A black rectangular surface of area ' $A$ ' emits energy ' $E$ ' per second at $27^{\circ} \mathrm{C}$. If…

A black rectangular surface of area ' $A$ ' emits energy ' $E$ ' per second at $27^{\circ} \mathrm{C}$. If length and breadth is reduced to $(1 / 3)^{\text {rd }}$ of its initial value and temperature is raised to $327^{\circ} \mathrm{C}$ then energy emitted per second becomes
  1. $\frac{20 \mathrm{E}}{9}$
  2. $\frac{8 E}{9}$
  3. $\frac{16 \mathrm{E}}{9}$
  4. $\frac{4 \mathrm{E}}{9}$

Solution

$\begin{aligned} & \mathrm{E}=\sigma \mathrm{AT}^4 \\ & \mathrm{E}^{\prime}=\sigma \mathrm{A}^{\prime} \mathrm{T}^{\prime 4} \\ & \mathrm{~A}=\mathrm{L} \times \mathrm{B} \\ & \mathrm{A}^{\prime}=\frac{\mathrm{L}}{3} \times \frac{\mathrm{B}}{3}=\frac{\mathrm{A}}{9} \\ & \mathrm{~T}=27^{\circ} \mathrm{C}=300 \mathrm{~K} \\ & \mathrm{~T}^{\prime}=327^{\circ} \mathrm{C}=600 \mathrm{~K} \\ & \frac{\mathrm{E}^{\prime}}{\mathrm{E}}=\frac{\mathrm{A}^{\prime}}{\mathrm{A}}\left(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}\right)^4=\frac{1}{9}(2)^4 \\ & \therefore \mathrm{E}^{\prime}=\frac{16 \mathrm{E}}{9}\end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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