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Q.

Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surfaces 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 shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A, by 1.00 μ m. If the temperature of A is 5802 K.

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a

The temperature of B is 1934 K

b

λB=1.5μm

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

(a,b) According to stefan’s law.E=eAσT4⇒EA=eAAσTA4 and EB=eBAσTB4∵EA=EB∴eATA4=eBTB4⇒TB=eAeBTA414=181×(5802)414⇒TB=1934K And, from Wien's law λA×TA=λB×TB⇒λAλB=TBTA⇒λB−λAλB=TA−TBTA⇒1λB=5802−19345802=38685802⇒λB=1.5μm
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