## Q. 2.SP.1

The specific weight of water at ordinary pressure and temperature is $62.4 \mathrm{lb} / \mathrm{ft}^3$ . The specific gravity of mercury is 13.56. Compute the density of water and the specific weight and density of mercury.

## Verified Solution

$\begin{array}{rlr}\rho_{\text {water }} & =\frac{\gamma_{\text {water }}}{g}=\frac{62.4 \mathrm{lb} / \mathrm{ft}^3}{32.2 \mathrm{ft} / \mathrm{sec}^2}=1.938 \mathrm{slugs} / \mathrm{ft}^3 \\\\\gamma_{\text {mercury }} & =s_{\text {mercury }} \gamma_{\text {water }}=13.56(62.4)=846 \mathrm{lb} / \mathrm{ft}^3 \\\\\rho_{\text {mercury }} & =s_{\text {mercury }} \rho_{\text {water }}=13.56(1.938)=26.3 \mathrm{slugs} / \mathrm{ft}^3 \end{array}$