# Uncertainty in the volume

1. Sep 7, 2013

### lep11

1. The problem statement, all variables and given/known data
As you eat your way through a bag of chocolate chip cookies, you observe that each cookie is a circular disk with a diameter of 8.50 ± 0.02 cm and a thickness of 0.050 ± 0.005 cm. (a) Find the average volume of a cookie and the uncertainty in the volume. (b) Find the ratio of the diameter to the thickness and the uncertainty in this ratio.

2. Relevant equations
V=pi*r2*h
d=8.50 +/- 0.02 cm → r=4.25+-0.01 cm

3. The attempt at a solution
a.) average volume V=pi*r^2*h=pi*(4.25cm)2*0.050cm≈2.8cm3

volume of the smallest cookie V=pi*(4.24cm)2*0.045cm=2.54152...cm3
volume of the biggest cookie V=pi*(4.26cm)2*0.055cm=3.1356...cm3
uncertainty= (biggest - smallest)/2=0.30cm3
b.) ratio of the diameter to the thickness= 8.50cm/0.050cm=170
uncertainty=biggest ratio-smallest ratio=8.52cm/0.045cm-8.48cm/0.055cm=35
Please check my answers. I am not sure what is meant with that uncertainty thing.

Last edited: Sep 7, 2013
2. Sep 7, 2013

### Staff: Mentor

The idea is that an actual value of a given quantity must lie between the biggest and smallest values and the assumption is made that it will most likely be right in the middle of that range. Half the spread should be "to the left" and half should be "to the right" of the average value. So, divide the total range by two to yield the uncertainty. In other words for f ±Δf you want:
$$Δf = \frac{f_{big} - f_{small}}{2}$$

In part b, check the thickness value you've used for the "biggest" term.

3. Sep 7, 2013

### lep11

So now the volume could be expressed as V=2.8 +-0.30cm3 ?

4. Sep 7, 2013

### Staff: Mentor

V = 2.8 ± 0.3 cm3

Don't add a '0' past the digit you've rounded! That would imply another digit of accuracy, but you've already disposed of it by rounding.

5. Sep 7, 2013

6. Sep 7, 2013

### Staff: Mentor

You wrote:
Shouldn't that be 0.045 cm?

7. Sep 7, 2013

### lep11

Yes it should! A typo. Thanks. Is b otherwise correct or should it be 35/2?

Last edited: Sep 7, 2013
8. Sep 7, 2013

### Staff: Mentor

Not correct. Redo the calculation and remember what I said about dividing the result in half.

9. Sep 7, 2013

### lep11

I appreciate your help. However, I still don't know what was the whole exercise for.

Last edited: Sep 7, 2013
10. Sep 7, 2013

### Staff: Mentor

The idea of uncertainty values is to establish a range over which the results of calculations may lie when they are derived from imperfect data. Every measurement has an error associated with it -- no measurement is ever perfectly accurate.

You will use uncertainty calculations a LOT in the sciences, particularly if you have to do labs and the associated lab reports.