Question

the pitch of a screw Guage is 0.5mm and the head scale is divided in 100 parts. what is the L. C of the screw Guage?

Answer 1

0.0005 m , but in my lab pitch is 1 , and least cound of my screw gauge is 0.001 . dividei it by 2 , you will get 0.005

Answer 2

Pitch of the Screw Guage = 0.5 mm

Number of Divisions on the circular scale = 100

Least Count of the Screw Guage ( L.C.)

$= \frac{pitch \: of \: the \: screw \: guage}{number \: of \: divisions \: on \: the \: circular \: scale } \\ \\ l.c. = \frac{0.05}{100} = 0.005 \\ \\ least \: count \: (l.c.) = 0.005mm$

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