Q.

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 isa×102W/m2 . The value of 'a' will be

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answer is 84.

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Detailed Solution

P=10% of 110W=10100×110W11WI1I2=P4πr12P4πr22=114π11125=114π×2425=264π×102=84×102W/m2 

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