Question

In a screw gauge, the pitch of the screw is 0.5 mm and the number of circular scale divisions are 50, then the least count is ______ $\times {10}^{-2}$mm.

Answer 1

$\begin{array}{cc}\text{for Screw gauge :}& \\ \text{Least count}& =\frac{\text{Pitch}}{\text{No. of circular scale divisions}}\\ & =\frac{0.5mm}{50}\\ & ={10}^{-2}mm.\end{array}$