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JEE Main 2021 (27 Jul, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2021 (27 Jul, Shift 1)

Q44JEE Main 2021 · 27 JUL · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

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 = PitchTotal divisions on circular scale\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:

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Answer: (a)

  1. Pitch =5 mm5=1 mm= \dfrac{5\text{ mm}}{5} = 1\text{ mm}
  2. LC =1 mm50=0.02 mm=0.002 cm= \dfrac{1\text{ mm}}{50} = 0.02\text{ mm} = 0.002\text{ cm}, not 0.001 cm0.001\text{ cm}, so A is wrong.
  3. R is the correct formula for least count.
  4. So the answer is (a) A is not correct but R is correct.
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