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About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation :at a distance of 1m from the bulb ? |
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Answer» <P> SOLUTION :Here, power of RADIATION = 5% of ELECTRICAL power`=100xx(5)/(100)` `THEREFORE P=5W` According to formula, `I=(P)/(A)` `therefore I=(P)/(4pi r^(2))` `therefore I=(5)/(4xx3.14xx(1)^(2)) ""(because r=1m)` `therefore I=0.3981 Wm^(-2)~~ 0.4 Wm^(-2)` |
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