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On the head scale of any screw guage, th...

On the head scale of any screw guage, there are 50 divisions marked. If the thread difference of the screw is 1 mm then least count of screw guage will be

A

0.50 mm

B

0.02 mm

C

0.002 mm

D

0.05 mm

Text Solution

Verified by Experts

The correct Answer is:
C
Least count `= ("Thread difference")/("Number of divisions on head scale")`
`= (1)/(50) = 0.02 mm`
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