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# A surface of area 1m2 kept perpendicular to the sun rays absorbs 1.8MJ of solar energy in one hour.(a) What is the amount of electrical energy produced when a solar panel of area 5m2 is exposed perpendicular to the sun rays for 5h. Take the efficiency of solar panel as 30%.(b) Calculate the power generated in kW.

A
13.5 MJ; 0.75 kW
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B
1.5 MJ; 0.75 kW
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C
13.5 MJ; 1.75 kW
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D
1.5 MJ; 1.75 kW
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Solution

## The correct option is A 13.5 MJ; 0.75 kWThe amount of energy absorbed by 1m2 area =1.8MJh−1=1.8×106Jh−1The area of solar panel =5m2.∴ The total amount of energy absorbed by solar panel in 1 hour =5×1.8×106Jh−1=9×106Jh−1∴ The amount of energy absorbed in 5h=5×9×106J=45×106J∴ Efficiency of conversion =30%.∴ Amount of electric energy produced =30% [45×106]J =30100×[45×106]J=1.5×9×106J =13.5×106J=13.5MJ.∴ Power generated (p)=Et=energyunittime=13.5×106J5hour =13.5×106J5×60×60s=13.536×5×104Js−1 =13.5180×104W=13.518×103W=0.75kW∴ Power generated =0.75kW.  Suggest Corrections  0      Similar questions
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