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PHYSICS PAPER 5 QUESTION 2

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I have a problem with the uncertainties when they're in a square root form and the questions that follow after the graph.Anyone please?
 
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There are usually 2 ways of finding uncertainties.
First is by using the formula of uncertainty
Second is by finding the maximum value or the minimum value and finding the difference between your value or finding the difference between the max and min and then dividing by 2 (just like how you find for the gradient |worst - best| or |worst(steepest) - worst(shallowest)| / 2.
In this paper to find the %uncertainty in u, you could use both. I’ll show you both the methods.

Method 1 - Uncertainty formula
To find u you used u = T/ (gradient x 2)^2
u = (1/4) T/ gradient^2
(1/4) is a constant so we don”t consider that
Now the formula
(delta u) / u = ((delta T) /T) + 2((delta gradient) /gradient) Note: When values are multiplied or divided, their uncertainties will be added and the values will also be multiplied by their power like here where we used gradient^2, we had to multiply the uncertainty of gradient by 2
Then you can just multiply the RHS by 100 because %uncertainty = ((delta u) /u) x 100

Method 2 - Max-value
u = (1/4) T/ gradient^2
Value of u will be max when T will be max and gradient will be min.
Max u = (1/4) (30 + 3) / (141 - 5.8)^2 Note: The gradient values are my values
Now you take the mod of the difference between these values and substitute in the %uncertainty formula.
 
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For the question where you have to find the %U in r
When you rearrange the formula given, you get
r = root(u / ( p x pi) )
here we will not consider p and pi because they are constants and they don’t have any uncertainties
so according to method 1
delta r / r = (1/2) delta u /u Remember that the we multiply by the power
if you calculate further you will see that %U of r is half that of u.

With method 2
Again find the max value and use that to find the uncertainty

Note: You will get different answers from both methods but both will be correct until you show your whole working. The reason for it being different is that uncertainties are not the same on both the sides of the value
Example if a has value 250 and uncertainty 10 that does not mean it is 10 on both the sides. The uncertainty is the average uncertainty on the both the sides. like it may be 14 on the left side and 6 on the right but the average is 10. therefore method 1 gives a much accurate answer but you can get the same answer with method 2 only if you use |value(max) - value(min)| / 2 but not with |value( max or min) - value|.
 
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Can some one tell me how to find gradient and y-intercept in the beginning of Q2.... the graph they plotted....:eek:
 
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Can some one tell me how to find gradient and y-intercept in the beginning of Q2.... the graph they plotted....:eek:
For the gradient, you should draw a right-angled triangle from your chosen two points in the graph. (The triangle should cover half of the line) It is best to pick the point that intersect the lines that are provided in the graph. (Not the x and y axis!).

For y-intercept, you simply choose a point, make an equation based from the gradient that you have found (y-y1=m(x-x1)), and from there, after forming the equation, input x=0. There you go the y-intercept.

Hope this helps!
 
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