JEE Main 2022PhysicsExperimental PhysicsMediumMCQ

JEE Main 2022 — Experimental Physics Question with Solution

JEE Main 2022 (29 Jul Shift 1)

Question

In an experiment to find out the diameter of wire using screw gauge, the following observation were noted:

(a) Screw moves 0.5 mm on main scale in one complete rotation
(b) Total divisions on circular scale =50
(c) Main scale reading is 2.5 mm
(d) 45th  division of circular scale is in the pitch line
(e) Instrument has 0.03 mm negative error Then the diameter of wire is :

Choose an option

Show full solutionCorrect option: C
Correct answer
C2.98 mm

Step-by-step explanation

From the data given in the question,

Pitch of the screw gauge=0.5 mm.

Total divisions on the circular scalen=50.

Therefore, the least count of the screw gauge is LC=pitchn=0.550=0.01 mm

Now, MSR=2.5 mm, CSR=45

Diameter reading =MSR+LC×CSR- zero error

⇒d=2.5+0.45--0.03

⇒d=2.98 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 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.