Two rain drops of same radius $r$ falling with terminal velocity $V$ merge and form a bigger drop with radius $R$. The terminal velocity of the bigger drop is:

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 $Pa \ s$):

$A$ sphere of radius $r$ and density $\rho$ is dropped from a height $h$. When it falls into water,it attains a terminal velocity. If the coefficient of viscosity of water is $\eta$,then $h =$

$A$ metal ball of radius $9 \times 10^{-4} \ m$ and density $10^4 \ kg/m^3$ falls freely under gravity through a distance $h$ and enters a tank of water. Considering that the metal ball has constant velocity in water,the value of $h$ is [coefficient of viscosity of water $= 8.1 \times 10^{-4} \ Pa \cdot s, g = 10 \ m/s^2$,density of water $= 10^3 \ kg/m^3$]. (in $m$)

$A$ steel ball is dropped in a viscous liquid. The distance of the steel ball from the top of the liquid is shown in the graph below. The terminal velocity of the ball is closest to .......... $m/s$.

$A$ rain drop of diameter $1 \ mm$ falls with a terminal velocity of $0.7 \ ms^{-1}$ in air. If the coefficient of viscosity of air is $2 \times 10^{-5} \ Pa \cdot s$, the viscous force on the rain drop is