Q148 / 150 · MHT CET 2024 (03 May Shift 1) · Physics
Two spherical black bodies of radii ' $R_1$ ' and ' $\mathrm{R}_2$ ' and with surface temperature ' $\mathrm{T}_1$ ' and ' $\mathrm{T}_2$ ' respectively radiate the same power. The ratio of ' $R_1$ ' to ' $R_2$ ' will be
- A$\left(\frac{T_2}{T_1}\right)^4$
- Correct answerB$\left(\frac{T_2}{T_1}\right)^2$
- C$\left(\frac{T_1}{T_2}\right)^4$
- D$\left(\frac{T_1}{T_2}\right)^2$
Solution
For any two bodies of radii $R_1$ and $R_2$, kept at temperatures $T_1$ and $T_2$, the power radiated or rate of loss of heat by them can be given as,
$\frac{\mathrm{Q}_1}{\mathrm{Q}_2}=\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^2 \times\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^4$
In case of non-spherical bodies, $\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^2$ can be replaced by $\left(\frac{A_1}{A_2}\right)$ where $A_1, A_2$ are areas of the given bodies.
$\therefore \quad$ For same power,
$\begin{aligned}
&\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^2=\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^4 \\
& \therefore \quad\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)=\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^2
\end{aligned}$
Hence, the correct answer is option: 'B'.