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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
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{2}}\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^2$
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:
$$
\begin{aligned}
& \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\sigma 4 \pi \mathrm{r}_1^2 \mathrm{~T}_1^4}{\sigma 4 \pi \mathrm{r}_2^2 \mathrm{~T}_2^4} \\
& \frac{1}{2}=\frac{\mathrm{r}_1^2 \mathrm{~T}_1^4}{\mathrm{r}_2^2 \mathrm{~T}_2^4} \\
& \frac{\mathrm{r}_1^2}{\mathrm{r}_2^2}=\frac{1}{2} \frac{\mathrm{T}_2^4}{\mathrm{~T}_1^4} \\
& \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{1}{\sqrt{2}}\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^2
\end{aligned}
$$
$\therefore \quad$ 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:
$$
\begin{aligned}
& \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\sigma 4 \pi \mathrm{r}_1^2 \mathrm{~T}_1^4}{\sigma 4 \pi \mathrm{r}_2^2 \mathrm{~T}_2^4} \\
& \frac{1}{2}=\frac{\mathrm{r}_1^2 \mathrm{~T}_1^4}{\mathrm{r}_2^2 \mathrm{~T}_2^4} \\
& \frac{\mathrm{r}_1^2}{\mathrm{r}_2^2}=\frac{1}{2} \frac{\mathrm{T}_2^4}{\mathrm{~T}_1^4} \\
& \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{1}{\sqrt{2}}\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^2
\end{aligned}
$$
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