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Saturday, 31 October

JEE Main 2021 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 24 February 2021, Shift 2, Physics Q30

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

A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is −5-5 dB per km and cable length is 20 km. The power received at receiver is 10−x10^{-x} W. The value of xx is __________. [Gain in dB =10log⁡10(PoPi)= 10\log_{10}\left(\frac{P_o}{P_i}\right)]

Official answer

8

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.