$A$ 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

$A$ 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 $g = 10 \,m / s^2$.)

$A$ large drop of oil (density $0.8 \, g/cm^3$ and viscosity $\eta_0$) floats up through a column of another liquid (density $1.2 \, g/cm^3$ and viscosity $\eta_L$). Assuming that the two liquids do not mix,the velocity with which the oil drop rises will depend on.

The terminal velocity of a copper ball of radius $5\,mm$ falling through a tank of oil at room temperature is $10\,cm\,s^{-1}$. If the viscosity of oil at room temperature is $0.9\,kg\,m^{-1}s^{-1}$,the viscous drag force is:

$A$ liquid disturbed by stirring comes to rest after some time due to its property of

$A$ spherical liquid drop of radius $r$ acquires the terminal velocity $v_1$ when falling through a gas of viscosity $\eta$. Now the drop is broken into $64$ identical droplets and each droplet acquires terminal velocity $v_2$ falling through the same gas. The ratio of terminal velocities $v_1/v_2$ is . . . . . . .