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Asseretion $A:$ If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5\, {mm}$ and there are $50$ total divisions on circular scale, then least count is $0.001\, {cm}$.
Reason $R:$ Least Count $=\frac{\text { Pitch }}{\text { Total divisions on circular scale }}$
In the light of the above statements, choose the most appropriate answer from the options given below:
Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$
$A$ is not correct but $R$ is correct.
Both $A$ and $R$ are correct and $R$ is NOT the correct explanation of $A$.
${A}$ is correct but ${R}$ is not correct.
Solution
Least count $=\frac{\text { Pitch }}{\text { total division on circular scale }}$
In $5$ revolution, distance travel, $5 \,mm$
In $1$ revolution, it will travel $1 \,mm$.
So least count $=\frac{1}{50}=0.02$