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Question: A spherical solid ball of volume \(V\)is made of a material of density \({\rho _1}\). It is falling ...

A spherical solid ball of volume VVis made of a material of density ρ1{\rho _1}. It is falling through a liquid of density ρ2{\rho _2} (ρ2<ρ1)\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 vv,that is Fviscous=−kv2 {F_{viscous}} = - k{v^{2\,}} (k>0)(k > 0). What is the terminal speed of the ball ?
A. Vg(ρ1−ρ2)k\sqrt {\dfrac{{Vg\left( {{\rho _1} - {\rho _2}} \right)}}{k}}
B. Vgρ1k\dfrac{{Vg{\rho _1}}}{k}
C. Vgρ1k\sqrt {\dfrac{{Vg{\rho _1}}}{k}}
D. Vg(ρ2<ρ1)k\dfrac{{Vg\left( {{\rho _2} < {\rho _1}} \right)}}{k}

Explanation

Solution

Hint- The condition for terminal velocity is that weight is equal to sum of buoyant force and viscous force.
wball=Fbuoyant+Fviscous{w_{ball}} = {F_{buoyant}} + {F_{viscous}}
Where wball{w_{ball}}is the weight of the ball , Fbuoyant{F_{buoyant}} is the buoyant force and Fviscous{F_{viscous}}is the viscous force
wball=Vρ1g{w_{ball}} = V{\rho _1}g
Fbuoyant=Vρ2g{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.
wball=Fbuoyant+Fviscous{w_{ball}} = {F_{buoyant}} + {F_{viscous}} (1)
Where wball{w_{ball}}is the weight of the ball , Fbuoyant{F_{buoyant}} is the buoyant force and Fviscous{F_{viscous}}is the viscous force

wball=mg =Vρ1g  {w_{ball}} = mg \\\ = V{\rho _1}g \\\
Since, Density is mass divided by volume.ρ=mV\rho = \dfrac{m}{V}
Buoyant force Fbuoyant=Vρ2g{F_{buoyant}} = V{\rho _2}g .since buoyant force is equal to the weight of the liquid displaced
Given viscous force Fviscous=−kv2 {F_{viscous}} = - k{v^{2\,}}
Substitute all the values in equation(1)
wball=Fbuoyant+Fviscous Vρ1g=Vρ2g+kv2   {w_{ball}} = {F_{buoyant}} + {F_{viscous}} \\\ V{\rho _1}g = V{\rho _2}g + k{v^{2\,}} \\\
Solve this equation to find terminal velocity vv

Vρ1g=Vρ2g+kv2  Vρ1g−Vρ2g=kv2  v2=(Vρ1g−Vρ2gk) v=Vρ1g−Vρ2gk =Vg(ρ1−ρ2)k  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.