Step 1: Let $P$ be the principal at any time $t$. According to the problem, principal increases continuously at the rate of $r\%$ per year. Therefore, the differential equation is:

$$\frac{dP}{dt} = \frac{r}{100} \times P$$

Step 2: Separating the variables, we get:

$$\frac{dP}{P} = \frac{r}{100}\, dt$$

Integrating both sides:

$$\log P = \frac{rt}{100} + C_1$$

which gives:

$$P = C \times e^{rt/100}$$

Step 3: Using initial condition: when $t = 0$, $P = 100$. Substituting in the equation:

$$100 = C \times e^{0} = C$$

Therefore, $C = 100$, and the equation becomes:

$$P = 100 \times e^{rt/100}$$

Step 4: Given that principal doubles in 10 years. So when $t = 10$, $P = 200$. Substituting:

$$200 = 100 \times e^{r \times 10/100}$$

which gives:

$$2 = e^{r/10}$$

Step 5: Taking logarithm on both sides:

$$\log_e 2 = \frac{r}{10}$$

Given $\log_e 2 = 0.6931$, we get:

$$r = 10 \times 0.6931 = 6.931$$

∴ $r = \mathbf{6.931\%}$ per year.