JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Units & Measurement

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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 is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm.

Reason R: Least Count

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

In the light of the above statement, choose the most appropriate answer from the options given below:

(A) $A$ is correct but $R$ is not correct.
(B) $A$ is not correct but $R$ is correct.✓ Correct
(C)Both $A$ and $R$ are correct and $R$ is NOT the correct explanation of $A$ .
(D)Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$ .

Answer: (B)

Solution

The Reason R correctly states the formula for least count. In 5 rotations, the distance is 5 mm, so pitch = 1 mm per rotation. Least count =

\frac{1 mm}{50} = 0.02

mm = 0.002 cm, not 0.001 cm as stated in Assertion A. Therefore, A is incorrect but R is correct.

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Units & Measurement

Solution

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