At $$t = 0$$, a body of mass $$100$$ g starts moving under the influence of a force $$(5\hat{i} + 10\hat{j})$$ N. After $$2$$ s, its  position is $$(2x\hat{i} + 5y\hat{j})$$ m. The ratio $$x : y$$ is ______.

The mass of the body is given in grams, so first convert it into SI units:
$$m = 100\,\text{g} = 0.1\,\text{kg}$$

The constant force acting on the body is
$$\mathbf{F} = 5\hat{i} + 10\hat{j}\,\text{N}$$

Using Newton’s second law $$\mathbf{F}=m\mathbf{a}$$, the acceleration components are

$$a_x = \frac{5}{0.1} = 50\,\text{m\,s}^{-2}$$
$$a_y = \frac{10}{0.1} = 100\,\text{m\,s}^{-2}$$

The question does not state any initial velocity, so we take the body to start from rest at $$t = 0$$:
$$u_x = 0,\; u_y = 0$$.

For motion with constant acceleration, the displacement after time $$t$$ in each direction is
$$s = ut + \tfrac12 a t^2$$.

Time interval: $$t = 2\,\text{s}$$.

Along the $$x$$-axis:
Displacement is given as $$2x$$, so
$$2x = \tfrac12 a_x t^2$$
$$2x = \tfrac12 \times 50 \times (2)^2$$
$$2x = 0.5 \times 50 \times 4 = 100$$
$$x = 50$$.

Along the $$y$$-axis:
Displacement is given as $$5y$$, so
$$5y = \tfrac12 a_y t^2$$
$$5y = \tfrac12 \times 100 \times (2)^2$$
$$5y = 0.5 \times 100 \times 4 = 200$$
$$y = 40$$.

Therefore, the required ratio is
$$x : y = 50 : 40 = 5 : 4$$.

Hence the correct choice is:
Option D which is: $$5:4$$