In an experiment,a small steel ball falls through a liquid at a constant speed of $10\, cm/s$. If the steel ball is pulled upward with a force equal to twice its effective weight,how fast will it move upward? ......... $cm/s$

Eight identical drops of water falling through air with a uniform terminal velocity of $10 \ cm/s$ combine to form a single big drop. The terminal velocity of the big drop will be: (in $cm/s$)

State Stokes' law. By using it,deduce the expression for:
$(i)$ Initial acceleration of a smooth sphere.
$(ii)$ Equation of terminal velocity of a sphere falling freely through a viscous medium.
$(iii)$ Explain: Upward motion of bubbles produced in a fluid.

$A$ spherical solid ball of volume $V$ is made of a material of density $\rho_1$. It is falling through a liquid of density $\rho_2$ $(\rho_2 < \rho_1)$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$,i.e.,$F_{viscous} = -kv^2$ $(k > 0)$. The terminal speed of the ball is:

$A$ metal sphere of mass $m$ and density $\sigma_{1}$ falls with terminal velocity through a container containing liquid. The density of the liquid is $\sigma_{2}$. The viscous force acting on the sphere is

Two metal spheres are falling through a liquid of density $2.5 \times 10^3 \ kg/m^3$ with the same uniform speed. The density of the material of the first sphere and the second sphere is $11.5 \times 10^3 \ kg/m^3$ and $8.5 \times 10^3 \ kg/m^3$ respectively. The ratio of the radius of the first sphere to that of the second sphere is