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Q.
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 wire is :
JEE MainJEE Main 2019Physical World, Units and Measurements
Solution:
Least count = $\frac{\text{Pitch}}{\text{Number} \; \text{of} \;\text{divisio} \;\text{on} \;\text{circular} \;\text{scale} }$
$\begin{aligned}
L C &=\frac{P}{ N } \\
L C &=5 \mu m =5 \times 10^{-6} m . \\
&=10^{-3} m . \\
\Rightarrow & N=\frac{P}{L C}=\frac{10^{-3}}{5 \times 10^{-6}} \\
\Rightarrow & N=200 .
\end{aligned}$