Question

One main scale division of a vernier caliper is '$a$' cm and $n^{th}$ division of the vernier scale coincides with ${\left(n-1\right)}^{th}$ division of the main scale. The least count of the calipers in $mm$ is:

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

Answer 1

The correct option is B

$\raisebox{1ex}{$10a$}\!\left/ \!\raisebox{-1ex}{$n$}\right.$

Step 1: Given data:

One main scale division of a vernier caliper = '$a$'

$n^{th}$ division of the vernier scale coincides with = ${\left(n-1\right)}^{th}$ division of the main scale

Step 2: Formula used:

$n$ VSD = $\left(n-1\right)$ MSD

Where MSD → Main scale division

VSD → Vernier scale division

L⋅C=$1$ MSD – $1$ VSD

Where LC → Least count

Step 3. Vernier caliper:

It is a kind of measuring device that is used for the measurement of linear dimensions. It is also used for the measurement of the diameters of round objects with the help of the measuring jaws.

Step 4: Calculations:

We will use the given data and putting in above for calculations:

$n$ VSD = $\left(n-1\right)$ MSD

$1$  VSD= $\left[\raisebox{1ex}{$\left(n-1\right)$}\!\left/ \!\raisebox{-1ex}{$n$}\right.\right]$ MSD

Now L⋅C=$1$ MSD – $1$ VSD

=$1$ MSD- $\left[\raisebox{1ex}{$\left(n-1\right)$}\!\left/ \!\raisebox{-1ex}{$n$}\right.\right]$MSD

=$1$ MSD-$1$ MSD+ MSD$/n$

= MSD$/n$

=$\left(\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$n$}\right.\right)$cm

=$\left(\raisebox{1ex}{$10a$}\!\left/ \!\raisebox{-1ex}{$n$}\right.\right)$ mm

Hence, option (b) is correct.