Nearly 10% of the power of a $110 \mathrm{~W}$ light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of $1 \mathrm{~m}$ from the bulb to a distance of $5 \mathrm{~m}$ is $a \times 10^{-2} \mathrm{~W} / \mathrm{m}^{2}$. The value of 'a' will be _________.
Answer (integer)
84
Solution
$\mathbf{P}^{\prime}=10 \%$ of $110 \mathbf{W}$
<br/><br/>$=\frac{10}{100} \times 110 \mathrm{~W}$
<br/><br/>$=11 \mathrm{~W}$
<br/><br/>$\mathrm{I}_1-\mathrm{I}_2=\frac{\mathrm{P}^{\prime}}{4 \pi \mathrm{r}_1^2}-\frac{\mathrm{P}^{\prime}}{4 \pi \mathrm{r}_2^2}$
<br/><br/>$=\frac{11}{4 \pi}\left[\frac{1}{1}-\frac{1}{25}\right]$
<br/><br/>$=\frac{11}{4 \pi} \times \frac{24}{25}$
<br/><br/>$=\frac{264}{\pi} \times 10^{-2}=84 \times 10^{-2} \mathrm{~W} / \mathrm{m}^2$
About this question
Subject: Physics · Chapter: Electromagnetic Waves · Topic: Properties of EM Waves
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