Correct Option is (A) = 9
Given,
\(r_1 = R\)
\(r_2 = \frac{R}{3}\)
The terminal velocity \(v = \frac{2}{9} \frac{r^2 (\rho - \sigma)g}{\eta}\)
where,
r = radius of falling body
\(\eta\) = cofficient of viscosity
g = gravity constant
\(\rho\) = density of the falling body
\(\sigma\) = density of fluid
\(v_1 \propto r^2\)
\(\frac{v_1}{v_2} = \frac{r_1{^2}}{r_2{^2}}\)
\(\frac{v_1}{v_2} \frac{R^2}{\left(\frac{R}{3}\right)^2}\)
\(\frac{v_1}{v_2} = \frac{9R^2}{R^2}\)
\(\frac{v_1}{v_2} = \frac{9}{1}\) \(\Rightarrow 9\)
