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JEE Advanced 2010Paper 1
Exam
JEE Advanced
Year
2010
Paper
Paper 1
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Advanced 2010 · Paper 1 · Physics · Thermal Properties of Matter

Two spherical bodies A (radius 6 ~cm ) and B (radius 18 ~cm ) are...

Two spherical bodies AA (radius 6 cm6 \mathrm{~cm} )andBand B (radius 18 cm18 \mathrm{~cm} ) are at temperatures T1T_1 and T2T_2, respectively. The maximum intensity in the emission spectrum of AA is at 500 nm500 \mathrm{~nm} and in that of BB is at 1500 nm1500 \mathrm{~nm}. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by AA to that of BB ?

Correct answer

9

Step-by-step solution

λm∝1T∴λAλB=TBTA=5001500=13E∝T4A (where A= surface area =4\piR2 ) ∴E∝T4R2EAEB=(TATB)4(RARB)2=(3)4(618)2=9\begin{aligned}& \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 T⁴ A \text{ (where }A=\text{ surface area }\\ & =4 \piR^2 \text{ ) }\\ & \therefore E \propto T⁴ R² \\ & \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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