A mass of $$10$$ kg is suspended vertically by a rope of length $$5$$ m from the roof. A force of $$30$$ N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $$\alpha = \tan^{-1}(x \times 10^{-1})$$. The value of $$x$$ is ______.
(Given, $$g = 10$$ m s$$^{-2}$$)

We begin by noting that a mass of $$10$$ kg hangs from a rope of length $$5$$ m, and that a horizontal force of $$30$$ N is applied at the midpoint of the rope. Let $$T_1$$ denote the tension in the upper half and $$T_2$$ the tension in the lower half, while $$\alpha$$ and $$\beta$$ are the angles these halves make with the vertical.

Since there is no horizontal force below the midpoint, the lower half supports only the weight of the mass, so it hangs vertically and thus $$\beta = 0$$. Consequently, $$T_2 = mg = 10 \times 10 = 100$$ N.

At the midpoint, three forces are in equilibrium: the tension $$T_1$$ along the upper rope at an angle $$\alpha$$ to the vertical, the downward tension $$T_2 = 100$$ N, and the applied horizontal force $$F = 30$$ N. From horizontal equilibrium we have $$T_1 \sin\alpha = 30$$ ... (i), and from vertical equilibrium we have $$T_1 \cos\alpha = T_2 = 100$$ ... (ii).

Dividing equation (i) by equation (ii) gives $$\tan\alpha = \frac{30}{100} = \frac{3}{10} = 3 \times 10^{-1}$$, so $$\alpha = \tan^{-1}(3 \times 10^{-1})$$.

Comparing this expression for $$\alpha$$ with the given form $$\alpha = \tan^{-1}(x \times 10^{-1})$$ shows that $$x = 3$$.

The answer is $$\boxed{3}$$.