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JEE Main 202127 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
27 July
Shift
Morning Shift
Subject
Physics
Chapter
Experimental Physics

JEE Main 2021 · 27 July · Morning Shift · Physics · Experimental Physics

AssertionA: If in five complete rotations of the circular scale,...

AssertionA:AssertionA: If in five complete rotations of the circular scale, the distance travelled on the main scale of the screw gauge is 5mm5mm and there are 5050 total divisions on a circular scale, then the least count is 0.001cm.0 . 001cm .

ReasonR:ReasonR:
Least Count =PitchTotal divisions on circular scale= \dfrac{\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.

Options

  1. AAA is not correct but RR is correct.
  2. BBoth AA and RR are correct and RR is the correct explanation of A.A .
  3. CAA is correct but RR is not correct.
  4. DBoth AA and RR are correct and RR is NOT the correct explanation of A.A .

Correct answer

A. AA is not correct but RR is correct.

Step-by-step solution

Least count =Pitchtotal division on circular scale= \dfrac{\text{Pitch}}{\text{total division on circular scale}}
In 55 revolution, distance travel, 5mm.5mm .

In 11 revolution, it will travel 1mm.1mm .
So the least count =150=0.02= \dfrac{1}{50} = 0 . 02

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