JEE Main 2022PhysicsExperimental PhysicsMediumMCQ

JEE Main 2022Experimental Physics Question with Solution

JEE Main 2022 (26 Jul Shift 2)

Question

In a Vernier Caliper 10 divisions of Vernier scale is equal to the 9 divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and 4th Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to 1 mm. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and 6th  Vernier scale division exactly. coincides with the main scale reading. The diameter of the spherical body will be:

Choose an option

Show full solutionCorrect option: C
Correct answer
C3.10 cm

Step-by-step explanation

Given that 1 M.S.D=1 mm

And 9 M.S.D =10 V.S.D

1 V.S.D=0.9M.S.D=0.9 mm

L.C of vernier caliper =1-0.9=0.1 mm=0.01 cm

Zero error =10-4×0.1 mm=0.6 mm as at zero the, the zero of the vernier is shifted towards left zero error is negative. Therefore, zero error=-0.6 mm.

Reading =M.S.R+V.S.R-  Zero error

Reading=3+6×0.01--0.06

Reading=3+0.06+0.06

Reading=3.12 cm

The closest option is 3.10 cm

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Experimental Physics chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2022, 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.