NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.6) Exercise 13.6

Exercise 13.6

1. The circumference of the base of cylindrical vessel is $\text{132 cm}$ and its height is $\text{25 cm}$. How many litres of water can it hold? $\left( \text{1000 c}{{\text{m}}^{\text{3}}}\text{ = 1 litre} \right)$. $\left[ \text{Assume  }\!\!\pi\!\!\text{  =}\frac{\text{22}}{\text{7}} \right]$

Ans:

Given the height of the vessel $\text{h = 25 cm}$

Let us assume the radius be $\text{r}$.

The circumference of the vessel $\text{= 132 cm}$

$\therefore \text{2 }\!\!\pi\!\!\text{ r = 132 cm}$

$\Rightarrow \text{r = }\left( \frac{\text{132  }\!\!\times\!\!\text{  7}}{\text{2  }\!\!\times\!\!\text{  22}} \right)\text{ cm}$

$\text{r = 21 cm}$

The volume of the cylindrical vessel $\text{V =  }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{21} \right)}^{\text{2}}}\text{  }\!\!\times\!\!\text{  25}$

$\Rightarrow \text{V = 34650 c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\frac{\text{34650}}{\text{1000}}\text{ litres}$

$\Rightarrow \text{V = 34}\text{.65 litres}$

Therefore, the cylindrical vessel can hold $\text{34}\text{.65 litres}$ of water.

2. The inner diameter of a cylindrical wooden pipe is $\text{24 cm}$ and its outer diameter is $\text{28 cm}$. The length of the pipe is $\text{35 cm}$. Find the mass of the pipe, if $\text{1 c}{{\text{m}}^{\text{3}}}$ of wood has a mass of $\text{0}\text{.6 g}$. $\left[ \text{Assume  }\!\!\pi\!\!\text{  =}\frac{\text{22}}{\text{7}} \right]$

Ans:

Given the inner diameter of cylindrical pipe $\text{= 24 cm}$

So, the inner radius $\text{r = }\frac{\text{24}}{\text{2}}\text{ = 12 cm}$

Given the outer diameter of cylindrical pipe $\text{= 28 cm}$

So, the outer radius $\text{R = }\frac{\text{28}}{\text{2}}\text{ = 14 cm}$

The height of the wooden pipe $\text{h = 35 cm}$.

The volume of the cylindrical pipe $\text{V =  }\!\!\pi\!\!\text{ }\left( {{\text{R}}^{\text{2}}}\text{ - }{{\text{r}}^{\text{2}}} \right)\text{h}$

$\Rightarrow \text{V = }\left[ \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }\left( \text{1}{{\text{4}}^{\text{2}}}\text{ - 1}{{\text{2}}^{\text{2}}} \right)\text{  }\!\!\times\!\!\text{  35} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left( \text{110  }\!\!\times\!\!\text{  52} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 5720 c}{{\text{m}}^{\text{3}}}$

It is given that the mass of $\text{1 c}{{\text{m}}^{\text{3}}}$ wood $\text{= 0}\text{.6 g}$

So, the mass of $\text{5720 c}{{\text{m}}^{\text{3}}}$ wood $\text{= }\left( \frac{\text{5720  }\!\!\times\!\!\text{  0}\text{.6}}{\text{1000}} \right)\text{ kg = 3}\text{.432 kg}$

Therefore, the mass of the pipe is $\text{3}\text{.432 kg}$.

3. A soft drink is available in two packs − (i) a tin can with a rectangular base of length $\text{5 cm}$ and width $\text{4 cm}$, having a height of $\text{15 cm}$ and (ii) a plastic cylinder with circular base of diameter $\text{7 cm}$ and height $\text{10 cm}$. Which container has greater capacity and by how much? $\left[ \text{Assume  }\!\!\pi\!\!\text{  =}\frac{\text{22}}{\text{7}} \right]$

Ans:

We will calculate for the cuboidal tin can.

Length of tin can $\text{l = 5 cm}$

Breadth of tin can $\text{b = 4 cm}$

Height of tin can $\text{h = 15 cm}$

The capacity of a tin can ${{\text{V}}_{1}}\text{ = l  }\!\!\times\!\!\text{  b  }\!\!\times\!\!\text{  h}$

$\Rightarrow {{\text{V}}_{1}}\text{ = }\left( \text{5  }\!\!\times\!\!\text{  4  }\!\!\times\!\!\text{  15} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow {{\text{V}}_{1}}\text{ = 300 c}{{\text{m}}^{\text{3}}}$

We will calculate for the plastic cylinder with circular base.

