Question

A spherical solid ball of volume V is 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., $F_{viscous}=-k{v}^2(k>0)$. The terminal speed of the ball is.

• A. $\frac{Vg({\rho }_1-{\rho }_2)}{k}$
• B. $\sqrt{\frac{Vg{\rho }_1}{k}}$
• C. $\frac{Vg{\rho }_1}{k}$
• D. $\sqrt{\frac{Vg({\rho }_1-{\rho }_2)}{k}}$

Answer 1

The correct option is D

$\sqrt{\frac{Vg({\rho }_1-{\rho }_2)}{k}}$

The forces acting on the ball are gravity force, buoyancy force and viscous force. When ball acquires terminal speed, it is in dynamic equilibrium, let terminal speed of ball be $v_T$. So

$V{\rho }_{2}g+k{v}_T^2=V{\rho }_{1}g$

$v_T=\sqrt{\frac{V({\rho }_1-{\rho }_2)g}{k}}$