Two balls A and B have radii in the ratio 1:2. What will be the ratio of their terminal velocities in a liquid?

Let the radius of the smaller ball $=r$

the radius of the bigger ball$=2r$

Density of the ball's material$={\rho }_b$

Density of liquid$=\rho$

Terminal velocity of the balls is v1 and v2 for the smaller and bigger respectively. 

Volume of the smaller ball $V=\frac{4}{3}\pi r^3$

The mass of the ball will be $={\rho }_{b}V$

Volume of the bigger ball $=\frac{4}{3}\pi (2r{)}^3=8V$

The mass of the ball will be$=8{\rho }_{b}V$

Upthrust on the smaller ball$=\rho V_g$

Viscous drag force on smaller ball $=6\pi \eta r{v}_1$

Weight of the smaller ball $={\rho }_b{V}_g$

Now, since the balls are moving with terminal velocity, the total upward force will balance its weight 

$\rho V_g+6\pi \eta r{v}_1={\rho }_b{V}_g----1.Upthrustonthebiggerball$= 8 \rho V_g$Viscousdragforceonbiggerball$= 6\pi \eta 2rv_2$Weightofthebiggerball$= 8\rho _bV_g$Similarly,$8\rho V_g+6\pi\eta 2rv_2=8\rho _bV_g$----2.Dividingeqn.1byeqn.2$\dfrac{ρVg+6πηrv1}{8ρVg+12πηrv2}=\dfrac{1}{8}$=>$8\rho V_g+12\pi \eta rv_2=8\rho V_g+48\pi \eta rv_1$=>$12\pi \eta rv_2=48\pi \eta rv_1$=>$v_2=4v_1$=>$\dfrac{v_1}{v_2}=\dfrac{1}{4}$Thus,theratiooftheirterminalvelocity$= 1:4$$