# Absolute Uncertainty of a rule ?Watch

#1
Now, I've been told that the uncertainty of :

a ruler is 0.5 mm

a vernier calliper is 0.005 cm

a micrometer is 0.005 mm

I do this by determining the least count and dividing it by half.

Is this correct ?

However, in the attached question I am first asked to measure the diameter of a mass (just a weight).

The markscheme then goes on to say that I must use 0.1 cm in the calculation of my percentage uncertainty, why ?

This is exactly twice the value of the uncertainty above.
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#2
bump
0
7 years ago
#3
The mark scheme says 0.1mm
A vernier calliper normally measures to the nearest 0.1mm (0.01cm)
Take a look at this, it might help.
1
#4
(Original post by Stonebridge)
x
Ah, yes. Thank you. That cleared up a few other issues too.

My school supplies us with two types of micrometers.

I know the uncertainty on one is smaller than the other. How do I determine the uncertainty ?

I know the least count is used somehow....
0
7 years ago
#5
You can usually tell what the uncertainty is by noting the change on the main scale when you rotate the drum once, and noting how many divisions there are on the scale on the drum.
It's explained on the sheet I posted in part b)
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#6
I've been told that the least count for a vernier calliper is 1 MSD - 1 VSD.

The picture below is of a vernier calliper I own.

1 MSD = 1mm

1 VSD = 2 mm

1 -2 = -1 mm..... what have I done wrong ?

(Original post by Stonebridge)
c
0
7 years ago
#7
The picture shows that the vernier scale has 20 divisions.
(The one in the sheet I posted has 10.)
This means that on yours, 1mm on the main scale can be divided up into 20 divisions.
This means that the smallest reading is 1/20 mm. That is 0.05mm
(The one on my sheet was 1/10 mm or 0.1mm)

Readings on yours would normally be given to ± 0.05mm
0
#8
(Original post by Stonebridge)
The picture shows that the vernier scale has 20 divisions.
(The one in the sheet I posted has 10.)
This means that on yours, 1mm on the main scale can be divided up into 20 divisions.
This means that the smallest reading is 1/20 mm. That is 0.05mm
(The one on my sheet was 1/10 mm or 0.1mm)

Readings on yours would normally be given to ± 0.05mm
So if asked for the percentage uncertainty when using my calliper it would be :

0.05/ (reading in mm) * 100 correct ?

I came across this method on the net to determine the least count :

39 MSD coincides with 20 VSD

(39-20)/20 = 0.95

1 - 0.95 = 0.05 mm
0
7 years ago
#9
Yes. To both.
% uncertainty would be () x 100%
0
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