Let the principal be \(p\) rupees and the annual rate of interest be \(r\%\). Therefore, the simple interest for 2 years is: \[ \frac{p \times 2 \times r}{100} \text{ rupees} \] According to the question: \[ \frac{p \times 2 \times r}{100} = 4000 \Rightarrow pr = 200000 \quad \text{(i)} \] Now, the compound amount for 2 years is: \[ p\left(1 + \frac{r}{100}\right)^2 \] According to the question: \[ p\left(1 + \frac{r}{100}\right)^2 - p = 4100 \Rightarrow p\left[\left(1 + \frac{r}{100}\right)^2 - 1\right] = 4100 \] Expanding the square: \[ p\left[\cancel{1} + \frac{2r}{100} + \frac{r^2}{10000} - \cancel{1}\right] = 4100 \Rightarrow \frac{pr}{100}\left(2 + \frac{r}{100}\right) = 4100 \] Substituting \(pr = 200000\) from equation (i): \[ \frac{200000}{100} \left(2 + \frac{r}{100}\right) = 4100 \Rightarrow 2 + \frac{r}{100} = \frac{4100}{2000} \] \[ \frac{r}{100} = \frac{41}{20} - 2 = \frac{1}{20} \Rightarrow r = \frac{1}{20} \times 100 = 5 \] Substituting \(r = 5\) into equation (i): \[ 5p = 200000 \Rightarrow p = 40000 \] ∴ The required principal is ₹40,000 and the rate of interest is 5%.