Question

Assertion A : If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5mm$ and there are $50$ total divisions on circular scale, then least count is $0.001cm$.
Reason R : Least Count $=\frac{\text{ Pitch }}{\text{ Total divisionson circular scale }}$ In the light of the above statements, choose the most appropriate answer from the options given below :

• A. A is not correct but R is correct.
• B. Both A and R are correct and R is the correct explanation of A.
• C. A is correct but R is not correct.
• D. Both A and R are correct and R is NOT the correct explanation of A.

Answer 1

Answer: (A)

Least Count $=\frac{\text{ Pitch }}{\text{ Total divisionson circular scale }}$
In $5$ revolution, distance travel, $5mm$
In $1$ revolution, it will travel $1mm$.
So least count $=\frac{1}{50}=0.02$