Question

A spherical solid ball of volume $V$ is made up 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 $v,i.e.F_v=-k{v}^2,(k>0).$ The terminal speed of the ball is

• A. $\sqrt{\frac{({\rho }_1-{\rho }_2)gV}{k}}$
• B. $\sqrt{\frac{({\rho }_1-{\rho }_2)gV}{2k}}$
• C. $\sqrt{\frac{({\rho }_1+{\rho }_2)gV}{k}}$
• D. $\sqrt{\frac{({\rho }_1+{\rho }_2)gV}{2k}}$

Answer 1

The correct option is A $\sqrt{\frac{({\rho }_1-{\rho }_2)gV}{k}}$

Forces acting on the ball,
$1.$ Buoyant force, $F_b={\rho }_{2}gV$ $[$upward$]$
$2.$ Viscous force, $F_v=k{v}^2$ $[$upward$]$
$3.$ Weight, $F_g={\rho }_{1}gV$ $[$downward$]$
We know that, at terminal velocity, net force is zero.
So, $F_b+F_v=F_g$
$\Rightarrow {\rho }_{2}gV+k{v}^2={\rho }_{1}gV$
$\Rightarrow k{v}^2=({\rho }_1-{\rho }_2)gV$
$\Rightarrow v=\sqrt{\frac{({\rho }_1-{\rho }_2)gV}{k}}$