Three tennis balls are just packed in a cylindrical jar. If radius of each ball is r, volume of air inside the jar is
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According to a famous geometric theorem by Archimedes, the volume of a sphere is exactly \(\frac{2}{3}\) of the volume of its circumscribed cylinder.
This means the three balls occupy exactly \(\frac{2}{3}\) of the jar's volume, leaving \(\frac{1}{3}\) of the jar's volume as air space:
\[ V_{\text{air}} = \frac{1}{3} \times V_{\text{cylinder}} = \frac{1}{3} \times 6\pi r^3 = 2\pi r^3 \]
Remembering this ratio saves you from doing long calculations!
Step 1: Recall the fixed volume ratio between a sphere and its snug cylinder. A sphere that fits exactly inside a cylinder (same radius, height equal to diameter) always occupies $\frac{2}{3}$ of that cylinder's volume, this is a standard geometric fact. Step 2: Apply it to this jar. Here the jar's height is exactly filled by the three balls, so together they occupy $\frac{2}{3}$ of the jar's volume, leaving $\frac{1}{3}$ as air. Step 3: Compute the jar's volume and take one-third. $V_{\text{cylinder}}=\pi r^2(6r)=6\pi r^3$, so $V_{\text{air}}=\frac13\times6\pi r^3=2\pi r^3$, matching option (A). \[ \boxed{2\pi r^3} \]
