The divisions on the main scale of a screw gauge are 1 mm apart and the screw of the spindle advances by 5 main scale divisions when the spindle is given 5 complete rotations . How many divisions are to be provided on the circular scale for the least count of the instrument is to be 1 `mu m` ? What changes are requied of the number if divisions can be only 500 for the same least count?

Explanation

Least count = `(1 MSD)/(N)`
`=(0.5 mm)/(10) =0.05 mm`
Diameter =x +1y
9.75 mm =(x+0.05y) mm
=x+ 0.05 y
Since y `lt ` 10 and 0.05 y `lt` 0.5. 9.75 can be
written as 9.5 + 0.25
(9.5 mm) + (0.25 mm) = (xx mm ) +(0.05 y mm)
since 1 MSD =0.5 mm
The MSR `=(9.5)/(0.5) =10 " and " 0.05 y0.25`
(b) 9.75 =(x+ly) -0.35
=9.75 +0.35
=10.10
=10+ 0.1
x=10 =MSR `=(10)/(0.5) =20`
0.05 =0.1
`y= (0.1)/(0.05) =2 (=VCD)`
MSR =20 ,VCD =2