Metal balls

Four metal balls with a diameter of 5 cm are placed in a measuring cylinder with an inner diameter of 10 cm. What is the smallest water volume to be poured into the cylinder so that all balls are below the water level?

Correct answer:

V =  470.9871 cm³

Step-by-step explanation:

$$\begin{gathered} D=10\text{ cm } \\ R=D\mathrm{/}2=10\mathrm{/}2=5\text{ cm } \\ n=4 \\ d=5\text{ cm } \\ r=d\mathrm{/}2=5\mathrm{/}2={\displaystyle \frac{5}{2}}=2.5\text{ cm } \\ {V}_1={\displaystyle \frac{4}{3}}\cdot \pi \cdot r^3={\displaystyle \frac{4}{3}}\cdot 3.1416\cdot 2.5^3\doteq 65.4498{\text{cm}}^3 \\ {h}_1={\displaystyle \frac{\sqrt{3}}{2}}\cdot (2\cdot r)={\displaystyle \frac{\sqrt{3}}{2}}\cdot (2\cdot 2.5)\doteq 4.3301\text{ cm } \\ h=h_1+2\cdot r=4.3301+2\cdot 2.5\doteq 9.3301\text{ cm } \\ {V}_2=\pi \cdot R^2\cdot h=3.1416\cdot 5^2\cdot 9.3301\doteq 732.7865{\text{cm}}^3 \\ V=V_2-n\cdot V_1=732.7865-4\cdot 65.4498=470.9871{\text{cm}}^3 \end{gathered}$$

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