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 $cm/s$)

Two identical drops of water are falling through air with steady velocity $V$. If the two drops coalesce to form a single drop,what is the new terminal velocity of the single drop?

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6 \,cm \,s^{-1}$. If they coalesce to form one big drop,what will be the terminal speed of the bigger drop? (Neglect the buoyancy of the air)

$A$ ball of mass $m$ and radius $r$ is gently released in a viscous liquid. The mass of the liquid displaced by it is $m'$ such that $m > m'$. The terminal velocity is proportional to

$A$ tiny spherical oil drop carrying a net charge $q$ is balanced in still air, with a vertical uniform electric field of strength $\frac{81}{7} \pi \times 10^5 \,V / m$. When the field is switched $OFF$, the drop is observed to fall with terminal velocity $2 \times 10^{-3} \,m / s$. Here $g=9.8 \,m / s^2$, viscosity of air is $\eta = 1.8 \times 10^{-5} \,N s / m^2$ and density of oil is $\rho = 900 \,kg / m^3$. The magnitude of $q$ is

What is the velocity $v$ of a metallic ball of radius $r$ falling in a tank of liquid at the instant when its acceleration is one-half that of a freely falling body? (The densities of metal and of liquid are $\rho$ and $\sigma$ respectively,and the viscosity of the liquid is $\eta$).