About `6%` of the power of a 100 W light bulb is converted ot 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.)

Options

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)`

Answer

C

Explanation

Here , power of bulb `=100 W`
As intensity, `I = ("Power of visible light")/("area")`
`= (100 xx 6//100)/(4pi(8)^(2)) = 7.2 xx 10^(-3) W m^(-2)`