Two uniform solid balls of same density and of radii $r$ and $2r$ are dropped in air and fall vertically downwards. The terminal velocity of the ball with radius $r$ is $1 \, cm \, s^{-1}$. Find the terminal velocity of the ball of radius $2r$ (neglect buoyant force on the balls). ........... $cm \, s^{-1}$

Write the equation of terminal velocity.

$A$ spherical solid ball of volume $V$ is made up of a material of density $\rho$. It is falling through a liquid of density $\sigma$ $(\sigma < \rho)$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of the terminal speed $v_{T}$,$F = -K v_{T}^2$ $(K > 0)$. Then,the terminal speed of the ball is ($g =$ acceleration due to gravity).

$A$ spherical ball of radius $1 \text{ mm}$ and density $10.5 \text{ g/cc}$ is dropped in glycerine of coefficient of viscosity $9.8 \text{ poise}$ and density $1.5 \text{ g/cc}$. Viscous force on the ball when it attains constant velocity is $3696 \times 10^{-x} \text{ N}$. The value of $x$ is (Given, $g = 9.8 \text{ m/s}^2$ and $\pi = 22/7$)

Two solid spheres of the same metal but of mass $M$ and $8M$ fall simultaneously in a viscous liquid. If their terminal velocities are $v$ and $nv$, then the value of $n$ is:

Small water droplets of radius $0.01 \,mm$ are formed in the upper atmosphere and falling with a terminal velocity of $10 \,cm/s$. Due to condensation, if $8$ such droplets are coalesced to form a larger drop, the new terminal velocity will be ........... $cm/s$.