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JEE Main 2012 (AIEEE) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2012 (AIEEE)

Q62JEE Main 2012 · AIEEEMedium

Least Count, Vernier Callipers & Screw Gauge

A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading: 58.5 degree

Vernier scale reading: 09 divisions

Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data is

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Answer: (c)

  1. 3030 VSD =29= 29 MSD, so LC =130 MSD=0.5∘30=160∘= \dfrac{1}{30}\text{ MSD} = \dfrac{0.5^\circ}{30} = \dfrac{1}{60}^\circ
  2. Angle =58.5∘+9×160∘=58.5∘+0.15∘= 58.5^\circ + 9\times \dfrac{1}{60}^\circ = 58.5^\circ + 0.15^\circ
  3. So the answer is (c) 58.65∘58.65^\circ.
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