The pitch and the number of circular scale divisions in a screw gauge with least count 0.02 mm are respectively

Answer (3)

We know that,

Least count of screw gauge

$\displaystyle \begin{aligned}&=\frac{Pitch}{Number\:of\:circular\:scale\:division}\\&=\frac {1\:mm} {50}=0.02\:mm\end{aligned}$

Therefore the pitch and no. of circular scale divisions are 1 mm and 50 respectively.