Discussion
Calculate the shape factor for the sphere diameter 'd' inside a cubical box length, l =d
Answer: A
The answer is \(\pi/6\).
All radiation leaving the sphere (surface 1) reaches the box (surface 2), so \(F_{12} = 1\). The non-trivial shape factor is the one from the box to the sphere, found by reciprocity, \(A_{1}F_{12} = A_{2}F_{21}\):
\[ \begin{aligned} F_{21} &= \frac{A_{1}}{A_{2}} = \frac{\pi d^{2}}{6 l^{2}} \\ &= \frac{\pi}{6}\quad (l = d) \end{aligned} \]
Since \(F_{12} = 1\) is not offered, the key reads the question as asking for this factor. \(\pi/3\) would follow from taking only three faces of the cube, and \(4\pi/3\) and \(3\pi/2\) exceed one, which no shape factor can.
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