JEE Main 2022PhysicsThermal Properties of MatterMediumNumerical

JEE Main 2022 — Thermal Properties of Matter Question with Solution

JEE Main 2022 (29 Jul Shift 2)

Question

Nearly 10% of the power of a 110 W light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of 1 m from the bulb to a distance of 5 m is a×10-2 W m-2. The value of 'a' will be

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Show full solutionCorrect answer: 84
Correct answer
84

Step-by-step explanation

Power of visible radiation is P'=10100×110 W =11 W

Change in average intensity of visible radiation is  ∆Iradiation=Iradiation1-Iradiation2

⇒I1-I2=P'4πr12-P'4πr22

=114π11-125  =114π×2425

=264π×10-2=84×10-2 W m-2

Hence, the value of a=84.

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About this question

This is a previous-year question from JEE Main 2022, covering the Thermal Properties of Matter 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.