Q21E z - University Physics with Modern Physics

Question

In Wagner’s opera Das Rheingold, the goddess Freia is ransomed for a pile of gold just tall enough and wide enough to hide her from sight. Estimate the monetary value of this pile. The density of gold is$\mathbf{19}\mathbf{.}\mathbf{3}\mathbf{ }\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$ 19.3 g / cm², and take its value to be about $\mathbf{\$}\mathbf{10}$ per gram.

Step-by-Step Solution

Answer

The value of gold will be around $100$ million dollars.

Step 1: Density of a substance

Here,

Density of a substance is the ratio of the mass to that of the volume of the substance. Density is a quantity that describe how closely the atoms are packed of the system. 

The SI unit of density is . Kg / m³

Here density of gold is given  19.3 g / cm³

Convert the given mass density into dollars per cubic feet.

Step 2: Calculation of density in dollar per cubic feet

Let us Convert the given mass density into dollars per cubic feet

Because each conversion factor has a value of $1$ they may be squared or cubed without affecting the results. $$\begin{gathered} =19.3\frac{\mathrm{g}}{{\mathrm{cm}}^3}\times \frac{\$10}{1 \mathrm{g}}\times {\left(\frac{2.54}{1 \mathrm{m}}\right)}^3\times {\left(\frac{12 \mathrm{m}}{1\mathrm{ft}}\right)}^3 \\ =5.47 \times 106 \raisebox{1ex}{$\$$}\!\left/ \!\raisebox{-1ex}{$f{t}^3$}\right. \end{gathered}$$

To calculate the ransom value, multiply the monetary density by Freia's volume (roughly that of a rectangle box                   $5 ft \times 2 ft \times 2 ft$). 

$\begin{array}{rcl}& & 5.48 \times {10}^6\frac{\$}{f{t}^3}\times 5 ft \times 2 ft \times 2 ft\\ & =& \$ {10}^8\end{array}$

Therefor the value of gold will be around $100$ million dollars