In a bank, principal increases continuously at the rate of

Let p, t, and r represent the principal, time, and rate of interest respectively.

It is given that the principal increases continuously at the rate of r% per year.

$\Rightarrow \frac{d p}{d t}=\left(\frac{r}{100}\right) p$

$\Rightarrow \frac{d p}{p}=\left(\frac{r}{100}\right) d t$

Integrating both sides, we get:

$\int \frac{d p}{p}=\frac{r}{100} \int d t$

$\Rightarrow \log p=\frac{r t}{100}+k$

$\Rightarrow p=e^{\frac{r t}{100}+k}$                ...(1)

It is given that when $t=0, p=100$.

$\Rightarrow 100=e^{k} \ldots(2)$

Now, if $t=10$, then $p=2 \times 100=200$.

Therefore, equation (1) becomes:

$200=e^{\frac{r}{10}+k}$

$\Rightarrow 200=e^{\frac{r}{10}} \cdot e^{k}$

$\Rightarrow 200=e^{\frac{r}{10}} \cdot 100$              ...(2)

$\Rightarrow e^{\frac{r}{10}}=2$

$\Rightarrow \frac{r}{10}=\log _{e} 2$

$\Rightarrow \frac{r}{10}=0.6931$

$\Rightarrow r=6.931$

Hence, the value of r is 6.93%.