# Ex.13.4 Q1 Surface Areas and Volumes Solution - NCERT Maths Class 10

## Question

A drinking glass is in the shape of a frustum of a cone of height \(14\,\rm{cm.}\) The diameters of its two circular ends are \(4\,\rm{cm}\) and \(2\,\rm{cm.}\) Find the capacity of the glass.

## Text Solution

**What is Known?**

A drinking glass is in the shape of a frustum of a cone of height is \(14\, \rm{cm}.\)

The diameter of its two circular ends are \(4\, \rm{cm}\) and \(2\,\rm{cm}.\)

**What is Unknown?**

The capacity of the glass

**Reasoning:**

Draw a figure to visualize the shape better

Since the glass is in the shape of a frustum of a cone

Therefore, the capacity of the glass \(=\) Volume of frustum of a cone

We will find the capacity of the glass by using formulae

Volume of frustum of a cone\(\begin{align} = \frac{1}{3}\pi h\left( {r_1^2 + r_2^2 + {r_1}{r_2}} \right)\end{align} \)

where *\(r1, r2\)* and *\(h\)* are the radii and height of the frustum of the cone respectively.

**Steps:**

Height of glass,\(h = 14cm\)

Radius of the larger base,\(\begin{align} {r_1} = \frac{{4cm}}{2} = 2cm\end{align} \)

Radius of the smaller base,\(\begin{align} {r_2} = \frac{{2cm}}{2} = 1cm\end{align} \)

The capacity of the glass \(=\) Volume of frustum of a cone

\[\begin{align}&= \frac{1}{3}\pi h\left( {r_1^2 + r_2^2 + {r_1}{r_2}} \right)\\&= \frac{1}{3} \times \frac{{22}}{7} \times 14cm \times \left( {{{\left( {2cm} \right)}^2} + {{\left( {1cm} \right)}^2} + 2cm \times 1cm} \right)\\&= \frac{{44}}{3}cm \times \left( {4c{m^2} + 1c{m^2} + 2c{m^2}} \right)\\&= \frac{{44}}{3}cm \times 7c{m^2}\\&= \frac{{308}}{3}c{m^2}\\&= 102\frac{2}{3}c{m^2}\end{align}\]

Therefore, the capacity of the glass is \(\begin{align}102\frac{2}{3}\,\rm{cm^3}\end{align}\)