1. Visible power emitted by the bulb

Nearly $10\%$ of the bulb's power is converted to visible radiation.

$P_{\text{visible}} = 0.10 \times 110 = 11\,\text{W}$

2. Intensity at distance $r$

Assuming the bulb radiates uniformly in all directions, intensity at distance $r$ is

$I = \frac{P}{4\pi r^2}$

So,

• At $r=1\,\text{m}$:

$I_1 = \frac{11}{4\pi(1)^2} = \frac{11}{4\pi}$

• At $r=5\,\text{m}$:

$I_5 = \frac{11}{4\pi(5)^2} = \frac{11}{100\pi}$

3. Change in intensity

$\Delta I = I_1 - I_5 = \frac{11}{4\pi} - \frac{11}{100\pi}$

Taking LCM:

$\Delta I = \frac{275 - 11}{100\pi} = \frac{264}{100\pi} = \frac{66}{25\pi}$

Now using $\pi \approx \frac{22}{7}$,

$\Delta I = \frac{66}{25\times \frac{22}{7}} = \frac{66\times 7}{550} = 0.84\,\text{W/m}^2$

4. Match with the given form

Given,

$\Delta I = a \times 10^{-2}\,\text{W/m}^2$

So,

$0.84 = 84 \times 10^{-2}$

Hence,

$a=84$

5. Comparison with stored answer

Stored correct answer = $84$

Our derived answer also = $84$.

So the answer agrees with the stored answer.