because that's how I've done it in every year I practised. :S

 

[smzimran]

I never did it!