In a screw gauge, the zero of mainscale coincides with fifth division of circular scale in figure (i). The circular division of screw gauge are `50`. It moves `0.5 mm` on main scale In one rotation. The diameter of the ball in figure (ii) is

Options

A. (a) `2.25mm`
B. (b) `2.20mm`
C. ( c ) `1.20mm`
D. (d) `1.25mm`

Answer

C

Explanation

( c ) `Least count = (0.5)/(50) = 0.01mm`
Zero error `= 5 xx L.C = 5 xx 0.01mm = 0.05mm`
`Diameter of ball = [Reading on main scale] + [Reading on circular scale xx L.C] - Zero error`
`= 0.5 xx 2 + 25 xx 0.01 - 0.05 = 1.20mm`