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Terminal velocity is defined as the highest velocity attained by a body that is falling through a fluid. It is observed when the sum of drag force and buoyant force becomes equal to the downward gravitational force that is acting on the body. The acceleration of the body is zero as the net force acting on the body is zero.

Viscous drag or viscous force acts in the direction opposite to the motion of the body in the fluid. According to Stokeâ€™s law, the magnitude of the opposing viscous drag increases with the increasing velocity of the body.

As the body falls through a medium, its velocity goes on increasing due to the force of gravity acting on it. Thus, the opposing viscous or drag force, which acts upwards, also goes on increasing. A stage reaches when the true weight of the body is just equal to the sum of the upward thrust due to buoyancy, that is, upthrust, and the upward viscous drag. At this stage, there is no net force to accelerate the moving body. Hence it starts falling with a constant velocity, which is called terminal velocity.

Let ${{\rho }_{o}}$ be the density of the material of the spherical body, ball of radius $r$ and $\rho $ be the density of the medium.

True weight of the body is given as,

$W=\text{Volume}\times \text{Density}\times g$

Volume of spherical body is, $V=\dfrac{4}{3}\pi {{r}^{3}}$

Where,

$r$ is the radius of the ball

Therefore,

$W=\dfrac{4}{3}\pi {{r}^{3}}{{\rho }_{o}}g$

Where,

$r$ is the radius of the ball

${{\rho }_{o}}$ is the density of the material of the ball

Now,

Letâ€™s say ${{F}_{T}}$ is the weight of the liquid displaced

${{F}_{T}}=\text{Volume of liquid displaced }\times \text{ Density }\times g$

Or,

${{F}_{T}}=\dfrac{4}{3}\pi {{r}^{3}}\rho g$

Where,

$r$ is the radius of the ball

$\rho $ is the density of the material

$g$ is the acceleration due to gravity

If $v$ is the terminal velocity of the body, then according to Stokeâ€™s law,

Upward viscous drag is given as,

${{F}_{V}}=6\pi \eta rv$

Where,

$r$ is the radius of the ball

$\eta $ is the viscosity of the liquid

When the body acquires terminal velocity,

${{F}_{T}}+{{F}_{V}}=W$

Putting values,

$\begin{align}

& {{F}_{T}}=\dfrac{4}{3}\pi {{r}^{3}}\rho g \\

& {{F}_{V}}=6\pi \eta rv \\

& W=\dfrac{4}{3}\pi {{r}^{3}}{{\rho }_{o}}g \\

\end{align}$

We get,

$\begin{align}

& \dfrac{4}{3}\pi {{r}^{3}}\rho g+6\pi \eta rv=\dfrac{4}{3}\pi {{r}^{3}}{{\rho }_{o}}g \\

& 6\pi \eta rv=\dfrac{4}{3}\pi {{r}^{3}}\left( {{\rho }_{o}}-\rho \right)g \\

& v=\dfrac{2{{r}^{2}}\left( {{\rho }_{o}}-\rho \right)g}{9\eta } \\

\end{align}$

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