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JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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Assertion A: If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge in 5 mm and there are 50 total divisions on circular scale, then least count of screw gauge is 0.002 cm.
Reason R: Least count

=\frac{\text { Pitch }}{\text { Total divisions on circular scale }}

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Correct option: C

Both A and R are true and R is the correct explanation of A

Option C is correct because both the assertion and reason are true, and the reason correctly explains how the least count of a screw gauge is calculated.

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