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) As we know,

Pitch = distance moved ahead in 1 revolution

Given,

Distance covered in two revolutions = 1 mm

Therefore, we get,

$\text {Pitch} = \dfrac{1}{2} \text{ mm} = 0.5 \text{ mm} \\[0.5em]$

Hence, pitch of the screw gauge = 0.5 mm

(ii) As we know,

$\text{Least Count} = \dfrac {\text{Pitch}}{\text{Total no. of div on circular head}}\\[0.5em]$

Given,

Pitch = 0.5 mm

Total number of divisions on circular scale = 50

Substituting the values in the formula above we get,

$\text {Least count} = \dfrac{0.5}{50} \\[0.5em] \text {Least count} = 0.01 \text { mm} \\[0.5em]$

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