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# Ex.13.6 Q5 Surface Areas and Volumes Solution - NCERT Maths Class 9

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## Question

It costs $$\rm Rs\; 2200$$ to paint the inner curved surface of a cylindrical vessel $$10\; \rm m$$ deep. If the cost of painting is at the rate of $$\rm Rs\; 20$$ per $$\rm m^2$$, find.

(i) inner curved surface area of the vessel,

(ii) radius of the base,

(iii) capacity of the vessel.

Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-6 | Question 5

## Text Solution

Reasoning:

The ratio between the total cost and the rate per $$\rm m^2$$ will give the inner curve surface area. Curve surface area of a cylinder is $$2\pi rh$$ if the height of the cylinder is $$h$$ and base radius is $$r$$.

What is known?

Total cost to paint and cost per $$\rm {m^2}$$. And height of the cylindrical vessel.

What is unknown?

Inner curved surface area.

Steps:

Total cost to paint $$=2200$$

Cost of paint per $$\rm {m^2}$$ $$=20$$

$$\therefore$$ Surface area \begin{align} = \frac{{2200}}{{200}} \end{align} \begin{align} = 110\,\, \rm cm^2 \end{align}

What is unknown?

Radius of the base.

Steps:

Inner curved surface =\begin{align}110\,\, \rm cm^2 \end{align}

$$\text{Height} \;(h) = 10\; \rm m$$

\begin{align}2\pi rh &= 110 \\ 2 \times \frac{{22}}{7} \times r \times 10 &= 110 \\r &= \frac{{110 \times 7}}{{22 \times 2 \times 10}}\\ &= \frac{7}{4}\\ &= 1.75\,\,\, \rm cm \end{align}

What is unknown?

Capacity of the vessel.

Steps:

Capacity of the vessel = Volume of the vessel = \begin{align}\,\pi {r^2}h \end{align}

Radius (r)\begin{align}\, = 1.75\,\,\,\rm cm \end{align}

Height (h)\begin{align}\, = 10\,\, \rm cm \end{align}

Capacity

= $$\,2 \times \frac{{22}}{7} \times 1.75 \times 1.75 \times 10$$

\begin{align} = 96.25\,\, \rm cm^3 \end{align}

The capacity of the vessel is \begin{align}\, = 96.25\,\, \rm cm^3 \end{align}

Answer:

Inner curved surface area\begin{align}\,110\,\, \rm cm^2 \end{align}

Radius of the box $$=1.75\; \rm m$$

Capacity of the vessel \begin{align}\, = 96.25\,\, \rm cm^3 \end{align}

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