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Two spherical black bodies of radii `R_(1)` and `R_(2)` and with surface temperature `T_(1)` and `T_(2)` respectively radiate the same power. `R_(1)//R_(2)` must be equal toA. `((T_(2))/(T_(1)))^(2)`B. `((T_(2))/(T_(1)))^(4)`C. `((T_(1))/(T_(2)))^(2)`D. `((T_(1))/(T_(2)))^(4)` |
Answer» Correct Answer - A Radius of Ist body = `r_(1)` radius of Iind body = `r^_(2)` Temperature of Ist body = `T_(1)` Temperature of Iind body = `T_(2)` The emissivity of a black body or radiated power is given `"by, "E=AsigmaT^(4)xxtpropAT^(4)` or `Aprop(E)/(T^(4))` (where,A is the surface area of the spherical black body) As in the condition of question , the power radiated by Ist and IInd body is same. `"Hence " Aprop(1)/(T^(4))` So , `(A_(1))/(A_(2))=((T_(2))/(T_(1)))^(4)rArr(4pir_(1)^(2))/(4pir_(2)^(2))=((T_(2))/(T_1))^(4)` `"Thus "(r_(1))/(r_(2))=((T_(2))/(T_(1)))^(2)` |
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