# Exercise E13.6 Surface Areas and Volumes NCERT Solutions Class 9

## Chapter 13 Ex.13.6 Question 1

The circumference of the base of a cylindrical vessel is \(132\; cm\) and its height is \(25\; cm\). How many litres of water can it hold? \(\begin{align}(1000\,c{m^3} = 1\,l) \end{align}\)

**Solution**

**Video Solution**

**Reasoning:**

Volume of a cylinder of base radius \(r\) and height is \(V = \pi r^{2} h\).

**What is known?**

Circumference of the base and the height.

**What is unknown?**

Litres of water the cylindrical vessel can hold.

**Steps:**

Since the base of a cylindrical vessel is a circle, the circumference is \(2 \pi r=132 \mathrm{cm}\) (given)

\(\begin{align} 2\pi r &= 132\\2 \times \frac{{22}}{7} \times r &= 132\\r &= \frac{{132 \times 7}}{{2 \times 22}}\\ &= 21\,\,cm \end{align}\)

Radius \(\begin{align}(r) = 21\, \rm{cm} \end{align}\)

Height \(\begin{align}(h) = 25\, \rm{cm}\end{align}\)

Capacity of the cylindrical vessel \(=\) Volume of the cylindrical vessel \( =𝜋𝑟^2ℎ\)

\(\begin{align}&= \frac{{22}}{7} \times 21 \times 21 \times 25 \\&= 34650\,c{m^3} \end{align}\)

Answer should be in litres

\(\begin{align} 1000\,c{m^3} &= 1\,\,l\\34650\,c{m^3} &= x\,\,l\\x &= \frac{{34650}}{{1000}}\\& = 34.65\,l \end{align}\)

**Answer:**

Capacity of the cylindrical vessel\(\begin{align}= 34.65\,l \end{align}\)

## Chapter 13 Ex.13.6 Question 2

The inner diameter of a cylindrical wooden pipe is \(24 \;cm\) and its outer diameter is \(28 \;cm\). The length of the pipe is \(35\ cm\). Find the mass of the pipe, if \(\begin{align}1\,\,c{m^3} \end{align}\) of wood has a mass of \(0.6\; g\).

**Solution**

**Video Solution**

**Reasoning:**

Since the cylindrical wooden pipe is made up of two concentric circles at the top and bottom, we have to find the volume of both the cylinders.

**Diagram**

**What is known?**

Diameter of the inner and outer cylinder. Length of the cylinder.

**What is unknown?**

Mass of the pipe if \(1 \mathrm{cm}^{3}\) of pipe has mass \(6\; g\).

**Steps:**

Volume of the outer cylinder \(V_{1}=\pi R^{2} h\)

Outer diameter *\((2R) = 28\; cm \)*

Outer radius \(\begin{align}(R)=\frac{28}{2}=14 \mathrm{cm}\end{align}\)

Height \((h) = 35\ cm\)

Volume of outer cylinder

\(\begin{align}V_{1}&=\frac{22}{7} \times 14 \times 14 \times 35\\&=21560 \;\rm{cm}^{3}\end{align}\)

Volume of inner cylinder \(V_{2}=\pi r^{2} h\)

Inner diameter \(\begin{align} \left( {2r} \right) = 24\,cm. \end{align}\)

Inner radius \(\begin{align} (r) = \frac{{24}}{2} = 12\,cm \end{align}\)

Height \((h) = 35\ cm\)

Volume of outer cylinder

\(\begin{align} {V_2} &= \frac{{22}}{7} \times 12 \times 12 \times 35 \\&= 15840\,c{m^3} \end{align}\)

Volume of the wood used \(=\) Outer volume \(-\) Inner volume

\(\begin{align} &= 21560 - 15840 \\ & = 5720\;{\rm{cm}}^{\rm{3}} \\ \end{align}\)

Mass of \(\mathrm{1cm}^{3} \text { pipe }=0.6 \mathrm{g}\)

Therefore,

Mass of \(5720 \,{cm}^{3}\) pipe

\(\begin{align} &={ 0}{\rm{.6 \times 5720}}\\ &= 3432\,\,g\\&\quad\rm OR\\ &= 3.432\,\,kg \end{align}\)

**Answer:** Mass of the pipe \(= 3.432\ kg\).

## Chapter 13 Ex.13.6 Question 3

A soft drink is available in two packs – (i) a tin can with a rectangular base of length \(5 \;cm\) and width \(4\; cm\), having a height of \(15\ cm\) and (ii) a plastic cylinder with circular base of diameter \(7\ cm\) and height \(10\ cm\). Which container has greater capacity and by how much?