Height of plastic cylinder $\text{H = 10 cm}$

The diameter of circular base $\text{= 7 cm}$

So, the radius $\text{r = }\frac{\text{7}}{\text{2}}\text{ = 3}\text{.5 cm}$

The capacity of plastic cylinder ${{\text{V}}_{\text{2}}}\text{ =  }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow {{\text{V}}_{\text{2}}}\text{ = }\left( \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{3}\text{.5} \right)}^{\text{2}}}\text{  }\!\!\times\!\!\text{  10} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow {{\text{V}}_{\text{2}}}\text{ = }\left( \text{11  }\!\!\times\!\!\text{  35} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow {{\text{V}}_{\text{2}}}\text{ = 385 c}{{\text{m}}^{\text{3}}}$

From both the values of capacities we found that ${{\text{V}}_{\text{2}}}\text{  }{{\text{V}}_{\text{1}}}\text{ }$.

The difference in capacity $\text{= }\left( \text{385 - 300} \right)\text{ c}{{\text{m}}^{\text{3}}}\text{ = 85 c}{{\text{m}}^{\text{3}}}$

Therefore, we can say that the capacity of plastic cylinder is greater than the tin can by $\text{85 c}{{\text{m}}^{\text{3}}}$.

4. If the lateral surface of a cylinder is $\text{94}\text{.2 c}{{\text{m}}^{\text{2}}}$ and its height is $\text{5 cm}$, then find

i.radius of its base

Ans:

We are given the height of cylinder $\text{h = 5 cm}$.

Let us assume the radius be $\text{r}$.

The curved surface area of cylinder $\text{A = 94}\text{.2 c}{{\text{m}}^{\text{2}}}$.

But we know the curved surface area of cylinder $\text{= 2 }\!\!\pi\!\!\text{ rh}$

$\therefore \text{2 }\!\!\pi\!\!\text{ rh = 94}\text{.2 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \left( \text{2  }\!\!\times\!\!\text{  3}\text{.14  }\!\!\times\!\!\text{  r  }\!\!\times\!\!\text{  5} \right)\text{ cm = 94}\text{.2 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{r = }\frac{\text{94}\text{.2}}{\text{31}\text{.4}}\text{ cm}$

$\Rightarrow \text{r = 3 cm}$

Therefore, the radius of the base is $\text{3 cm}$.

ii. its volume. $\left[ \text{Use  }\!\!\pi\!\!\text{  = 3}\text{.14} \right]$

Ans:

The volume of the cylinder $\text{V =  }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\left( \text{3}\text{.14  }\!\!\times\!\!\text{  }{{\left( \text{3} \right)}^{\text{2}}}\text{  }\!\!\times\!\!\text{  5} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{V = 141}\text{.3 c}{{\text{m}}^{\text{3}}}$

Therefore, the volume of the cylinder is $\text{141}\text{.3 c}{{\text{m}}^{\text{3}}}$.

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

i.Inner curved surface area of the vessel 

Ans:

It is given that it requires $\text{Rs}\text{. 20}$ to paint an area of $\text{1 }{{\text{m}}^{\text{2}}}$.

So, it requires $\text{Rs}\text{. 2200}$ to paint an area of $\left( \frac{\text{1}}{\text{20}}\text{  }\!\!\times\!\!\text{  2200} \right)\text{ }{{\text{m}}^{\text{2}}}\text{ = 110 }{{\text{m}}^{\text{2}}}$

Therefore, the inner surface area is $\text{110 }{{\text{m}}^{\text{2}}}$.

ii. Radius of the base 

Ans:

Let us assume the radius of the vessel be $\text{r}$.

The height of the vessel is $\text{h = 10 m}$.

The surface area of the vessel is $\text{110 }{{\text{m}}^{\text{2}}}$.

$\therefore \text{2 }\!\!\pi\!\!\text{ rh = 110 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \left[ \text{2  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  r  }\!\!\times\!\!\text{  10} \right]\text{ m = 110 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{r = }\frac{\text{7}}{\text{4}}\text{ m}$

$\Rightarrow \text{r = 1}\text{.75 m}$

Therefore, the radius of the base is $\text{1}\text{.75 m}$.

iii. Capacity of the vessel

Ans:

The volume of the vessel $\text{V =  }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\left[ \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{1}\text{.75} \right)}^{\text{2}}}\text{  }\!\!\times\!\!\text{  10} \right]\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 96}\text{.25 }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left( \text{96}\text{.25  }\!\!\times\!\!\text{  1000} \right)\text{ litres}$

$\Rightarrow \text{V = 96250 litres}$

Therefore, the capacity of the vessel is $\text{96250 litres}$.

6. The capacity of a closed cylindrical vessel of height $\text{1 m}$ is $\text{15}\text{.4 litres}$. How many square metres of metal sheet would be needed to make it? $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

Given height of cylindrical vessel $\text{h = 1 m}$

Let us assume the radius be $\text{r}$.

