$A$ gas bubble of $2 \ cm$ diameter rises through a liquid of density $1.75 \ g \ cm^{-3}$ with a fixed speed of $0.35 \ cm \ s^{-1}$. Neglect the density of the gas. The coefficient of viscosity of the liquid is (in $\text{poise}$)

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:

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 $.5$)

$A$ copper ball of radius $3.0 \,mm$ falls in an oil tank of viscosity $1 \,kg / ms$. Then, the terminal velocity of the copper ball will be (Density of oil $= 1.5 \times 10^3 \,kg / m^3$, Density of copper $= 9 \times 10^3 \,kg / m^3$ and $g = 10 \,m / s^2$.)

Eight identical small drops of water are falling down vertically through a medium,each with terminal velocity $v$. If they combine to form a single drop,then its terminal velocity will be (in $v$)