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

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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.

 Video Solution
Surface Areas And Volumes
Ex 13.4 | Question 1

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,\(h = 14 \rm cm\)

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

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

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

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

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

  
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