JEE Main 2015 · 10 Apr · Shift 1 · Physics Q30

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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 :

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 :

0.56 cm

0.59 cm

0.53 cm

0.52 cm

Answer and worked solution

Official answer: B

Least count: $0.1-0.09=0.01$ cm. Readings $0.58$ , $0.54$ and $0.56$ cm, mean $0.56$ cm. Correcting the zero error of $-0.03$ :
$0.56-(-0.03)=0.59\text{ cm}.$

Topics:Vernier callipers and screw gauge

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JEE Main 2015 · 10 Apr · Shift 1, Physics question 30