To solve this problem, we need to understand the relationship between the viscous force, terminal velocity, and heat production.$\\$When a small sphere falls through a viscous liquid, it experiences a viscous force given by Stokes' law:$\\ F_{viscous} = 6 \pi \eta r v \\$where $\eta$ is the viscosity of the liquid, $r$ is the radius of the sphere, and $v$ is the velocity of the sphere.$\\$At terminal velocity $v_t,$ the net force on the sphere is zero, so the gravitational force is balanced by the viscous force:$\\ mg = 6 \pi \eta r v_t \\$The terminal velocity $v_t$ can be expressed as:$\\ v_t = \frac{mg}{6 \pi \eta r} \\$The rate of production of heat due to viscous force is given by the power dissipated by the force:$\\ P = F_{viscous} \cdot v_t \\$Substitute $F_{viscous} = 6 \pi \eta r v_t$ into the power equation:$\\ P = (6 \pi \eta r v_t) \cdot v_t \\ P = 6 \pi \eta r v_t^2 \\$Substitute the expression for $v_t: \\ v_t = \frac{mg}{6 \pi \eta r} \\ P = 6 \pi \eta r \left(\frac{mg}{6 \pi \eta r}\right)^2 \\ P = 6 \pi \eta r \cdot \frac{m^2 g^2}{(6 \pi \eta r)^2} \\ P = \frac{m^2 g^2}{6 \pi \eta r} \\$Since $m$ is the mass of the sphere, it can be expressed in terms of its volume and density:$\\ m = \frac{4}{3} \pi r^3 \rho \\$where $\rho$ is the density of the sphere. Substitute $m$ into the power equation:$\\ P = \frac{\left(\frac{4}{3} \pi r^3 \rho\right)^2 g^2}{6 \pi \eta r} \\ P = \frac{\left(\frac{16}{9} \pi^2 r^6 \rho^2\right) g^2}{6 \pi \eta r} \\ P = \frac{16 \pi r^5 \rho^2 g^2}{54 \eta} \\ P \propto r^5 \\$Thus, the rate of production of heat when the sphere attains its terminal velocity is proportional to $r^5. \\$This corresponds to Option 1.