The volume of cylindrical vessel $\text{= 15}\text{.4 litres = 0}\text{.0154 }{{\text{m}}^{\text{3}}}$

$\therefore \text{ }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{h = 0}\text{.0154}$

$\Rightarrow \left( \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\text{r}}^{\text{2}}}\text{  }\!\!\times\!\!\text{  1} \right)\text{ m = 0}\text{.0154 }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{r = 0}\text{.07 m}$

The total surface area of the vessel $\text{A = 2 }\!\!\pi\!\!\text{ r }\left( \text{h + r} \right)$

$\Rightarrow \text{A = }\left[ \text{2  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  0}\text{.07 }\left( \text{1 + 0}\text{.07} \right) \right]\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{0}\text{.44  }\!\!\times\!\!\text{  1}\text{.07} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 0}\text{.4708 }{{\text{m}}^{\text{2}}}$

Therefore, the amount of metal sheet required to make the cylindrical vessel is $\text{0}\text{.4708 }{{\text{m}}^{\text{2}}}$.

7. A lead pencil consists of a cylinder of wood with solid cylinder of graphite filled in the interior. The diameter of the pencil is $\text{7 mm}$ and the diameter of the graphite is $\text{1 mm}$. If the length of the pencil is $\text{14 cm}$, find the volume of the wood and that of the graphite. $\left[ \text{Assume  }\!\!\pi\!\!\text{  =}\frac{\text{22}}{\text{7}} \right]$

Ans:

Given the diameter of the pencil $\text{= 7 mm}$

So, the radius of pencil ${{\text{r}}_{\text{1}}}\text{ = }\frac{\text{7}}{\text{2}}\text{  }\!\!\times\!\!\text{  }\frac{\text{1}}{\text{10}}\text{ cm = 0}\text{.35 cm}$

The diameter of graphite $\text{= 1 mm}$

So, the radius of graphite ${{\text{r}}_{\text{2}}}\text{ = }\frac{\text{1}}{\text{2}}\text{  }\!\!\times\!\!\text{  }\frac{\text{1}}{\text{10}}\text{ cm = 0}\text{.05 cm}$

The height of the pencil $\text{h = 14 cm}$

The volume of wood in the pencil $\text{V =  }\!\!\pi\!\!\text{ }\left( \text{r}_{\text{2}}^{\text{2}}\text{ - r}_{\text{1}}^{\text{2}} \right)\text{h}$

$\Rightarrow \text{V =}\left[ \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }\left\{ {{\left( \text{0}\text{.35} \right)}^{\text{2}}}\text{ - }{{\left( \text{0}\text{.05} \right)}^{\text{2}}} \right\}\text{  }\!\!\times\!\!\text{  14} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left[ \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }\left\{ \text{0}\text{.1225 - 0}\text{.0025} \right\}\text{  }\!\!\times\!\!\text{  14} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left( \text{44  }\!\!\times\!\!\text{  0}\text{.12} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 5}\text{.28 c}{{\text{m}}^{\text{3}}}$

The volume of graphite $\text{{V}' =  }\!\!\pi\!\!\text{ r}_{\text{2}}^{\text{2}}\text{h}$

$\Rightarrow \text{{V}' =}\left[ \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{0}\text{.05} \right)}^{\text{2}}}\text{  }\!\!\times\!\!\text{  14} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{{V}' =}\left( \text{44  }\!\!\times\!\!\text{  0}\text{.0025} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{{V}' = 0}\text{.11 c}{{\text{m}}^{\text{3}}}$

Therefore, the volume of wood is $\text{5}\text{.28 c}{{\text{m}}^{\text{3}}}$ and the volume of graphite is $\text{0}\text{.11 c}{{\text{m}}^{\text{3}}}$.

8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter $\text{7 cm}$. If the bowl is filled with soup to a height of $\text{4 cm}$, how much soup the hospital has to prepare daily to serve $\text{250}$ patients? $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

It is given that the diameter of cylindrical bowl $\text{= 7 cm}$

So, the radius $\text{r = }\frac{\text{7}}{\text{2}}\text{ = 3}\text{.5 cm}$

Let $\text{h}$ be the height of the soup in the bowl and $\text{h = 4 cm}$.

The volume of the soup in the bowl $\text{V =  }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\left[ \text{ }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{3}\text{.5} \right)}^{\text{2}}}\text{  }\!\!\times\!\!\text{  4} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left[ \text{ 11  }\!\!\times\!\!\text{  3}\text{.5  }\!\!\times\!\!\text{  4} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 154 c}{{\text{m}}^{\text{3}}}$

So, the total volume of the soup given to $250$ patients $\text{= }\left( \frac{\text{250  }\!\!\times\!\!\text{  154}}{\text{1000}} \right)\text{ litres = 38}\text{.5 litres}$

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.6

Opting for the NCERT solutions for Ex 13.6 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.6 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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