(d) 50%
Let the sum be Rs x. Then, Amount =Rs \(\frac{9X}{4}\)
n = 2, r = ?
\(\therefore\) \(\frac{9X}{4}\) = x\(\big(1+\frac{r}{100}\big)^2\)
\(\Rightarrow\) \(\frac{9}{4}\) = \(\big(1+\frac{r}{100}\big)^2\) \(\Rightarrow\) \(\big(\frac{3}{2}\big)^2\) = \(\big(1+\frac{r}{100}\big)^2\)
\(\Rightarrow\) \(\frac{r}{100}\) = \(\frac{3}{2}\) - 1 = \(\frac{1}{2}\) \(\Rightarrow\) r = 50% p.a.