Question

castor oil at 20°c has a coefficient of viscosity 2·42Ns/m^2 ana a density 940kg/m^3.
calculate the terminal velocity of a steel ball of radius 2.0mm falling under gravity in the oil, taking the density of steel as 7800kg/m^3

Answer 1

The formula for terminal velocity of a ball of radius r, of material of density $\rho_{s}$, falling in a material of density $\rho_{l}$, and coefficient of viscosity $\eta$ is given by:

$v_{t} = \frac{2}{9} \frac{r^2 * (\rho_{s} - \rho_{l}) * g}{\eta}$

Substituting the given values, we get:

$v_{t} = 0.025 m/s$

Note #1: The value of g is taken to be $10m/s^2$.

Note #2: However, the calculation with the given data seems tough. If the viscosity was 3.43, however, it would become significantly easy, and the answer then, would be:

$v_{t} = 0.0177... m/s, or  \frac{16}{9} * 10^-^2 m/s$

For the derivation of the initial formula, kindly refer to my answer to:

https://brainly.in/question/8357895

Answer 2

so the terminal velocity of steel ball is 0.0251 m/s

given

temperature of castor oil = 20°c

coefficient of viscosity = 2.42 Ns/$m^{2}$

density = 940kg/$m^{3}$.

radius of steel ball = 2mm =2/1000 m

density of steel = 7800kg/  $m^{3}$

to find

terminal velocity of steel ball falling under gravity in the oil = v

solution

• terminal velocity is the maximum velocity that is attatined by a body falling through the medium.

• when a body falls through a medium the velocity of its motion changes throughout but after some time it become constant and this constant velocity is also called terminal velocity

• this happens when gravitational force acting downward equals the buoyant force acting upward plus the viscous drag force acting upward

• terminal velocity formula(v) :

$\frac{2 r^{2} g (p-s) }{9 n}$

where r is the radius of the spherical body

g is acceleration due to gravity

n is the viscosity of the medium

p is the density of the falling body

s is the  fluid medium density

• so putting all the values we get

v  =  $\frac{2 X (0.002)^{2} X 10 X ( 7800 - 940) }{9 X (2.42)}$

v   =  $\frac{0.5488}{21.78}$

v  =  0.0251 m/s

so the terminal velocity of steel ball is 0.0251 m/s

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