The rectangular surface of a black body at a temperature of $127^{o} C$ emits energy at the rate of E per second. If the area of the surface is reduced to 1/4th of the initial value and the temperature is raised to $327^{o} C$ , the rate of emission of energy will become

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$\frac{81}{64} E$

$\frac{9}{16} E$

$\frac{3}{8} E$

$\frac{81}{16} E$

answer is D.

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Detailed Solution

$(Q)_{Blackbody} = A σ T^{4} t \Rightarrow \frac{Q}{t} \propto P = A σ T^{4}$
Breadth are halved so area becomes one fourth.
$\begin{aligned} \Rightarrow \frac{P_{1}}{P_{2}} = \frac{A_{1}}{A_{2}} \times {(\frac{T_{1}}{T_{2}})}^{4} \Rightarrow \frac{A_{1}}{(A_{1} / 4)} \times (\frac{273 + 327}{273 + 127}) \\ \Rightarrow P_{2} = \frac{81}{64} E \end{aligned}$