Question

An inclosure is formed by three closed surface (1),(2) and (3) where $F_{11}=0.3,F_{12}=0.4$ so what will be value of $F_{31}$ where area of surface (3) is two times as that of surface (1)?

• A. 0.6
• B. 0.45
• C. 0.15
• D. 0.05

Answer 1

The correct option is C 0.15

for inclosures,
$F_{11}+F_{22}+F_{13}=1$
$0.3+0.4+F_{13}=1$
Reciprocatory law, $A_1{F}_{13}=A_3{F}_{31}$
$F_{31}=F_{13}\left(\frac{A_1}{A_3}\right)$
$F_{31}=0.03\times \frac{1}{2}$
$F_{31}=0.15$