question_answer

A spherical solid ball of volume Vis made of a material of density \[{{\rho }_{1}}\]. It is falling through a liquid of density \[{{\rho }_{2}}({{\rho }_{2}}<{{\rho }_{1}})\]. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed v, i.e., \[{{F}_{\text{viscous}}}=-k{{v}^{2}}(k>0)\]. The terminal speed of the ball is

A) \[\sqrt{\frac{Vg({{\rho }_{1}}-{{\rho }_{2}})}{k}}\]

B) \[\frac{Vg{{\rho }_{1}}}{k}\]

C) \[\sqrt{\frac{Vg{{\rho }_{1}}}{k}}\]       

D)        \[\frac{Vg({{\rho }_{1}}-{{\rho }_{2}})}{k}\]

Correct Answer: A

Solution :

The condition for terminal speed \[({{\text{v}}_{t}})\] is Weight = Buoyant force \[+\] Viscous force \[\therefore \,\,\,\,\,V{{\rho }_{1}}g=V{{p}_{2}}g+k\text{v}_{t}^{2}\] \[\therefore \,\,\,\,\,\,{{\text{v}}_{t}}=\sqrt{\frac{Vg({{\rho }_{1}}-{{\rho }_{2}})}{k}}\]