The Student Room Group

How do you add uncertainties?

The volume V of a cylinder of height h and radius r is given by the expression
V = πr2h.
In a particular experiment, r is to be determined from measurements of V and h. The uncertainties
in V and in h are as shown below.
V  7 %
h  3 %
The approximate uncertainty in r is
A. 10 %.
B. 5 %.
C. 4 %.
D. 2 %.

The correct answer is B, but I have no idea how to get there. Thank you!:wink:
Reply 1
Uncertainty is defined as half the range of the readings or, if the readings are the same, the instrument precision. So basically uncertainty is half the range of the repeat values. I've only spent 30 mins on uncertainty in class this is all i can come up with for now lol. Ill try and help tomorrow
Relative uncertainities add up when dividing and multiplying (see formula, 2 and other values have no uncertainty). Absolute uncertainties sum up when adding or whatever - stands for.

I'm fairly certain it should be A. Dividing by 2 shouldn't reduce the relative uncertainty, only the absolute one.

EDIT: oh that formula had a square in there not 2. Uncertainty is multiplied with the exponent and the square root exponent is 1/2.
DYB
The volume V of a cylinder of height h and radius r is given by the expression
V = πr2h.
In a particular experiment, r is to be determined from measurements of V and h. The uncertainties
in V and in h are as shown below.
V  7 %
h  3 %
The approximate uncertainty in r is
A. 10 %.
B. 5 %.
C. 4 %.
D. 2 %.

The correct answer is B, but I have no idea how to get there. Thank you!:wink:



Volume of a cylinder = πr²h [area of crossection x height]

rearrange for r and you get
= V/(πh)
r = √(V/πh)

The rule is

you add percentage uncertainties when you multiply of divide quantities.

if a quantity is squared you double the uncertainty, if cubed it is 3 times etc. i.e. multiply by the index. Square root is index half so divide by 2.



Uncertainty in r will be (7% + 3%)/2
Answer B is correct
Reply 4
I dont think its B, it should be C.

You need to add the uncertainties in quadrature (like Pythagoras) , then devide by two.

The uncertainty in r = (1/2) * sqrt[(7^2) + (3^2)] = 3.807 = 4%
Thus I believe C is the correct answer.
Chris454
I dont think its B, it should be C.

You need to add the uncertainties in quadrature (like Pythagoras) , then devide by two.

The uncertainty in r = (1/2) * sqrt[(7^2) + (3^2)] = 3.807 = 4%
Thus I believe C is the correct answer.


I suggest you check this out
http://www.rit.edu/cos/uphysics/uncertainties/Uncertaintiespart2.html#powers

It's not a standard deviation type approach that is required.
Reply 6
Stonebridge
I suggest you check this out
http://www.rit.edu/cos/uphysics/uncertainties/Uncertaintiespart2.html#powers

It's not a standard deviation type approach that is required.


I see what you mean. If DYB's sure that B is correct then I guess it must be these type of absolute errors he's using.
Reply 7
@ Stonebridge

I see what you mean. If DYB's sure that B is correct then it must be these type of absolute errors being used. Sry for confusing :smile:
Reply 8
uncertainties regardless if they are being subtracted or added in the equation are always added:

V=πr2h.
7 = (ignore constant) r2(3)
7-3 = r2
-even though its squared, you do
7-3 / 2 = 2

isnt it D ? :s-smilie:
Glee
uncertainties regardless if they are being subtracted or added in the equation are always added:

V=πr2h.
7 = (ignore constant) r2(3)
7-3 = r2
-even though its squared, you do
7-3 / 2 = 2

isnt it D ? :s-smilie:


No. The standard approach to doing these (up to A Level at least!) is...
If the quantities are added or subtracted, the actual errors are added.
If the quantities are multiplied or divided, the percentage errors are added.
If a quantity is raised to a power (positive or negative), the percentage error is multiplied by the positive value that power.
In the example given the quantities are raised to the power 1/2 and -1/2
Original post by Stonebridge
No. The standard approach to doing these (up to A Level at least!) is...
If the quantities are added or subtracted, the actual errors are added.
If the quantities are multiplied or divided, the percentage errors are added.
If a quantity is raised to a power (positive or negative), the percentage error is multiplied by the positive value that power.
In the example given the quantities are raised to the power 1/2 and -1/2

Yeah, but.. why is it done that way? I have no intuition as to why.
Original post by moomin valley
Yeah, but.. why is it done that way? I have no intuition as to why.


This is 10 years old…
Physics changed a lot in 10 years i see