A small metal sphere of density is dropped into a tall cylinder containing a viscous liquid of density (where > ). The sphere eventually reaches a terminal velocity v_T . The acceleration of the sphere at the exact instant its velocity is half of its terminal velocity is

Step-by-step solution

Let the volume of the sphere be V . The mass of the sphere is m = V . At terminal velocity v_T , the net force on the sphere is zero. The downward gravitational force is balanced by the upward buoyant force and the upward viscous force: F_ vT = F_g - F_b F_ vT = V g - V g = V( - )g According to Stokes' law, the viscous force is directly proportional to the velocity of the sphere ( F_v v ). Therefore, when the velocity is v = v_T 2 , the viscous force is half of its maximum value: F_v = 1 2 F_ vT = 1 2 V( - )g Now,