NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volume

Exercise (13.1)

1. A plastic box $\text{1}\text{.5}$ $\text{m}$ long, $\text{1}\text{.25}$ $\text{m}$ wide and $\text{65}$ $\text{cm}$ deep, is to be made. It is open at the top. Ignoring the thickness of the plastic sheet, determine:

(i) The area of the sheet required for making the box.

Ans:

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We are given the length $\left( \text{l} \right)$ of the box $\text{= 1}\text{.5}$ $\text{m}$.

The breadth $\left( \text{b} \right)$ of box = $\text{1}\text{.25}$ $\text{m}$.

The depth $\left( \text{h} \right)$ of box = $\text{0}\text{.65}$ $\text{m}$.

We are also told that the box is open at the top. So, we can write the area of sheet required

So, area of sheet required

$\text{= 2lh+2bh+lb}$

$\text{= }\left[ \left( \text{2 }\times\text{ 1}\text{.5 }\times\text{ 0}\text{.65} \right)\text{+}\left( \text{2 }\times\text{ 1}\text{.25 }\times\text{ 0}\text{.65} \right)\text{+}\left( \text{1}\text{.5 }\times\text{ 1}\text{.25} \right) \right]$

$\text{= 1}\text{.95+1}\text{.625+1}\text{.875}$

$\text{= 5}\text{.45}$

The area of the sheet required is $\text{5}\text{.45}$ ${{\text{m}}^{\text{2}}}$.

(ii) The cost of the sheet for it, if a sheet measuring $\text{1}$ ${{\text{m}}^{\text{2}}}$ costs $\text{Rs}\text{.}$ $\text{20}$.

Ans: It is given that the cost of sheet per ${{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 20}$

Cost of sheet of $\text{5}\text{.45}$ ${{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \text{20 }\times\text{ 5}\text{.45} \right)\text{ = Rs}\text{. 109}$

So, the cost of the sheet required is $\text{Rs}\text{.}$ $\text{109}$.

2. The length, breadth and height of a room are $\text{5}$ $\text{m}$, $\text{4}$ $\text{m}$ and $\text{3}$ $\text{m}$ respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of $\text{Rs}\text{.}$ $\text{7}\text{.50}$ per ${{\text{m}}^{\text{2}}}$.

Ans:

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We are given the following: 

The length $\left( \text{l} \right)$ of the room $\text{= 5}$ $\text{m}$.

The breadth $\left( \text{b} \right)$ of the room $\text{= 4}$ $\text{m}$.

The height $\left( \text{h} \right)$ of the room $\text{= 3}$ $\text{m}$.

We have to whitewash the walls of the room and also to the ceiling.

The area to be whitewashed 

$\text{=}$ Area of walls $\text{+}$ Area of ceiling

$\text{= 2lh+2bh+lb}$

$\text{= }\left[ \text{2 }\times\text{ 5 }\times\text{ 3+2 }\times\text{ 4 }\times\text{ 3+5 }\times\text{ 4} \right]$ ${{\text{m}}^{\text{2}}}$

$\text{= }\left( \text{30+24+20} \right)$ ${{\text{m}}^{\text{2}}}$

$\text{= 74}$ ${{\text{m}}^{\text{2}}}$

It is given that the cost of whitewashing per ${{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 7}\text{.50}$

So, the cost of whitewashing $\text{74}$ ${{\text{m}}^{\text{2}}}$ area 

$\text{= Rs}\text{. }\left( \text{7}\text{.5 }\times\text{ 74} \right)$

$\text{= Rs}\text{. 555}$

The cost of whitewashing the walls of given room is $\text{Rs}\text{. 555}$.

3. The floor of a rectangle hall has a perimeter $\text{250}$ $\text{m}$. If the cost of painting the four walls at the rate of $\text{Rs}\text{.}$ $\text{10}$ per ${{\text{m}}^{\text{2}}}$ is $\text{Rs}\text{.}$ $\text{15000}$, find the height of the hall.

(Hint: Area of four walls $\text{=}$ Lateral surface area)

Ans:

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Let us assume the following:

The length of the hall $\text{= l}$ $\text{m}$

The breadth of the hall $\text{= b}$ $\text{m}$

The height of the hall $\text{= h}$ $\text{m}$

So, the area of the four walls will be 

$\text{= 2lh+2bh}$

$\text{= 2}\left( \text{l+b} \right)\text{h}$

We are given the perimeter of the floor of the hall $\text{= 250}$ $\text{m}$. So, we can write the perimeter of the floor $\text{= 2}\left( \text{l+b} \right)\text{ = 250}$ $\text{m}$.

