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

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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
Ex exercise-13-6 | Question 5

Text Solution


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.


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.


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.


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


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

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

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


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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