Assertion A : If in five complete rotations of the circular scale, the distance travelled on the main scale of the screw gauge is 5 mm and there are 50 total divisions on a circular scale, then the least count is 0 . 001 cm . Reason R : Least Count = Pitch Total divisions on circular scale In the light of the above statements, choose the most appropriate answer from the options given below.

Question

Assertion A : If in five complete rotations of the circular scale, the distance travelled on the main scale of the screw gauge is 5 mm and there are 50 total divisions on a circular scale, then the least count is 0 . 001 cm . Reason R : Least Count = Pitch Total divisions on circular scale In the light of the above statements, choose the most appropriate answer from the options given below., A is not correct but R is correct., Both A and R are correct and R is the correct explanation of A ., A is correct but R is not correct., Both A and R are correct and R is NOT the correct explanation of A .

MARKS App · Accepted Answer

Least count = Pitch total division on circular scale In 5 revolution, distance travel, 5 mm . In 1 revolution, it will travel 1 mm . So the least count = 1 50 = 0 . 02

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