**Solution**

**Video Solution**

**Reasoning:**

Volume of the cuboid is *\(lbh\)* and volume of cylinder in\(\begin{align}\,\pi {r^2}h \end{align}\)

**What is known?**

Measurements of cuboid can. Measurements of cylindrical can.

**What is unknown?**

Which can is big and how much** **?

**Steps:**

**Diagram **

Tin Can:

\(\text{Length} \; (l) = 5\ cm \)

\(\text{Breadth}\; (b) = 4\ cm\)

\(\text{Height}\; (h) = 15\ cm\)

\(\text{Capacity = Volume} = l b h\).

\[\begin{align}&= 5 \times 4 \times 15 \\&= 300\,\,c{m^3} \\ \end{align}\]

Plastic Cylinder:

\(\text{Diameter} (2r) =7\; cm\)

Radius\(\begin{align} \,(r) = \frac{7}{2} \end{align}\)cm

\(\text{Height}\; (h) =10\; cm\)

\(\begin{align}\text{Capacity} &= \rm{Volume} = \pi {r^2}h\ \\ &= \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2} \times 10\\ &= 385\,\,\,c{m^3} \end{align}\)

Clearly, the plastic cylinder has greater capacity than the tin container.

Difference\(\begin{align} = \,385 - 300 = 85\,c{m^3} \end{align}\)

**Answer:**

The plastic cylindrical can have more capacity than the Tin Can by\(\begin{align}\, = 85\,\,c{m^3} \end{align}\).

## Chapter 13 Ex.13.6 Question 4

If the lateral surface of a cylinder is \(92.4\,\,\rm cm^2 \) and its height is \(5\; \rm cm\), then find

(i) radius of its base.

(ii) its volume. (Use \(\pi = 3.14\))

**Solution**

**Video Solution**

**Reasoning:**

Lateral surface area of a cylinder is \(2\pi rh \).

Volume of cylinder is \( \pi {r^2}h.\)

**What is known?**

Lateral surface area of the cylinder and its height.

**What is unknown?**

Base radius.

**Steps:**

Lateral surface area of a cylinder having \(r\) as radius and height \(h\) is \( 2\pi rh \)

Lateral surface area \( = 94.2\, \rm cm^2 \)

\(\text{Height}\; h = 5\; \rm cm\)

\(\begin{align}2\pi rh &= 94.2\\2 \times 3.14 \times r \times 5 &= 94.2\\r &= \frac{{94.2}}{{5 \times 3.14 \times 2}} \\ & = 3\,\,\rm cm \end{align}\)

**What is unknown?**

Volume of the cylinder.

**Steps:**

\(\text{Radius}\; r = 3\; \rm cm\)

\(\text{Height}\; h = 5\; \rm cm\)

Volume of cylinder

\[\begin{align} & =\pi r^{2} h \\ & =3.14 \times 3 \times 3 \times 5 \\ & =141.3 \, \mathrm{cm}^{3}\end{align} \]

**Answer:**

\(\text{Radius of its base} = 3\; \rm cm\)

Its volume \( = 141.3\,\, \rm cm^3 \)

## Chapter 13 Ex.13.6 Question 5

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.

**Solution**

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

## Chapter 13 Ex.13.6 Question 6

The capacity of a closed cylindrical vessel of height \(1 \; \rm m\) is \(15.4\) litres. How many square metres of metal sheet would be needed to make it?

**Solution**

**Video Solution**

**Reasoning:**

Surface area total of a closed cylinder is \(2\pi rh + 2\pi {r^2} \).

\(2\pi rh \) corresponds to the Curved Surface Area of the cylinder and \(2\pi {r^2} \) corresponds to the area of the circular lids of the cylindrical vessel in order to make it a closed container. Volume of cylinder \(\pi {r^2}h \).

**What is known?**

Capacity and height of the cylinder.

**What is unknown?**

Metal sheet needed to make it.

