Question

For a ball falling in a iiquid with constant velocity, ratio of resistance force due to the liquid to that due to gravity is

• A. $\frac{2a^2\rho g}{9xv}$
• B. $\frac{2a^2\rho g}{9{\eta }^2}$
• C. $\frac{2a^2(\rho -\sigma )g}{9\eta }$
• D. none of these

Answer 1

The correct option is A $\frac{2a^2\rho g}{9xv}$

Force due to liquid (stokes law):

$f=6\pi xrv$

Force due to gravity:

$\begin{array}{c}f=mg\\ =\rho .\frac{4\pi r^3}{3}.g\end{array}$

Ratio =

$\begin{array}{c}\frac{4\rho .\pi r^2.g}{3.6\pi xr.v}\\ =\frac{2a^2\rho g}{9xv}\end{array}$

$\therefore \frac{2a^2\rho g}{9xv}$

so,

Option $A$ is correct answer.