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JEE Main 2025 (4 Apr, Shift 2) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2025 (4 Apr, Shift 2)

Q14JEE Main 2025 · 4 APR · SHIFT 2Medium

Least Count, Vernier Callipers & Screw Gauge

For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm. The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm):

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Answer: (a)

  1. 1 MSD =15 cm300=0.05 cm= \dfrac{15\text{ cm}}{300} = 0.05\text{ cm}
  2. 2525 VSD =24= 24 MSD, so LC =1 MSD−1 VSD=125 MSD= 1\text{ MSD} - 1\text{ VSD} = \dfrac{1}{25}\text{ MSD}
  3. LC =0.0525=0.002 cm= \dfrac{0.05}{25} = 0.002\text{ cm}. So the answer is (a).
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