**Steps:**

\(\begin{align}\text{Capacity} &= \rm{15.4 \, liters}\\ V &= \frac{{15.4}}{{1000}} = .0154\,\,\rm {m^3} \end{align}\)

\(\begin{align}\rm{Volume} &= \pi {r^2}h\\ 0.0154 &= \frac{{22}}{7} \times {r^2} \times 1 \\{r^2} &= \frac{{.0154 \times 7}}{{22}}\\{r^2} &= .0049\\r &= \sqrt {.0049} \\ &= 0.07\,\, \rm m \end{align}\)

Curved surface area

\[\begin{align} & = 2\pi rh + 2\pi {r^2} \\ & = 2\pi r(r + h) \\ & = 2 \times \frac{{22}}{7} \times 0.07 \times (0.07 + 1) \\ & = 0.4708\,\, \rm {m^2} \end{align}\]

**Answer:**

\(0.47\, \rm {m^2} \) of metal sheet would be needed**.**

## Chapter 13 Ex.13.6 Question 7

A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is \(7 \; \rm mm\) and the diameter of the graphite is \(1\; \rm mm\). If the length of the pencil is \(14 \; \rm cm\), find the volume of the wood and that of the graphite.

**Solution**

**Video Solution**

**Reasoning:**

Volume of cylinder \( \pi {r^2}h \)

**What is known?**

Diameter of the pencil, diameter of graphite and length of the pencil.

**What is unknown?**

Volume of the wood and that if graphite.

**Steps:**

For cylinder graphite.

Diameter \((2r) = 1 \; \rm mm\)

Radius \(\begin{align}(r) = \frac{1}{2}\,\, \rm mm \end{align}\)

Length of the pencil \((h) = 14 \; \rm cm = 140 \;\rm mm\)

\(\begin{align}\text{Capacity}&= \text{Volume} \\ &=\,\pi {r^2}h\\ &= \frac{{22}}{7} \times \frac{1}{2} \times \frac{1}{2} \times 140\\ &= 110\,\, \rm mm^3 \end{align}\)

\(\begin{align} \rm{In} \,\,c{m^3} &= \frac{{110}}{{10 \times 10 \times 10}} = 0.11\,\, \rm cm^3 \end{align}\)

For cylinder of wood:

To find the volume of the wood:

Total volume of the pencil \(-\) Volume of graphite

\(\begin{align}\pi {R^2}h - \pi {r^2}h \pi h({R^2} - {r^2})\end{align}\)

Diameter of pencil \((2R) = 7 \; \rm mm\)

Radius \((r) = \frac{7}{2} \rm mm \)

Length of the pencil \((h) = 14 \; \rm cm = 140 \; \rm mm\)

Volume of wood \(\begin{align}=\,\pi h({R^2} - {r^2}) \end{align}\)

\[\begin{align} &= \frac{{22}}{7} \times 140 \times [{(\frac{7}{2})^2} - {(\frac{1}{2})^2}]\\ &= \frac{{22}}{7} \times 140 \times [\frac{{49}}{4} - \frac{1}{2}]\\ &= 5280\,\, \rm mm^3\\ &= \frac{{5280}}{{10 \times 10 \times 10}}\, \rm cm^3 \\ &= 5.28\,\, \rm cm^3 \end{align}\]

**Answer:**

Volume of the wood \( = 5.28\,\, \rm cm^3 \)

Volume of graphite \( = 0.11\,\,\, \rm cm^3 \)

## Chapter 13 Ex.13.6 Question 8

A patient in a hospital is given soup daily in a cylindrical bowl of diameter \(7 \; \rm cm\). If the bowl is filled with soup to a height of \(4 \; \rm cm\), how much soup the hospital has to prepare daily to serve \(250\) patients?

**Solution**

**Video Solution**

**Reasoning:**

Volume of cylinder \(\begin{align} = \pi {r^2}h \end{align}\). So, the amount of soup to be prepared will be the product of volume of the bowl (with \(h = 4\; \rm cm\)) and total number of patients.

**What is known?**

Diameter of the bowl and length of the soup, number of patients**.**

**What is unknown?**

Total soup to be prepared.

**Steps:**

Volume of soup bowl \(= \pi {r^2}h \)

Diameter \((2R) = 7 \; \rm cm\)

Radius \((r) = \frac{7}{2}\,\,\, \rm cm \)

Height \((h) = 4\; \rm cm\)

\(\therefore \) Volume of soup in one bowl

\[\begin{align} &= \pi {r^2}h \\ &= \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2} \times 4\\ &= 154\,\, \rm cm^3 \end{align}\]

Volume of soup for \(1\) patient \( = 154\,\, \rm cm^3 \)

Volume of soup for \(250\) patients

\[\begin{align} &= 154 \times 250\\ & = 38500\,\, \rm cm^3 \end{align}\]

**Answer:**

Volume of soup to be prepared for \(250\) patients is

\(\begin{align} = 38500\,\, \rm cm^3 \end{align}\) or \(38.5\) litres [as \(1000\) \( \rm cm^3 \) \(= 1\) litre]