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Physics: Post your doubts here!

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sagar65265

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sma786 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_21.pdf

Question 1 part c(ii).. the markscheme ( http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_ms_21.pdf ) says that for area's uncertaininty, we have to multiply 2 as well.. Why do we multiply 2?
Click to expand...

Supposing we have two numbers defined as

a = 5.0 ± 0.2

b = 3.0 ± 0.6

and we have to multiply them together, thus giving us a result of

a * b = (5.0 ± 0.2)(3.0 ± 0.6)

In this case, we convert their absolute uncertainties into percentage uncertainties before multiplying, and this gives us

a * b = (5.0 ± 4.0%)(3.0 ± 20%)

and so we write our answer as 15.0 ± 24%, because whenever we multiply two numbers, we add their percentage uncertainties together and write that percentage uncertainty for the final product.

If we now extend this to the case where we square a particular number, say we square a, then we get

a^2 = a * a = (5.0 ± 0.2)(5.0 ± 0.2)
= (5.0 ± 4.0%)(5.0 ± 4.0%)
= (25 ± 8.0%)

This 8.0% value we have obtained by adding up the percentage uncertainties in the products, and writing that value as the percentage uncertainty for the final product. Note, however, we could have obtained the same value by multiplying the percentage uncertainty in a by the power we were raising a to, i.e. 2.

Suppose we cube a, we get a similar argument:

a^3 = a * a * a= (5.0 ± 0.2)(5.0 ± 0.2)(5.0 ± 0.2)
= (5.0 ± 4.0%)(5.0 ± 4.0%)(5.0 ± 4.0%)
= (125 ± 12%)

Again, we could have gotten this percentage uncertainty by multiplying the percentage uncertainty in a (=4.0%) by the power we were raising a to (=3), which for accuracy's sake we can see is 12%, the same as above.

Suppose we raise a to a power of 0.5, i.e. we take the square-root of a. The same principle applies here:

a^0.5 = (5.0 ± 0.2)^0.5
= (5.0 ± 4.0%)^0.5
= (2.23 ± 2.o%)
Where 2.23 is the square root of 5 to 3 significant figures, and the percentage uncertainty has been obtained by multiplying the percentage uncertainty in a (=4.0%) by the power we were raising a to (=0.5) to give us 4.0% * 0.5 = 2.0%.

In the question, you are given the diameter and asked to calculate the area. The formula for doing so is given by

Area = π(d^2)/4

where d is the diameter. In this case, there is no uncertainty in π and no uncertainty in 4, so the uncertainty in the Area is equal to the uncertainty caused by squaring the value of d. So, according to what we've seen above, the % uncertainty in the final value is equal to the %uncertainty in d * the power it has been raised to. So, since it is squared in the formula, we get

% uncertainty in Area = 2 * percentage uncertainty in d.

Hope this helped!

Good Luck for all your exams!
 
Suchal Riaz

Suchal Riaz

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Haya Ahmed said:
hmmm okay I got the idea now ! .. so in the question it is asking for , draw the stationary wave at time (t + 5.0 ms) so it means we have to add 1/4 to the 1 cycle right ??.. then why did we draw the straight line it isn't 2/4 or 4/4 it is 5/4 !!
Click to expand...
it was 5 ms which is =1/4T
it was initially at the 1/4th from my diagram so add 1/4 we get 2/4th stage. :)
 
sma786

sma786

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K
sagar65265 said:
Supposing we have two numbers defined as

a = 5.0 ± 0.2

b = 3.0 ± 0.6

and we have to multiply them together, thus giving us a result of

a * b = (5.0 ± 0.2)(3.0 ± 0.6)

In this case, we convert their absolute uncertainties into percentage uncertainties before multiplying, and this gives us

a * b = (5.0 ± 4.0%)(3.0 ± 20%)

and so we write our answer as 15.0 ± 24%, because whenever we multiply two numbers, we add their percentage uncertainties together and write that percentage uncertainty for the final product.

If we now extend this to the case where we square a particular number, say we square a, then we get

a^2 = a * a = (5.0 ± 0.2)(5.0 ± 0.2)
= (5.0 ± 4.0%)(5.0 ± 4.0%)
= (25 ± 8.0%)

This 8.0% value we have obtained by adding up the percentage uncertainties in the products, and writing that value as the percentage uncertainty for the final product. Note, however, we could have obtained the same value by multiplying the percentage uncertainty in a by the power we were raising a to, i.e. 2.

Suppose we cube a, we get a similar argument:

a^3 = a * a * a= (5.0 ± 0.2)(5.0 ± 0.2)(5.0 ± 0.2)
= (5.0 ± 4.0%)(5.0 ± 4.0%)(5.0 ± 4.0%)
= (125 ± 12%)

Again, we could have gotten this percentage uncertainty by multiplying the percentage uncertainty in a (=4.0%) by the power we were raising a to (=3), which for accuracy's sake we can see is 12%, the same as above.

Suppose we raise a to a power of 0.5, i.e. we take the square-root of a. The same principle applies here:

a^0.5 = (5.0 ± 0.2)^0.5
= (5.0 ± 4.0%)^0.5
= (2.23 ± 2.o%)
Where 2.23 is the square root of 5 to 3 significant figures, and the percentage uncertainty has been obtained by multiplying the percentage uncertainty in a (=4.0%) by the power we were raising a to (=0.5) to give us 4.0% * 0.5 = 2.0%.

In the question, you are given the diameter and asked to calculate the area. The formula for doing so is given by

Area = π(d^2)/4

where d is the diameter. In this case, there is no uncertainty in π and no uncertainty in 4, so the uncertainty in the Area is equal to the uncertainty caused by squaring the value of d. So, according to what we've seen above, the % uncertainty in the final value is equal to the %uncertainty in d * the power it has been raised to. So, since it is squared in the formula, we get

% uncertainty in Area = 2 * percentage uncertainty in d.

Hope this helped!

Good Luck for all your exams!
Click to expand...

Thabkyou for the fantastic reply :eek: <3
 
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