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NEET 2018 · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from NEET 2018

Q57NEET 2018Medium

Least Count, Vernier Callipers & Screw Gauge

A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of –0.004 cm, the correct diameter of the ball is

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Answer: (d)

  1. Observed reading =MSR+CSR×LC=0.5 cm+25×0.001 cm=0.525 cm= \text{MSR} + \text{CSR}\times\text{LC} = 0.5\text{ cm} + 25\times 0.001\text{ cm} = 0.525\text{ cm}
  2. Correct reading == observed −- zero error =0.525−(−0.004)= 0.525 - (-0.004)
  3. So the answer is (d) 0.529 cm0.529\text{ cm}.
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