Question

One main scale division of a Vernier calliper is ${}^{\prime }{a}^{\prime }\text{cm}$ and the $n^{th}$ division of the Vernier scale coincides with the $(n-1{)}^{th}$ division of the main scale. The least count of the calliper (in $\text{mm}$) is:

• A. $\left(\frac{n-1}{10n}\right)a$
• B. $\frac{10a}{n}$
• C. $\frac{10na}{(n-1)}$
• D. $\frac{10a}{(n-1)}$

Answer 1

The correct option is B $\frac{10a}{n}$

$MSD\to$ Main scale division
$VSD\to$ Vernier scale division
$LC\to$ Least count

Given,
$nVSD=(n-1)MSD$
$\Rightarrow 1VSD=\left(\frac{n-1}{n}\right)MSD$

$L.C=1MSD-1VSD$
$=1MSD-\left(\frac{n-1}{n}\right)MSD$
$=1MSD-1MSD+\frac{MSD}{n}$
$=\frac{MSD}{n}$
$=\frac{a}{n}\text{cm}$
$=\frac{10a}{n}\text{mm}$
Hence, option (b) is correct.