Three discs x, y and z having radii 2 m, 2 | vedclass

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(B) According to Wien's displacement law,the temperature $T$ is inversely proportional to the wavelength of maximum emission: $T \propto \frac{1}{\lam

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$P_z$ is maximum

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(B) According to Wien's displacement law,the temperature $T$ is inversely proportional to the wavelength of maximum emission: $T \propto \frac{1}{\lambda_{\max }}$.From Stefan-Boltzmann law,the power radiated $P$ by a blackbody is given by $P = \sigma A T^4$,where $A$ is the surface area and $\sigma$ is the Stefan-Boltzmann constant.Since $A = \pi r^2$ for a disc (assuming radiation from both sides,$A = 2\pi r^2$),we have $P \propto r^2 T^4$.Substituting $T \propto \frac{1}{\lambda_{\max }}$,we get $P \propto \frac{r^2}{\lambda_{\max }^4}$.Given radii $r_x = 2 \ m, r_y = 2 \ m, r_z = 6 \ m$ and wavelengths $\lambda_x = 3 \ \mu m, \lambda_y = 4 \ \mu m, \lambda_z = 5 \ \mu m$:$P_x \propto \frac{2^2}{3^4} = \frac{4}{81} \approx 0.049$$P_y \propto \frac{2^2}{4^4} = \frac{4}{256} = 0.0156$$P_z \propto \frac{6^2}{5^4} = \frac{36}{625} = 0.0576$Comparing the values,$P_z > P_x > P_y$. Since the question asks for the maximum power,$P_z$ is the maximum.

Column-$I$	Column-$II$
$(a)$ Cup of hot tea in room	$(i)$ Forced convection
$(b)$ Substance kept near fire	$(ii)$ Natural convection
	$(iii)$ Conduction
	$(iv)$ Radiation

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In Column-$I$,system surroundings and in Column-$II$ modes of heat transfer are given below. Match the following:

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$(a)$ Cup of hot tea in room $(i)$ Forced convection

$(b)$ Substance kept near fire $(ii)$ Natural convection

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