Question

The terminal speed of a sphere of gold (density = $19.5kg/{m^3}$) is $0.2m/s$ in a viscous liquid (density= $1.5kg/{m^3}$), find the terminal speed of a sphere of silver (density = $10.5kg/{m^3}$) of the same size in the same liquid
A. 0.4m/s
B. 0.133m/s
C. 0.1 m/s
D. 0.2m/s

Answer 1

Hint: The terminal velocity of a body is the speed at which the viscous force and buoyant force together balance its weight so that no net force acts on the body. The relation for terminal velocity is $v = \dfrac{2}{9}\dfrac{{(\sigma - \rho ){r^2}g}}{\eta }$

Complete step-by-step answer:
We know that when a sphere moves with a velocity $v$ through a medium of viscosity $\eta$, It experiences an opposing Force given by
$F = 6\pi r\eta v$ where $r$ is the radius of the sphere.
The terminal velocity of a body can be obtained by equating this viscous force with the apparent weight of the body in the fluid.
For a body falling downwards in a fluid, The net downward force on the body is:
Weight of body $-$ buoyant force
$F_d = mg-V \rho g$ where $\rho$ is the density of the fluid and $V$ is the volume of the sphere.
At terminal velocity, this downward force and the upward force due to viscosity balance each other.
So we can equate the forces as: $mg - V\rho g = 6\pi \eta rv$

$v = \dfrac{{mg - V\rho g}}{{6\pi \eta r}}$

We can now replace $m$ in the equation as $V\sigma g$ where V is the volume of sphere and $\sigma$ its density to get:
$v = \dfrac{{V(\sigma - \rho )g}}{{6\pi \eta r}}$. (1)
In the question, we are asked to compare the terminal velocities of two spheres of different densities with all other parameters fixed.
So we can say from equation (1) that :
$\dfrac{v}{{\sigma - \rho }} = constant$
$\dfrac{{{v_1}}}{{{\sigma _1} - \rho }} = \dfrac{{{v_2}}}{{{\sigma _2} - \rho }}$
Substituting the given values of ${v_1}$,${v_2}$, $\rho$, ${\sigma _1}$ and ${\sigma _2}$ into the equation gives :
$\dfrac{{0.2}}{{19.5 - 1.5}} = \dfrac{{{v_2}}}{{10.5 - 1.5}}$
${v_2} = \dfrac{{0.2}}{{19.5 - 1.5}}(10.5 - 1.5)$
${v_2} = \dfrac{{(0.2)(9)}}{{18}}$
${v_2} = 0.1m/s$

This is the required answer.

Note: Be careful while using $v = \dfrac{{(V\sigma - \rho )g}}{{6\pi \eta r}}$ for other calculations. The $V$ in equations is also dependent on $r$ and hence, terminal velocity is not inversely proportional to $r$. We may substitute $V = \dfrac{4}{3}\pi {r^3}$ and obtain a general relation: $v = \dfrac{2}{9}\dfrac{{(\sigma - \rho ){r^2}g}}{\eta }$