JEE Main 2019PhysicsExperimental PhysicsMediumMCQ

JEE Main 2019 — Experimental Physics Question with Solution

JEE Main 2019 (09 Jan Shift 2)

Question

The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.

The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is:

Choose an option

Show full solutionCorrect option: C
Correct answer
C5.725 mm

Step-by-step explanation

least count =0.5 mm100=0.005 mm

Reading =5.5 mm+48-3×0.005 mm

=5.725 mm.

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About this question

This is a previous-year question from JEE Main 2019, 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.