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 ${\rho }_2<{\rho }_1$. 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 ofthe ball is :

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

Answer 1

Answer: (C)

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

At terminal velocity, the weight of the ball is balance by the viscous force and the buoyant force upwards.

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

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