Measurements and Experimentation — Question 10

(i) As we know,

Pitch = distance moved ahead in 1 revolution

Given,

Distance covered in two revolutions = 1 mm

∴, 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