Errors in measurementHard
Question
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 measurementsfor a ball are given as :
If the zero error is − 0.03 cm, then mean corrected diameter is :
If the zero error is − 0.03 cm, then mean corrected diameter is :
Options
A.0.53 cm
B.0.56 cm
C.0.59 cm
D.0.52 cm
Solution
Least count = 0.01 cm
d1 = 0.5 + 8 × 0.01 + 0.03 = 0.61 cm
d2 = 0.5 + 4 × 0.01 + 0.03 = 0.57 cm
d3 = 0.5 + 6 × 0.01 + 0.03 = 0.59 cm
Mean diameter =
= 0.59 cm
d1 = 0.5 + 8 × 0.01 + 0.03 = 0.61 cm
d2 = 0.5 + 4 × 0.01 + 0.03 = 0.57 cm
d3 = 0.5 + 6 × 0.01 + 0.03 = 0.59 cm
Mean diameter =
= 0.59 cmCreate a free account to view solution
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