IIT-JEE 2010 · Paper 1 · Physics Q78

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PhysicsInteger+3Question 78 of 84

Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures $T_1$ and $T_2$ , respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B?

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Answer and worked solution

Official answer: 9

Temperatures (Wien's law). $\lambda_mT=$ const, so $\dfrac{T_1}{T_2}=\dfrac{1500}{500}=3$ .

Power (Stefan's law). $P\propto r^2T^4$ :
$\frac{P_A}{P_B}=\left(\frac6{18}\right)^2\times3^4=\frac19\times81=9.$

Answer: 9

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

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IIT-JEE 2010 · Paper 1, Physics question 78