Question

A table-tennis ball has a diameter of $\mathbf{\text{3.80\hspace{0.17em}cm}}$ and average density of $\mathbf{\text{0.0840\hspace{0.17em}}}\raisebox{1ex}{$\mathbf{\text{g}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{\text{cm}}}^{\mathbf{\text{3}}}$}\right.$. What force is required to hold it completely submerged under water?

Short answer

The required force to hold it completely submerged under water is $F=\text{0.258\hspace{0.17em}N}$.

Step-by-step solution

Given Data:

The diameter of the table tennis ball, $d=\text{3.8\hspace{0.17em}cm}$

The radius,$r=1.9\times {10}^{-2}\text{\hspace{0.17em}m}$

The density of ball,${\rho }_b=0.084\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{${\text{cm}}^{\text{3}}$}\right.$

A concept:

When an object is partially or completely submerged in a fluid, the fluid exerts an upward force on the object called buoyancy. According to Archimedes' principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object.

$B={\rho }_{Fluid}g{V}_{disp}$

Here, $V_{disp}$the volume of fluid, $g$is the acceleration due to gravity having a value $9.8\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.$, and ${\rho }_{Fluid}$ is the density of the fluid.

Determine what force is required to hold it completely submerged under water:

You know the density of water is,

${\rho }_w=1\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{${\text{cm}}^{\text{3}}$}\right.$

As you can see that ${\rho }_w>{\rho }_b$. So the ball will float over water, to submerge it into water extra force is required, Let it be $F$.

Now balancing forces on ball,

$$\begin{gathered} w+F=F_b \\ {\rho }_{b}Vg+F={\rho }_{w}Vg \end{gathered}$$

$\begin{array}{rcl}F& =& {\rho }_{w}Vg-{\rho }_{b}Vg\\ & =& ({\rho }_w-{\rho }_b)Vg\end{array}$

Substitute known values in the above equation.

$\begin{array}{rcl}F& =& (1-0.084)\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\times \frac{4}{3}\pi r^3\times 9.8\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\\ & =& (1-0.084)\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{m}}^{\text{3}}}$}\right.\text{×}\frac{4}{3}\times 3.14\times {(1.9\times {10}^{-2}\text{m})}^3\times 9.8\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\\ & =& 0.258\text{N}\end{array}$

Hence, the required force is to hold it completely submerged under water is $0.258\text{N}$.