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JEE Main 2011 (AIEEE) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2011 (AIEEE)

Q63JEE Main 2011 · AIEEEMedium

Least Count, Vernier Callipers & Screw Gauge

A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading: 0 mm

Circular scale reading: 52 divisions

Given that 1 mm on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is:

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Answer: (a)

  1. LC =1 mm100=0.01 mm= \dfrac{1\text{ mm}}{100} = 0.01\text{ mm}
  2. Diameter =0+52×0.01=0.52 mm= 0 + 52\times 0.01 = 0.52\text{ mm}
  3. So the answer is (a) 0.052 cm0.052\text{ cm}.
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