JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Heat & Thermodynamics

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?

Answer: 9

Solution

Wien's displacement law:

λ_{ma x} T = b

(constant)

For body A:

λ_{ma x, A} = 500

nm, so

T_{1} = \frac{b}{500}

For body B:

λ_{ma x, B} = 1500

nm, so

T_{2} = \frac{b}{1500}

Therefore:

\frac{T _{1}}{T _{2}} = \frac{1500}{500} = 3

, i.e.,

T_{1} = 3 T_{2}

Stefan-Boltzmann law: Power radiated

P = σ A T^{4} = σ (4 π r^{2}) T^{4}

For body A:

P_{A} = σ \cdot 4 π (6)^{2} \cdot T_{1}^{4} = 144 π σ T_{1}^{4}

For body B:

P_{B} = σ \cdot 4 π (18)^{2} \cdot T_{2}^{4} = 1296 π σ T_{2}^{4}

Ratio:

\frac{P _{A}}{P _{B}} = \frac{144 T _{1}^{4}}{1296 T _{2}^{4}} = \frac{144}{1296} \times (\frac{T _{1}}{T _{2}})^{4} = \frac{1}{9} \times 3^{4} = \frac{81}{9} = 9

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer

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JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Heat & Thermodynamics

Solution

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