Two capacitors, each having capacitance $$40 \mu F$$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $$K$$ such that the equivalence capacitance of the system became $$24 \mu F$$. The value of $$K$$ will be:

Two capacitors of 40 μF each are connected in series. One capacitor is filled with a dielectric of constant K, and the equivalent capacitance becomes 24 μF.

When a dielectric of constant $$K$$ is inserted in one capacitor:

$$C_1 = 40K \; \mu F$$ and $$C_2 = 40 \; \mu F$$

$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$$

$$\frac{1}{24} = \frac{1}{40K} + \frac{1}{40}$$

$$\frac{1}{24} = \frac{1}{40}\left(\frac{1}{K} + 1\right)$$

$$\frac{40}{24} = \frac{1}{K} + 1$$

$$\frac{5}{3} = \frac{1}{K} + 1$$

$$\frac{1}{K} = \frac{5}{3} - 1 = \frac{2}{3}$$

$$K = \frac{3}{2} = 1.5$$

Hence, the correct answer is Option A: 1.5.