Q103 / 150 · MHT CET 2023 (11 May Shift 1) · Physics

Two spherical black bodies of radii ' $r_1$ ' and ' $r_2$ ' at temperature ' $T_1$ ' and ' $T_2$ ' respectively radiate power in the ratio $1: 2$ Then $r_1: r_2$ is

  • A
    $\frac{1}{2}\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^4$
  • Correct answer
    B
    $\frac{1}{\sqrt{2}}\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^2$
  • C
    $2\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^4$
  • D
    $2\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^2$

Solution

Power Radiated by black body, $\mathrm{P}=\sigma \mathrm{AT}^4$ $\therefore \quad$ For first black body: $$ \mathrm{P}_1=\sigma 4 \pi \mathrm{r}_1^2 \mathrm{~T}_1^4 $$ $\therefore \quad$ For second black body: $$ \mathrm{P}_2=\sigma 4 \pi \mathrm{r}_2{ }^2 \mathrm{~T}_2{ }^4 $$ $\therefore \quad$ The ratio will be: $$ \begin{aligned} & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\sigma 4 \pi \mathrm{r}_1^2 \mathrm{~T}_1^4}{\sigma 4 \pi \mathrm{r}_2^2 \mathrm{~T}_2^4} \\ & \frac{1}{2}=\frac{\mathrm{r}_1^2 \mathrm{~T}_1^4}{\mathrm{r}_2^2 \mathrm{~T}_2^4} \\ & \frac{\mathrm{r}_1^2}{\mathrm{r}_2^2}=\frac{1}{2} \frac{\mathrm{T}_2^4}{\mathrm{~T}_1^4} \\ & \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{1}{\sqrt{2}}\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^2 \end{aligned} $$

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

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