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JEE Main 2015 (10 Apr, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2015 (10 Apr, Shift 1)

Q60JEE Main 2015 · 10 APR · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

Diameter of a steel ball is measured using a Vernier callipers which has divisions of 0.1 cm on its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main scale. Three such measurements for a ball are given as:

Figure

If the zero error is −- 0.03 cm, then mean corrected diameter is:

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Answer: (b)

  1. LC =0.1−0.09=0.01= 0.1 - 0.09 = 0.01 cm
  2. Readings: 0.5+8(0.01)=0.580.5 + 8(0.01) = 0.58, 0.540.54, 0.560.56 cm; mean =0.56= 0.56 cm
  3. Zero error −0.03-0.03 cm, so corrected =0.56−(−0.03)=0.59= 0.56 - (-0.03) = 0.59 cm
  4. So the answer is (b) 0.590.59 cm.
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