If 50 Vernier divisions are equal to 49 main scale divisions of a...

The number of divisions on the vernier scale $= 50$

Main scale division $( MSD ) = 0 . 5mm$

As, $50$ divisions on the vernier scale coincide with $49$ divisions on the main scale,

$VSD = \dfrac{49}{50}MSD$

The minimum inaccuracy in the measurement of distance by a vernier scale,

= Least count of the vernier scale

$= 1MSD – 1VSD$

$= 1MSD - \dfrac{49}{50}MSD$

$= \dfrac{1}{50}MSD = \dfrac{1}{50} \times 0.5mm$

$= 0 . 01mm$.