Chapter 1, Problem 1.25P

To determine

The radius of a solid aluminum sphere that balance a solid iron sphere on an equal arm balance.

Answer to Problem 1.25P

The radius of a solid aluminum sphere is $\text{2}\text{.86}\text{cm}$ that balance a solid iron sphere on an equal arm balance.

Explanation of Solution

The mass of one cubic meter aluminum is $2.70\times {10}^3\text{kg}$ and the mass of one cubic meter iron is $7.86\times {10}^3\text{kg}$. The radius of the solid iron sphere is $2.0\text{cm}$.

Write the formula to calculate the density of aluminum sphere

    ${\rho }_{\text{alu}}=\frac{m_1}{V_1}$

Here, ${\rho }_{\text{alu}}$ is the density of the aluminum sphere, $m_{\text{1}}$ is the mass of one cubic meter aluminum sphere and $V_1$ is the volume of the aluminum sphere.

Substitute $2.70\times {10}^3\text{kg}$ for $m_{\text{1}}$ and $1.0{\text{m}}^3$ for $V_1$ to find ${\rho }_{\text{alu}}$.

    $\begin{array}{c}{\rho }_{\text{alu}}=\frac{2.70\times {10}^3\text{kg}}{1.0{\text{m}}^3}\\ =2.70\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Write the formula to calculate the density of iron sphere

    ${\rho }_{\text{iron}}=\frac{m_2}{V_2}$

Here, ${\rho }_{\text{iron}}$ is the density of the iron sphere, $m_{\text{2}}$ is the mass of one cubic meter iron sphere and $V_2$ is the volume of the iron sphere.

Substitute $7.86\times {10}^3\text{kg}$ for $m_{\text{1}}$ and $1.0{\text{m}}^3$ for $V_1$ to find ${\rho }_{\text{iron}}$.

    $\begin{array}{c}{\rho }_{\text{iron}}=\frac{7.86\times {10}^3\text{kg}}{1.0{\text{m}}^3}\\ =7.86\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Write the formula to calculate the mass of an aluminum sphere

    $m_{\text{alu}}={\rho }_{\text{alu}}\cdot V_{\text{alu}}$

Here, $m_{\text{alu}}$ is the mass of the aluminum sphere, $V_{\text{alu}}$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of an aluminum sphere

    $V_{\text{alu}}=\frac{4}{3}\pi r_{\text{alu}}^3$

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

Substitute $\frac{4}{3}\pi r_{\text{alu}}^3$ for $V_{\text{alu}}$.

    $\begin{array}{c}{m}_{\text{alu}}={\rho }_{\text{alu}}\cdot \frac{4}{3}\pi r_{\text{alu}}^3\\ =\frac{4}{3}\pi r_{\text{alu}}^3{\rho }_{\text{alu}}\end{array}$                                                    (I)

Write the formula to calculate the mass of a solid iron sphere of radius $2.0\text{cm}$

    $m_{\text{iron}}={\rho }_{\text{iron}}\cdot V_{\text{iron}}$

Here, $m_{\text{iron}}$ is the mass of the aluminum sphere, $V_{\text{iron}}$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of iron sphere

    $V_{\text{iron}}=\frac{4}{3}\pi r_{\text{iron}}^3$

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

Substitute $\frac{4}{3}\pi r_{\text{iron}}^3$ for $V_{\text{iron}}$.

    $\begin{array}{c}{m}_{\text{iron}}={\rho }_{\text{iron}}\cdot \frac{4}{3}\pi r_{\text{iron}}^3\\ =\frac{4}{3}\pi r_{\text{iron}}^3{\rho }_{\text{iron}}\end{array}$                                                    (II)

Since both sphere must balance to each other on an equal arm balance. So, they both have equal mass.

Equating equation (I) and equation (II),

    $\begin{array}{c}{m}_{\text{alu}}=m_{\text{iron}}\\ \frac{4}{3}\pi r_{\text{alu}}^3{\rho }_{\text{alu}}=\frac{4}{3}\pi r_{\text{iron}}^3{\rho }_{\text{iron}}\\ r_{\text{alu}}^3{\rho }_{\text{alu}}=r_{\text{iron}}^3{\rho }_{\text{iron}}\\ r_{\text{alu}}=\sqrt[3]{\frac{{\rho }_{\text{iron}}}{{\rho }_{\text{alu}}}}\cdot r_{\text{iron}}\end{array}$

Conclusion:

Substituting $2.0\text{cm}$ for $r_{\text{iron}}$, $7.86\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{iron}}$ and $2.70\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{alu}}$ in the above equation to find $r_{\text{alu}}$.

    $\begin{array}{c}{r}_{\text{alu}}=\left(\sqrt[3]{\frac{7.86\times {10}^3\text{kg}/{\text{m}}^{\text{3}}}{2.70\times {10}^3\text{kg}/{\text{m}}^{\text{3}}}}\right)\cdot \left(2.0\text{cm}\right)\\ =1.427\times 2.0\text{cm}\\ =\text{2}\text{.86}\text{cm}\end{array}$

Therefore, the radius of a solid aluminum sphere is $\text{2}\text{.86}\text{cm}$ that balance a solid iron sphere on an equal arm balance.