1.

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`

Answer» Correct Answer - C
`LC = ("Pitch")/("No of" CSD)`
so pitch `= 0.005 xx 200 = 1.00mm`


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