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When Vernier Scale Divisions are smaller than Main Scale Divisions, the least count is given by $1 \text{MSD} - 1 \text{VSD}$.

But what if say, $3 \text{VSD}=5 \text{MSD}$ ?

Do we write the least count as $1 \text{VSD} - 1 \text{MSD}$ ?I am having difficulty in visualizing the situation. Can someone please explain it with the help of diagrams?

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  • $\begingroup$ Practically, does the situation of a Vernier scale division being bigger than the main scale division ever arise? $\endgroup$ Commented Apr 24, 2017 at 2:45

1 Answer 1

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The least count is $\frac{1}{3}$MSD.

If 3VSD = 5MSD then 1VSD =$\frac{5}{3}$MSD. The Vernier scale division is just short of 2MSD. They differ by $\frac{1}{3}$MSD. Therefore the least count is $\frac{1}{3}$MSD.

Graphically it looks like this (for zero distance):

Main scale
0  1  2  3  4  5  6  7  8  9  10
|  |  |  |  |  |  |  |  |  |  |
|    |    |    
0    1/3  2/3  
Vernier scale

A measurement of $5\frac{2}{3}$ would look like this:

Main scale
0  1  2  3  4  5  6  7  8  9  10
|  |  |  |  |  |  |  |  |  |  |
                 |    |    |    
                 0    1/3  2/3  
Vernier scale

In general you can find the least count by finding the minimum nonzero difference between any integer multiple of the VSD and any integer multiple of the MSD. $$\text{least count}=\min_{\text{non zero}}(|n\cdot MSD-m\cdot VSD|) \quad m,n\in \mathbb{N}$$ If you fill in $m=n=1$, you obtain your original equation.

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  • $\begingroup$ So, if 8 vsd = 5 msd then why is least count not 3/8 mm . why is it 1/8 mm? $\endgroup$
    – maveric
    Commented Mar 5, 2019 at 6:17

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