JEE Main 2023PhysicsExperimental PhysicsEasyNumerical

JEE Main 2023Experimental Physics Question with Solution

JEE Main 2023 (30 Jan Shift 1)

Question

In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by 0.5 mm on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while 46th division the circular scale coincide with the reference line. The diameter of the wire is ______ ×10-2 mm.

Enter your answer

Show full solutionCorrect answer: 220
Correct answer
220

Step-by-step explanation

Least count of screw gauge is=PitchNo. of circular divisions

=0.5 mm100

Least count, LC=5×10-3 mm

The zero of circular scale lies 6 divisions below the line of graduation. So, zero error is positive.

Positive zero error =MSR+CSR(LC)

=0 mm+65×10-3 mm

Diameter of wire, d=MSR+CSR (LC)Positive zero error

=4×0.5 mm+465×10-3-65×10-3 mm

=2 mm+40×5×10-3 mm=2.2 mm

Hence, diameter of wire is 220×10-2 mm.

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 2023, 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.