JEE Advanced 2015 · Paper 1 · Physics Q5

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PhysicsInteger+4Question 5 of 60

Two spherical stars $A$ and $B$ emit blackbody radiation. The radius of $A$ is 400 times that of $B$ and $A$ emits $10^4$ times the power emitted from $B$ . The ratio $\left(\dfrac{\lambda_A}{\lambda_B}\right)$ of their wavelengths $\lambda_A$ and $\lambda_B$ at which the peaks occur in their respective radiation curves is

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Answer and worked solution

Official answer: 2

Stefan's law. $P\propto R^2T^4$ :
$10^4=400^2\left(\frac{T_A}{T_B}\right)^4\implies\left(\frac{T_A}{T_B}\right)^4=\frac1{16}\implies\frac{T_A}{T_B}=\frac12.$

Wien's law. $\lambda_{max}\propto1/T$ , so $\dfrac{\lambda_A}{\lambda_B}=2$ .

Answer: 2

Topics:Radiation and Stefan's law Wien's displacement law

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JEE Advanced 2015 · Paper 1, Physics question 5