IIT-JEE 2012 · Paper 1 · Physics Q2

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Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures $2T$ and $3T$ respectively. The temperature of the middle (i.e. second) plate under steady state condition is

\left(\dfrac{65}{2}\right)^{1/4}T

\left(\dfrac{97}{4}\right)^{1/4}T

\left(\dfrac{97}{2}\right)^{1/4}T

(97)^{1/4}T

Answer and worked solution

Official answer: C

Steady state of the middle plate. It absorbs radiation from both neighbours and emits from both faces:
$\sigma(2T)^4+\sigma(3T)^4=2\sigma T_2^4.$

Solve. $T_2^4=\dfrac{16+81}2T^4=\dfrac{97}2T^4$ , so $T_2=\left(\dfrac{97}2\right)^{1/4}T$ .

(C)

Topics:Radiation and Stefan's law

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IIT-JEE 2012 · Paper 1, Physics question 2