Experimental Physics

JEE Main level, single correct, medium

60 of 70

Question 60 of 70

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 for a ball are given as: S.No. MS (cm) VS divisions 1. 0.5 8 2. 0.5 4 3. 0.5 6
If the zero error is - 0.03 cm, then mean corrected diameter is:

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Answer: (C) 0.59cm0.59cm

L.C. of Vernier calipers

=1mainscaledivisionTotaldivisonsvernierscale=\dfrac{1mainscaledivision}{Totaldivisonsvernierscale}

=0.110=0.01cm=\dfrac{0.1}{10}=0.01cm

Reading of Vernier calipers

=M.S.R.+(L.C)×vsdivisions.=M.S.R.+\left(L.C\right)\times vsdivisions.

d1=0.5+0.01×8{d}_{1}=0.5+0.01\times 8

d2=0.5+0.01×4{d}_{2}=0.5+0.01\times 4

d3=0.5+0.01×6{d}_{3}=0.5+0.01\times 6

∴∴ Measured diameter are respectively

0.58cm0.54cm,0.56cm0.58cm0.54cm,0.56cm

∴averagediameter=0.58+0.54+0.563∴averagediameter=\dfrac{0.58+0.54+0.56}{3}

=1.683=0.56=\dfrac{1.68}{3}=0.56

Negative zero error is added in the average diameter to get the corrected diameter

∴correcteddiameter=0.56−(−0.03)∴correcteddiameter=0.56-(-0.03)

=0.56+0.03=0.59cm=0.56+0.03=0.59cm

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