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

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

Solution

$\mathrm{E}=\sigma \mathrm{AT}^{4}$ $\therefore \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}} \cdot\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}$ $\mathrm{~A}_{2}=\frac{\mathrm{A}_{1}}{9} \quad \therefore \frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\frac{1}{9}$ $\mathrm{~T}_{1}=27+273=300 \mathrm{k}$ $\mathrm{T}_{2}=327+273=600 \mathrm{k}$ $\therefore \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{600}{300}=2$ $\therefore \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{1}{9} \times(2)^{4}=\frac{16}{9}$ $\therefore \mathrm{E}_{2}=\frac{16}{9} \mathrm{E}_{1}=\frac{16}{9} \mathrm{E}$

Asked in: MHT CET 2020 (14 Oct Shift 2)

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