1.Units, Dimensions and Measurement
medium

Asseretion $A:$ 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 $=\frac{\text { Pitch }}{\text { Total divisions on circular scale }}$

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

A

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

B

$A$ is not correct but $R$ is correct.

C

Both $A$ and $R$ are correct and $R$ is NOT the correct explanation of $A$.

D

${A}$ is correct but ${R}$ is not correct.

(JEE MAIN-2021)

Solution

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 $=\frac{1}{50}=0.02$

Standard 11
Physics

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