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:

The terminal velocity $v$ of a spherical ball of lead of radius $R$ falling through a viscous liquid varies with $R$ such that

An iron sphere of mass $20 \times 10^{-3} \ kg$ falls through a viscous liquid with terminal velocity $0.5 \ ms^{-1}$. The terminal velocity (in $ms^{-1}$) of another iron sphere of mass $54 \times 10^{-2} \ kg$ is (in $.5$)

Two small spherical metal balls,having equal masses,are made from materials of densities $\rho_{1}$ and $\rho_{2}$ (where $\rho_{1} = 8 \rho_{2}$) and have radii of $1 \; mm$ and $2 \; mm$,respectively. They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals $\eta$ and whose density is $0.1 \rho_{2}$. The ratio of their terminal velocities would be:

Small water droplets of radius $0.01 \,mm$ are formed in the upper atmosphere and falling with a terminal velocity of $10 \,cm/s$. Due to condensation, if $8$ such droplets are coalesced to form a larger drop, the new terminal velocity will be ........... $cm/s$.