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

Least count of screw gauge
`LC= ("Pitch")/("Number of division on circular scale")=(0.5)/(50)`
`=0.01mm = 0.001cm`
Negative zero error `= -5xxLC`
`= -5xx0.01mm = -0.05mm`
Measured thickness of alminium sheet is
`trianglex = MSR+ CSR xx LC - "zero error"`
`triangle x =0.5mm+25xx0.01-(-0.05)`
`triangle x= 0.80mm`