If the term independent of $$x$$ in the expansion of $$\left(\sqrt{a}x^2 + \frac{1}{2x^3}\right)^{10}$$ is 105, then $$a^2$$ is equal to :

We need to find the term independent of $$x$$ in the expansion of $$\left(\sqrt{a}x^2 + \frac{1}{2x^3}\right)^{10}$$.

The general term using the Binomial Theorem is:

$$T_{r+1} = \binom{10}{r}(\sqrt{a}x^2)^{10-r}\left(\frac{1}{2x^3}\right)^r$$

$$= \binom{10}{r}(\sqrt{a})^{10-r} \cdot \frac{1}{2^r} \cdot x^{2(10-r)-3r}$$

$$= \binom{10}{r}\frac{a^{(10-r)/2}}{2^r} \cdot x^{20-5r}$$

For the term independent of $$x$$: $$20 - 5r = 0 \Rightarrow r = 4$$.

The term independent of $$x$$ is:

$$T_5 = \binom{10}{4}\frac{a^{3}}{2^4} = 210 \cdot \frac{a^3}{16} = \frac{210a^3}{16} = \frac{105a^3}{8}$$

Setting this equal to 105:

$$\frac{105a^3}{8} = 105$$

$$a^3 = 8$$

$$a = 2$$

Therefore $$a^2 = 4$$.

The correct answer is Option 2: 4.