If the terminal speed of a sphere of gold (density $\rho_g = 19.5 \times 10^3 \ kg/m^3$) is $0.2 \ m/s$ in a viscous liquid (density $\rho_L = 1.5 \times 10^3 \ kg/m^3$),find the terminal speed of a sphere of silver (density $\rho_s = 10.5 \times 10^3 \ kg/m^3$) of the same size in the same liquid. (in $m/s$)

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6 \,cm \,s^{-1}$. If they coalesce to form one big drop,what will be the terminal speed of the bigger drop? (Neglect the buoyancy of the air)

$A$ spherical solid ball of volume $V$ is made up of a material of density $\rho$. It is falling through a liquid of density $\sigma$ $(\sigma < \rho)$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of the terminal speed $v_{T}$,$F = -K v_{T}^2$ $(K > 0)$. Then,the terminal speed of the ball is ($g =$ acceleration due to gravity).

$A$ metal sphere of mass $m$ and density $\sigma_{1}$ falls with terminal velocity through a container containing liquid. The density of the liquid is $\sigma_{2}$. The viscous force acting on the sphere is

In an experiment to calculate the coefficient of viscosity of a liquid by measuring terminal velocity, it was observed that a ball of mass $(\frac{\pi}{2}) \ gm$ and radius $0.5 \ cm$ takes $5 \ s$ to fall steadily through a height of $50 \ cm$ inside a long column of liquid of density $1.2 \ gm/cc$. The coefficient of viscosity will be given as: (in $\text{poise}$)

What is the velocity $v$ of a metallic ball of radius $r$ falling in a tank of liquid at the instant when its acceleration is one-half that of a freely falling body? (The densities of metal and of liquid are $\rho$ and $\sigma$ respectively,and the viscosity of the liquid is $\eta$).