Two long parallel surfaces each of emissivity 0.7 are maintained at different temperatures and accordingly have radiation heat exchange between them. It is desired to reduce 75% of the radiant heat transfer by inserting thin parallel shields of emissivity 1 on both sides. The number of shields should be

A. One

B. Two

C. Three

D. Four

Answer: Option C

Solution(By Examveda Team)

$$\frac{{{\text{Q with shield}}}}{{{\text{Q without shield}}}} = \frac{1}{{{\text{n}} + 1}}.$$
N = number of parallel shields
0.25n + 0.25 = 1
0.25n = 0.75
n = 3.