MHT CET · Physics · Thermal Properties of Matter
Two bodies A and B at temperatures ' \(\mathrm{T}_1\) ' \(\mathrm{K}\) and ' \(\mathrm{T}_2\) ' \(\mathrm{K}\) respectively have the same dimensions. Their emissivities are in the ratio \(1: 3\). If they radiate the same amount of heat per unit area per unit time, then the ratio of their temperatures \(\left(T_1: T_2\right)\) is
- A \(1: 3\)
- B \(3^{1 / 4}: 1\)
- C \(9^{1 / 4}: 1\)
- D \(81: 1\)
Answer & Solution
Correct Answer
(B) \(3^{1 / 4}: 1\)
Step-by-step Solution
Detailed explanation
From Stefan - Boltzmann's law
\(\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{e}\left(\sigma \mathrm{AT}^4\right)\)
Given \(\mathrm{A}\) and \(\frac{\mathrm{dQ}}{\mathrm{dt}}\) are same for both the bodies
\(\begin{gathered}
\Rightarrow \mathrm{e}_1 \mathrm{~T}_1^4=\mathrm{e}_2 \mathrm{~T}_2^4 \\
\therefore \quad\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^4=\frac{\mathrm{e}_2}{\mathrm{e}_1}=\frac{3}{1} \\
\Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\sqrt[4]{3}}{1}=\frac{3^{\frac{1}{4}}}{1}
\end{gathered}\)
\(\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{e}\left(\sigma \mathrm{AT}^4\right)\)
Given \(\mathrm{A}\) and \(\frac{\mathrm{dQ}}{\mathrm{dt}}\) are same for both the bodies
\(\begin{gathered}
\Rightarrow \mathrm{e}_1 \mathrm{~T}_1^4=\mathrm{e}_2 \mathrm{~T}_2^4 \\
\therefore \quad\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^4=\frac{\mathrm{e}_2}{\mathrm{e}_1}=\frac{3}{1} \\
\Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\sqrt[4]{3}}{1}=\frac{3^{\frac{1}{4}}}{1}
\end{gathered}\)
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