The least count of the main scale of a screw gauge is $1\, mm$. The minimum number of divisions on its circular scale required to measure $5\,\mu m$ diameter of a wire is
JEE MAIN 2019, Medium
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Least count $= \frac{{\text{Pitch}}}{{\text{No.of divisions on circular scale}}}$
$5 \times {10^{ - 6}} = \frac{{{{10}^{ - 3}}}}{N}$
$N = 200$
$5 \times {10^{ - 6}} = \frac{{{{10}^{ - 3}}}}{N}$
$N = 200$
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