Now we can rewrite the area of four walls as the following.

Area of four walls $\text{= 2}\left( \text{l+b} \right)\text{h = 250h}$ ${{\text{m}}^{\text{2}}}$

The cost of painting per ${{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 10}$

So, the cost of painting $\text{250h}$ ${{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \text{250h }\times\text{ 10} \right)\text{ = Rs}\text{. 2500h}$

But in the question it is given that the cost of painting the area of four walls is $\text{Rs}\text{. 15000}$

$\therefore \text{ 2500h = 15000}$

\[\text{h=6}\]

Therefore, the height of the rectangular hall is $\text{6}$ $\text{m}$.

4. The paint in a certain container is sufficient to paint an area equal to $\text{9}\text{.375}$ ${{\text{m}}^{\text{2}}}$. How many bricks of dimensions $\text{22}\text{.5 cm }\times \text{10 cm }\times \text{7}\text{.5 cm}$ can be painted out of this container?

Ans:

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We will find the total surface area of one brick.

Total surface area of one brick

$\text{= 2}\left( \text{lb + bh + hl} \right)$

$\text{= }\left[ \text{2}\left( \text{22}\text{.5  }\times\text{  10 + 10  }\times\text{  7}\text{.5 + 7}\text{.5  }\times\text{  22}\text{.5} \right) \right]$ $\text{c}{{\text{m}}^{\text{2}}}$

$\text{= 2}\left( \text{225 + 75 + 168}\text{.75} \right)$ $\text{c}{{\text{m}}^{\text{2}}}$

$\text{= }\left( \text{2  }\times\text{  468}\text{.75} \right)$ $\text{c}{{\text{m}}^{\text{2}}}$

$\text{= 937}\text{.5}$ $\text{c}{{\text{m}}^{\text{2}}}$

We are given that the paint in the container can paint out an area $\text{= 9}\text{.375}$ ${{\text{m}}^{\text{2}}}$ $\text{= 93750}$ ${{\text{m}}^{\text{2}}}$

Let us assume that $\text{n}$ bricks can be painted out of the container.

So, the total area of $\text{n}$ bricks $\text{= }\left( \text{n  }\times\text{  937}\text{.5} \right)$ $\text{c}{{\text{m}}^{\text{2}}}$ $\text{= 937}\text{.5n}$ $\text{c}{{\text{m}}^{\text{2}}}$

We can now equate the following:

$\text{93750 = 937}\text{.5n}$

$\text{n = 100}$

Therefore, we can paint out $\text{100}$ bricks by the paint of the container. 

5. A cubical box has each edge $\text{10}$ $\text{cm}$ and another cuboidal box is $\text{12}\text{.5}$ $\text{cm}$ long, $\text{10}$ $\text{cm}$ wide and $\text{8}$ $\text{cm}$ high.

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i. Which box has the greatest lateral surface area and by how much?

Ans: We are given the following:

Edge of the cubical box $\text{= 10 cm}$

Length $\left( \text{l} \right)$ of the cuboidal box $\text{= 12}\text{.5 cm}$ 

Breadth $\left( \text{b} \right)$ of the cuboidal box $\text{= 10 cm}$ 

Height $\left( \text{h} \right)$ of the cuboidal box $\text{= 8 cm}$ 

Now we will calculate the lateral surface area of both the boxes.

Lateral surface area of cubical box $\text{= 4}{{\left( \text{edge} \right)}^{\text{2}}}$

$\text{= 4}{{\left( \text{10 cm} \right)}^{\text{2}}}$

$\text{= 400 c}{{\text{m}}^{\text{2}}}$

Lateral surface area of cuboidal box $\text{= 2}\left[ \text{lh + bh} \right]$

$\text{= }\left[ \text{2}\left( \text{12}\text{.5  }\times\text{  8 + 10  }\times\text{  8} \right) \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= }\left( \text{2  }\times\text{  180} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= 360 c}{{\text{m}}^{\text{2}}}$

We can clearly see that the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box.

