To solve this problem, we need to analyze the motion of both the disc and the sphere as they roll down the inclined plane.$\\$Both objects are rolling without slipping, so we need to consider both translational and rotational motion.$\\$The total kinetic energy of a rolling object is the sum of its translational and rotational kinetic energies:$\\ K_{total} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \\$where $m$ is the mass, $v$ is the linear velocity, $I$ is the moment of inertia, and $\omega$ is the angular velocity.$\\$For rolling without slipping, the condition $v = r \omega$ holds, where $r$ is the radius of the object.$\\$Let's consider the moment of inertia for each object:$\\$• For a disc, $I_{disc} = \frac{1}{2} m r^2 \\$• For a sphere, $I_{sphere} = \frac{2}{5} m r^2 \\$Using the rolling condition $v = r \omega,$ we can express $\omega$ as $\omega = \frac{v}{r}. \\$Substitute $\omega$ into the kinetic energy equation:$\\ K_{total} = \frac{1}{2} m v^2 + \frac{1}{2} I \left(\frac{v}{r}\right)^2 \\$For the disc:$\\ K_{total, disc} = \frac{1}{2} m v^2 + \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 \\ K_{total, disc} = \frac{1}{2} m v^2 + \frac{1}{4} m v^2 \\ K_{total, disc} = \frac{3}{4} m v^2 \\$For the sphere:$\\ K_{total, sphere} = \frac{1}{2} m v^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \\ K_{total, sphere} = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 \\ K_{total, sphere} = \frac{7}{10} m v^2 \\$The gravitational potential energy at the top of the incline is converted into kinetic energy at the bottom.$\\$For both objects, the potential energy is $mgh,$ where $h$ is the height of the incline.$\\$Equating potential energy to total kinetic energy:$\\ mgh = K_{total} \\$For the disc:$\\ mgh = \frac{3}{4} m v_{disc}^2 \\ gh = \frac{3}{4} v_{disc}^2 \\ v_{disc}^2 = \frac{4}{3} gh \\$For the sphere:$\\ mgh = \frac{7}{10} m v_{sphere}^2 \\ gh = \frac{7}{10} v_{sphere}^2 \\ v_{sphere}^2 = \frac{10}{7} gh \\$Since the sphere has a larger $v^2$ term, it will have a greater velocity at the bottom of the incline.$\\$Thus, the sphere reaches the bottom first.$\\$Therefore, the correct option is Option 4: Sphere.