Two small spheres of radii r and 4r fall through a viscous liquid with the same terminal velocity. The ratio between the viscous forces acting on them is

Question

Two small spheres of radii r and 4r fall through a viscous liquid with the same terminal velocity. The ratio between the viscous forces acting on them is (A)  1:2  (B)  4:1  (C)  1:16  (D)  1:4

Tardigrade Admin · Accepted Answer

The magnitude of the viscous force depends on the shape and size of the body, its speed and the viscosity of the fluid. Stokes established that if a sphere of radius

r

moves with velocity

v

through a fluid of viscosity

η

, the viscous force opposing the motion of the sphere

F = 6 π ηr v

\frac{F _{1}}{F _{2}} = \frac{r _{1}}{r _{2}}

[As

η =

constant and

v_{1} = v_{2}

]

\frac{F _{1}}{F _{2}} = \frac{r}{4 r}

Q. Two small spheres of radii $r$ and $4 r$ fall through a viscous liquid with the same terminal velocity. The ratio between the viscous forces acting on them is

$1 : 2$

$4 : 1$

$1 : 16$

$1 : 4$

Q. Two small spheres of radii r and 4r fall through a viscous liquid with the same terminal velocity. The ratio between the viscous forces acting on them is

1:2

4:1

1:16

1:4

Q. Two small spheres of radii $r$ and $4 r$ fall through a viscous liquid with the same terminal velocity. The ratio between the viscous forces acting on them is

$1 : 2$

$4 : 1$

$1 : 16$

$1 : 4$