In a screw gauge, 5 complete rotations of the screw cause if to move a linear distance of `0.25 cm`. There are `100` circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is :

Options

A. `0.4300 cm`
B. `0.3150 cm`
C. `0.0430 cm`
D. `0.2150 cm`

Answer

D

Explanation

It moves a distance of 0.25cm in five rotations and there are 100 divisions on circular scale : hence
Least count of screw gauge `=(0.25)/(5xx100^(-4))cm`
`=5xx10^(-4)cm`
One rotation cover linear distance
`=(0.25cm)/(5)=0.05cm`
Reading of thickness of wire = main scale division `xx n+` circular Scale division `xx` least count
`=4xx0.05cm +30xx5xx10^(-4)cm`
`=(0.2+0.0150)cm=0.2150cm`.