A black sphere has radius R whose rate of rad | vedclass

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(C) According to Stefan-Boltzmann Law,the rate of radiation (power) $P$ from a black body is given by $P = \sigma A T^4$,where $A$ is the surface area

Vedclass · Accepted Answer

$\frac{81 E}{4}$

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(C) According to Stefan-Boltzmann Law,the rate of radiation (power) $P$ from a black body is given by $P = \sigma A T^4$,where $A$ is the surface area and $T$ is the absolute temperature.For a sphere,the surface area $A = 4 \pi R^2$.Thus,$E = \sigma (4 \pi R^2) T^4$.When the radius becomes $R' = R/2$ and the temperature becomes $T' = 3T$,the new rate of radiation $E'$ is:$E' = \sigma (4 \pi (R/2)^2) (3T)^4$$E' = \sigma (4 \pi R^2 / 4) (81 T^4)$$E' = \frac{81}{4} \sigma (4 \pi R^2) T^4$Since $E = \sigma (4 \pi R^2) T^4$,we have $E' = \frac{81}{4} E$.

$A$ black sphere has radius $R$ whose rate of radiation is $E$ at temperature $T$. If the radius is made $R/2$ and the temperature $3T$,the rate of radiation will be:

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