Two drops of equal radius are falling with a terminal velocity of $5 \, cm/s$. If they coalesce to form a single larger drop,what will be the terminal velocity of the new drop?

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

The velocity of a small ball of mass $m$ and density $\rho$ when dropped in a container filled with glycerine of density $\sigma$ becomes constant after some time. The viscous force acting on the ball in the final stage is

$A$ lead sphere of mass $m$ falls with a terminal velocity $V$ in a viscous liquid. With what terminal velocity will another lead sphere of mass $8m$ fall in the same liquid?

Two drops of same radius are falling through air with steady velocity of $v \ cm/s$. If the two drops coalesce,what would be the terminal velocity?

$A$ solid metal sphere released in a vertical liquid column has attained terminal velocity in the downward direction. The magnitudes of viscous force,buoyant force,and gravitational force acting on it are $F_{v}$,$F_{B}$,and $F_{W}$ respectively. Then,the correct relation between them is: