Two metal spheres are falling through a liquid of density $3 \times 10^3 \ kg/m^3$ with the same uniform speed. The material densities of sphere $1$ and sphere $2$ are $8 \times 10^3 \ kg/m^3$ and $11 \times 10^3 \ kg/m^3$ respectively. The ratio of their radii is

Eight equal drops of water are falling through air with a steady speed of $10\,cm/s$. If the drops coalesce,the new velocity is $.........\,cm/s$.

$A$ spherical ball of radius $1 \times 10^{-4} \,m$ and density $10^5 \,kg/m^3$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering the water the velocity of the ball does not change, then the value of $h$ is approximately: (The coefficient of viscosity of water is $9.8 \times 10^{-6} \,N s/m^2$) (in $\,m$)

The terminal velocity of a copper ball of radius $5\,mm$ falling through a tank of oil at room temperature is $10\,cm\,s^{-1}$. If the viscosity of oil at room temperature is $0.9\,kg\,m^{-1}s^{-1}$,the viscous drag force is:

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 (in $cm \ s^{-1}$)? (Neglect the buoyancy of the air)

The terminal velocity of a liquid drop of radius $r$ falling through air is $v$. If two such drops are combined to form a bigger drop,the terminal velocity with which the bigger drop falls through air is (ignore any buoyant force due to air).