A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two laws of the screw gauge are brought in contact, the `45^(th)` division coincides with the main scale is barely visible. What is the thickness of the sheet is the main scale reading is 0.5 mm and the `25^(th)` division coincides with the main scale line ?

Options

A. `0.50 mm`
B. `0.75 mm`
C. `0.80 mm`
D. `0.70 mm`

Explanation

Leas count `=(0.5)/(50)=(1)/(100)=0.01 mm`
Zero error `=-5xx0.01=0.05 mm`
Reading `=0.5+(25)/(100)=0.75 mm`
`:.` Thickness `=0.75-(-0.05)=0.80 mm`