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Physics · Atomic Physics

JEE Main 2024 — 6 April, Shift 2 — Question 44

The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is : (Given e=1.6×10−19C\mathrm{e}=1.6 \times 10^{-19} \mathrm{C} )

  1. Option A:

    31.25×101731.25 \times 10^{17}

    Correct
  2. Option B:

    6.25×10186.25 \times 10^{18}

  3. Option C:

    6.25×10176.25 \times 10^{17}

  4. Option D:

    1.25×10191.25 \times 10^{19}

Answer: A

Step-by-step solution

Power (P)=(\mathrm{P})= V.I ⇒110=(220)(I)\Rightarrow 110=(220)(I)

⇒I=0.5 A\Rightarrow \mathrm{I}=0.5 \mathrm{~A}

Now, I=n⋅etI=\frac{\mathrm{n} \cdot \mathrm{e}}{\mathrm{t}}

⇒0.5=(nt)(1.6×10−19)\Rightarrow 0.5=\left(\frac{\mathrm{n}}{\mathrm{t}}\right)\left(1.6 \times 10^{-19}\right)

⇒nt=0.51.6×10−19\Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=\frac{0.5}{1.6 \times 10^{-19}}

⇒nt=31.25×1017\Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=31.25 \times 10^{17}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
6 April, Shift 2
Subject
Physics
Chapter
Atomic Physics
Topic
Photon Theory of Light and Radiation Pressure

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