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

We have a 110 W light bulb, and nearly 10% of its power is converted to visible radiation. So the power of visible radiation is $$P = 0.10 \times 110 = 11$$ W.

The intensity of radiation at a distance $$r$$ from a point source is given by $$I = \frac{P}{4\pi r^2}$$.

At a distance $$r_1 = 1$$ m: $$I_1 = \frac{11}{4\pi (1)^2} = \frac{11}{4\pi}$$

At a distance $$r_2 = 5$$ m: $$I_2 = \frac{11}{4\pi (5)^2} = \frac{11}{4\pi \times 25} = \frac{11}{100\pi}$$

The change in intensity is: $$\Delta I = I_1 - I_2 = \frac{11}{4\pi} - \frac{11}{100\pi} = \frac{11}{\pi}\left(\frac{1}{4} - \frac{1}{100}\right) = \frac{11}{\pi} \times \frac{25 - 1}{100} = \frac{11 \times 24}{100\pi}$$

$$\Delta I = \frac{264}{100\pi} = \frac{264}{314.159} \approx 0.8403$$ W/m$$^2$$ $$= 84.03 \times 10^{-2}$$ W/m$$^2$$.

Hence the value of $$a = 84$$.

Hence, the correct answer is 84.