The difference between the lateral surface areas $\text{= 400 c}{{\text{m}}^{\text{2}}}\text{ - 360 c}{{\text{m}}^{\text{2}}}$

$\text{= 40 c}{{\text{m}}^{\text{2}}}$

Hence, the lateral surface area of cubical box is greater than the lateral surface area of cuboidal box by $\text{40 c}{{\text{m}}^{\text{2}}}$.

ii. Which box has the smaller total surface area and by how much?

Ans:

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We will calculate the total surface area of both the boxes.

Total surface area of cubical box $\text{= 6}{{\left( \text{edge} \right)}^{\text{2}}}$

$\text{= 6}{{\left( \text{10 cm} \right)}^{\text{2}}}$

$\text{= 600 c}{{\text{m}}^{\text{2}}}$

Total surface area of cuboidal box $\text{= 2}\left[ \text{lh + bh + lb} \right]$

$\text{= 2}\left[ \text{12}\text{.5  }\times\text{  8 + 10  }\times\text{  8 + 12}\text{.5  }\times\text{  10} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= 2}\left[ \text{100 + 80 + 125} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= 610 c}{{\text{m}}^{\text{2}}}$

We can clearly see that the total surface area of the cubical box is smaller than the total surface area of the cuboidal box.

The difference between the total surface areas $\text{= 610 c}{{\text{m}}^{\text{2}}}\text{ - 600 c}{{\text{m}}^{\text{2}}}$

$\text{= 10 c}{{\text{m}}^{\text{2}}}$

Hence, the total surface area of cubical box is smaller than the total surface area of cuboidal box by $\text{10 c}{{\text{m}}^{\text{2}}}$.

6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is $\text{30 cm}$ long, $\text{25 cm}$ wide and $\text{25 cm}$ high.

(i) What is the area of the glass?

Ans:

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We are given the following:

Length $\left( \text{l} \right)$ of the greenhouse $\text{= 30 cm}$ 

Breadth $\left( \text{b} \right)$ of the greenhouse $\text{= 25 cm}$ 

Height $\left( \text{h} \right)$ of the greenhouse $\text{= 25 cm}$ 

Now we will calculate the total surface area of the greenhouse.

Total surface area of the green house made of glass

$\text{= 2}\left[ \text{lb + lh + bh} \right]$

$\text{= 2}\left[ \text{30  }\times\text{  25 + 30  }\times\text{  25 + 25  }\times\text{  25} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= 2}\left[ \text{750 + 750 + 625} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= }\left( \text{2  }\times\text{  2125} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= 4250 c}{{\text{m}}^{\text{2}}}$

Hence, the area of glass is $\text{4250 c}{{\text{m}}^{\text{2}}}$.

(ii) How much of tape is needed for all the $\text{12}$ edges?

Ans:

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From the figure we can see that the total length of the tape required will be the total length of all the edges.

Total length of tape $\text{= 4}\left( \text{l + b + h} \right)$

$\text{= }\left[ \text{4}\left( \text{30 + 25 + 25} \right) \right]\text{ cm}$

$\text{= 320 cm}$

Hence, the total length of tape required is $\text{320 cm}$.

7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions \[\text{25 cm  }\times\text{  20 cm  }\times\text{  5 cm}\] and the smaller of dimensions \[\text{15 cm  }\times\text{  12 cm  }\times\text{  5 cm}\]. For all the overlaps, $\text{5  }\%\text{ }$ of the total surface area is required extra. If the cost of the cardboard is $\text{Rs}\text{. 4}$ for $\text{1000 c}{{\text{m}}^{\text{2}}}$, find the cost of cardboard required for supplying $\text{250}$ boxes of each kind.  

Ans:

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We are given the following:

Length $\left( \text{L} \right)$ of the bigger box $\text{= 25 cm}$ 

Breadth $\left( \text{B} \right)$ of the bigger box $\text{= 20 cm}$ 

Height $\left( \text{H} \right)$ of the bigger box $\text{= 5 cm}$ 

Total surface area of the bigger box

$\text{= 2}\left[ \text{LB + LH + BH} \right]$

$\text{= }\left[ \text{2}\left( \text{25  }\times\text{  20 + 25  }\times\text{  5 + 20  }\times\text{  5} \right) \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= }\left[ 2\left( \text{500 + 125 + 100} \right) \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= }\left( \text{2  }\times\text{  725} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= 1450 c}{{\text{m}}^{\text{2}}}$

We have to find the extra area due to overlapping.

