Chapter 15, Problem 25PQ

A hollow copper (ρ_Cu = 8.92 × 10³ kg/m³) spherical shell of mass m = 0.950 kg floats on water with its entire volume below the surface. a. What is the radius of the sphere? b. What is the thickness of the shell wall?

To determine

The radius of the sphere.

Answer to Problem 25PQ

The radius of the sphere is $6.10\text{cm}$.

Explanation of Solution

Since the sphere is in equilibrium condition so that weight of the spherical shell is equal to the buoyant force of the water.

  $F_g=F_B$                                                                                   (I)

Here, $F_g$ is the force due to gravity and $F_B$ is the buoyant force.

Write the expression for force due to gravity.

  $F_g=mg$                                                                                  (II)

Here, $m$ is the mass of the copper spherical shell and $g$ is the acceleration due to gravity.

Write the expression for buoyant force.

  $F_B={\rho }_{\text{water}}Vg$                                                                          (III)

Here, ${\rho }_{\text{water}}$ is the density of the water and $V$ is the volume of the spherical shell.

Write the expression for volume of the spherical shell.

  $V=\frac{4}{3}\pi r_{\text{outer}}^3$                                                                           (IV)

Here, $r_{\text{outer}}$ is the outer radius of the sphere.

Conclusion:

Substitute the equation (II) and (III) in the equation (I).

  $mg={\rho }_{\text{water}}Vg$

Substitute equation (IV) in the above equation to solve for .

  $\begin{array}{c}mg={\rho }_{\text{water}}\frac{4}{3}\pi r_{\text{outer}}^{3}g\\ m={\rho }_{\text{water}}\frac{4}{3}\pi r_{\text{outer}}^3\\ r_{\text{outer}}^3=\frac{3m}{4\pi {\rho }_{\text{water}}}\\ r_{\text{outer}}={\left(\frac{3m}{4\pi {\rho }_{\text{water}}}\right)}^{1/3}\end{array}$

Substitute $0.950\text{kg}$ for $m$ and $1000{\text{kg/m}}^3$ for ${\rho }_{\text{water}}$ in the above equation to find $r_{\text{outer}}$.

  $\begin{array}{c}{r}_{\text{outer}}={\left(\frac{3\left(0.950\text{kg}\right)}{4\left(3.14\right)\left(1000{\text{kg/m}}^3\right)}\right)}^{1/3}\\ =0.0610\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ =6.10\text{cm}\end{array}$

Therefore, the radius of the sphere is $6.10\text{cm}$.

To determine

The thickness of the shell wall.

Answer to Problem 25PQ

The thickness of the shell wall is $2.37\times {10}^{-3}\text{m}$.

Explanation of Solution

Write the expression for mass of the ball.

  $m={\rho }_{\text{Cu}}V$

Here, $m$ is the mass of the ball, ${\rho }_{\text{Cu}}$ is the density of the copper, and $V$ is the volume of the spherical shell.

Rearrange the above equation for volume of the inner and outer radius of the sphere.

  $m={\rho }_{\text{Cu}}\left(\frac{4}{3}\pi r_o^3-\frac{4}{3}\pi r_i^3\right)$                                                                        (V)

Here, $r_o$ is the outer radius of the sphere and $r_i$ is the inner radius of the sphere.

Write the expression for thickness of the shell wall.

  $\Delta r=r_o-r_i$                                                                                             (VI)

Here, $\Delta r$ is the thickness of the shell wall.

Conclusion:

Substitute $8.92\times {10}^3{\text{kg/m}}^3$ for ${\rho }_{\text{Cu}}$, $0.950\text{kg}$ for $m$, and $0.0610\text{m}$ for $r_o$ in the equation (V).$\begin{array}{c}\left(0.950\text{kg}\right)=\left(8.92\times {10}^3{\text{kg/m}}^3\right)\left(\frac{4}{3}\pi \right)\left[{\left(0.0610\text{m}\right)}^3-r_i^3\right]\\ \frac{0.950\text{kg}}{8.92\times {10}^3{\text{kg/m}}^3}=\left(4.187\right)\left[{\left(0.0610\text{m}\right)}^3-r_i^3\right]\\ 1.065\times {10}^{-4}{\text{m}}^3=9.503\times {10}^{-4}{\text{m}}^3-4.187r_i^3\\ 1.065\times {10}^{-4}{\text{m}}^3-9.503\times {10}^{-4}{\text{m}}^3=-4.187r_i^3\end{array}$

Solve the above equation to find $r_i$.

  $\begin{array}{c}-8.438\times {10}^{-4}{\text{m}}^3=-4.187r_i^3\\ r_i^3=2.015\times {10}^{-4}{\text{m}}^3\\ r_i={\left(2.015\times {10}^{-4}{\text{m}}^3\right)}^{1/3}\\ =5.86\text{cm}\end{array}$

Substitute $5.86\text{cm}$ for $r_i$ and $6.10\text{cm}$ for $r_o$ in the equation (VI) to find $\Delta r$.

  $\begin{array}{c}\Delta r=6.10\text{cm}-5.86\text{cm}\\ =0.24\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =2.37\times {10}^{-3}\text{m}\\ \approx \text{2}\text{.4}\times {10}^{-3}\text{m}\end{array}$

Therefore, the thickness of the shell wall is $2.37\times {10}^{-3}\text{m}$.