A) e = a = 0.0735
B) e = a = 0.0435
C) e = a = 0.0535
D) e = a = 0.0235
Correct Answer: A
Solution :
Using the relation \[E=\frac{Q}{At}\] \[=\frac{ms\,\theta }{At}=\frac{4.2\times 420\times 1\times {{10}^{-2}}}{5\times {{10}^{-2}}\times 12}=29.4W/{{m}^{2}}\] Emissive and absorptive powers are given by \[e=\frac{{{E}_{a}}}{{{E}_{b}}}=\frac{29.4}{400}\] 0(Here,\[{{E}_{b}}=400\,W/{{m}^{2}}\]) = 0.0735You need to login to perform this action.
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