Question

In a vernier callipers, each $cm$ on the main scale is divided into $20$ equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be..... $\times 10^{-2}mm$.

Answer 1

Answer: 5

$20MSD=1cm$
$1MSD=\frac{1}{20}cm$
$10VSD=9MSD$
$1VSD=\frac{9}{10}MSD$
$=\frac{9}{10}\times \frac{1}{20}cm$
$1VSD=\frac{9}{200}cm$
$VC=1MSD-1VSD$
$=\frac{1}{20}cm-\frac{9}{200}cm$
$=\frac{1}{200}\times 10mm$
$VC=5\times 10^{-2}mm$