2012-I-12

$\begin{array}{cl}
= & \dfrac{1}{3} \pi (48)^2 (96) \\
= & 73~728\pi\mbox{ cm}^3
\end{array}$

1. The volume of the milk in the vessel

   $\begin{array}{cl}
   = & \dfrac{1}{2} \times \dfrac{4}{3}\pi (60)^3 \\
   = & 144~000\pi\mbox{ cm}^3
   \end{array}$

2. Let $h\mbox{ cm}$ and $b\mbox{ cm}$ be the height and the upper base radius of the frustum immersed in the milk respectively. Therefore, we have

   $\begin{array}{rcl}
   h & = & \sqrt{60^2-48^2} \\
   & = & 36
   \end{array}$

   Also,

   $\begin{array}{rcl}
   \dfrac{b}{48} & = & \dfrac{96-36}{96} \\
   b & = & 30
   \end{array}$

   Therefore, the volume of the frustum

   $\begin{array}{cl}
   = & 73~728\pi-\dfrac{1}{3}\pi (30)^2(96-36) \\
   = & 55~728\pi \mbox{ cm}^3
   \end{array}$

   Hence, the volume of the milk remaining in the vessel

   $\begin{array}{cl}
   = & 144~000\pi-55~728\pi \\
   = & 277~314.667\mbox{ cm}^3 \\
   = & 0.277314667 \mbox{ m}^3 \\
   < & 0.3 \mbox{ m}^3 \end{array}$

   Therefore, I don’t agree the claim.