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(B) According to Stefan-Boltzmann's law,the rate of emission of radiation $E$ from a metal surface is given by $E = \sigma A T^4$.Taking the ratio for

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$\left(\frac{81}{64}\right) E \ Js^{-1}$

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(B) According to Stefan-Boltzmann's law,the rate of emission of radiation $E$ from a metal surface is given by $E = \sigma A T^4$.Taking the ratio for two states,we have $\frac{E_2}{E_1} = \left(\frac{A_2}{A_1}ight) \left(\frac{T_2}{T_1}ight)^4$.Given:$A_1 = 8 \ cm 	imes 4 \ cm = 32 \ cm^2$$A_2 = 4 \ cm 	imes 2 \ cm = 8 \ cm^2$$T_1 = 127 + 273 = 400 \ K$$T_2 = 327 + 273 = 600 \ K$Substituting these values into the ratio formula:$\frac{E_2}{E_1} = \left(\frac{8}{32}ight) \left(\frac{600}{400}ight)^4$$\frac{E_2}{E_1} = \left(\frac{1}{4}ight) \left(\frac{3}{2}ight)^4 = \frac{1}{4} 	imes \frac{81}{16} = \frac{81}{64}$.Therefore,$E_2 = \left(\frac{81}{64}ight) E \ Js^{-1}$.

$A$ rectangular metal plate $8 \ cm \times 4 \ cm$ at $127^{\circ} C$ emits $E \ Js^{-1}$. If both length and breadth are halved and the temperature is raised to $327^{\circ} C$,the rate of emission is

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