Answer:
[tex]V=\dfrac{2}{9\ \mu}R^2g(\rho_d-\rho_f)[/tex]
Explanation:
Given that
Radius =R
[tex]Density\ of\ droplet=\rho_d[/tex]
[tex]Density\ of\ fluid=\rho_f[/tex]
When drop let will move downward then so
[tex]F_{net}=F_{weight}-F_{b}-F_d[/tex]
Fb = Bouncy force
Fd = Drag force
We know that
[tex]F_b=\dfrac{4\pi }{3}R^3\ \times \rho_f\times g[/tex]
[tex]F_{weight}=\dfrac{4\pi }{3}R^3\ \times \rho_d\times g[/tex]
[tex]F_{d}=6\pi \mu\ R\ V[/tex]
μ=Dynamic viscosity of fluid
V= Terminal velocity
So at the equilibrium condition
[tex]F_{net}=F_{weight}-F_{b}-F_d[/tex]
[tex]0=F_{weight}-F_{b}-F_d[/tex]
[tex]F_{weight}=F_{b}+F_d[/tex]
[tex]\dfrac{4\pi }{3}R^3\ \times \rho_d\times g=\dfrac{4\pi }{3}R^3\ \times \rho_f\times g+6\pi \mu\ R\ V[/tex]
So
[tex]V=\dfrac{2}{9\ \mu}R^2g(\rho_d-\rho_f)[/tex]
This is the terminal velocity of droplet.
