Diameter 7648

The mug has the shape of a cylinder with a height of 60.7 mm. There is two dl of water in it. If we dip a ball with a diameter of 40 cm into the water, the water will not overflow. What is the minimum diameter of the cup?

Correct answer:

D =  80 cm

Step-by-step explanation:

$$\begin{gathered} h=60.7\text{mm}\to \text{cm}=60.7:10\text{cm}=6.07\text{cm} \\ {V}_1=2\text{dl}\to {\text{cm}}^3=2\cdot 100{\text{cm}}^3=200{\text{cm}}^3 \\ r=40\text{cm} \\ {V}_2=\frac{4}{3}\cdot \pi \cdot r^3=\frac{4}{3}\cdot 3.1416\cdot 40^3\doteq 268082.5731{\text{cm}}^3 \\ V=V_1+V_2=200+\frac{45842120}{171}=\frac{200\cdot 171}{171}+\frac{45842120}{171}=\frac{34200}{171}+\frac{45842120}{171}=\frac{34200+45842120}{171}=\frac{45876320}{171}\doteq 268282.5731{\text{cm}}^3 \\ S=V/h=\frac{45876320}{171}/\frac{607}{100}=\frac{45876320}{171}:\frac{607}{100}=\frac{45876320}{171}\cdot \frac{100}{607}=\frac{45876320\cdot 100}{171\cdot 607}=\frac{4587632000}{103797}\doteq 44198.1175{\text{cm}}^2 \\ S=\pi \cdot D_1^2/4 \\ {D}_1=\sqrt{4\cdot S/\pi }=\sqrt{4\cdot 44198.1175/3.1416}\doteq 237.2231\text{cm} \\ {D}_1<2r \\ D=2\cdot r=2\cdot 40=80\text{cm} \end{gathered}$$

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