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A rectangular metal plate \(8 \mathrm{~cm} \times 4 \mathrm{~cm}\) at \(127^{\circ} \mathrm{C}\) emits \(E \mathrm{Js}^{-1}\). If both length and breadth are halved and the temperature is raised to \(327^{\circ} \mathrm{C}\), the rate of emission is
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Verified Answer
The correct answer is:
\(\left(\frac{81}{64}\right) E \mathrm{Js}^{-1}\)
According to Stefan-Boltzmann's law, rate of emission of radiation from metal surface is given as,
\(\begin{aligned}
E & =\sigma A T^4 \\
\Rightarrow \quad \frac{E_2}{E_1} & =\left(\frac{A_2}{A_1}\right)\left(\frac{T_2}{T_1}\right)^4 \quad \ldots (i)
\end{aligned}\)
\(\begin{aligned}
\text{Given, } A_1 & =8 \mathrm{~cm} \times 4 \mathrm{~cm} \\
& =3.2 \times 10^{-3} \mathrm{~m}^2 \\
T_1 & =127+273=400 \mathrm{~K} \\
A_2 & =\frac{8}{2} \mathrm{~cm} \times \frac{4}{2} \mathrm{~cm} \\
& =4 \times 2 \mathrm{~cm}^2=8 \times 10^{-4} \mathrm{~m}^2 \\
T_2 & =327+273=600 \mathrm{~K}
\end{aligned}\)
Putting these values in Eq. (i), we have
\(\begin{aligned}
& \frac{E_2}{E_1}=\left(\frac{8 \times 10^{-4}}{3.2 \times 10^{-3}}\right)\left(\frac{600}{400}\right)^4 \\
& =\frac{1}{4} \times \frac{81}{16} \\
& \Rightarrow \quad \frac{E_2}{E_1}=\frac{81}{64} \\
& \Rightarrow \quad E_2=\frac{81}{64} E_1=\left(\frac{81}{64}\right) E \mathrm{Js}^{-1}\left[\because E_1=E \mathrm{Js}^{-1}\right] \\
\end{aligned}\)
\(\begin{aligned}
E & =\sigma A T^4 \\
\Rightarrow \quad \frac{E_2}{E_1} & =\left(\frac{A_2}{A_1}\right)\left(\frac{T_2}{T_1}\right)^4 \quad \ldots (i)
\end{aligned}\)
\(\begin{aligned}
\text{Given, } A_1 & =8 \mathrm{~cm} \times 4 \mathrm{~cm} \\
& =3.2 \times 10^{-3} \mathrm{~m}^2 \\
T_1 & =127+273=400 \mathrm{~K} \\
A_2 & =\frac{8}{2} \mathrm{~cm} \times \frac{4}{2} \mathrm{~cm} \\
& =4 \times 2 \mathrm{~cm}^2=8 \times 10^{-4} \mathrm{~m}^2 \\
T_2 & =327+273=600 \mathrm{~K}
\end{aligned}\)
Putting these values in Eq. (i), we have
\(\begin{aligned}
& \frac{E_2}{E_1}=\left(\frac{8 \times 10^{-4}}{3.2 \times 10^{-3}}\right)\left(\frac{600}{400}\right)^4 \\
& =\frac{1}{4} \times \frac{81}{16} \\
& \Rightarrow \quad \frac{E_2}{E_1}=\frac{81}{64} \\
& \Rightarrow \quad E_2=\frac{81}{64} E_1=\left(\frac{81}{64}\right) E \mathrm{Js}^{-1}\left[\because E_1=E \mathrm{Js}^{-1}\right] \\
\end{aligned}\)
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