Physics · 2024
JEE Main · 6 April 2024, Shift 2 · Q43
The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is: (Given e =1.6 × 10^-19 C )
The number of electrons flowing per second in the filament of a $\displaystyle 110$ W bulb operating at $\displaystyle 220$ V is : (Given $\displaystyle \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ )
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle 31.25 \times 10^{17}$
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JEE Main 2024 Physics question, with the answer from NTA’s final answer key. Where our answers come from.