In a bank, principal increases at the rate of $\mathrm{r} \%$ per annum. Find the value of $\mathrm{r}$ if ' 100 double itself in 10 years. (Given $\left.\log _{\mathrm{e}} 2=0.6931\right)$

Home » In a bank, principal increases at the rate of per annum. Find the value of if ‘ 100 double itself in 10 years. (Given

In a bank, principal increases at the rate of $\mathrm{r} \%$ per annum. Find the value of $\mathrm{r}$ if ‘ 100 double itself in 10 years. (Given $\left.\log _{\mathrm{e}} 2=0.6931\right)$

In a bank, principal increases at the rate of $\mathrm{r} \%$ per annum. Find the value of $\mathrm{r}$ if ‘ 100 double itself in 10 years. (Given $\left.\log _{\mathrm{e}} 2=0.6931\right)$

Solution:

It is given that: $\frac{d p}{d t}=\left(\frac{r}{100}\right) \times p$
Here, $p$ is the principal, $r$ is the rate of interest per annum and $t$ is the time in years.
On solving the differential equation we obtain,
$\begin{array}{l} \frac{d p}{p}=\left(\frac{r}{100}\right) d t \\ \Rightarrow \int \frac{d p}{p}=\int \frac{r}{100} d t \\ \Rightarrow \log p=\frac{r t}{100}+c \\ \Rightarrow p=e^{\frac{r t}{100}+e} \end{array}$
As the principal doubles itself in 10 years, therefore
Suppose the initial interest be $\mathrm{pl}$ (for $\mathrm{t}=0$ ), after 10 years $\mathrm{p} 1$ becomes $2 \mathrm{p} 1$ .
As a result, $p 1=e^{c}$ for $(\mathrm{t}=0) \ldots$ (i)
$p=2 p 1=e^{\frac{r(10)}{200}}, e^{c \cdots(\mathrm{ii})}$
Substituting equation(i) in equation(ii), we obtain,
$\begin{array}{l} \Rightarrow 2 \mathrm{p} 1=e^{\frac{r}{10}}, p 1 \\ \Rightarrow 2=e^{\frac{r}{10}} \\ \Rightarrow \log 2=\frac{r}{10} \\ \Rightarrow r=10 \log 2 \end{array}$
$\Rightarrow r=6.931$
$\therefore$ Rate of interest $=6.931$