Question

A table-tennis ball has a diameter of $\mathbf{3}\mathbf{.}\mathbf{80}\mathit{c}\mathit{m}$ and average density of $\mathbf{0}\mathbf{.}\mathbf{0840}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$. What force is required to hold it completely submerged under water?

Short answer

The force is required to hold it completely submerged under water:

$F=0.257N$

Step-by-step solution

Step 1:

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

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

Where$V_{disp}$ the weight of fluid is displaced at and${\rho }_{Fluid}$ is the density of the fluid.

Step 2:

Given:

Diameteroftabletennisball$d=3.8cm$

Radius:$r=1.9\times {10}^{-2}m$

Density of ball${\rho }_b=0.084g/c{m}^3$

We know density of water${\rho }_w=1g/c{m}^3$

As we 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,

$$\begin{gathered} w+F=F_b \\ {\rho }_{b}Vg+F={\rho }_{w}Vg..............\left(\because v=volume\right) \\ F={\rho }_{w}Vg-{\rho }_{b}Vg \\ F=\left({\rho }_w-{\rho }_b\right)Vg \\ F=\left(1-0.084\right)\frac{4}{3}\pi r^3\times 9.8 \\ F=\left(1-0.084\right)\frac{4}{3}\times \frac{22}{7}\times 1.9^3\times 9.8 \\ F=0.257N \end{gathered}$$