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+ <meta name="description" content="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?">
+ <meta name="author" content="Aliaksandr Melnichenka">
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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <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:
+$$\frac{dV}{dt} = 4 \pi r^2 v=126\;sm^3$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$\frac{dV}{dt} = 4 \pi r^2 v$$
+ </p>
+
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