The terminal velocity $(v_{t})$ of a spherical raindrop depends on the radius $(r)$ of the raindrop as:

Two small spherical metal balls,having equal masses,are made from materials of densities $\rho_{1}$ and $\rho_{2}$ (where $\rho_{1} = 8 \rho_{2}$) and have radii of $1 \; mm$ and $2 \; mm$,respectively. They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals $\eta$ and whose density is $0.1 \rho_{2}$. The ratio of their terminal velocities would be:

$Assertion :$ Falling raindrops acquire a terminal velocity.
$Reason :$ $A$ constant force in the direction of motion and a velocity dependent force opposite to the direction of motion,always result in the acquisition of terminal velocity.

What will be the approximate terminal velocity of a rain drop of diameter $1.8 \times 10^{-3} \ m$, when density of rain water $\approx 10^{3} \ kg \ m^{-3}$ and the coefficient of viscosity of air $\approx 1.8 \times 10^{-5} \ N \ s \ m^{-2}$ (in $m \ s^{-1}$)? (Neglect buoyancy of air)

$A$ copper ball of radius $3.0 \,mm$ falls in an oil tank of viscosity $1 \,kg / ms$. Then, the terminal velocity of the copper ball will be (Density of oil $= 1.5 \times 10^3 \,kg / m^3$, Density of copper $= 9 \times 10^3 \,kg / m^3$ and $g = 10 \,m / s^2$.)

$A$ 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: