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MHT CET · Physics · Thermal Properties of Matter

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. A \(\frac{20 \mathrm{E}}{9}\)
  2. B \(\frac{8 E}{9}\)
  3. C \(\frac{16 \mathrm{E}}{9}\)
  4. D \(\frac{4 \mathrm{E}}{9}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{16 \mathrm{E}}{9}\)

Step-by-step Solution

Detailed explanation

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