Given,
Principal = Rs.1000
Amount = Rs.1102.50
Time = 2 years
Let rate of interest = R % per annum
So,
A \(={P}[({1}+\frac{R}{100})^T]\)
\(={1000}({1}+\frac{R}{100})^2\) = 1102.50
\(=({1}+\frac{R}{100})^2\) \(=\frac{1102.50}{1000}\) \(=\frac{4410}{4000}\) \(=(\frac{21}{20})^2\)
\(=1+\frac{R}{100}\) \(=\frac{21}{20}\)
\(=\frac{R}{100}\) \(=\frac{21}{20}-1\) \(=\frac{1}{20}\)
= R \(=\frac{1}{20}\times100\) = 5%
Hence,
Rate of interest = 5 % per annum