General Questions | JEE 2012 | Discussion Page

Three very large plates of the same area are kept parallel and close to each other. They are considered 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 conditions is

$\left(\frac{65}{2}\right)^{\frac{1}{4}}T$

$\left(\frac{97}{4}\right)^{\frac{1}{4}}T$

$\left(\frac{97}{2}\right)^{\frac{1}{4}}T$

$\left(97\right)^{\frac{1}{4}}T$

Answer:

Option C

Explanation:

Let temperature of middle plate in steady state is $T_{0}$

$Q_{1}=Q_{2}$

Q= net rate of heat flow

$\therefore$ $\sigma A(3T)^{4}-\sigma AT_{0}^{4}=\sigma AT_{0}^{4}-\sigma A(2T)^{4}$

Solving this equation , we get

$T_{0}=\left(\frac{97}{2}\right)^{\frac{1}{4}}T$

Discussion Forum

Answer:

Explanation: