Nearly 10 % of power of a 110,W light bulb is converted to visible radiation. change in average inte

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8.Electromagnetic waves

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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

$84$

(JEE MAIN-2022)

Solution

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

$=\frac{10}{100} \times 110\,W$

$=11\,W$

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

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

$=\frac{11}{4 \pi} \times \frac{24}{25}$

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

Standard 12

Physics

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