To solve this problem, we need to understand the forces acting on the ball when it reaches terminal velocity.$\\$At terminal velocity, the net force acting on the ball is zero. This means the gravitational force is balanced by the buoyant force and the viscous force.$\\$Let's denote:$\\$• $M$ as the mass of the ball.$\\$• $d$ as the density of the ball.$\\$• $d_2$ as the density of glycerine.$\\$• $g$ as the acceleration due to gravity.$\\$The gravitational force acting on the ball is given by:$\\ F_{gravity} = Mg \\$The buoyant force acting on the ball is given by Archimedes' principle:$\\ F_{buoyant} = V \cdot d_2 \cdot g \\$where $V$ is the volume of the ball. Since the ball is spherical, $V = \frac{M}{d}. \\$Substituting the volume, we get:$\\ F_{buoyant} = \frac{M}{d} \cdot d_2 \cdot g \\$At terminal velocity, the viscous force $F_{viscous}$ balances the net force:$\\ F_{viscous} = F_{gravity} - F_{buoyant} \\$Substitute the expressions for $F_{gravity}$ and $F_{buoyant}: \\ F_{viscous} = Mg - \frac{M}{d} \cdot d_2 \cdot g \\$Simplify the expression:$\\ F_{viscous} = Mg \left(1 - \frac{d_2}{d}\right) \\$Given that the ball reaches terminal velocity, the density of the ball is twice that of glycerine:$\\ d = 2d_2 \\$Substitute $d = 2d_2$ into the expression for $F_{viscous}: \\ F_{viscous} = Mg \left(1 - \frac{d_2}{2d_2}\right) \\ F_{viscous} = Mg \left(1 - \frac{1}{2}\right) \\ F_{viscous} = Mg \cdot \frac{1}{2} \\$Therefore, the viscous force acting on the ball is $\frac{Mg}{2}$, which corresponds to Option 1.