Fluid Mechanics

JEE Main level, single correct, medium

38 of 192

Question 38 of 192

A small ball of mass MM and density ρ\rho is dropped in a viscous liquid of density ρ0{\rho }_{0}. After some time, the ball falls with a constant velocity. What is the viscous force on the ball?

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Answer: (C) F=Mg(1−ρ0ρ)F=Mg\left(1-\dfrac{{\rho }_{0}}{\rho }\right)

The buoyant force is given by
B=Vρ0gB=V{\rho }_{0}g
where VV is the volume of the ball. The weight of the ball
W=Mg=VρgW=Mg=V\rho g, is acting in the opposite direction of the buoyant force . Hence, at terminal velocity
Mg−B=FvMg-B={F}_{v}
⇒B+Fv=Mg\Rightarrow B+{F}_{v}=Mg
⇒Fv=Vρg−Vρ0g⇒Fv=ρVg(1−ρ0ρ)⇒Fv=Mg(1−ρ0ρ)\Rightarrow {F}_{v}=V\rho g-V{\rho }_{0}g \Rightarrow {F}_{v}=\rho Vg(1-\dfrac{{\rho }_{0}}{\rho }) \Rightarrow {F}_{v}=Mg(1-\dfrac{{\rho }_{0}}{\rho })

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