Q. 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
A) $\left( \frac{65}{2} \right)^{\frac{1}{4}}T$
B) $\left( \frac{97}{4} \right)^{\frac{1}{4}}T$
C) $\left( \frac{97}{2} \right)^{\frac{1}{4}}T$
D) $(97)^{\frac{1}{4}}T$
Explanation
(c) [image:11_image65.emf]
In steady state energy absorbed by middle plate is equal to energy released by middle plate
$\sigma A(3T)^{4} - \sigma A(T'')^{4} = \sigma A(T'')^{4} - \sigma A(2T)^{4}$
$(3T)^{4} - \left( T^{''} \right)^{4} = (T'')^{4} - (2T)^{4}$
$2(T'')^{4} - (16 + 81)T^{4}$
$T'' = \left( \frac{97}{2} \right)^{1/4}T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.