Q.

Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength λB corresponding to maximum spectral radiancy in the radiation from B is greater than the wavelength I, corresponding to maximum spectral radiancy in the radiation from A by 1.00 μm. If the temperature of A is 5802 K

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a

the temperature of B is 1934 K

b

λB=1⋅5 μm

c

the temperature of B is 11604 K

d

the temperature of B is 2901 K

answer is A.

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Detailed Solution

Power radiated P=EA=eσT4ANow   eATA4=eBTB4      ∵P1=P2or    TB=eAeB1/4×TA=0⋅010⋅811/4×5802      =13×5802  or  TB=1934KAccording to Wien's displacement law,     λmT= constant So   λATA=λBTB  or  λB/λA=TA/TBor   λBλA=58021934  or  λA=λB/3Further,  λB−λA=1μmor    λB−λB3=1μm  or  λB=1⋅5 μm
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