Chapter 10, Problem 101RQ

Interpretation Introduction

Interpretation:

The density of water of $\text{lb in}{\text{.}}^2$ (pounds per square inch) and the mass of

water in pounds present in the given column is to be calculated.

Concept Introduction:

The conversion of units is used in resolving a numerical problem that involves the conversion of one kind of unit into another one. Before applying for any conversion, it is necessary to know the relationship between the two kinds of conversions.

The conversion of a unit is determined by using a simple methodology:

$\left(\text{given quantity}\right)\times \left(\text{conversion method}\right)=\left(\text{desired quantity}\right)$.

The conversion of the height of the column of water from feet $\left(\text{ft}\right)$ to inches $\left(\text{in}\text{.}\right)$ is taken place with the help of the given relationship:

$\begin{array}{l}1\text{ ft}=12\text{ in}\text{.}\\ \text{Hence, the conversion factor is as:}\\ \frac{12\text{}\text{in}\text{.}}{1\text{}\text{ ft}}\end{array}$

The conversion of $\text{mL}$ to ${\text{cm}}^3$ in the relationship is:

$\begin{array}{l}1{\text{ cm}}^3=1\text{ mL}\\ \text{Hence, conversion factor is as:}\\ \frac{1\text{ mL}}{1{\text{ cm}}^3}\end{array}$

The conversion of centimeters $\left(\text{cm}\right)$ to inches $\left(\text{in}\text{.}\right)$ the relationship is:

$\begin{array}{l}2.54\text{ cm}=\text{ 1 in}\text{.}\\ \text{Hence, conversion factor is as:}\\ \frac{\text{1 in}\text{.}}{2.54\text{ cm}}\end{array}$