Question:
One main scale division of a vernier callipers is ' $\mathrm{a}$ ' $\mathrm{cm}$ and $\mathrm{n}^{\text {th }}$ division of the vernier scale coincide with $(\mathrm{n}-1)^{\mathrm{th}}$ division of the main scale. The least count of the callipers in $\mathrm{mm}$ is :
Correct Option: 4
Solution:
(4)
$(\mathrm{n}-1) \mathrm{a}=\mathrm{n}\left(\mathrm{a}^{\prime}\right)$
$\mathrm{a}^{\prime}=\frac{(\mathrm{n}-1) \mathrm{a}}{\mathrm{n}}$
$\therefore$ L. $\mathrm{C}=1 \mathrm{MSD}-1 \mathrm{VSD}$
$=\left(\mathrm{a}-\mathrm{a}^{\prime}\right) \mathrm{cm}$
$=a-\frac{(n-1) a}{n}$
$=\frac{n a-n a+a}{n}=\frac{a}{n} \mathrm{~cm}$
$=\left(\frac{10 a}{n}\right) m m$