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A star (P) behaves like a perfectly black body emitting radiant energy at temperature ' $\mathrm{T}$ '. Another star (Q) also behaves like perfectly black body emitting radiant energy at temperature ' $\mathrm{T} / 4$ ' and has radius eight times the radius of star $(\mathrm{P})$. The ratio of radiant energy emitted by $(\mathrm{P})$ to that by $(\mathrm{Q})$ is
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The correct answer is:
1:1
Concept: According to the Stefan-Boltzmann law energy radiated per unit area (j)
$\mathrm{j}=\sigma \mathrm{T}^4$
Where $\sigma$ is the proportionality constant and $\mathrm{T}$ is temperature radiant energy $E=j A$, where $A$ is the surface black body.
Given,
For $(\mathrm{a})$ blackbody $\mathrm{P}$, radius $=\mathrm{R}$ and temperature $=\mathrm{T}$
(b) blackbody $\mathrm{Q}$, radius $=8 \mathrm{R}$ and temperature $=\frac{\mathrm{T}}{4}$
$\therefore \frac{E_{\mathrm{P}}}{\mathrm{E}_{\mathrm{Q}}}=\frac{\sigma \mathrm{T}^4 4 \pi \mathrm{R}^2}{\sigma\left(\frac{\mathrm{T}}{4}\right)^4 4 \pi(8 \mathrm{R})^2}=\frac{4^4}{8^2}=\frac{(8)^2}{8^2}=1$
$\mathrm{j}=\sigma \mathrm{T}^4$
Where $\sigma$ is the proportionality constant and $\mathrm{T}$ is temperature radiant energy $E=j A$, where $A$ is the surface black body.
Given,
For $(\mathrm{a})$ blackbody $\mathrm{P}$, radius $=\mathrm{R}$ and temperature $=\mathrm{T}$
(b) blackbody $\mathrm{Q}$, radius $=8 \mathrm{R}$ and temperature $=\frac{\mathrm{T}}{4}$
$\therefore \frac{E_{\mathrm{P}}}{\mathrm{E}_{\mathrm{Q}}}=\frac{\sigma \mathrm{T}^4 4 \pi \mathrm{R}^2}{\sigma\left(\frac{\mathrm{T}}{4}\right)^4 4 \pi(8 \mathrm{R})^2}=\frac{4^4}{8^2}=\frac{(8)^2}{8^2}=1$
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