Chapter 11, Problem 84QAP

To determine

(a)

Height of the cube of gold won in lottery and is it possible for the person to take it to home?

Answer to Problem 84QAP

The height is $71.017\text{ cm}$ and yes, it is possible to take it at home.

Explanation of Solution

Given:

Price of lottery  = $146.9\text{ million euros = 211}\text{.8 million U}\text{.S}\text{. Dollars}$

Rate of gold  = $\text{\$953 per try ounce}$

Rate of silver  = $\$14.16\text{ per troy ounce}$

Relation= $\text{31}\text{.1035 g}$

Density of gold  = $\text{19}{\text{.3 g/cm}}^3$

Density of silver  = $10.5{\text{g/cm}}^3$

Formula used:

  $\begin{array}{l}\text{Density(d) = }\frac{\text{mass(m)}}{\text{Volume(V)}}\\ \text{Volume of cube = }{\left(\text{side}\right)}^3\end{array}$

Calculation:

Substituting the values,

  $\begin{array}{l}\text{Density(d) = }\frac{\text{mass(m)}}{\text{Volume(V)}}\\ \text{Volume of cube = }{\left(\text{side}\right)}^3\end{array}$

  $\begin{array}{l}\$211.8\times {10}^6\text{ = }\$2.118\times {10}^8\\ \text{1 Troy ounce = \$953}\\ \text{Number of troy ounces}\\ \text{N = }\frac{2.118\times {10}^8}{\text{953}}\\ \text{N = }222245.5404\text{ try ounces}\\ \text{1 Troy ounce = 31}\text{.1035 g}\\ 222245.5404\text{ try ounces =}222245.5404\times \text{ 31}\text{.1035 g}\\ \text{Therefore mass = 6912614}\text{.166 g}\\ \text{Density(d) = }\frac{\text{mass(m)}}{\text{Volume(V)}}\\ \text{Volume(V) = }\frac{\text{mass(m)}}{\text{Density(d)}}\\ \text{Volume(V) = }\frac{\text{6912614}\text{.166 g}}{\text{19}{\text{.3 g/cm}}^3}\\ \text{Volume(V) = }358166.5372{\text{ cm}}^3\\ \text{Now,}\\ \text{Volume of cube = }{\left(\text{side}\right)}^3\\ \text{side = }\sqrt[3]{358166.5372{\text{ cm}}^3}\\ \text{side = }71.017\text{ cm}\end{array}$

Conclusion:

The flow rate is $\text{0}\text{.11}\times {10}^{-3}{\text{ m}}^3\text{/s}$

To determine

(b)

Calculate the height of the silver cube with the same lottery amount.

Answer to Problem 84QAP

The height of the silver cube will be $353.856\text{ cm}$

Explanation of Solution

Given:

Price of lottery  = $146.9\text{ million euros = 211}\text{.8 million U}\text{.S}\text{. Dollars}$

Rate of gold  = $\text{\$953 per try ounce}$

Rate of silver  = $\$14.16\text{ per troy ounce}$

Relation= $\text{31}\text{.1035 g}$

Density of gold  = $\text{19}{\text{.3 g/cm}}^3$

Density of silver  = $10.5{\text{g/cm}}^3$

Formula used:

  $\begin{array}{l}\text{Density(d) = }\frac{\text{mass(m)}}{\text{Volume(V)}}\\ \text{Volume of cube = }{\left(\text{side}\right)}^3\end{array}$

Calculation:

Substituting the values,

  $\begin{array}{l}\text{Density(d) = }\frac{\text{mass(m)}}{\text{Volume(V)}}\\ \text{Volume of cube = }{\left(\text{side}\right)}^3\end{array}$

  $\begin{array}{l}\$211.8\times {10}^6\text{ = }\$2.118\times {10}^8\\ \text{1 Troy ounce = }\$14.16\\ \text{Number of troy ounces}\\ \text{N = }\frac{2.118\times {10}^8}{14.16}\\ \text{N = }14957627.1186\text{ try ounces}\\ \text{1 Troy ounce = 31}\text{.1035 g}\\ 14957627.1186\text{ try ounces =}14957627.1186\times \text{ 31}\text{.1035 g}\\ \text{Therefore mass = 465234555}\text{.0833 g}\\ \text{Density(d) = }\frac{\text{mass(m)}}{\text{Volume(V)}}\\ \text{Volume(V) = }\frac{\text{mass(m)}}{\text{Density(d)}}\\ \text{Volume(V) = }\frac{\text{465234555}\text{.0833 g}}{\text{10}{\text{.5 g/cm}}^3}\\ \text{Volume(V) = 44308052}{\text{.8650 cm}}^3\\ \text{Now,}\\ \text{Volume of cube = }{\left(\text{side}\right)}^3\\ \text{side = }\sqrt[3]{\text{44308052}{\text{.8650 cm}}^3}\\ \text{side = }353.856\text{ cm}\end{array}$

Conclusion:

Thus, the height of the silver cube will be $353.856\text{ cm}$