Question

if the density of water is $62.4 lb/ft^{3}$ and the density of gold is $1208 lb/ft^{3}$, what is the specific gravity of gold?
$15.23 lb/ft^{3}$
$19.36 lb/ft^{3}$
$19.36$
$15.23$

Explanation:

Step1: Recall the formula for specific gravity

Specific gravity \(SG=\frac{
ho_{substance}}{
ho_{water}}\), where \(
ho_{substance}\) is the density of the substance and \(
ho_{water}\) is the density of water.

Step2: Substitute the given values

Given \(
ho_{gold} = 1208\space lb/ft^{3}\) and \(
ho_{water}=62.4\space lb/ft^{3}\). Then \(SG=\frac{1208}{62.4}\).

Step3: Calculate the value

\(SG=\frac{1208}{62.4}\approx19.36\) (since specific gravity is a dimensionless quantity, we discard the units of \(lb/ft^{3}\) for the final result)

Answer:

19.36