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Two spherical starts A and B emit black body radiation. The radius of A is 400 times that of B and A emits `10^(4)` times the power emitted from B. The ratio `(lambda_(A)//lambda_(B))` of their wavelengths `lambda_(A)` and `lambda_(B)` at which the peaks oc cur in their respective radiation curves is :

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The correct Answer is:
B
Here, `R_(A)= 400R_(B), E_(A)= 10^(4)E_(B)`
As `E= omegaT^(4)A=omegaT^(4)4 piR^(2), i.e., E prop T^(4)R^(2)`
` :. (E_A)/(E_(B))=(T_(A)^(4)R_(A)^(2))/(T_(B)^(4)R_(B)^(2)) or 10^(4)=(T_(A)/T_(B))^(4)xx(400)^(2)`
or `(T_(A))/(T_(B))=[(10^(4))/((400)^(2))]^(1/4)=1/2`
From Wien's displacement law
`(lambda_(A))/(lambda_(B))= (T_(B))/(T_(A))= 2/1`
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