# Ex.13.2 Q7 Surface Areas and Volumes - NCERT Maths Class 9

## Question

The inner diameter of a circular well is $$3.5\, \rm{m}\,.$$ It is $$10 \,\rm{m}$$ deep. Find

1. Its inner curved surface area
2. The cost of plastering this curved surface at the rate of $$40$$ per\begin{align}\rm{{m^2}} \end{align}

Video Solution
Surface Areas And Volumes
Ex 13.2 | Question 7

## Text Solution

Reasoning:

The curved surface area of a right circular cylinder of base radius and height h is \begin{align}2\pi rh \end{align}

What is the known?

1.  The inner diameter and the depth of the well.
2.  Cost of plastering the curved surface area per\begin{align}\rm{{m^2}} \end{align}

1.Its inner curved surface area

What is the unknown?

The inner curved surface area.

Steps:

Diameter $$= 2r = 3.5 \rm{m}$$

\begin{align} r &= \frac{{3.5}}{2}\,\,\rm{m}\\height = h &= 10\,\,\rm{m}\end{align}

The Curved surface area
\begin{align}&= 2\pi rh\\& = 2 \times \frac{{22}}{7} \times \frac{{3.5}}{2} \times 10\\& = 110\,\,{m^2} \end{align}

2. Cost of plastering the curved surface area per \begin{align}\rm{{m^2}} \end{align}.

What is the unknown?

Cost of plastering the curved surface area.

Steps:

Cost of plastering \begin{align}\,1\,\rm{{m^2}} = Rs\,\,40 \end{align}

Cost of plastering

\begin{align}110\,\,\rm{{m^2}} = 110 \times 40 = Rs\,\,4400 \end{align}

Its inner curved surface \begin{align} = 110\,\,\rm{{m^2}} \end{align}

Cost of plastering the curved surface

$$= \rm{Rs}\, 4400$$

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