Thickness of a thin sheet of aluminium is measured with a screw gauge of pitch of 0.5 mm and having 50 divisions on circular scale . Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the `45^(th)` division coincides with the main scale line left to the zero mark. What will be the thickness (in mm ) of the sheet if the main scale reading is 0.5 mm and the 30th division of circular scale coincides with the main scale line?

Answer

`00.85 `

Explanation

Main Scale Reading
(M.S. R.) = 0.5 mm Circular Scale Division (C.S. D.) = `30^(th)` Number of divisions on circular scale = 50
Pitch of screw = 0.5 mm
`therefore` L.C. of screw gauge
` = (0.5)/(50) = 0.01 mm`
`therefore` zero error
`= -5 xx L.C. = -0.05 mm`
`therefore` Zero correction = `+0.05 mm`
Observed reading
`= 0.5mm + (30 xx 0.01) mm `
`= 0.80 mm `
Corrected reading
`= 0.80mm + 0.05 mm `
`= 0.85 mm `