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Saturday, 31 October

JEE Main 2021 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 26 February 2021, Shift 1, Physics Q26

Question 26 of 90 in this shift, Physics question 26 of 30, Section B.

A radiation is emitted by 1000 W bulb and it generates an electric field and magnetic field at P, placed at a distance of 2 m. The efficiency of the bulb is 1.25%. The value of peak electric field at P is x×10−1x\times10^{-1} V/m. Value of xx is __________. (Rounded-off to the nearest integer) [Take ε0=8.85×10−12 C2 N−1 m−2\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^2\,N^{-1}\,m^{-2}}, c=3×108 ms−1]\mathrm{c}=3\times10^{8}\ \mathrm{ms^{-1}}]

Official answer

137

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.