JEE Main 2021PhysicsExperimental PhysicsMediumMCQ

JEE Main 2021 — Experimental Physics Question with Solution

JEE Main 2021 (25 Feb Shift 1)

Question

The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72nd  division on circular scale coincides with the reference line. The radius of the wire is

Choose an option

Show full solutionCorrect option: B
Correct answer
B0.82 mm

Step-by-step explanation

Least count =1 mm100=0.01 mm

zero error =+8×LC=+0.08 mm

True reading (Diameter)

=1 mm+72×LC- (Zero error)

=1 mm+72×0.01 mm-0.08 mm

=1.72 mm-0.08 mm

=1.64 mm

therefore, radius =1.642=0.82 mm.

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About this question

This is a previous-year question from JEE Main 2021, 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.