JEE Main 2022PhysicsExperimental PhysicsMediumNumerical

JEE Main 2022 — Experimental Physics Question with Solution

JEE Main 2022 (26 Jun Shift 1)

Question

In a vernier callipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be _____ ×10-2 mm

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Show full solutionCorrect answer: 5
Correct answer
5

Step-by-step explanation

Vernier constant is defined as the difference between the value of one main scale division and one vernier scale division. It is also known as the least count of vernier calliper.

The main scale division of the given vernier calliper is MSD=1 cm20=0.5 mm.

LC=MSDn=0.5 mm10=0.05 mm=5×10-2 mm

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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.