A boy inflates a balloon. With a ball radius of 10 cm, the rate of increase in the radius is 1 \frac{mm}{s}. How much air does the boy exhale every second?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#1.2">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.2.12.$ A boy inflates a balloon. With a ball radius of $10$ cm, the rate of increase in the radius is $1$ $\frac{mm}{s}$. How much air does the boy exhale every second?
    </p>

    <h3>Solution</h3>
    <p>
        The volume of the ball is defined by the expression 
        $$V = \frac{4}{3} \pi r^3$$
        Then, the rate of change of the hugram or the volume of air that the boy exhales every second:

$$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$$
Given $v = \frac{dr}{dt},$ we obtain:
$$\frac{dV}{dt} = 4 \pi r^2 v=126\;sm^3$$

    <h4>Answer</h4>
    <p>
        $$\frac{dV}{dt} = 4 \pi r^2 v$$
    </p>


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