Chapter 1, Problem 43GP

To determine

To Find: The diameter of a 1 − ton rock.

Answer to Problem 43GP

  $d=2.74\text{ ft}$

Explanation of Solution

Given info:

Density of rock, ${\rho }_r=3{\rho }_W=3\times 62{\text{ lb/ft}}^3$

Mass of the rock, $M=1\text{ ton}$

Formula used:

  ${\rho }_r=\frac{M}{V}\mathrm{...............}\text{(i)}$

Where,

  ${\rho }_r=$ Density of rock.

  $M=$ Mass of rock.

  $V=$ Volume of rock.

Volume of spherical rock:

  $V=\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3\mathrm{.............}\text{(ii)}$

Where,

  $d=$ diameter of spherical rock.

Calculation:

  $1\text{ ton}=2000\text{ lb}$

Therefor, volume of rock can be calculated from equation (i),

i.e.

  $V=\frac{M}{{\rho }_r}$

Plugging in the given values:

  $\begin{array}{c}V=\frac{1\text{ ton}}{3\times 62{\text{ lb/ft}}^3}\\ =\frac{2000\text{ lb}}{3\times 62{\text{ lb/ft}}^3}\\ =10.75{\text{ ft}}^3\end{array}$

The diameter of the rock can be calculated from equation (ii),

i.e.

  $\begin{array}{c}V=\frac{4}{3}\pi \times \frac{d^3}{8}\\ d^3=\frac{6V}{\pi }\\ d={\left(\frac{6V}{\pi }\right)}^{\frac{1}{3}}\end{array}$

Plugging in the given values:

  $\begin{array}{c}d={\left(\frac{6\times 10.75{\text{ ft}}^3}{3.14}\right)}^{\frac{1}{3}}\\ =2.74\text{ ft}\end{array}$

Conclusion:

Hence, the diameter of the rock: $2.74\text{ ft}$