Question

If an oil drop of density 0.95 X 103 Kg/m3 and radius 10-4 cm is falling in air whose density is 1.3 Kg/m3. Coefficient of viscosity is 18 X 10 -6. Calculate the terminal speed of the drop.

Answer 1

Answer:

ok other will answer be given by other one

Answer 2

The terminal speed of the drop is $1.171\times10^-^4m/s$.

Explanation:

Terminal velocity id given as

$v=\frac{2}{9} (\frac{d_o-d_a}{\eta}) r^2g$

Here $\eta$ is the coefficient of viscosity, r is the radius of the drop, $d_0$ is the density of the oil, d_a is the density of the air and g is the acceleration due to gravity.

Given  $d_o=0.95\times10^3kg/m^3$, $d_a=1.3kg/m^3$, $\eta=18\times10^-^6$ and $r=10^-^4cm=10^-^6m$.

substitute the given values with the value of g = 9.8 m/s^2 we get

$v=\frac{2}{9} (\frac{0.95\times10^3kg/m^3-1.3kg/m^3}{18\times10^-^6}) (10^-^6m)^2\times10m/s^2$

$v=1.171\times10^-^4m/s$

Thus, the terminal speed of the drop is $1.171\times10^-^4m/s$.

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