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JEE Main 2023 (30 Jan, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2023 (30 Jan, Shift 1)

Q29JEE Main 2023 · 30 JAN · SHIFT 1Difficult

Least Count, Vernier Callipers & Screw Gauge

In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by 0.5 mm0.5\text{ mm} on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while 4646th division the circular scale coincide with the reference line. The diameter of the wire is ______________ ×10−2 mm\times 10^{-2}\text{ mm}.

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

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Answer: 220

  1. LC =0.5100=0.005 mm= \dfrac{0.5}{100} = 0.005\text{ mm}; zero 6 divisions below line: zero error =+6×0.005=+0.03 mm= +6\times 0.005 = +0.03\text{ mm}
  2. Linear scale reading =4×0.5=2 mm= 4\times 0.5 = 2\text{ mm}; observed reading =2+46×0.005=2.23 mm= 2 + 46\times 0.005 = 2.23\text{ mm}
  3. Diameter =2.23−0.03=2.20 mm=220×10−2 mm= 2.23 - 0.03 = 2.20\text{ mm} = 220\times 10^{-2}\text{ mm}
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