A screw gauge with a pitch of 0.5 mm and a circular scale with 50 division is used to measure the thickness of a thin sheet of aluminium. Before starting the measurement, it is found that when the two jaws of the screw are brought in constant, 45th division coincides with the main scale line and that the zero of main scale line and that the zero of main scale is barely visible. What is the thickness of the sheet, if the main scale reading is 0.5 mm and 25th division coincides with the main scalel line ?

Options

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

Answer

C

Explanation

Least count `= ("pitch")/("no. of division on circular scale")=(0.5"mm")/(50)`
`L.C = 0.001 mm`
`-ve` zero error `= -5 xx L.C = -0.005 mm`
Measured value =
main scale reading + screw gauge reading - zero error
`=0.5 mm+[25xx0.001-(-0.05)]mm`
`=0.8 mm`