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MHT CET 202623 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
23 April
Shift
Evening Shift
Subject
Physics
Chapter
Thermal Properties of Matter

MHT CET 2026 · 23 April · Evening Shift · Physics · Thermal Properties of Matter

Three very large plates of same area are kept parallel (in vertical...

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 2T2T and 3T3T respectively. The temperature of the middle (i. e. second) plate under steady state condition is

Options

  1. A(97)1/4 T(97)^{1/4}\,T
  2. B(972)1/4T\left(\dfrac{97}{2}\right)^{1/4}T
  3. C(974)1/4T\left(\dfrac{97}{4}\right)^{1/4}T
  4. D(652)1/4T\left(\dfrac{65}{2}\right)^{1/4}T

Correct answer

B. (972)1/4T\left(\dfrac{97}{2}\right)^{1/4}T

Step-by-step solution

Let the temperature of the middle plate be T2T_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:

σ(T14−T24)+σ(T34−T24)=0\sigma(T_1^4 - T_2^4) + \sigma(T_3^4 - T_2^4) = 0

T14+T34=2T24T_1^4 + T_3^4 = 2T_2^4

Given T1=2TandT3=3TT_1 = 2T and T_3 = 3T, substituting these values:

(2T)4+(3T)4=2T24(2T)^4 + (3T)^4 = 2T_2^4

16T4+81T4=2T2416T^4 + 81T^4 = 2T_2^4

97T4=2T2497T^4 = 2T_2^4

T24=972T4T_2^4 = \dfrac{97}{2}T^4

T2=(972)1/4TT_2 = \left(\dfrac{97}{2}\right)^{1/4}T

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