A sphere of mass \(M\) is dropped into glycerin. The density of glycerin is half the density of the sphere. After some t…

ExplanationKey Concept:

At terminal velocity (constant velocity), the net force acting on the sphere is zero. The forces acting on the sphere are:

1. Gravitational Force (\({F}_{g}\)): Downward force due to the weight of the sphere. \({F}_{g}=Mg\)
2. Buoyant Force (\({F}_{b}\)​): Upward force due to the displaced glycerin. \({F}_{b}={\rho }_{\text{glycerin}}Vg\) where \({\rho }_{\text{glycerin}}\)is the density of glycerin, \(V\) is the volume of the sphere, and \(g\) is the acceleration due to gravity.
3. Viscous Force (\({F}_{v}\)​): Upward force due to the viscosity of glycerin. This balances the remaining downward force at terminal velocity.

From equilibrium at terminal velocity:

\({F}_{v}={F}_{g}−{F}_{b}\)​

Step 1: Relationship Between Densities

The density of glycerin (\({\rho }_{\text{glycerin}}\)) is half the density of the sphere (\({\rho }_{\text{sphere}}\)):

\({\rho }_{\text{glycerin}}=\frac{{\rho }_{\text{sphere}}}{2}\)​​

The mass \(M\) of the sphere is related to its density and volume:

\(M={\rho }_{\text{sphere}}V\)

Substitute \({\rho }_{\text{glycerin}}=\frac{{\rho }_{\text{sphere}}}{2}\)​​ into \({F}_{b}\):

\({F}_{b}=\frac{{\rho }_{\text{sphere}}}{2}Vg=\frac{M}{2}g\)

Step 2: Calculate the Viscous Force

Using \({F}_{v}={F}_{g}−{F}_{b}\):

\({F}_{v}=Mg−\frac{M}{2}g\)

Simplify:

\({F}_{v}=\frac{M}{2}g\)

Final Answer:

The viscous force acting on the sphere is:

\(\frac{Mg}{2}\)​​