JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
Nearly \(10 \%\) of the power of a \(110\,W\) light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of \(1\, m\) from the bulb to a distance of \(5\,m\) is \(a \times 10^{-2}\,W / m ^{2}\). The value of ' \(a\) ' will be.
- A \(80\)
- B \(29\)
- C \(54\)
- D \(84\)
Answer & Solution
Correct Answer
(D) \(84\)
Step-by-step Solution
Detailed explanation
\(P ^{\prime}=10 \%\) of \(110\,W\) \(=\frac{10}{100} \times 110\,W\) \(=11\,W\) \(I _{1}- I _{2}=\frac{ P ^{\prime}}{4 \pi r _{1}^{2}}-\frac{ P ^{\prime}}{4 \pi r _{2}^{2}}\) \(=\frac{11}{4 \pi}\left[\frac{1}{1}-\frac{1}{25}\right]\) \(=\frac{11}{4 \pi} \times \frac{24}{25}\)…
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