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. First and third plates are maintained at absolute temperatures 2T and 3T respectively. Temperature of the middle plate in steady state is

Options

A. `((65)/(2))^((1)/(4))T`
B. `((97)/(4))^((1)/(4))T`
C. `((97)/(2))^((1)/(4))T`
D. `(97)^((1)/(4))T`

Answer

C

Explanation

Let assume temperature of middle plate is `T_(0)`. As temperatures are constant,
`therefore` Heat gained by middle surface = Heat lost by middle surface.
So, `[sigma A(3T)^(4)-T_(0)^(4)]=sigma A[T_(0)^(4)-(2T)^(4)]`
`rArr" "T_(0)^(4)=(97)/(2) T^(4) rArr T_(0) =((97)/(2))^(1//4) T`