2018-I-14

$\begin{array}{cl}
= & \pi \times 8^2 \times 64 \\
= & 4096\pi \text{ cm}^3
\end{array}$

$\begin{array}{rcl}
\dfrac{r}{20} & = & \dfrac{h}{60} \\
r & = & \dfrac{h}{3}
\end{array}$

Hence, by the result of (a), we have

$\begin{array}{rcl}
\dfrac{1}{3} \pi r^2 \times h & = & 4096\pi \\
\dfrac{1}{3} \pi \times \left(\dfrac{h}{3} \right)^2 \times h & = & 4096\pi \\
h^3 & = & 110\ 592 \\
h & = & 48
\end{array}$

Therefore, the depth of water in the vessel is $48\text{ cm}$.

$\begin{array}{cl}
= & \dfrac{1}{3} \pi (20)^2 \times 60 \\
= & 8000\pi \text{ cm}^3
\end{array}$

The total volume of the metal sphere and the water

$\begin{array}{cl}
= & \dfrac{4}{3} \pi (14)^3 + 4096\pi \\
= & 7754\dfrac{2}{3}\pi \text{ cm}^3 \\
< & 8000\pi \text{ cm}^2 \end{array}$

Therefore, the water will not overflow.