Question 638: Thermodynamics & KTG

Question

Two spherical black bodies have radii ' $R_1$ ' and ' $\mathrm{R}_2$ '. Their surface temperatures are ' $\mathrm{T}_1$ ' and ' $T_2$ '. If they radiate same power, then $\frac{R_2}{R_1}$ is

Accepted Answer

$$ \frac{Q_1}{Q_2}=\left(\frac{R_1}{R_2}ight)^2 	imes\left(\frac{T_1}{T_2}ight)^4 $$
$$ \begin{aligned} &	herefore \quad 	ext { for same power, }\ &\begin{array}{ll}  & \left(\frac{\mathrm{R}_1}{\mathrm{R}_2}ight)^2=\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}ight)^4 \ 	herefore & \left(\frac{\mathrm{R}_1}{\mathrm{R}_2}ight)=\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}ight)^2 \ & \left(\frac{\mathrm{R}_2}{\mathrm{R}_1}ight)=\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}ight)^2 \end{array} \end{aligned} $$

Answer

$\frac{T_2}{T_1}$

Answer

$\frac{T_1}{T_2}$

Answer

$\left(\frac{T_2}{T_1}\right)^2$

Answer

$\left(\frac{T_1}{T_2}\right)^2$

Thermodynamics & KTG

Two spherical black bodies have radii ' R_1 ' and ' \mathrm{R}_2 '. Their surface temperatures are ' \mathrm{T}_1 ' and ' T_2 '. If they radiate same power, then \frac{R_2}{R_1} is

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