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JEE Main 2022 (26 Jul, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2022 (26 Jul, Shift 1)

Q34JEE Main 2022 · 26 JUL · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

A screw gauge of pitch 0.5 mm0.5\text{ mm} is used to measure the diameter of uniform wire of length 6.8 cm6.8\text{ cm}, the main scale reading is 1.5 mm1.5\text{ mm} and circular scale reading is 7. The calculated curved surface area of wire to appropriate significant figures is:

[Screw gauge has 50 divisions on its circular scale]

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Answer: (b)

  1. LC =0.550=0.01 mm= \dfrac{0.5}{50} = 0.01\text{ mm}; d=1.5+7×0.01=1.57 mm=0.157 cmd = 1.5 + 7\times 0.01 = 1.57\text{ mm} = 0.157\text{ cm}
  2. Curved surface area =πdL=3.14×0.157×6.8≈3.35 cm2= \pi d L = 3.14\times 0.157\times 6.8 \approx 3.35\text{ cm}^2
  3. Least significant figures in data are 2 (6.86.8), so area =3.4 cm2= 3.4\text{ cm}^2. So the answer is (b).
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