Sphere 1 with a diameter of 0.1 m is completely enclosed by another sphere 2 of diameter 0.4 m. The view factor F₁₂ is

The correct option is 1.0.

Analysis of the Given Image:
The attached image shows two concentric circles representing the cross-section of the two spheres:
1. An inner sphere labeled as 1 with an arrow indicating a dimension of 0.1 (its diameter or radius is 0.1 m).
2. An outer sphere labeled as 2 with a longer arrow indicating a dimension of 0.4 (its diameter or radius is 0.4 m).
Because sphere 1 is completely enclosed by sphere 2, they form a two-surface radiation enclosure.

Step-by-Step Derivation:

1. Concept of View Factor:
The view factor (or shape factor) $F_{ij}$ is defined as the fraction of radiation leaving surface i that is directly intercepted by surface j.

2. Summation Rule for Enclosures:
For any surface i in an N-surface enclosure, the sum of view factors to all surfaces in the enclosure (including itself) must equal 1:
$\sum _{j=1}^N{F}_{ij}=1$
For a two-surface enclosure consisting of surface 1 and surface 2, the summation rule for surface 1 is:
$F_{11}+F_{12}=1$

3. Self-View Factor of the Inner Sphere ($F_{11}$):
The inner sphere (surface 1) has a convex outer surface. A convex surface cannot see itself, meaning that no radiation leaving surface 1 can directly strike surface 1. Thus:
$F_{11}=0$

4. Calculating $F_{12}$:
Substituting $F_{11}=0$ into the summation rule:
$0+F_{12}=1$
$F_{12}=1.0$

This means 100% of the radiation leaving the convex inner sphere 1 is intercepted by the enclosing outer sphere 2, regardless of their respective diameters.