Two equal drops are falling through air with a steady velocity of $5\, cm/s$. If two drops coalesce to form one drop,then the new terminal velocity will be:

An air bubble of radius $r$ rises steadily through a liquid of density $\rho$ with velocity $v$. The coefficient of viscosity of the liquid is:

$A$ small steel ball is dropped from a height of $1.5 \,m$ into a glycerine jar. The ball reaches the bottom of the jar $1.5 \,s$ after it was dropped. If the retardation in the glycerine is $2.66 \,m/s^2$, the height of the glycerine in the jar is about (acceleration due to gravity $g = 9.8 \,m/s^2$) (in $\,m$)

$A$ liquid drop of mass $m$ and radius $r$ is falling from a great height. Its terminal velocity is proportional to ............

If the terminal speed of a sphere $A$ [density $\rho_A = 7.5 \ kg \ m^{-3}$] is $0.4 \ ms^{-1}$ in a viscous liquid [density $\rho_L = 1.5 \ kg \ m^{-3}$],the terminal speed of sphere $B$ [density $\rho_B = 3 \ kg \ m^{-3}$] of the same size in the same liquid is: (in $ms^{-1}$)

$A$ solid steel ball of diameter $3.6 \ mm$ acquires a terminal velocity of $2.45 \times 10^{-2} \ m/s$ while falling under gravity through an oil of density $925 \ kg/m^3$. Take the density of steel as $7825 \ kg/m^3$ and $g$ as $9.8 \ m/s^2$. The viscosity of the oil in $SI$ units is: