# Ex.13.7 Q2 Surface Areas and Volumes Solution - NCERT Maths Class 9

## Question

Find the capacity in litres of a conical vessel with

(i) radius $$7\rm\, cm,$$ slant height $$25\rm\, cm$$

(ii) height $$12\rm\, cm,$$ slant height $$13\rm\, cm$$

Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-7 | Question 2

## Text Solution

Reasoning:

Capacity of a conical vessel is nothing but the Volume of the cone \begin{align} = \frac{1}{3}\pi {r^2}h \end{align}

What is  known?

Radius $$(r) = 7\rm\, cm$$

Slant height $$(1) = 25\rm\, cm$$

What is  unknown?

Capacity of the vessel in litres.

Steps:

Capacity of the conical vessel $$=$$ volume of cone \begin{align} = \frac{1}{3}\pi {r^2}h \end{align}

Where \begin{align}h = \sqrt {{l^2} - {r^2}} \end{align}

Radius \begin{align}(r) = 7\rm \,\rm\,cm \end{align}

Slant height \begin{align}(l) = 25\rm\,\rm \,cm \end{align}

\begin{align}\therefore h &= \sqrt {{{(25)}^2} - {7^2}} \\ &= \sqrt {(25 + 7)(25 - 7)}\\ {\rm{ [ (}}{{\rm{a}}^2} - {b^2}) &= (a + b)(a - b)]\\ &= \sqrt {32 \times 18} \\ &=\! \sqrt {2 \!\times\! 2\! \times\! 2\! \times \!2\! \times \!2 \!\times\! 3 \!\times \!3\! \times\! 2} \\ &= \sqrt {576} \\ &= 24\,\,\rm\,cm \end{align}

Volume
\begin{align}& = \frac{1}{3} \times \frac{{22}}{7} \times 7 \times 7 \times 24\\ &= 1232\,\,\rm\,c{m^3}\,\,or\,\,1.232\,\text{l} \end{align}

\begin{align} [\therefore 1000\,\,c{m^3} = 1\,l] \end{align}

What is  known?

Radius $$(r) = 12 \rm\,cm$$

Slant height \begin{align}(l) = 13\,\,\rm\,cm \end{align}

What is  unknown?

Capacity of the vessel in litres.

Steps:

Capacity of the conical vessel $$=$$ volume of cone \begin{align} = \frac{1}{3}\pi {r^2}h \end{align}

Height $$(h) = 12\rm\, cm$$

Slant height \begin{align}(l) = 13\,\,\rm\,cm \end{align}

\begin{align}&{r^2} + {h^2} = {l^2} \\ &{r^2} = {l^2} - {h^2} \\ &r=\sqrt{l^{2}-h^{2}} \\ &= \sqrt{(13)^{2}-(12)^{2}} \\ &=\sqrt{(13+12)(13-12)} \\&= \sqrt {25} \\ &= 5\,\,\rm\,cm \end{align}

Capacity

\begin{align}&= \frac{1}{3} \times \frac{{22}}{7} \times 5 \times 5 \times 12\\&= \frac{{2200}}{7}\\& = \frac{2200}{7000}\\& = \frac{11}{35}l \end{align}

Capacity $$=1.232\,\, l$$
Capacity \begin{align}=\frac{{11}}{{35}} \,l \end{align}