To solve this problem, we need to match the gravitational terms with their respective dimensional formulas.$\\$Let's analyze each term:$\\$(a) Gravitational constant (G):$\\$The gravitational constant is used in Newton's law of gravitation:$\\ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \\$Here, $F$ has dimensions $[MLT^{-2}], \\ m_1$ and $m_2$ have dimensions $[M], \\$and $r$ has dimensions $[L]. \\$Rearranging for $G,$ we get:$\\ G = \frac{F \cdot r^2}{m_1 \cdot m_2} \\$Thus, the dimensions of $G$ are:$\\ [M^{-1}L^3T^{-2}] \\$(b) Gravitational potential energy:$\\$Gravitational potential energy is given by $U = mgh, \\$where $m$ is mass $[M], \\ g$ is acceleration due to gravity $[LT^{-2}], \\$and $h$ is height $[L]. \\$Thus, the dimensions of gravitational potential energy are:$\\ [ML^2T^{-2}] \\$(c) Gravitational potential:$\\$Gravitational potential $V$ is defined as the potential energy per unit mass:$\\ V = \frac{U}{m} \\$Thus, the dimensions of $V$ are:$\\ [ML^2T^{-2}] \cdot [M^{-1}] = [L^2T^{-2}] \\$(d) Gravitational intensity:$\\$Gravitational intensity (or field) $g$ is the force per unit mass:$\\ g = \frac{F}{m} \\$Thus, the dimensions of $g$ are:$\\ [MLT^{-2}] \cdot [M^{-1}] = [LT^{-2}] \\$Now, we match the terms with their dimensional formulas:$\\$(a) Gravitational constant (G) $\rightarrow (ii) [M^{-1}L^3T^{-2}] \\$(b) Gravitational potential energy $\rightarrow (iv) [ML^2T^{-2}] \\$(c) Gravitational potential $\rightarrow (i) [L^2T^{-2}] \\$(d) Gravitational intensity $\rightarrow (iii) [LT^{-2}] \\$Therefore, the correct option is:$\\$Option 2: (a)->(ii), (b)->(iv), (c)->(i), (d)->(iii)