Viscosity and Stoke's Law and Terminal Velocity Questions in English

(D) When a spherical object falls through a viscous fluid, it experiences three forces: the gravitational force $(F_g)$ acting downwards, the buoyant force $(F_b)$ acting upwards, and the viscous drag force $(F_v)$ acting upwards.
As the velocity of the sphere increases, the viscous force increases. Terminal velocity is reached when the net force on the sphere becomes zero, meaning the object stops accelerating.
At this point, the downward force is balanced by the sum of the upward forces:
$F_g = F_b + F_v$
Therefore, the buoyant force plus the viscous force becomes equal to the force due to gravity.

(D) When a small metal sphere falls through a viscous liquid, it experiences three forces: gravitational force $(mg)$ acting downwards, buoyant force $(F_B)$ acting upwards, and viscous drag force $(F_v = 6\pi\eta rv)$ acting upwards.
The net force is $F_{net} = mg - F_B - 6\pi\eta rv = ma$.
Initially, the velocity $(v)$ is zero, so the viscous drag is zero, and the acceleration is maximum. As the velocity increases, the viscous drag force increases, causing the net force and acceleration to decrease.
Eventually, the net force becomes zero when the viscous drag balances the effective weight, and the sphere attains a constant velocity known as terminal velocity. This behavior is represented by a curve that starts from the origin, increases with a decreasing slope, and approaches a horizontal asymptote, which corresponds to graph $(d)$.