JEE Main 2015PhysicsExperimental PhysicsMediumMCQ

JEE Main 2015 — Experimental Physics Question with Solution

JEE Main 2015 (10 Apr Online)

Question

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:

Choose an option

Show full solutionCorrect option: C
Correct answer
C0.59 cm

Step-by-step explanation

L.C. of Vernier calipers

= 1 main scale divisionTotal divisons vernier scale  

=0.110=0.01 cm

Reading of Vernier calipers

=M.S.R. +L.C×vs divisions.

d1=0.5+0.01×8

d2=0.5+0.01×4

d3=0.5+0.01×6

∴    Measured diameter are respectively

0.58 cm     0.54 cm,    0.56 cm

∴   average diameter = 0.58 + 0.54 + 0.563 

=1.683=0.56

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

∴   corrected diameter=0.56-(-0.03)

=0.56+0.03=0.59 cm

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About this question

This is a previous-year question from JEE Main 2015, covering the Experimental Physics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.