Question

In a bank, principal increases continuously at the rate of $r$ per year. Find the value of $r\%$ if $Rs.100$ double itself in $10$ years $({\log }_{e}2=0.6931).$

Answer 1

Let $p$ be the principle
Given: Principle increases at the rate $r\%$ per year
$\frac{dp}{dt}=r\%\times p$
$\frac{dp}{dt}=\frac{r}{100}\times p$
$\frac{dp}{p}=\frac{r}{100}\times dt$
Integrating both sides, we get
$\int \frac{dp}{p}=\frac{r}{100}\int dt$
$\log p=\frac{rt}{100}+\log c$
$\log p-logc=\frac{rt}{100}$
$\log \frac{p}{c}=\frac{rt}{100}$
$\frac{p}{c}=e^{\frac{rt}{100}}...\left(i\right)$

$\begin{array}{ccc}\text{time}& t=0& t=10\\ \text{Principal(p)}& 100& 200\end{array}$

Substituting $t=0,p=100in\left(i\right),$
$\frac{100}{c}=e^{\frac{r\times 0}{100}}$

$\frac{100}{c}=e^0$
$c=100$

Substituting value of $c$ in (i),
$\frac{P}{100}=e^{\frac{rt}{100}}...\left(ii\right)$
Also, it given that $Rs.100$ will double itselft in $10$ years
Substituting $t=10,p=200in\left(ii\right),$
$\frac{200}{100}=e^{\frac{10r}{100}}$
$2=e^{\frac{r}{10}}$

Taking $\log$both sides,
$\log 2=\frac{r}{10}$

$0.6931=\frac{r}{10}$
$r=6.931\%$
Hence, rate of interest $r=6.931\%$