The value of $$\frac{(0.13)^2 + (0.21)^2}{(0.39)^2 + 81(0.07)^2} \div \frac{(2.4)^4 + 3 \times (11.52) + 9}{(2.4)^6 + 6(2.4)^4 + 3 \times (17.28)}$$ lies between:

Given that

$$\frac{(0.13)^2 + (0.21)^2}{(0.39)^2 + 81(0.07)^2} \div \frac{(2.4)^4 + 3 \times (11.52) + 9}{(2.4)^6 + 6(2.4)^4 + 3 \times (17.28)}$$

$$\Rightarrow \frac{(0.13)^2 + (0.21)^2}{(0.39)^2 + 81(0.07)^2} \times  \frac{(2.4)^6 + 6(2.4)^4 + 3 \times (17.28)}{(2.4)^4 + 3 \times (11.52) + 9}$$

$$\Rightarrow \frac{0.0169+ 0.0441}{0.1521 + 81\times 0.0049} \times \frac{191.10+ 33.1776 + 51.84}{5.76 + 34.56 + 9}$$

$$\Rightarrow \frac{0.061}{0.549} \times \frac{276.1176}{49.32}$$

$$\Rightarrow \frac{16.84317}{27.07668}$$

$$\Rightarrow 0.6222$$

then It lies between 0. 6 and 0.7

therefore Option (C) 0.6 and 0.7 Ans