Question

A semi-circular sheet of metal of diameter$28\mathrm{cm}$ is bent to form an open conical cup. Find the capacity of the cup.

Answer 1

Step 1. Calculate the radius of the open conical cup.

The diameter of the semi-circular sheet is $28\mathrm{cm}$.

$\therefore$ The radius of the semi-circular sheet $\left(r\right)=\frac{28}{2}=14cm$

Since the semi-circular sheet is bent to form an open conical cup. Thus, the slant height of a conical cup$\left(l\right)$ is equal to the radius of semicircular sheet $\left(r\right)$.

$\therefore l=r=14\mathrm{cm}$

Since the circumference of the base of a cone is equal to the Circumference of a Semicircle.

$$\begin{gathered} \therefore 2\mathrm{\pi R}=\mathrm{\pi r} \\ \Rightarrow \mathrm{R}=\frac{\mathrm{r}}{2} \\ \Rightarrow \mathrm{R}=\frac{14}{2}\left[\because \mathrm{r}=14\mathrm{cm}\right] \\ \Rightarrow \mathrm{R}=7\mathrm{cm} \end{gathered}$$, Where $R$ is the radius of the cone.

Step 2. calculate the height of the open conical cup.

The height $\left(h\right)$ of the circular cone is computed as,

$$\begin{gathered} h=\sqrt{l^2-R^2} \\ \Rightarrow =\sqrt{{\left(14\right)}^2-{\left(7\right)}^2} \\ \Rightarrow =\sqrt{196-49} \\ \Rightarrow =\sqrt{147}\mathrm{cm} \end{gathered}$$

Step 3. Calculate the capacity of the cup.

The capacity is equivalent to the volume of the cone.

$$\begin{gathered} \therefore V=\frac{1}{3}{\mathrm{\pi R}}^2\mathrm{h} \\ \Rightarrow =\frac{1}{3}\times \frac{22}{7}\times 7^2\times \sqrt{147} \\ \Rightarrow =\frac{154}{3}\times \sqrt{147} \\ \Rightarrow =622.38{\mathrm{cm}}^3 \end{gathered}$$

Hence, the capacity of the cup is $622.38{\mathrm{cm}}^3$.