A screw gauge with a pitch of `0.5mm` and a circular scale with `50` divisions is used to measure the thicknes of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the jaws of the screw gauge are brought in cintact, the `45^(th)` division coincide with the main scale line and the zero of the main scale is barely visible. what is the thickness of the sheet if the main scale readind is `0.5 mm` and the `25th` division coincide with the main scale line?

Options

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

Answer

B

Explanation

Screw gauge has negative zero error
least count of screw gauge
`LC=("Pitch")/("Number of divisions on circular scale")`
`LC=(0.5mm)/(50)=0.01mm`
zero error`=(45-50)xx0.01mm=-0.05mm`
Thickness of sheets=Main Scale reading `+`(cuircular scale reading `xx1C`)`-`zero error
`=0.5+(25xx0.01)-(-0.05)`
`=0.5+0.30=0.80mm`