Question

The head scale of a screw gauge has 200 divisions. Its head advances its 1 mm for each 2 complete rotations of its head. Find its least count.

• A. 0.0025 mm
• B. 0.056 mm
• C. 0.025 mm
• D. 0.0342 mm

Answer 1

The correct option is A 0.0025 mm

Head of the screw gauge advances 1 mm for two complete rotations.

So, pitch of the screw gauge $P=\frac{1mm}{2}=0.5mm$

Number of divisions $N=200$

So, Least count $L.C.=\frac{P}{N}=\frac{0.5mm}{200}=0.0025mm$