$125$ 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 $\,m/s$)

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 $m/s$)

An air bubble of radius $1 \ cm$ rises from the bottom portion through a liquid of density $1.5 \ g/cc$ at a constant speed of $0.25 \ cm \ s^{-1}$. If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (in $Pa \ s$):

An air bubble of diameter $6\,mm$ rises steadily through a solution of density $1750\,kg/m^3$ at the rate of $0.35\,cm/s$. The coefficient of viscosity of the solution (neglect density of air) is $..........\,Pa\cdot s$ (given,$g = 10\,m/s^2$).

$A$ water drop of radius $1\,\mu m$ falls in a situation where the effect of buoyant force is negligible. The coefficient of viscosity of air is $1.8 \times 10^{-5}\,Nsm^{-2}$ and its density is negligible as compared to that of water $10^{3}\,kgm^{-3}$. The terminal velocity of the water drop is . . . . . . $\times 10^{-6}\,ms^{-1}$. (Take acceleration due to gravity $= 10\,ms^{-2}$)

$A$ copper ball of radius $r$ travels with a uniform speed $v$ in a viscous fluid. If the ball is changed with another ball of radius $2r$,then the new uniform speed will be