In a screw gauge, $5$ complete rotations of circular scale give $1.5\, mm$ reading on linear scale. Circular scale has $50$ divisions. Least count of the screw gauge is
A$0.006 \,mm$
B$0.003\, mm$
C$0.015\, mm$
D$0.03\, mm$
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A$0.006 \,mm$
a Pitch $=\frac{1.5}{5}=0.3 \mathrm{mm}$
Least count $=\frac{0.3}{50}=0.006 \mathrm{mm}$
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