Q135 / 160 · AP EAMCET 2015 · Physics

The ratio energies of emitted radiation by a black body at $600 \mathrm{~K}$ and $933 \mathrm{~K}$ when the surrounding temperature is $300 \mathrm{~K}$

  • A
    $\frac{5}{16}$
  • B
    $\frac{7}{16}$
  • Correct answer
    C
    $\frac{3}{16}$
  • D
    $\frac{9}{16}$

Solution

$\begin{aligned} & \text { } \frac{E_1}{E_2}=\frac{T_1^4-T_S^4}{T_2^4-T_S^4}\left[\because E=\sigma A T^4\right] \\ & \Rightarrow \frac{E_1}{E_2}=\frac{(600)^4-(300)^4}{(900)^4-(300)^4}=\frac{3}{16}\end{aligned}$

Hence, the correct answer is option: 'C'.

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