A spherical ball of density ρ and radius 0.003 m is dropped into a tube containing a viscous fluid. If the viscosity of the fluid =1.260 N m -2 and its density is ρL= ρ / 2=1260 kg m -3, then find the terminal speed (in cm / s ) of the ball. (g=. Acceleration due to gravity .=10 m s -2)

Question

A spherical ball of density ρ and radius 0.003 m is dropped into a tube containing a viscous fluid. If the viscosity of the fluid =1.260 N m -2 and its density is ρL= ρ / 2=1260 kg m -3, then find the terminal speed (in cm / s ) of the ball. (g=. Acceleration due to gravity .=10 m s -2) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

M a=F text net =M g-B-6 π η r v At v=vT, a=0 ⇒ M g-B=6 π η r vT ⇒ vT=((4/3) π r3[ρ-(ρ/2)] g/6 π r η)=2 m / s

Q. A spherical ball of density ρ and radius 0.003m is dropped into a tube containing a viscous fluid. If the viscosity of the fluid =1.260Nm−2 and its density is ρL​= ρ/2=1260kgm−3, then find the terminal speed (in cm/s ) of the ball. (g= Acceleration due to gravity =10ms−2)

Ma=Fnet ​=Mg−B−6πηrv At v=vT​,a=0 ⇒Mg−B=6πηrvT​ ⇒vT​=6πrη34​πr3[ρ−2ρ​]g​=2m/s

$M a = F_{net} = M g - B - 6 π ηr v$
At $v = v_{T}, a = 0$
$\Rightarrow M g - B = 6 π ηr v_{T}$
$\Rightarrow v_{T} = \frac{\frac{4}{3} π r ^{3} [ ρ - \frac{ρ}{2} ] g}{6 π rη} = 2 m / s$