Step 1: Understanding the Concept:
This problem describes a scenario of continuous compounding. When a quantity increases continuously at a proportional rate, it models exponential growth. The rate of change of the principal $P$ with respect to time $t$ is directly proportional to the current principal.
Step 2: Key Formula or Approach:
1. Differential equation for continuous growth: $\frac{dP}{dt} = \frac{x}{100} P$.
2. The solution is the exponential growth model: $P(t) = P_0 e^{rt}$, where $r = x/100$ is the continuous growth rate as a decimal and $P_0$ is the initial principal.
Step 3: Detailed Explanation:
Let $P$ represent the principal at any given time $t$ in years.
The continuous rate of increase is given by:
\[ \frac{dP}{dt} = \left(\frac{x}{100}\right) P \]
Separate variables to solve the differential equation:
\[ \frac{dP}{P} = \left(\frac{x}{100}\right) dt \]
Integrate both sides:
\[ \int \frac{dP}{P} = \int \left(\frac{x}{100}\right) dt \]
\[ \log_e P = \frac{x}{100} t + C \]
Convert to exponential form:
\[ P(t) = e^{\frac{x}{100} t + C} = e^C \cdot e^{\frac{xt}{100}} \]
Let $e^C = P_0$, representing the initial principal at $t=0$. Thus, $P(t) = P_0 e^{\frac{xt}{100}}$.
We are given that the initial principal $P_0$ is ₹100.
The problem states it doubles in 10 years. So at $t = 10$, $P(10) = 2 \times 100 = 200$.
Substitute these values into the growth model:
\[ 200 = 100 e^{\frac{x \cdot 10}{100}} \]
Divide by 100:
\[ 2 = e^{\frac{x}{10}} \]
Take the natural logarithm ($\log_e$) of both sides to isolate the exponent:
\[ \log_e(2) = \frac{x}{10} \]
We are given the approximation $\log_e(2) = 0.6931$.
\[ 0.6931 = \frac{x}{10} \]
Multiply by 10 to solve for $x$:
\[ x = 10 \times 0.6931 = 6.931 \]
The value of $x$ is approximately $6.93$. The question frames $x$ as the percentage value itself, so the rate is $6.93%$.
Step 4: Final Answer:
The rate $x$ is $6.93%$.