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About 6% of the power of a 100 W light b...

About `6%` of the power of a 100 W light bulb is converted to visible radiation. The average intensity of visible radiation at a distance of 8 m is (Assume that the radiation is emitted isotropically and neglect reflection.)

A

`3.5xx10^(-3)"W m"^(-2)`

B

`5.1xx10^(-3)"W m"^(-2)`

C

`7.2xx10^(-3)"W m"^(-2)`

D

`2.3xx10^(-3)"W m"^(-2)`

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The correct Answer is:
To solve the problem step by step, we will calculate the average intensity of visible radiation emitted by a 100 W light bulb at a distance of 8 m, given that 6% of the power is converted to visible radiation. ### Step 1: Calculate the Effective Power of Visible Radiation The total power of the light bulb is 100 W, and only 6% of this power is converted to visible radiation. \[ \text{Effective Power} = 6\% \text{ of } 100 \text{ W} = \frac{6}{100} \times 100 = 6 \text{ W} \] ### Step 2: Calculate the Area of the Sphere at 8 m Since the radiation is emitted isotropically, we can consider the light spreading out uniformly in all directions, forming a sphere. The formula for the surface area \(A\) of a sphere is given by: \[ A = 4 \pi r^2 \] Where \(r\) is the radius (distance from the bulb). Here, \(r = 8 \text{ m}\). \[ A = 4 \pi (8)^2 = 4 \pi (64) = 256 \pi \text{ m}^2 \] ### Step 3: Calculate the Intensity of Visible Radiation Intensity \(I\) is defined as the power per unit area. Therefore, we can calculate the intensity of the visible radiation using the effective power and the area we just calculated. \[ I = \frac{\text{Power}}{\text{Area}} = \frac{6 \text{ W}}{256 \pi \text{ m}^2} \] ### Step 4: Simplify the Expression Now, we can calculate the numerical value of the intensity. \[ I = \frac{6}{256 \pi} \approx \frac{6}{804.25} \approx 0.00748 \text{ W/m}^2 \] ### Step 5: Express in Scientific Notation To express the intensity in scientific notation: \[ I \approx 7.48 \times 10^{-3} \text{ W/m}^2 \] ### Final Answer The average intensity of visible radiation at a distance of 8 m is approximately: \[ I \approx 7.48 \times 10^{-3} \text{ W/m}^2 \] ---

To solve the problem step by step, we will calculate the average intensity of visible radiation emitted by a 100 W light bulb at a distance of 8 m, given that 6% of the power is converted to visible radiation. ### Step 1: Calculate the Effective Power of Visible Radiation The total power of the light bulb is 100 W, and only 6% of this power is converted to visible radiation. \[ \text{Effective Power} = 6\% \text{ of } 100 \text{ W} = \frac{6}{100} \times 100 = 6 \text{ W} \] ...
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