Question

In a bank principal increases at the rate of r% per year. Find the value of r if ₹100 double itself in 10 years (log_e 2 = 0.6931).

Answer 1

Let P be the principal at any instant t.
Given:
\[\frac{dP}{dt} = \frac{r}{100}P\]
\[ \Rightarrow \frac{dP}{P} = \frac{r}{100}dt\]
Integrating both sides, we get
\[\int\frac{dP}{P} = \int\frac{r}{100}dt\]
\[ \Rightarrow \log P = \frac{rt}{100} + C . . . . . . (1)\]
\[\text{ Initially, i . e . at t = 0, let }P = P_0 . \]
\[\text{ Putting }P = P_0 ,\text{ we get }\]
\[\log P_0 = C, \]
\[\text{ Putting }C = \log P_0\text{ in }(1), \text{ we get }\]
\[\log P = \frac{rt}{100} + \log P_0 \]
\[ \Rightarrow \log \frac{P}{P_0} = \frac{rt}{100}\]
\[\text{ Substituting }P_0 = 100, P = 2 P_0 = 200\text{ and }t = 10 \text{ in }(2), \text{ we get }\]
\[\log 2 = \frac{r}{10}\]
\[ \therefore r = 10 \log 2\]
\[ = 10 \times 0 . 6931\]
\[ = 6 . 931\]