Learn

Saturday, 31 October

JEE Main 2021 · PhysicsNumerical answerNumerical valueMediumMulti-step

JEE Main 25 February 2021, Shift 2, Physics Q28

Question 28 of 90 in this shift, Physics question 28 of 30, Section B.

The peak electric field produced by the radiation coming from the 8 W bulb at a distance of 10 m is x10μ0cπ Vm\frac{x}{10}\sqrt{\frac{\mu_0 c}{\pi}}\ \frac{V}{m}. The efficiency of the bulb is 10% and it is a point source. The value of xx is ____________.

Official answer

2

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.