Ans: (a) $60\text{ mm}$ (b) $36\text{ mm}$ (c) Yes
- Let $OX= R\text{ mm}$.
$\begin{array}{rcl}
\pi R^2 \times \dfrac{288^\circ}{360^\circ} & = & 2880\pi \\
R^2 & = & 3600 \\
R & = & 60
\end{array}$Therefore, the length of $OX$ is $60\text{ mm}$.
- Let $r\text{ mm}$ be the base radius of the container.
$\begin{array}{rcl}
\pi \times r \times 60 & = & 2880\pi \\
r & = & 48
\end{array}$Therefore, the height of the container
$\begin{array}{cl}
= & \sqrt{60^2 – 48^2} \\
= & 36\text{ mm}
\end{array}$ - The volume of the container
$\begin{array}{cl}
= & \dfrac{1}{3} \pi \times (48)^2 \times 36 \\
= & 86~858.753~69 \text{ mm}^3 \\
= & 86.858~753~69 \text{ cm}^3 \\
< & 150\text{ cm}^3 \end{array}$Therefore, the water will overflow.
