Solve the infinite geometric series equation sum_r=1^infinity (m)/(2^r−1) = 10 for m.

The given series is $\sum_{r=1}^{\infty} \frac{m}{2^{r-1}} = 10$.
This is an infinite geometric series. Let's write out the first few terms to identify the first term and the common ratio.
For $r=1$, the term is $\frac{m}{2^{1-1}} = \frac{m}{2^0} = m$.
For $r=2$, the term is $\frac{m}{2^{2-1}} = \frac{m}{2^1} = \frac{m}{2}$.
For $r=3$, the term is $\frac{m}{2^{3-1}} = \frac{m}{2^2} = \frac{m}{4}$.
So the series is $m + \frac{m}{2} + \frac{m}{4} + \dots$

The first term is $a = m$.
The common ratio is $k = \frac{\text{second term}}{\text{first term}} = \frac{m/2}{m} = \frac{1}{2}$.

The sum $S$ of an infinite geometric series is given by the formula $S = \frac{a}{1-k}$, provided that $|k| < 1$.
In this case, $a = m$ and $k = \frac{1}{2}$. Since $|\frac{1}{2}| < 1$, the series converges.
We are given that the sum $S = 10$.

Divide both sides by 2:

The value of $m$ is 5.

The correct option is B.

The final answer is $\boxed{\text{B}}$.