Step 1:

From the information given,let $p$ be the principle, $t$ be the time and $r$ be the rate of interest respectively

Using the information in the tool box $\large\frac{dp}{dt}= (\large\frac{r}{100})$$p$

Now seperating the variables

we get $\large\frac{dp}{p}=(\large\frac{r}{100})$$dt$

Step 2:

Integrating on both sides we get

$\int\large\frac{dp}{p}= (\large\frac{r}{100})\int dt$

$ \log p=\large\frac{rt}{100}$$+k$

Or $p=e^{(\Large\frac{rt}{100})+k}$

Step 3:

But it is given when $t=0,p=100$.Hence $100= e^k$

If $t=10$, then $p= 2(100) = 200$

Therefore $200= e^{[(\large\frac{r}{10})+k]} $

Or $e^{(\large\frac{r}{10}) }. e^k$

but $e^k = 100$.

Hence $200 = e^{(\Large\frac{r}{10})}$$ . 100$

$e^{(\large\frac{r}{10})} = 2$

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

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

$r=6.931$

Hence the value of $r=6.93%$