In Millikan's oil drop experiment, a charged drop falls with terminal velocity $V$. If an electric field $E$ is applied in a vertically upward direction, then it starts moving in an upward direction with terminal velocity $2V$. If the magnitude of the electric field is decreased to $E/2$, then the terminal velocity will become

Eight identical drops of same radii are falling through air with steady velocity $6 \ cm/s$. If these drops coalesce,then the terminal velocity would be .......... $cm/s$.

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: $A$ spherical body of radius $(5 \pm 0.1) \ mm$ having a particular density is falling through a liquid of constant density. The percentage error in the calculation of its terminal velocity is $4\,\%$.
Reason $R$: The terminal velocity of the spherical body falling through the liquid is inversely proportional to its radius.
In the light of the above statements,choose the correct answer from the options given below:

$A$ spherical ball of radius $1 \times 10^{-4} \,m$ and density $10^4 \,kg \,m^{-3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If the velocity of the ball does not change after entering the water, then the value of $h$ is:

Eight equal drops of water,each of radius $r = 2 \ mm$,are falling through air with a terminal velocity of $16 \ cm/s$. The eight drops combine to form a single big drop. The terminal velocity of the bigger drop will be $.... \ cm/s$.

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 . . . . . . $cm/s$.