To solve this problem, we need to determine the minimum velocity required for a body to complete a vertical loop of radius $R. \\$Consider the forces acting on the body at the top of the loop. At this point, the gravitational force and the centripetal force must be balanced.$\\$The gravitational force acting on the body is $mg,$ where $m$ is the mass of the body and $g$ is the acceleration due to gravity.$\\$The centripetal force required to keep the body moving in a circle is $\frac{mv^2}{R},$ where $v$ is the velocity of the body at the top of the loop.$\\$For the body to just complete the loop, the centripetal force must be equal to the gravitational force at the top of the loop:$\\ \frac{mv^2}{R} = mg \\$Simplifying, we get:$\\ v^2 = gR \\$Thus, the minimum velocity at the top of the loop is $v = \sqrt{gR}. \\$However, to find the minimum velocity at the bottom of the loop, we need to consider the conservation of energy.$\\$At the bottom of the loop, the body has kinetic energy and potential energy. At the top of the loop, it has only potential energy.$\\$Let $v_0$ be the velocity at the bottom of the loop. The total mechanical energy at the bottom is:$\\ \frac{1}{2}mv_0^2 \\$At the top of the loop, the total mechanical energy is:$\\ \frac{1}{2}mv^2 + 2mgR \\$Using conservation of energy, we equate the energies at the bottom and top of the loop:$\\ \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + 2mgR \\$Substitute $v^2 = gR$ into the equation:$\\ \frac{1}{2}mv_0^2 = \frac{1}{2}m(gR) + 2mgR \\$Simplify:$\\ \frac{1}{2}mv_0^2 = \frac{1}{2}mgR + 2mgR \\ \frac{1}{2}mv_0^2 = \frac{5}{2}mgR \\$Solve for $v_0^2: \\ v_0^2 = 5gR \\$Thus, the minimum velocity at the bottom of the loop is $v_0 = \sqrt{5gR}. \\$Therefore, the correct answer is Option 2: $\sqrt{5gR}.$