Question. A screw gauge has least count $0.01\ \text{mm}$ and there are 50 divisions in its circular scale. The pitch of the screw gauge is:
a. $0.01\ \text{mm}$
b. $0.25\ \text{mm}$
c. $0.5\ \text{mm}$
d. $1.0\ \text{mm}$

Answer. c. $0.5\ \text{mm}$

Solution. Least count (LC) $=\dfrac{\text{pitch}}{\text{number of circular divisions}}$.

Therefore

$
\text{pitch}=\text{LC}\times N=0.01\ \text{mm}\times 50=0.5\ \text{mm}.
$