Two perfectly black parallel plates A and B of equal area facing each other are maintained at constant temperatures $T_{A}$ and $T_{B} (T_{A} > T_{B}) .$ 99 perfectly black and conducting identical screens of same area as plates are introduced between A and B, parallel to each other having equal space between them. Once the thermal equilibrium is reached, choose the INCORRECT statements :
(Take $σ (T_{A}^{4} - T_{B}^{4}) = 5000 \frac{W}{m^{2}};$ Ignore convection)

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Heat flux between screen 1 and 2 is greater than heat flux between screen 98 and 99.

Heat flux between screen 99 and plate B is $50 \frac{W}{m^{2}} .$

Effective heat flux between plates A and B reduces due to introduction of screens between them.

Heat flux between plate A and screen 1 is $5000 \frac{W}{m^{2}} .$

answer is A, B.

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Detailed Solution

$\begin{array}{l} q_{n e t . f}^{"} = σ (T_{A}^{4} - T_{1}^{4}) = σ (T_{1}^{4} - T_{2}^{4}) ..... = σ (T_{n - 1}^{4} - T_{n}^{4}) = σ (T_{n}^{4} - T_{B}^{4}) \\ \Rightarrow σ (T_{n}^{4} - T_{B}^{4}) = (n + 1) q_{n e t . f}^{"} = \frac{σ}{n + 1} (T_{A}^{4} - T_{B}^{4}) \\ \Rightarrow \frac{q_{n e t . f}^{"}}{q_{n e t . i}^{"}} = \frac{1}{n + 1} \Rightarrow q_{n e t . f}^{'} = \frac{5000}{99 + 1} = 50 W / m^{2} \end{array}$