Diameter of a steel ball is measured using a Vernier calipers 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

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Correct answer: C. 0 . 59cm

L.C. of Vernier calipers = (1mainscaledivision)/(Totaldivisonsvernierscale) = (0.1)/(10) = 0.01cm Reading of Vernier calipers = M . S . R . + (L . C) × vsdivisions . d _ 1 = 0.5 + 0.01 × 8 d _ 2 = 0.5 + 0.01 × 4 d _ 3 = 0.5 + 0.01 × 6 ∴ Measured diameter are respectively 0 .58 cm0 .54 cm, 0 .56 cm ∴ averagediameter = (0.58 + 0.54 + 0.56)/(3) = (1.68)/(3) = 0.56 Negative zero error is added in the average diameter to get the corrected diameter ∴ correcteddiameter = 0.56 - ( - 0.03 ) = 0.56 + 0.03 = 0.59cm

S.No.	MS (cm)	VS divisions
1.	0.5	8
2.	0.5	4
3.	0.5	6

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