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. `1.2 mm`
B. `1.25 mm`
C. `2.20 mm`
D. `2.25 mm`

Answer

A

Explanation

`"Least count" = ("Pitch")/("Number of division on circular scale") = ( 0.5)/( 50)`
`= 0.01 mm`
Now , diameter of ball `= (2 xx 0.5 ) + ( 25 - 5) (0.01) = 1.2 mm `