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?

Answer 1

Answer:

The million dollar cube will be 10.79 cm tall

Step-by-step explanation:

Let the sides of the gold cube representing the tallness of the cube  be s

volume of the gold cube V = s³

Then :

$\mathbf{s = \sqrt[3]{V} }$  --- (1)

Similarly; we know that , density ρ = mass m/Volume V

making volume the subject of the formula, then ;

V = m/ρ

Replacing that into (1) above , we have:

$\mathbf{s = \sqrt[3]{\dfrac{m}{\rho}} }$

Given that:

At the time, gold is selling for $1282 per troy ounce

∴ In a million-dollar cube of gold, the number of troy ounce that can be bought with a million dollar = 1000000/1282 = 780.03 troy ounce

Since 1.0000 troy ounce equals 31.1035 g

∴ 780.03 troy ounce = 780.03  × 31.1035 g

= 24261.66 g

From the tables 12.1 of Gold , the density of Gold is obtained as 19.3 g/cm³

∴

$\mathbf{s = \sqrt[3]{\dfrac{m}{\rho}} }$

$\mathbf{s = \sqrt[3]{\dfrac{24261.66}{19.3}} }$

$\mathbf{s = \sqrt[3]{1257.080829} }$

$\mathbf{s = 10.79 \ cm }}$

The million dollar cube will be 10.79 cm tall