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JEE Main 2022 (26 Jun, Shift 2) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2022 (26 Jun, Shift 2)

Q37JEE Main 2022 · 26 JUN · SHIFT 2Medium

Least Count, Vernier Callipers & Screw Gauge

A travelling microscope is used to determine the refractive index of a glass slab. If 40 divisions are there in 1 cm on main scale and 50 Vernier scale divisions are equal to 49 main scale divisions, then least count of the travelling microscope is __________ ×\times 10−-6 m.

Numerical answer type. Work it out, then open the answer.

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Answer: 5

  1. 1 MSD =1 cm40=0.025 cm=2.5×10−4 m= \dfrac{1\text{ cm}}{40} = 0.025\text{ cm} = 2.5\times 10^{-4}\text{ m}
  2. 5050 VSD =49= 49 MSD, so LC =150 MSD= \dfrac{1}{50}\text{ MSD}
  3. LC =2.5×10−450=5×10−6 m= \dfrac{2.5\times 10^{-4}}{50} = 5\times 10^{-6}\text{ m}, so the answer is 5.
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