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A black rectangular surface of area ' $a$ ' emits energy ' $E$ ' per second at $27^{\circ} \mathrm{C}$. If length and breadth is reduced to $\left(\frac{1}{3}\right)^{\text {rd }}$ of initial value and temperature is raised to $327^{\circ} \mathrm{C}$ then energy emitted per second becomes
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Verified Answer
The correct answer is:
$\frac{16 \mathrm{E}}{9}$
$$
\mathrm{E}=\mathrm{e} \sigma \mathrm{AT}^4 \text { and } \mathrm{A}=\ell \mathrm{b}
$$
$$
\frac{E^{\prime}}{E}=\frac{A^{\prime}(327+273)^4}{A(27+273)^4}
$$
$$
\frac{E^{\prime}}{E}=\frac{1}{9}\left(\frac{600}{300}\right)^4
$$
$$
\begin{aligned}
& \therefore \mathrm{E}^{\prime}=\frac{1}{9} \times(2)^4 \times \mathrm{E} \\
& \Rightarrow \mathrm{E}^{\prime}=\frac{16 \mathrm{E}}{9}
\end{aligned}
$$
\mathrm{E}=\mathrm{e} \sigma \mathrm{AT}^4 \text { and } \mathrm{A}=\ell \mathrm{b}
$$
$$
\frac{E^{\prime}}{E}=\frac{A^{\prime}(327+273)^4}{A(27+273)^4}
$$
$$
\frac{E^{\prime}}{E}=\frac{1}{9}\left(\frac{600}{300}\right)^4
$$
$$
\begin{aligned}
& \therefore \mathrm{E}^{\prime}=\frac{1}{9} \times(2)^4 \times \mathrm{E} \\
& \Rightarrow \mathrm{E}^{\prime}=\frac{16 \mathrm{E}}{9}
\end{aligned}
$$
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