To solve this problem, we need to determine the height $h$ from which the body must be released to just complete a vertical circle of diameter $D. \\$Given:$\\$• Diameter of the circle $= D \\$• Radius of the circle $= \frac{D}{2} \\$For the body to just complete the vertical circle, it must have enough velocity at the top of the circle to maintain circular motion.$\\$At the top of the circle (point B), the gravitational force provides the necessary centripetal force.$\\$The condition for just completing the circle is:$\\ v_B^2 = g \cdot \frac{D}{2} \\$Using energy conservation between the initial point and the top of the circle:$\\ mgh = mgD + \frac{1}{2}mv_B^2 \\$Substitute $v_B^2 = g \cdot \frac{D}{2}: \\ mgh = mgD + \frac{1}{2}m(g \cdot \frac{D}{2}) \\$Simplify:$\\ mgh = mgD + \frac{mgD}{4} \\ mgh = \frac{4mgD}{4} + \frac{mgD}{4} \\ mgh = \frac{5mgD}{4} \\$Cancel $m$ from both sides:$\\ gh = \frac{5gD}{4} \\$Cancel $g$ from both sides:$\\ h = \frac{5D}{4} \\$Therefore, the height $h$ is equal to $\frac{5}{4}D. \\$This corresponds to Option 4.