Question

A screw gauge has a least count equal to 0.0005 mm. The spindle of the gauge advances 0.1 mm when the screw is turned through 2 revolutions. Find the number of divisions thimble of the screw gauge contains.

Answer 1

$\begin{array}{c}L\cdot C=0.0005mm\\ 2\text{revolution}\to 0.1mm\\ \text{1 revolution}\to \frac{0.1}{2}\\ \text{pitch}=0.05mm\\ L\cdot C=\frac{\text{pitch}}{\text{no. of division on circular scale}}\end{array}$

$\begin{array}{c}\Rightarrow \text{no of division on circular}\\ \text{scale}=\frac{\text{pitch}}{L\cdot C}\\ =\frac{0.05}{0.0005}\\ =100⟶\text{Ans.}\end{array}$