Learn

Saturday, 31 October

JEE Main 2021 · PhysicsNumerical answerNumerical valueHardMulti-step

JEE Main 26 August 2021, Shift 1, Physics Q26

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

The electric field in a plane electromagnetic wave is given by E⃗=200cos⁡[(0.5×103m)x−(1.5×1011rads×t)]Vmj^\vec{E}=200\cos\left[\left(\frac{0.5\times10^{3}}{\mathrm{m}}\right)x-\left(1.5\times10^{11}\frac{\mathrm{rad}}{\mathrm{s}}\times t\right)\right]\frac{\mathrm{V}}{\mathrm{m}}\hat{j} If this wave falls normally on a perfectly reflecting surface having an area of 100 cm2\mathrm{cm^2}. If the radiation pressure exerted by the E.M. wave on the surface during a 10 minute exposure is x109Nm2\frac{x}{10^9}\frac{\mathrm{N}}{\mathrm{m}^2}. Find the value of xx.

Official answer

354

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.