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Two spherical black bodies of radius ' $\mathrm{r}_{1}$ ' and ' $\mathrm{r}_{2}$ ' with surface temperature ' $\mathrm{T}_{1}$ ' and
${ }^{\prime} \mathrm{T}_{2}$ ' respectively, radiate same power, then $\mathrm{r}_{1}: \mathrm{r}_{2}$ is
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${ }^{\prime} \mathrm{T}_{2}$ ' respectively, radiate same power, then $\mathrm{r}_{1}: \mathrm{r}_{2}$ is
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Verified Answer
The correct answer is:
$\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}$
(A)
$4 \pi r_{1}^{2} T_{1}^{4}=4 \pi r_{2}^{2} T_{2}^{4}$
$\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{T_{2}^{4}}{T_{1}^{4}}$
$\frac{r_{1}}{r_{2}}=\frac{T_{2}^{2}}{T_{1}^{2}}$
$4 \pi r_{1}^{2} T_{1}^{4}=4 \pi r_{2}^{2} T_{2}^{4}$
$\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{T_{2}^{4}}{T_{1}^{4}}$
$\frac{r_{1}}{r_{2}}=\frac{T_{2}^{2}}{T_{1}^{2}}$
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