The circular scale of screw gauge has 50 divisions. It move 5 mm, when it is rotated 10 times. The least count of the screw gauge is

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0.01 mm

0.005 mm

0.05 mm

0.001 mm

answer is A.

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Detailed Solution

Pitch of the screw $= \frac{5 m m}{10} = 0.5 m m$
$L C = \frac{P . O . S}{N o . o f c i r c u l a r s c a l e d i v i s i o n s}$
$= \frac{0.5 m m}{50} = 0.01 m m$

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