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 ${\rho _2}$ $\left( {{\rho _2} < {\rho _1}} \right)$ Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$,that is ${F_{viscous}} = - k{v^{2\,}}$ $(k > 0)$. What is the terminal speed of the ball ?
A. $\sqrt {\dfrac{{Vg\left( {{\rho _1} - {\rho _2}} \right)}}{k}} $
B. $\dfrac{{Vg{\rho _1}}}{k}$
C. $\sqrt {\dfrac{{Vg{\rho _1}}}{k}} $
D. $\dfrac{{Vg\left( {{\rho _2} < {\rho _1}} \right)}}{k}$

Answer 1

Hint- The condition for terminal velocity is that weight is equal to sum of buoyant force and viscous force.
${w_{ball}} = {F_{buoyant}} + {F_{viscous}}$
Where ${w_{ball}}$is the weight of the ball , ${F_{buoyant}}$ is the buoyant force and ${F_{viscous}}$is the viscous force
${w_{ball}} = V{\rho _1}g$
\[{F_{buoyant}} = V{\rho _2}g\]

Step by step solution:
The condition for terminal velocity is that weight is equal to sum of buoyant force and viscous force.
${w_{ball}} = {F_{buoyant}} + {F_{viscous}}$ (1)
Where ${w_{ball}}$is the weight of the ball , ${F_{buoyant}}$ is the buoyant force and ${F_{viscous}}$is the viscous force

$
  {w_{ball}} = mg \\
   = V{\rho _1}g \\
 $
Since, Density is mass divided by volume.$\rho = \dfrac{m}{V}$
Buoyant force \[{F_{buoyant}} = V{\rho _2}g\] .since buoyant force is equal to the weight of the liquid displaced
Given viscous force ${F_{viscous}} = - k{v^{2\,}}$
Substitute all the values in equation(1)
$
  {w_{ball}} = {F_{buoyant}} + {F_{viscous}} \\
  V{\rho _1}g = V{\rho _2}g + k{v^{2\,}} \\
 $
Solve this equation to find terminal velocity $v$
\[
  V{\rho _1}g = V{\rho _2}g + k{v^{2\,}} \\
  V{\rho _1}g - V{\rho _2}g = k{v^{2\,}} \\
  {v^2} = \left( {\dfrac{{V{\rho _1}g - V{\rho _2}g}}{k}} \right) \\
  v = \sqrt {\dfrac{{V{\rho _1}g - V{\rho _2}g}}{k}} \\
   = \sqrt {\dfrac{{Vg\left( {{\rho _1} - {\rho _2}} \right)}}{k}} \\
 \]

So the answer is option A

Note: Terminal velocity is the maximum velocity attained by an object. It is attained when weight of the body is balanced by the viscous and the buoyant force. Here magnitude of the forces is taken into consideration. Thus when we substitute for the viscous force negative sign should not be taken only its magnitude should be taken.