Assertion 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 $=\frac{\text { Pitch }}{\text { Totaldivisionson circular scale }}$
In the light of the above statements, choose the most appropriate answer from the options given below :
Correct Option: 1,
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 $=\frac{1}{50}=0.02$
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