Saved Bookmarks
| 1. |
A black metal foil is warmed by radiation from a small sphere at temperature `T` and at a distance d it is found that the power received by the foil is `P` If both the temperature and the distance are doubled the power received by the foil will be .A. `16P`B. `4P`C. `2P`D. `P` |
|
Answer» `P_(Recived)=IA` `I="Intensity"` `"Intensity"=("Power of same")/("Area of sphere")` `(P_(0))/(4pir^(2))=(sigmaeA_(1)T_(0)^(4))/(4pir^(2))` `P=(sigmaeA_(1)T_(0)^(4))/(4pir^(2))A` [`A=` Area of foil] `P_(2)=(sigmaeA_(1)(2T_(0))^(4)A)/(4pi(2r^(2)))=4P` `A_(1)=` area of source |
|