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A screw gauge has a least count of 0.005...

A screw gauge has a least count of 0.005 mm and its head scale is divided into 200 equal division. The distance between consecutive threads on the screw is:

A

`0.25mm`

B

`0.5mm`

C

`1.00mm`

D

`2.00mm`

Text Solution

Verified by Experts

The correct Answer is:
C
`LC = ("Pitch")/("No of" CSD)`
so pitch `= 0.005 xx 200 = 1.00mm`
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