Learn

Saturday, 31 October

JEE Main 2022 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 29 July 2022, Shift 2, Physics Q52

Question 52 of 90 in this shift, Physics question 22 of 30, Section B.

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/m2a\times10^{-2}\ \mathrm{W/m^2}. The value of 'a' will be _______.

Official answer

84

NTA final key.

Chapter
Electromagnetic Waves
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.