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JEE Main 2020 (6 Sep, Shift 2) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2020 (6 Sep, Shift 2)

Q50JEE Main 2020 · 6 SEP · SHIFT 2Medium

Least Count, Vernier Callipers & Screw Gauge

A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings 5.50 mm, 5.55 mm, 5.45 mm, 5.65 mm. The average of these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The average diameter

of the pencil should therefore be recorded as:

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Answer: (a)

  1. The readings are given to 0.01 mm, so the result is kept to two decimal places.
  2. Mean 5.5375→5.545.5375 \to 5.54 mm; deviation 0.07395→0.070.07395 \to 0.07 mm
  3. So the answer is (a) (5.54±0.07)(5.54 \pm 0.07) mm.
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