Read the screw gauge shown below in the figure.

Given that circular scale has 100 divisions and in one complete rotation the screw advances by $1 \mathrm{mm}$.

L C $=\frac{\text{ Pitch }}{\text{ Number of divisions on circular scale }}$

$=\frac{1 \mathrm{mm}}{100}=0.01 \mathrm{mm}$

Linear scale reading =10 (pitch) $=10 \mathrm{mm}$

Circular scale reading $=n\left(\mathrm{LC}\right)=65\times 0.01$

$=0.65 \mathrm{mm}$

Total reading $=(10+0.65)\mathrm{mm}=10.65 \mathrm{mm}$