JEE Main 2024 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 6 April 2024, Shift 2, Physics Q43

Question 43 of 90 in this shift, Physics question 13 of 30, Section A.

The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is : (Given

e=1.6\times10^{-19}\ \mathrm{C}

)

(1) $6.25\times10^{17}$
(2) $1.25\times10^{19}$
(3) $31.25\times10^{17}$ Official answer
(4) $6.25\times10^{18}$

Official answer

Option 3

NTA final key.

Chapter: Current Electricity
Topic: Ohm's Law and Resistance
Idea tested: Electron Count Rate from Current

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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 42 of the 174 JEE Main shifts held from 2017 to 2026.