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In Mathematics / College | 2025-07-07

A beverage manufacturer performs a taste-test and discovers that people like their fizzy beverages best when the radius of the bubbles is about 0.5 mm. According to the formula below, what would be the volume of one of these bubbles?

[tex]r=\sqrt[3]{\frac{3 V}{4 \pi}}[/tex]

A. about [tex]$0.52 mm^3$[/tex]
B. about [tex]$0.35 mm^3$[/tex]
C. about [tex]$1.05 mm^3$[/tex]
D. about [tex]$0.49 mm^3$[/tex]

Asked by angv007

Answer (1)

Cube both sides of the equation: r 3 = 4 π 3 V ​ .
Isolate V: V = 3 4 π r 3 ​ .
Substitute r = 0.5 mm: V = 3 4 π ( 0.5 ) 3 ​ .
Calculate the volume: V ≈ 0.52 m m 3 .

Explanation

Understanding the Problem We are given the formula r = 3 4 π 3 V ​ ​ which relates the radius r of a bubble to its volume V . We are also given that the radius of the bubble is r = 0.5 mm. Our goal is to find the volume V of the bubble.

Isolating the Volume First, let's cube both sides of the equation to eliminate the cube root: r 3 = 4 π 3 V ​ Next, we want to isolate V . Multiply both sides of the equation by 4 π :
4 π r 3 = 3 V Finally, divide both sides by 3: V = 3 4 π r 3 ​

Substituting the Radius Now, substitute the given value of r = 0.5 mm into the equation: V = 3 4 π ( 0.5 ) 3 ​ V = 3 4 π ( 0.125 ) ​ V = 3 0.5 π ​ V ≈ 3 0.5 × 3.14159 ​ V ≈ 3 1.570795 ​ V ≈ 0.523598

Finding the Answer The volume of the bubble is approximately 0.523598 m m 3 . Looking at the answer choices, the closest value is 0.52 m m 3 .


Examples
Understanding the volume of bubbles is crucial in various applications, such as designing efficient aeration systems for wastewater treatment or optimizing the texture and taste of carbonated beverages. For instance, in wastewater treatment, smaller bubbles provide a larger surface area for oxygen transfer, enhancing the breakdown of pollutants. Similarly, in carbonated drinks, controlling bubble size affects the drink's fizziness and overall consumer experience. By applying the formula and principles used in this problem, engineers and scientists can fine-tune these processes for better performance and quality.

Answered by GinnyAnswer | 2025-07-07