Q 11-01-225JEE MainJEE Main 2019 (12 Jan, Shift 1)Easy
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:
Answer: (D) $200$
The least count must be $5\ \mu$m. Taking the pitch equal to the main-scale least count, $1$ mm:
$$N = \frac{1\ \text{mm}}{5\ \mu\text{m}} = \frac{1000}{5} = 200$$
Solution by Sreeraj P, M.Sc Physics