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(D) The power of the visible radiation emitted by the bulb is $P' = 10\% 	ext{ of } 110\,W = 0.10 	imes 110\,W = 11\,W$.The intensity $I$ at a dista

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$84$

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(D) The power of the visible radiation emitted by the bulb is $P' = 10\% 	ext{ of } 110\,W = 0.10 	imes 110\,W = 11\,W$.The intensity $I$ at a distance $r$ from a point source is given by $I = \frac{P'}{4\pi r^2}$.The intensity at $r_1 = 1\,m$ is $I_1 = \frac{11}{4\pi(1)^2} = \frac{11}{4\pi}$.The intensity at $r_2 = 5\,m$ is $I_2 = \frac{11}{4\pi(5)^2} = \frac{11}{100\pi}$.The change in intensity is $\Delta I = I_1 - I_2 = \frac{11}{4\pi} - \frac{11}{100\pi} = \frac{11}{\pi} \left( \frac{1}{4} - \frac{1}{100} ight)$.$\Delta I = \frac{11}{\pi} \left( \frac{25 - 1}{100} ight) = \frac{11}{\pi} 	imes \frac{24}{100} = \frac{264}{100\pi} = \frac{2.64}{\pi} \approx \frac{2.64}{3.14} \approx 0.84\,W/m^2$.Given $\Delta I = a 	imes 10^{-2}\,W/m^2$,we have $0.84 = a 	imes 10^{-2}$,which implies $a = 84$.

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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