JEE Main 2022PhysicsExperimental PhysicsMediumNumerical

JEE Main 2022 — Experimental Physics Question with Solution

JEE Main 2022 (29 Jun Shift 2)

Question

The Vernier constant of Vernier callipers is 0.1 mm and it has zero error of -0.05 cm. While measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5 . The corrected diameter will be _____×10-2  cm.

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Show full solutionCorrect answer: 180
Correct answer
180

Step-by-step explanation

Vernier constant is the least count of the vernier V.C=0.1 mm=0.01 cm

zero error =-0.05 cm

∴correction=0.05 cm

M.S.R.=1.7 cm

V.S.R.=5 V.C.=5×0.01=0.05 cm

The diameter of the sphere will be,

diameter=MSR+VSR+correction

1.7+0.05+0.05=1.8 cm

=180×10-2 cm

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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.