The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions. (i) What is the pitch of the screw gauge? (ii) What is the least count of the screw gauge?

(i) Pitch = distance moved ahead in 1 revolution

Given the distance covered in two revolutions = 1 mm

∴ we get,

Pitch = 1/2 = 0.5 mm

Hence, the pitch of the screw gauge = 0.5 mm

Pitch = 0.5 mm

Total number of divisions on a circular scale = 50

Substituting the values in the formula above we get,

$\text{Least count}=\; 0.5/50\; mm$

$\text{Least count}=0.01\text{ mm}$

Hence, the least count of the screw gauge = 0.01 mm