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

Answer

A

Explanation

Least count `=("Pitch")/("No of division on circular scale") `
`=(0.5 mm)/(50)`
`implies` LC = 0.01mm
`-` ve zero error `=-5 xx LC= -0.05` mm
Measured value = Main scale reading + Screw gauge reading - zero error `= 0.5 mm+ [25 xx 0.01 - (-0.05 mm)] = 0.80mm`.