MHT CET · Physics · Thermal Properties of Matter
Two spheres ' \(\mathrm{S}_{1}\) ' and ' \(\mathrm{S}_{2}\) ' have same radii but temperatures \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) respectively.
Their emissive power is same and emissivity is in the ratio \(1: 4\). Then the ratio of \(\mathrm{T}_{1}\)
to \(\mathrm{T}_{2}\) is
- A \(\sqrt{2}: 1\)
- B \(1: 2\)
- C \(2: 1\)
- D \(1: \sqrt{2}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{2}: 1\)
Step-by-step Solution
Detailed explanation
Emissive power \(P=\sigma \mathrm{e} T^{4}\)
\(
\begin{array}{l}
\therefore \sigma e_{1} T_{1}^{4}=\sigma e_{2} T_{2}^{4} \\
\frac{T_{1}^{4}}{T_{2}^{4}}=\frac{e_{2}}{e_{1}}=4 \\
\therefore \frac{T_{1}}{T_{2}}=\sqrt{2}
\end{array}
\)
\(
\begin{array}{l}
\therefore \sigma e_{1} T_{1}^{4}=\sigma e_{2} T_{2}^{4} \\
\frac{T_{1}^{4}}{T_{2}^{4}}=\frac{e_{2}}{e_{1}}=4 \\
\therefore \frac{T_{1}}{T_{2}}=\sqrt{2}
\end{array}
\)
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