Question

Calculate the least count of the screw gauge.

Answer 1

Least count:

1. The distance moved by the tip of the screw when cranked through one division of the head scale is defined as the screw gauge's least count.
2. The smallest division on the main scale will be $1\mathrm{mm}$ and it will be the pitch $\left(p\right)$ of the screw gauge. The number of circular divisions is $n=100$.
3. Mathematically, the least count ($\mathrm{LC}$) of the screw gauge can be represented as:

$\mathrm{LC}=\frac{\mathrm{Pitch}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{divisions}\mathrm{on}\mathrm{the}\mathrm{circular}\mathrm{scale}}...\left(1\right)$

Substitute the values in equation (1) as follows:

$$\begin{gathered} \mathrm{LC}=\frac{\mathrm{Pitch}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{divisions}\mathrm{on}\mathrm{the}\mathrm{circular}\mathrm{scale}} \\ =\frac{1\mathrm{mm}}{100} \\ ={10}^{-2}\mathrm{mm} \end{gathered}$$

Therefore, the least count of screw gauges is ${10}^{-2}\mathrm{mm}$.