The gravitational field in a region is given by: $$\vec{E} = (5N/kg)\hat{i} + (12N/kg)\hat{j}$$. If the potential at the origin is taken to be zero, then the ratio of the potential at the points (12 m, 0) and (0, 5 m) is :

The gravitational field is given by $$\vec{E} = (5 \text{N/kg}) \hat{i} + (12 \text{N/kg}) \hat{j}$$. The potential at the origin (0,0) is zero. We need to find the ratio of the potential at point (12 m, 0) to the potential at point (0, 5 m).

The gravitational potential $$V$$ is related to the gravitational field $$\vec{E}$$ by $$\vec{E} = -\nabla V$$. Therefore, the potential difference between two points is given by the line integral:

$$$ V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{r} $$$

Since the potential at the origin is zero, we can compute the potential at any point by integrating from (0,0) to that point. The field is conservative, so the path of integration does not matter. We will use straight-line paths for simplicity.

First, calculate the potential at point P(12 m, 0). We integrate along the x-axis from (0,0) to (12,0). Along this path, $$y = 0$$ and $$dy = 0$$, so $$d\vec{r} = dx \hat{i}$$. The dot product is:

$$$ \vec{E} \cdot d\vec{r} = (5 \hat{i} + 12 \hat{j}) \cdot (dx \hat{i}) = 5 dx $$$

The potential at P is:

$$$ V_P - V_{\text{origin}} = -\int_{(0,0)}^{(12,0)} \vec{E} \cdot d\vec{r} = -\int_{0}^{12} 5 dx $$$

Solving the integral:

$$$ -\int_{0}^{12} 5 dx = -5 \int_{0}^{12} dx = -5 \left[ x \right]_{0}^{12} = -5 (12 - 0) = -60 $$$

Since $$V_{\text{origin}} = 0$$, we have $$V_P = -60 \text{J/kg}$$.

Next, calculate the potential at point Q(0, 5 m). We integrate along the y-axis from (0,0) to (0,5). Along this path, $$x = 0$$ and $$dx = 0$$, so $$d\vec{r} = dy \hat{j}$$. The dot product is:

$$$ \vec{E} \cdot d\vec{r} = (5 \hat{i} + 12 \hat{j}) \cdot (dy \hat{j}) = 12 dy $$$

The potential at Q is:

$$$ V_Q - V_{\text{origin}} = -\int_{(0,0)}^{(0,5)} \vec{E} \cdot d\vec{r} = -\int_{0}^{5} 12 dy $$$

Solving the integral:

$$$ -\int_{0}^{5} 12 dy = -12 \int_{0}^{5} dy = -12 \left[ y \right]_{0}^{5} = -12 (5 - 0) = -60 $$$

Since $$V_{\text{origin}} = 0$$, we have $$V_Q = -60 \text{J/kg}$$.

The ratio of the potentials is:

$$$ \frac{V_P}{V_Q} = \frac{-60}{-60} = 1 $$$

Hence, the ratio is 1, which corresponds to Option B.

So, the answer is Option B.