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JEE Mains · Physics · STD 12 - Electromagnetic waves
  • JEE Mains 2022
  • Medium
  • Numerical

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.

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Correct Answer

(D) \(84\)

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Detailed explanation

\(P ^{\prime}=10 \%\) of \(110\,W\)

\(=\frac{10}{100} \times 110\,W\)

\(=11\,W\)

\(I _{1}- I _{2}=\frac{ P ^{\prime}}{4 \pi r _{1}^{2}}-\frac{ P ^{\prime}}{4 \pi r _{2}^{2}}\)

\(=\frac{11}{4 \pi}\left[\frac{1}{1}-\frac{1}{25}\right]\)

\(=\frac{11}{4 \pi} \times \frac{24}{25}\)

\(=\frac{264}{\pi} \times 10^{-2}=84 \times 10^{-2}\,W / m ^{2}\)

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