JEE Advanced · Physics · 12. Thermal Properties
Two spherical bodies \(A\) (radius \(6 \mathrm{~cm}\) ) and \(B\) (radius \(18 \mathrm{~cm}\) ) are at temperatures \(T_1\) and \(T_2\), respectively. The maximum intensity in the emission spectrum of \(A\) is at \(500 \mathrm{~nm}\) and in that of \(B\) is at \(1500 \mathrm{~nm}\). Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by \(A\) to that of \(B\) ?
- A 3
- B 6
- C 9
- D 12
Answer & Solution
Correct Answer
(C) 9
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \text { } \lambda_m \propto \frac{1}{T} \\
& \therefore \frac{\lambda_A}{\lambda_B}=\frac{T_B}{T_A}=\frac{500}{1500}=\frac{1}{3} \\
& E \propto T^4 A \text { (where } A=\text { surface area } \\
& =4 \pi R^2 \text { ) } \\
& \therefore E \propto T^4 R^2 \\
& \frac{E_A}{E_B}=\left(\frac{T_A}{T_B}\right)^4\left(\frac{R_A}{R_B}\right)^2 \\
& =(3)^4\left(\frac{6}{18}\right)^2=9
\end{aligned}
\)
\(\therefore\) Answer is 9 .
\begin{aligned}
& \text { } \lambda_m \propto \frac{1}{T} \\
& \therefore \frac{\lambda_A}{\lambda_B}=\frac{T_B}{T_A}=\frac{500}{1500}=\frac{1}{3} \\
& E \propto T^4 A \text { (where } A=\text { surface area } \\
& =4 \pi R^2 \text { ) } \\
& \therefore E \propto T^4 R^2 \\
& \frac{E_A}{E_B}=\left(\frac{T_A}{T_B}\right)^4\left(\frac{R_A}{R_B}\right)^2 \\
& =(3)^4\left(\frac{6}{18}\right)^2=9
\end{aligned}
\)
\(\therefore\) Answer is 9 .
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