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(A) According to Stefan-Boltzmann Law,the total emissive power $E$ of a black body is given by $E = \sigma T^4$,where $\sigma$ is the Stefan-Boltzmann

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$E_x$ is maximum

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(A) According to Stefan-Boltzmann Law,the total emissive power $E$ of a black body is given by $E = \sigma T^4$,where $\sigma$ is the Stefan-Boltzmann constant and $T$ is the absolute temperature.According to Wien's Displacement Law,the wavelength $\lambda_m$ corresponding to maximum intensity is inversely proportional to the absolute temperature $T$,i.e.,$T = \frac{b}{\lambda_m}$,where $b$ is Wien's constant.Substituting $T$ in the expression for $E$,we get $E \propto (\frac{1}{\lambda_m})^4 = \frac{1}{\lambda_m^4}$.Note that the emissive power $E$ of a black body depends only on its temperature,not on its surface area or radius.Given $\lambda_x = 200 \ nm$,$\lambda_y = 300 \ nm$,and $\lambda_z = 400 \ nm$.Since $E \propto \frac{1}{\lambda_m^4}$,we have:$E_x \propto \frac{1}{200^4}$$E_y \propto \frac{1}{300^4}$$E_z \propto \frac{1}{400^4}$Comparing the values,since $200  \frac{1}{300^4} > \frac{1}{400^4}$.Therefore,$E_x > E_y > E_z$,which means $E_x$ is the maximum.

Three black discs $x, y, z$ have radii $1 \ m, 2 \ m$ and $3 \ m$ respectively. The wavelengths corresponding to maximum intensity are $200 \ nm, 300 \ nm$ and $400 \ nm$ respectively. The relation between emissive power $E_x, E_y$ and $E_z$ is:

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