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Stokes' Law And Terminal Velocity - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Stokes' law & Terminal Velocity is considered one the most difficult concept.

  • 29 Questions around this concept.

Solve by difficulty

Spherical balls of radius R are falling in a viscous fluid of viscosity \eta with a velocity\nu. The retarding viscous force acting on the spherical ball is:

A spherical solid ball of volume V is made of a material of density \rho1. It is falling through a liquid of density \rho2 (\rho2  < \rho1). Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed \nu ,i.e.,F_{viscous}= -kv^{2}\left ( k> 0 \right ). The terminal speed of the ball is

If the terminal speed of a sphere of gold \left ( density = 19.5 kg/m^{3} \right ) is 0.2 m/s in a viscous liquid \left ( density = 1.5 kg/m^{3} \right ), find the terminal speed (in m/sec) of a sphere of silver \left ( density = 10.5 kg/m^{3} \right ) of the same size in the same liquid.

Concepts Covered - 1

Stokes' law & Terminal Velocity
  • Stokes' law-

When a body moves through a fluid then The fluid exerts a viscous force on the body to oppose its motion.

And according to Stokes' law, the magnitude of the viscous force depends on the shape and size of the body, its speed and the viscosity of the fluid.

So for the below figure

If a sphere of radius r moves with velocity v through a fluid of viscosity \eta

Then using  Stokes' law the viscous force (F) opposing the motion of the sphere is given by

$
F=6 \pi \eta r v
$


Where
$\eta$ - coefficient viscosity
$r$ - radius
$v-v e l o c i t y$

  • Terminal Velocity-

When the spherical body is dropped in a viscous fluid, it is first accelerated and then it's acceleration becomes zero and it attains a constant velocity and this constant velocity is known as terminal velocity.

For a  spherical body of radius r is dropped in a viscous fluid, The forces acting on it are shown in the below figure.

 

So Forces acting on the body are

1. Weight of Body (W)

$
W=m g=\frac{4}{3} \pi r^3 \rho g
$


Where $\rho \rightarrow$ density of body
2. Buoyant/ Thrust Force (T of $F_B$ )

$
T=F_B=\frac{4}{3} \pi r^3 \sigma g
$

where $\sigma \rightarrow$ density of fluid
3. Viscous force (F)

$
F=6 \pi \eta r v
$


So when the body attains terminal velocity the net force acting on the body is zero.
Apply force balance

$
\begin{aligned}
& F_B+F=W \\
& \rightarrow 6 \pi \eta r v+\frac{4}{3} \pi r^3 \sigma g=\frac{4}{3} \pi r^3 \rho g \\
& \rightarrow 6 \pi \eta r v=\frac{4}{3} \pi r^3 g(\rho-\sigma) \\
& \rightarrow v_t=\frac{2}{9} \frac{r^2(\rho-\sigma)}{\eta} g
\end{aligned}
$


Where $v_T=$ terminal velocity

                From this formula, we can say that

  • Terminal velocity depends on the radius of the sphere/body.
  • Greater the density of solid greater will be the terminal velocity     
  • Greater the density and viscosity of the fluid lesser will be the terminal velocity.      
  • If ρ > σ then Terminal velocity will be positive.

            I.e Spherical body attains constant velocity in a downward direction.     

  •  If ρ < σ then Terminal velocity will be negative.

         I.e Spherical body attains constant velocity in an upward direction.

  •  Terminal velocity graph

                  

 

 

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Stokes' law & Terminal Velocity

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Stokes' law & Terminal Velocity

Physics Part II Textbook for Class XI

Page No. : 264

Line : 5

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