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The least count of the main scale of a screw gauge is $1 \mathrm{~mm}$. The minimum number of divisions on its circular scale required to measure $5 \mu \mathrm{m}$ diameter of a wire is
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200
Least count of the main scale of screw gauge $=1 \mathrm{~mm}$
Least count of screw gauge $=$ Pitch/Number of division on the circular scale
$5 \times 10^{-6}=10^{-3} / \mathrm{N}$
$N=200$
Answer: (2) 200
Least count of screw gauge $=$ Pitch/Number of division on the circular scale
$5 \times 10^{-6}=10^{-3} / \mathrm{N}$
$N=200$
Answer: (2) 200
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