Chapter 10, Problem 72GP

To determine

To calculate: The height of the continent above the surrounding rock.

Answer to Problem 72GP

The height of the continent above the surrounding rock is found to be $5.3\text{km}$

Explanation of Solution

Given:

  ${\rho }_{continent}=2800{\text{kgm}}^{-3}$ ,

The density $\left({\rho }_{displacedmantle}\right)$ of rock $=3300{\text{kgm}}^{-3}$

  $$

The average thickness of the earth’s continental crust $=35\text{km}$ .

Formula used:

The buoyant force is equal to the mantle displaces weight. It must be equal to the continent’s weight. So, the formula is,

  $W=\rho \times g\times V$

Calculation:

The buoyant force is equal to the mantle displaces weight. It must be equal to the continent’s weight. Considered $h$ is the full continent height and y is the height of the continent above the surrounding rock. The force is given as,

  $\begin{array}{l}{W}_{displacedmantle}=W_{{}_{continent}}\\ Ah{\rho }_{displacedmantle}g=A\left(h-y\right){\rho }_{continent}g\\ \Rightarrow y=h\left(1-\frac{{\rho }_{continent}}{{\rho }_{displacedmantle}}\right)\\ =\left(35\text{km}\right)\times \left(1-\frac{\text{2800}\frac{\text{kg}}{{\text{m}}^{\text{3}}}}{\text{3300}\frac{\text{kg}}{{\text{m}}^{\text{3}}}}\right)\\ =5.3\text{km}\end{array}$

Conclusion:

The height of the continent above the surrounding rock is found to be $5.3\text{km}$ .