The half-target mass assigned to equal fragments contains
$$
N_f=\frac{\tfrac12(\pi\rho D_t^3/6)}{\pi\rho D_f^3/6}
=\frac{D_t^3}{2D_f^3}
$$
fragments. Their total <geometric cross-section> is consequently
$$
\boxed{\sigma_{{\rm tot},2}
=N_f\frac{\pi D_f^2}{4}
=\frac{\pi}{8}\frac{D_t^3}{D_f}}.
$$
Ignoring the negligible single largest fragment in $\sigma_{{\rm tot},1}$,
$$
\boxed{
\frac{\sigma_{{\rm tot},2}}{\sigma_{{\rm tot},1}}
\simeq
\frac{\alpha_f-3}{4-\alpha_f}
\frac{D_{\min,f}^{\alpha_f-3}D_{\max,f}^{4-\alpha_f}}{D_f}}.
$$
For example, if $D_f=D_{\min,f}$, this ratio is $(\alpha_f-3)(D_{\max,f}/D_{\min,f})^{4-\alpha_f}/(4-\alpha_f)$ and is large for a broad size range. This reflects the inverse-size <cross-section per unit mass> of equal-density spherical fragments: concentrating the mass at small $D_f$ creates more area.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 316 / 2 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-316/2/ii/solution