Question

You win the lottery and decide to impress your friends by exhibiting a million-dollar cube of gold. At the time, gold is selling for $1282 per troy ounce , and 1.0000 troy ounce equals 31.1035 g. How tall would your million-dollar cube be?

Short answer

The height of gold cubes purchased using million dollar is 10.79 cm .

Step-by-step solution

Identification of given data

The given data can be listed below,

• The selling price of gold is, p = 1282 dollar/troy ounce
• The value of one troy ounce is, 1troy ounce = 31.1035 g

Concept/Significance of volume of cube

A cube is made up of six squares that are all joined to make a solid. A cube's volume is always equal to the length of one side "cubed ," or L x L x L.

Determination of the height of million-dollar cube.

The troy bought with one million can be given by,

$\frac{1000000}{1282/\text{troy ounce}}=780.03\text{troy ounces}$

The mass of the gold is given by,

$\begin{array}{r}m=\frac{1000000}{1282}\times 31.1035\mathrm{g}\\ =24261.70\mathrm{g}\end{array}$

The volume of the cube of gold is given by

$V=\frac{m}{p}$ …(i)

Here, m is the mass of gold troy and $p$is he density of gold whose value is $19.3g/c{m}^3$.

The volume of cube can also be expressed as,

$V=a^3$ …(ii)

Here, a is the side of cube.

Equate equation (i) and (ii)

[A1]This should be done in next step 4.

$\begin{array}{r}{a}^3=\frac{m}{\rho }\\ a=\sqrt[3]{\frac{m}{\rho }}\end{array}$

Substitute values in the above equation.

$\begin{array}{r}a=\sqrt[3]{\frac{24261.70\mathrm{g}}{19.3\mathrm{g}/{\mathrm{cm}}^3}}\\ =10.79\mathrm{cm}\end{array}$

Thus, the height of gold cubes purchased using million dollar is 10.79 cm.