Q169 / 200 · MHT CET - Medical 2026 (23 April Shift 2) · Physics
Three very large plates of same area are kept parallel (in vertical plane) 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$(97)^{1/4}\,T$
- Correct answerB$\left(\dfrac{97}{2}\right)^{1/4}T$
- C$\left(\dfrac{97}{4}\right)^{1/4}T$
- D$\left(\dfrac{65}{2}\right)^{1/4}T$
Solution
Let the temperature of the middle plate be $T_2$.
In steady state, the net rate of heat received by the middle plate must be zero. The middle plate receives radiation from the first plate on one side and from the third plate on the other side. It also emits radiation from both of its sides.
According to Stefan-Boltzmann law, the net rate of heat transfer per unit area to the second plate is given by:
$\sigma(T_1^4 - T_2^4) + \sigma(T_3^4 - T_2^4) = 0$
$T_1^4 + T_3^4 = 2T_2^4$
Given $T_1 = 2T$ and $T_3 = 3T$, substituting these values:
$(2T)^4 + (3T)^4 = 2T_2^4$
$16T^4 + 81T^4 = 2T_2^4$
$97T^4 = 2T_2^4$
$T_2^4 = \dfrac{97}{2}T^4$
$T_2 = \left(\dfrac{97}{2}\right)^{1/4}T$
Answer: $\left(\dfrac{97}{2}\right)^{1/4}T$
In steady state, the net rate of heat received by the middle plate must be zero. The middle plate receives radiation from the first plate on one side and from the third plate on the other side. It also emits radiation from both of its sides.
According to Stefan-Boltzmann law, the net rate of heat transfer per unit area to the second plate is given by:
$\sigma(T_1^4 - T_2^4) + \sigma(T_3^4 - T_2^4) = 0$
$T_1^4 + T_3^4 = 2T_2^4$
Given $T_1 = 2T$ and $T_3 = 3T$, substituting these values:
$(2T)^4 + (3T)^4 = 2T_2^4$
$16T^4 + 81T^4 = 2T_2^4$
$97T^4 = 2T_2^4$
$T_2^4 = \dfrac{97}{2}T^4$
$T_2 = \left(\dfrac{97}{2}\right)^{1/4}T$
Answer: $\left(\dfrac{97}{2}\right)^{1/4}T$
Hence, the correct answer is option: 'B'.