Experiment #1 Error in Measurements Pre-lab
Questions
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
What is the uncertainty in a measurement made with a meter stick (1 mm
resolution)?
The uncertainty in a measurement made with a meter stick with 1 mm resolution would
be half of the smallest division of the instrument (see Exp1-3):
length  xx.x  0.5 mm
 length  0.5 mm

Calculate the uncertainty for the volume of the following cylinder:
Radius = 21.4 ± 0.2 cm
Height = 1.2 ± 0.1 cm
Hint: r2 = r · r
First, let’s define a few of the necessary equations and known variables we need (See
Exp1-7):
Known Variables:
h  1.2 cm
h  0.1 cm
r  21.4 cm
r  0.2 cm
We can calculate the volume of the cylinder by using the following:
 CYLINDER  r 2 h
 CYLINDER   21.4 cm2 1.2 cm  1726.5 cm3
The uncertainty of the volume is given by:
 r
 CYLINDER    
 CYLINDER

r

h 
r
h 
 r
  0.2 cm  0.1 cm 


 1726.5 cm 3   2
 21.4 cm  1.2 cm 

 



 CYLINDER  1726.5 cm 3  20.0093457   0.08333  1726.5 cm 3  0.1020249
 CYLINDER  553.4 cm 3
We could actually go one step further and determine the complete volume and
uncertainty of the cylinder:
 CYLINDER  1726.5  553.4 cm 3

What is the length reading from the following vernier caliper?
First we have to determine the number of whole millimeters. This is where the tenth’s of
millimeter’s 0 mark falls right after. In this case, the tenth’s of millimeter’s 0 mark falls
right after the 5 mm mark. So this is our 1’s of millimeters measurement. Next, we have
to find where one of the ten’s of millimeter’s lines matches exactly with one of the one’s
of millimeter’s lines. This occurs at 0.8 mm (the 8th line). (See Exp1-10)
Thus, the length reading from the micrometer is 5.8 mm. However, to be completely
correct, we must include the uncertainty of the measurement. This is half of the smallest
division of the instrument – in this case 0.05 mm.
length  5.8  0.05 mm
To go one step further, how would we convert this to cm?
10 mm  1 cm
Notice that if we divide both sides by 10 mm or 1 cm we get the identities:
10 mm
1 cm
1
10 mm
10 mm
or
10 mm
1 cm
1
1 cm
1 cm
Remember from mathematics that we can multiply anything by 1 and not change its
value. So choose the ratio that is most useful (that will cancel out the units you do not
want, and leave you with the units you DO want).
 1 cm 
  3.5 cm
35 mm  35 mm1  35 mm
 10 mm 
Note that the mm’s cancel (since one is in the numerator and one is in the denominator)
and we’re left with centimeters (which we want).
So, our answer in terms of centimeters is:
 1 cm 
  0.58 cm
5.8 mm  5.8 mm
 10 mm 
 1 cm 
  0.005 cm
0.05 mm  0.05 mm
 10 mm 
length  0.58  0.005 cm
For meters, the conversion is:
100 cm  1 m
Hence,
 1m 
  0.0058 m
0.58 cm  0.58 cm
 100 cm 
 1m 
  0.00005 m
0.005 cm  0.005 cm
 100 cm 
length  0.0058  0.00005 m
Finally, notice that all these lengths are the same, just represented in different scales.
length  0.58  0.005 cm  5.8  0.05 mm  0.0058  0.00005 m
Aside:
To convert square-units (or higher order) from one to another you follow the same
process, just remember to “get rid” of all the units that are being converted from:
Convert 13.45 cm2 to units of meters-squared:
 1 m  1 m 

  0.001345 m 2
13.45 cm 2  13.45 cm 2 
100
cm
100
cm






Notice that since we had centimeters-squared we had to use the identity two times to get
rid of both cm’s. If we had only used the identity one time, we’d have had something like
the following:
 1m 
  0.1345 cm  m
13.45 cm 2  13.45 cm 2 
100
cm





That’s a centimeter-meter (whatever that is), not a meter-squared. Notice we have a “leftover” cm that we didn’t get rid of, because only one of the two cm’s canceled out.
Convert 1726.5 cm3 to units of m3:
 1 m  1 m  1 m 


  0.0017265 m 3
1726.5 cm 3  1726.5 cm 3 
 100 cm  100 cm  100 cm 


