$A$ person moves in sea water at terminal velocity wearing an electronic digital watch. What is the effect on the measurement of time by the waterproof watch?

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).

Two metal spheres are falling through a liquid of density $2.5 \times 10^3 \ kg/m^3$ with the same uniform speed. The density of the material of the first sphere and the second sphere is $11.5 \times 10^3 \ kg/m^3$ and $8.5 \times 10^3 \ kg/m^3$ respectively. The ratio of the radius of the first sphere to that of the second sphere is

Eight identical small drops of water are falling down vertically through a medium,each with terminal velocity $v$. If they combine to form a single drop,then its terminal velocity will be (in $v$)

$A$ small spherical solid ball is dropped from a great height in a viscous liquid. Its journey in the liquid is best described in the diagram given below by the

Two solid spheres of the same metal but of mass $M$ and $8M$ fall simultaneously in a viscous liquid. If their terminal velocities are $v$ and $nv$, then the value of $n$ is: