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JEE Main 2024 (6 Apr, Shift 2) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2024 (6 Apr, Shift 2)

Q20JEE Main 2024 · 6 APR · SHIFT 2Medium

Least Count, Vernier Callipers & Screw Gauge

In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its 44th division coincides exactly with a certain division on main scale. If 50 vernier scale divisions equal to 49 main scale divisions and zero error in the instrument is 0.04 mm0.04\text{ mm} then how many main scale divisions are there in 1 cm1\text{ cm}?

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Answer: (d)

  1. 50 VSD=49 MSD⇒50\text{ VSD} = 49\text{ MSD} \Rightarrow LC =1 MSD50= \dfrac{1\text{ MSD}}{50}
  2. Zero error magnitude =4×LC=0.04= 4\times\text{LC} = 0.04 mm ⇒\Rightarrow LC =0.01= 0.01 mm
  3. 1 MSD =50×0.01=0.5= 50\times 0.01 = 0.5 mm
  4. Divisions in 1 cm =100.5=20= \dfrac{10}{0.5} = 20, answer (d).
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