KVPY SX SB Exam 19-Nov-2017 Question Paper

The planet should be in Thermal equilibrium with star.
Amount of heat energy emitted by start per second ${\mathit{E}}_1=\left(\mathit{\sigma }\mathit{T}{*}^4\right)\left(4\mathit{\pi }\mathit{R}{*}^2\right)$
∵ d = distance between star and planet
Hence, amount of energy reaching planet per unit area per second = $\frac{{\mathit{E}}_1}{4\mathit{\pi }{\mathit{d}}^2}$
$=\frac{\left(\mathit{\sigma }\mathit{T}{*}^4\right)\left(4\mathit{\pi }\mathit{R}{*}^2\right)}{4\mathit{\pi }{\mathit{d}}^2}$
$=\frac{\mathit{\sigma }\mathit{T}{*}^2\mathit{R}{*}^2}{{\mathit{d}}^2}$
Henceenergy received by planet per second
${\mathit{E}}_2=\left(\frac{\mathit{\sigma }\mathit{T}{*}^4\mathit{R}{*}^2}{{\mathit{d}}^2}\right)\mathit{\pi }{\mathit{R}}_{\mathit{P}}^2$
T = Temperature of planet then amount of
energy emitted by planet per second
${\mathit{E}}_3=\left(\mathit{\sigma }{\mathit{T}}^4\right)\left(4\mathit{\pi }{\mathit{R}}_{\mathit{P}}^2\right)$
For thermal equilibrium ${\mathit{E}}_3={\mathit{E}}_2$
⇒ $\left(\mathit{\sigma }{\mathit{T}}^4\right)\left(4\mathit{\pi }{\mathit{R}}_{\mathit{P}}^2\right)=\frac{\left(\mathit{\sigma }\mathit{T}{*}^4\mathit{R}{*}^2\right)}{{\mathit{d}}^2}\left(\mathit{\pi }{\mathit{R}}_{\mathit{P}}^2\right)$
⇒ ${\mathit{T}}^4=\frac{\mathit{T}{*}^4\mathit{R}{*}^2}{4{\mathit{d}}^2}$
⇒ $\mathit{T}=\mathit{T}*\sqrt{\frac{\mathit{R}*}{2\mathit{d}}}=\mathrm{fT}*$
⇒ $\mathrm{f}\propto \sqrt{\frac{\mathit{R}*}{\mathit{d}}}$
Hence correct answer is (A)