Q. A spherical solid ball of volume $V$ is made of a material 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$v,ie,F_{viscous} = - {kv}^{2}(k > 0)\rbrack.$ The terminal speed of the ball is
A) $\sqrt{\frac{Vg(\rho_{2} < \rho_{2})}{k}}$
B) $\frac{Vg\rho_{1}}{k}$
C) $\sqrt{\frac{Vg\rho_{1}}{k}}$
D) $\frac{Vg(\rho_{1} < \rho_{2})}{k}$
Explanation
(a) The force 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 is $v_{T}.$
So, $V\rho_{2}g + kv_{T}^{2} = V\rho_{1}g$
$v_{r} = \sqrt{\frac{V\left( \rho_{1} - \rho_{2} \right)g}{k}}$
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.