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________.

P = 10% of 110W = 11W

(Visible)

$I_1-I_2=\frac{\rho }{4\pi }\left(\frac{1}{r_1^2}-\frac{1}{r_2^2}\right)$

$=\frac{11}{4\times \frac{22}{7}}\left(\frac{1}{1}-\frac{1}{25}\right)$

$\frac{7}{8}\times \left(\frac{24}{25}\right)=a\times 10^{-2}W/m^2$

$\frac{28\times 24}{8}\times 10^{-2}=a\times 10^{-2}$

 ⇒ 84 × 10^-2 = a × 10^-2  ⇒ a = 84

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