Question

The terminal velocity of a steel ball $2\,mm$ in diameter falling through glycerin is $44 \times {10^{ - 2}}\,cm/s$ (Given that specific gravity of steel $= 8$ , specific gravity of glycerin a $1.3$ and viscosity of glycerin $8.3\,poise$ )

Answer 1

Terminal velocity is the velocity of the object that free falls on the fluid. Use the formula of the terminal velocity given below and substitute the known parameters or the derived parameters in the formula. The simplification of the equation provides the answer for the terminal velocity of the steel ball through the glycerin.

Formula used:
The formula of the terminal velocity is given by
$v = \dfrac{{2{r^2}\left( {s - 6g} \right)a}}{{9\eta }}$
Where $v$ is the terminal velocity of the steel ball, $r$ is the radius of the steel ball, $s$ is the specific gravity of the steel, $g$ is the specific gravity of the glycerin, $\eta$ is the viscosity of the glycerin and $a$ is the acceleration due to gravity.

Complete step by step answer:
Diameter of the steel ball, $D = 2\,mm = 2 \times {10^{ - 3}}\,m$
Specific gravity of steel, $s = 8$
Specific gravity of glycerin, $g = 1.3$
Viscosity of the glycerin, $\mu = 8.3 = 0.83\,deca\,poise$
By using the formula of the terminal velocity,
$v = \dfrac{{2{r^2}\left( {s - 6g} \right)a}}{{9\eta }}$
Substituting the known values in the above formula. The radius is obtained from the diameter, by dividing it by $2$ . $r = \dfrac{d}{2} = \dfrac{{2 \times {{10}^{ - 3}}\,m}}{2} = {10^{ - 3}}\,m$
$v = \dfrac{{2{{\left( {{{10}^{ - 3}}} \right)}^2}\left( {8 \times {{10}^3} \times 9.8 - 1.3 \times {{10}^3}} \right)}}{{9 \times 0.83}}$
By doing the basic arithmetic operation in the above step, we get
$v = \dfrac{{19.6 \times {{10}^{ - 6}} \times 6.7 \times {{10}^3}}}{{7.47}}$
By performing further simplification of the above step,
$v = 1.7 \times {10^{ - 2}}\,m{s^{ - 1}}$

Hence, The terminal velocity of the steel ball is $1.7 \times {10^{ - 2}}\,m{s^{ - 1}}$.

Note: Given that specific gravity of glycerin is $a1.3$, so the value of the specific gravity is calculated by multiplying the acceleration due to gravity with its value $\left( {9.8 \times 1.3} \right)$ is substituted in the simplification step of the above calculation.