Chapter 12.6, Problem 12PSC

To determine

To check: Whether the ice cream overflow or not when it melts into the cone.

Answer to Problem 12PSC

The ice cream does not overflow when it melts into the cone

Explanation of Solution

Given Information:

Diameter of a cone $\left(d_1\right)=4\text{cm}$

Height of a cone $\left(h\right)=9\text{cm}$

Diameter of a sphere $\left(d_2\right)=4\text{cm}$

Formula used:

Volume of a cone $\left(V_1\right)=\frac{1}{3}\pi {\left(r_1\right)}^{2}h$

  $r_1\text{and}h$ are the radius and height of a cone

Volume of a sphere $\left(V_2\right)=\frac{4}{3}\pi {\left(r_2\right)}^3$

  $r_2$ is the radius of a sphere

Calculation:

Diameter of a cone $\left(d_1\right)=4\text{cm}$

  $\therefore$ Radius of the cone $\left(r_1\right)=\frac{d_1}{2}=2\text{cm}$

Diameter of a sphere $\left(d_2\right)=4\text{cm}$

  $\therefore$ Radius of the cone $\left(r_2\right)=\frac{d_2}{2}=2\text{cm}$

We know that, Volume of a cone $\left(V_1\right)=\frac{1}{3}\pi {\left(r_1\right)}^{2}h$

As $r_1=2\text{cm}\text{and}h=9\text{cm}$

  $\begin{array}{l}\therefore V_1=\frac{1}{3}\times \frac{22}{7}\times 2^2\times 9\\ \therefore V_1=\frac{264}{7}\\ \therefore V_1=37.71\end{array}$

We know that, Volume of a sphere $\left(V_2\right)=\frac{4}{3}\pi {\left(r_2\right)}^3$

As $r_2=2\text{cm}$

  $\begin{array}{l}\therefore V_2=\frac{4}{3}\times \frac{22}{7}\times 2^3\\ \therefore V_2=\frac{704}{21}\\ \therefore V_2=33.52\end{array}$

So, we get $V_1>V_2$

Thus, the volume of the ice cream cone is greater than the volume of the single scoop ice cream.

Hence, the ice cream does not overflow when it melts into the cone.