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JEE Main 2021 (25 Feb, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2021 (25 Feb, Shift 1)

Q45JEE Main 2021 · 25 FEB · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72nd division on circular scale coincides with the reference line. The radius of the wire is:

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Answer: (c)

  1. LC =1100=0.01 mm= \dfrac{1}{100} = 0.01\text{ mm}; zero 8 divisions below reference line: zero error =+0.08 mm= +0.08\text{ mm}
  2. Observed reading =1+72×0.01=1.72 mm= 1 + 72\times 0.01 = 1.72\text{ mm}
  3. Diameter =1.72−0.08=1.64 mm= 1.72 - 0.08 = 1.64\text{ mm}
  4. Radius =0.82 mm= 0.82\text{ mm}. So the answer is (c).
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