To solve this problem, we will use the principle of conservation of angular momentum.$\\$Given:$\\$• Initial velocity $v_0$ at radius $R_0. \\$• Final radius $R = \frac{R_0}{2}. \\$The angular momentum is conserved, so:$\\ m v_0 R_0 = m v R \\$Substitute the final radius:$\\ m v_0 R_0 = m v \left(\frac{R_0}{2}\right) \\$Simplify to find the final velocity $v: \\ v_0 R_0 = v \frac{R_0}{2} \\ v = 2 v_0 \\$The initial kinetic energy $K_i$ is:$\\ K_i = \frac{1}{2} m v_0^2 \\$The final kinetic energy $K_f$ is:$\\ K_f = \frac{1}{2} m v^2 \\$Substitute the final velocity:$\\ K_f = \frac{1}{2} m (2 v_0)^2 \\ K_f = \frac{1}{2} m \cdot 4 v_0^2 \\ K_f = 2 m v_0^2 \\$Therefore, the final value of the kinetic energy is $2 m v_0^2. \\$This corresponds to Option 2.