To solve this problem, we need to analyze the motion of the object on an inclined plane.$\\$Given:$\\$• The angle of inclination $\theta_1 = 60^\circ$ for distance $x_1. \\$• The angle of inclination $\theta_2 = 30^\circ$ for distance $x_2. \\$• The initial velocity $u$ is the same in both cases.$\\$The object is projected up the inclined plane, so the acceleration due to gravity along the plane is $g \sin \theta. \\$Using the kinematic equation for motion along the inclined plane:$\\ v^2 = u^2 - 2a s \\$Since the object comes to rest at the maximum distance, $v = 0. \\$For the first case ($\theta_1 = 60^\circ$):$\\ 0 = u^2 - 2(g \sin 60^\circ) x_1 \\$Solve for $x_1: \\ x_1 = \frac{u^2}{2g \sin 60^\circ} \\$For the second case ($\theta_2 = 30^\circ$):$\\ 0 = u^2 - 2(g \sin 30^\circ) x_2 \\$Solve for $x_2: \\ x_2 = \frac{u^2}{2g \sin 30^\circ} \\$Now, find the ratio $x_1 : x_2: \\ \frac{x_1}{x_2} = \frac{\frac{u^2}{2g \sin 60^\circ}}{\frac{u^2}{2g \sin 30^\circ}} \\$Simplify the expression:$\\ \frac{x_1}{x_2} = \frac{\sin 30^\circ}{\sin 60^\circ} \\$Substitute the values of the sines:$\\ \sin 30^\circ = \frac{1}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2} \\ \frac{x_1}{x_2} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} \\ \frac{x_1}{x_2} = \frac{1}{\sqrt{3}} \\$Therefore, the ratio $x_1 : x_2$ is $1 : \sqrt{3}. \\$This corresponds to Option 3.