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

## 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.

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