Two spherical bodies A (radius 6 ,cm) and B ( | vedclass

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(A) According to Wien's displacement law,$\lambda_m T = 	ext{constant}$,so $\lambda_A T_A = \lambda_B T_B$.Given $\lambda_A = 500 \,nm$ and $\lambda_

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$9$

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(A) According to Wien's displacement law,$\lambda_m T = 	ext{constant}$,so $\lambda_A T_A = \lambda_B T_B$.Given $\lambda_A = 500 \,nm$ and $\lambda_B = 1500 \,nm$,we have $500 T_A = 1500 T_B$,which implies $T_A / T_B = 3$.The rate of total energy radiated by a black body is given by $P = \sigma A T^4 = \sigma (4 \pi r^2) T^4$.Therefore,the ratio of the power radiated by $A$ to that of $B$ is $\frac{P_A}{P_B} = \left( \frac{r_A}{r_B} ight)^2 \left( \frac{T_A}{T_B} ight)^4$.Substituting the values: $\frac{P_A}{P_B} = \left( \frac{6}{18} ight)^2 	imes (3)^4 = \left( \frac{1}{3} ight)^2 	imes 81 = \frac{1}{9} 	imes 81 = 9$.

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