Two electric lamps A and B radiate the same p | vedclass

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Question

(A) According to the Stefan-Boltzmann law,the power radiated by a body is given by $P = e A \sigma T^4$,where $e$ is the emissivity,$A$ is the surface

Vedclass · Accepted Answer

$(\frac{e_B}{e_A})^{1/4}$

Vedclass · Answer

(A) According to the Stefan-Boltzmann law,the power radiated by a body is given by $P = e A \sigma T^4$,where $e$ is the emissivity,$A$ is the surface area,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature.Since both lamps radiate the same power $(P_A = P_B)$ and have the same dimensions (same surface area $A$),we have:$e_A A \sigma T_A^4 = e_B A \sigma T_B^4$Canceling the common terms $A$ and $\sigma$,we get:$e_A T_A^4 = e_B T_B^4$Rearranging the terms to find the ratio $\frac{T_A}{T_B}$:$\frac{T_A^4}{T_B^4} = \frac{e_B}{e_A}$Taking the fourth root on both sides:$\frac{T_A}{T_B} = (\frac{e_B}{e_A})^{1/4}$

Two electric lamps $A$ and $B$ radiate the same power. Their filaments have the same dimensions and have emissivities $e_A$ and $e_B$. Their surface temperatures are $T_A$ and $T_B$. The ratio $\frac{T_A}{T_B}$ will be equal to :-

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