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 $.........\,cm/s$.

$A$ rain drop of radius $r$ is falling through air,starting from rest. The work done by all the forces on the drop,when it attains terminal velocity,is proportional to

$A$ sphere is dropped under gravity through a fluid of viscosity $\eta$. If the average acceleration is half of the initial acceleration,the time to attain the terminal velocity is ($\rho$ = density of sphere; $r$ = radius).

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

The graph between terminal velocity (along $y-$ axis) and square of radius (along $x-$ axis) of a spherical body of density $\rho$ allowed to fall through a fluid of density $\sigma$ is a

$125$ small water drops of same size fall through air with constant terminal velocity $4 \,cm/s$. They coalesce to form a big drop. The terminal velocity of the big drop is: (in $\,m/s$)