QJEEAdv15Ph12d2: JEE Advanced 2015 Physics - Class 11 JEEAdv15Ph12d2 Solution: Two spherical stars A and B emit blackbody radiation. The…

Power, $P = (\sigma {T^4}A) = \sigma {T^4}(4\pi {R^2})$

or, $P \propto {T^4}{R^2}$ ..... (i)

According to Wien's law,

$\lambda \propto {1 \over T}$

( $\lambda$ is the wavelength at which peak occurs)

$\therefore$ Eq. (i) will become,

$P \propto {{{R^2}} \over {{\lambda ^4}}}$

or, $\lambda \propto {\left[ {{{{R^2}} \over P}} \right]^{1/4}}$

$\Rightarrow {{{\lambda _A}} \over {{\lambda _B}}} = {\left[ {{{{R_A}} \over {{R_B}}}} \right]^{1/2}}{\left[ {{{{P_B}} \over {{P_A}}}} \right]^{1/4}}$

$= {[400]^{1/2}}{\left[ {{1 \over {{{10}^4}}}} \right]^{1/4}} = 2$