Two rain drops of same radius are falling through air each with a steady speed of $5 \,cm/s$. If the drops coalesce,the new steady velocity of the big drop will be

$A$ sphere is dropped under gravity through a fluid of viscosity $\eta$. If the average acceleration is half of the initial acceleration,the time to attain the terminal velocity is ($\rho$ = density of sphere; $r$ = radius).

Sixty-four rain drops of radius $1 \ mm$ each, falling down with a terminal velocity of $10 \ cm/s$, coalesce to form a bigger drop. The terminal velocity of the bigger drop is . . . . . . $cm/s$.

$A$ 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:

Two spheres of equal masses but radii $r_1$ and $r_2$ are allowed to fall in a liquid of infinite column. The ratio of their terminal velocities is

$A$ spherical body of radius $r$ and density $\sigma$ falls freely through a viscous liquid having density $\rho$ and viscosity $\eta$ and attains a terminal velocity $v_0$. The estimated maximum error in the quantity $\eta$ is: (Ignore errors associated with $\sigma, \rho$ and $g$, gravitational acceleration)