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QJEEM22Phe476: JEE Main 2022 Physics - Class 12 JEEM22Phe476 Solution: Nearly 10% of the power of a 110   W light bulb is…

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×10−2 W/m2. The value of 'a' will be _________.

Step-by-Step Verified Solution

P′=10%\mathbf{P}^{\prime}=10 \% of 110W110 \mathbf{W}

=10100×110 W=\frac{10}{100} \times 110 \mathrm{~W}

=11 W=11 \mathrm{~W}

I1−I2=P′4πr12−P′4πr22\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}

=114π[11−125]=\frac{11}{4 \pi}\left[\frac{1}{1}-\frac{1}{25}\right]

=114π×2425=\frac{11}{4 \pi} \times \frac{24}{25}

=264π×10−2=84×10−2 W/m2=\frac{264}{\pi} \times 10^{-2}=84 \times 10^{-2} \mathrm{~W} / \mathrm{m}^2

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