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Question

Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface area 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 μm. If the temperature of A is 5802 K

A
The temperature of B is 1934 K
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B
λB=1.5μm
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C
The temperature of B is 11604 K
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D
The temperature of B is 2901 K
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Solution

The correct options are
A The temperature of B is 1934 K
B λB=1.5μm
To calculate temperature of B

Power of body A=eAσT4A×(area)

Power of body B=eBσT4B×(area)

The two powers are equal,

eBσT4B×(area)=eAσT4A×(area)

or T4B=(eAeB)T4A T4B=(0.010.81)×(5802)4

or TB=1934K Option (a) is correct

d) According to Wien's displacement law,

λT=constant λATA=λBTB

or λAλB=TBTA or λAλB=19345802

or λAλB=13 or λB=3λA

Given: λBλA=1.0×106m

3λAλA=106 (or) 2λA=106

or λA=0.5×106 and λB=3×0.5×106

or λB=1.5×106m

Hence option (b) is correct

Options (c) and (d) are incorrect as option (a) is correct

Hence options (a) and (b) are correct.

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