Physics - Fluid Question with Solution | TestHub

When a small spherical body falls in a viscous liquid, it experiences three forces: its weight downwards, buoyant force upwards, and viscous drag force upwards.

The equation of motion is given by:
$mg-F_B-F_v=ma$

Using Stokes' law, $F_v=6\pi \eta rv$, the equation becomes:
$mg-\frac{4}{3}\pi r^3\sigma g-6\pi \eta rv=m\frac{dv}{dt}$

This can be written in the form:
$\frac{dv}{dt}=A-Bv$
where $A$ and $B$ are positive constants.

Integrating with the initial condition $v=0$ at $t=0$, we get:
$v(t)=\frac{A}{B}(1-e^{-Bt})$

This equation indicates that the velocity of the body initially increases with time, but the rate of increase (acceleration) continuously decreases. As $t\to \infty$, the velocity asymptotically approaches a constant maximum value known as the terminal velocity ($v_T=\frac{A}{B}$).

The first graph correctly represents this behavior, showing the velocity increasing and eventually leveling off to a constant value.

Answer: