MHT CET · Physics · Thermal Properties of Matter
Two spheres \(\mathrm{S}_1\) and \(\mathrm{S}_2\) have same radii but temperatures \(T_1\) and \(T_2\) respectively. Their emissive power is same and emissivity in the ratio 1:4. Then the ratio \(T_1: T_2\) is
- A \(2: 1\)
- B \(\sqrt{2}: 1\)
- C \(1: \sqrt{2}\)
- D \(1: 2\)
Answer & Solution
Correct Answer
(B) \(\sqrt{2}: 1\)
Step-by-step Solution
Detailed explanation
Given: \(\frac{\mathrm{Q}_1}{\mathrm{~A}_1 \mathrm{t}}=\frac{\mathrm{Q}_2}{\mathrm{~A}_2 \mathrm{t}}\)
But \(A_1=A_2\) as radii are same
\(\begin{array}{ll}
\therefore & \mathrm{Q}_1=\mathrm{Q}_2 \\
\therefore & \mathrm{e}_1 \sigma \mathrm{AT}_1^4=\mathrm{e}_2 \sigma \mathrm{AT}_2^4 \\
\therefore & \frac{\mathrm{~T}_1^4}{\mathrm{~T}_2^4}=\frac{\mathrm{e}_2}{\mathrm{e}_1}=\frac{4}{1} \\
\therefore & \frac{\mathrm{~T}_1}{\mathrm{~T}_2}=\left(\frac{4}{1}\right)^{\frac{1}{4}}=\frac{\sqrt{2}}{1}
\end{array}\)
But \(A_1=A_2\) as radii are same
\(\begin{array}{ll}
\therefore & \mathrm{Q}_1=\mathrm{Q}_2 \\
\therefore & \mathrm{e}_1 \sigma \mathrm{AT}_1^4=\mathrm{e}_2 \sigma \mathrm{AT}_2^4 \\
\therefore & \frac{\mathrm{~T}_1^4}{\mathrm{~T}_2^4}=\frac{\mathrm{e}_2}{\mathrm{e}_1}=\frac{4}{1} \\
\therefore & \frac{\mathrm{~T}_1}{\mathrm{~T}_2}=\left(\frac{4}{1}\right)^{\frac{1}{4}}=\frac{\sqrt{2}}{1}
\end{array}\)
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