Bronze and silver beads

combinatorics

Since you haven't edited the question for a month after I pointed out a somewhat trivial solution, I'll assume that you didn't intend to preclude it, so I'll post it as an answer:

Two weighings are enough: First put all beads on one side, then put only the bead of interest on one side. That gives you the mean weight per bead and the weight of the bead of interest. If it's bronze, it will deviate from the mean by $\frac7{13}\text g$; if it's silver, by $\frac6{13}\text g$.

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