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JEE Main 2022 (27 Jul, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2022 (27 Jul, Shift 1)

Q39JEE Main 2022 · 27 JUL · SHIFT 1Difficult

Least Count, Vernier Callipers & Screw Gauge

The one division of main scale of Vernier callipers reads 1 mm1\text{ mm} and 10 divisions of Vernier scale is equal to the 9 divisions on main scale. When the two jaws of the instrument touch each other, the zero of the Vernier lies to the right of zero of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies in between 4.1 cm4.1\text{ cm} and 4.2 cm4.2\text{ cm} and 66th Vernier division coincides scale division. The diameter of the bob will be ____________ ×\times 10−-2 cm.

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

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Answer: 412

  1. LC =1−0.9=0.1 mm=0.01 cm= 1 - 0.9 = 0.1\text{ mm} = 0.01\text{ cm}
  2. Vernier zero right of main zero: zero error =+4×0.01=+0.04 cm= +4\times 0.01 = +0.04\text{ cm}
  3. Observed reading =4.1+6×0.01=4.16 cm= 4.1 + 6\times 0.01 = 4.16\text{ cm}
  4. Diameter =4.16−0.04=4.12 cm=412×10−2 cm= 4.16 - 0.04 = 4.12\text{ cm} = 412\times 10^{-2}\text{ cm}
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