Question

a cylindrical can has a radius of 2 inches and its label says it contains liquid to a height of 12 inches. the height of the liquid in the can is actually $12\frac{1}{8}$ inches. by what percent is the actual volume of the liquid greater than the labeled volume of the liquid? (use 3.14 as the value of $\pi$.)

percentage excess = \\(\square\\)%
(type an integer or decimal rounded to the nearest tenth as needed )

Explanation:

Step1: Recall the volume formula for a cylinder

The volume \( V \) of a cylinder is given by \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. Since the radius is the same for both the labeled and actual volume, the percentage difference in volume will be the same as the percentage difference in height (because \( \pi r^2 \) is a constant factor for both volumes). So we can focus on the heights.

Step2: Find the difference in height

The labeled height \( h_{labeled} = 12 \) inches, and the actual height \( h_{actual} = 12\frac{1}{8}=12 + \frac{1}{8}=12.125 \) inches. The difference in height \( \Delta h=h_{actual}-h_{labeled}=12.125 - 12 = 0.125 \) inches.

Step3: Calculate the percentage excess

The formula for percentage excess is \( \text{Percentage Excess}=\frac{\Delta h}{h_{labeled}}\times 100\% \). Substituting the values, we have \( \frac{0.125}{12}\times 100\% \). First, calculate \( \frac{0.125}{12}\approx0.0104167 \), then multiply by 100% to get \( 0.0104167\times 100\%\approx1.04167\% \approx 1.0\% \) (rounded to the nearest tenth).

Answer:

\( 1.0 \)