40% of incident radiant energy on the surface of a thermally transparent body is reflected back. If the transmissivity of the body be 0.15, then the emissivity of surface is
A. 0.45
B. 0.55
C. 0.40
D. 0.75
Answer: Option A
Solution(By Examveda Team)
a + p + T = 1a + 0.40 + 0.15 = 1 (p = 0.40, T = 0.15)
a = 1 - 0.55
a = 0.45
This Question Belongs to Mechanical Engineering >> Heat And Mass Transfer
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