1. Tardigrade
  2. Question
  3. Physics
  4. 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 =( text Pitch / text Total divisionson circular scale ) In the light of the above statements, choose the most appropriate answer from the options given below :

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Q. 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 $=\frac{\text { Pitch }}{\text { Total divisionson circular scale }}$ In the light of the above statements, choose the most appropriate answer from the options given below :

JEE MainJEE Main 2021Physical World, Units and Measurements

Solution:

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