Extra area $\text{= }\left( \dfrac{\text{1450  }\times\text{  5}}{\text{100}} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= 72}\text{.5 c}{{\text{m}}^{\text{2}}}$

Now the total surface area of bigger box

$\text{= }\left( \text{1450 + 72}\text{.5} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= 1522}\text{.5 c}{{\text{m}}^{\text{2}}}$

So, the area of cardboard sheet required for $\text{250}$ boxes

$\text{= }\left( \text{1522}\text{.5  }\times\text{  250} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= 380625 c}{{\text{m}}^{\text{2}}}$

Similarly we will repeat the calculations for smaller box.

Length $\left( \text{l} \right)$ of the smaller box $\text{= 15 cm}$ 

Breadth $\left( \text{b} \right)$ of the smaller box $\text{= 12 cm}$ 

Height $\left( \text{h} \right)$ of the smaller box $\text{= 5 cm}$ 

Total surface area of the smaller box

$\text{= 2}\left[ \text{lb + lh + bh} \right]$

$\text{= }\left[ \text{2}\left( \text{15  }\times\text{  12 + 15  }\times\text{  5 + 12  }\times\text{  5} \right) \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= }\left[ 2\left( \text{180 + 75 + 60} \right) \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= }\left( \text{2  }\times\text{  315} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= 630 c}{{\text{m}}^{\text{2}}}$

We have to find the extra area due to overlapping.

Extra area $\text{= }\left( \dfrac{\text{630  }\times\text{  5}}{\text{100}} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= 31}\text{.5 c}{{\text{m}}^{\text{2}}}$

Now the total surface area of smaller box

$\text{= }\left( \text{630 + 31}\text{.5} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= 661}\text{.5 c}{{\text{m}}^{\text{2}}}$

So, the area of cardboard sheet required for $\text{250}$ boxes

$\text{= }\left( \text{661}\text{.5  }\times\text{  250} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= 165375 c}{{\text{m}}^{\text{2}}}$

Hence, the total cardboard sheet required

$\text{= }\left( \text{380625 + 165375} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\text{= 546000 c}{{\text{m}}^{\text{2}}}$

It is given that the cost of $\text{1000 c}{{\text{m}}^{\text{2}}}$ sheet $\text{= Rs}\text{. 4}$

So, the cost of  $\text{546000 c}{{\text{m}}^{\text{2}}}$ sheet $\text{= Rs}\text{. }\left( \dfrac{\text{546000  }\times\text{  4}}{\text{1000}} \right)$

$\text{= Rs}\text{. 2184}$

Therefore, the cost of cardboard sheet required to make $\text{250}$ boxes of each kind is $\text{Rs}\text{. 2184}$.

8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height $\text{2}\text{.5 m}$, with base dimensions $\text{4 m  }\times\text{  3 m}$?

Ans:

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We are given the following:

Length $\left( \text{l} \right)$ of the shelter $\text{= 4 m}$ 

Breadth $\left( \text{b} \right)$ of the shelter $\text{= 3 m}$ 

Height $\left( \text{h} \right)$ of the shelter $\text{= 2}\text{.5 m}$ 

A tarpaulin is made to cover the four sides and the top of the shelter.

The area of tarpaulin required

$\text{= 2}\left( \text{lh + bh} \right)\text{ + lb}$

$\text{= }\left[ \text{2}\left( \text{4  }\times\text{  2}\text{.5 + 3  }\times\text{  2}\text{.5} \right)\text{ + 4  }\times\text{  3} \right]\text{ }{{\text{m}}^{\text{2}}}$

$\text{= }\left[ \text{2}\left( \text{10 + 7}\text{.5} \right)\text{ + 12} \right]\text{ }{{\text{m}}^{\text{2}}}$

$\text{= 47 }{{\text{m}}^{\text{2}}}$

Hence, we would need $\text{47 }{{\text{m}}^{\text{2}}}$ of tarpaulin.

Exercise (13.2)

1. The curved surface area of a right circular cylinder of height $\text{14 cm}$ is $\text{88 c}{{\text{m}}^{\text{2}}}$. Find the diameter of the base of the cylinder. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$.

Ans: We are given the following:

Height $\left( \text{h} \right)$ of the cylinder $\text{= 14 cm}$

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Let us assume $\text{r}$ is the radius of base of cylinder

Curved surface area of the cylinder $\text{A = 88 c}{{\text{m}}^{\text{2}}}$

\[\Rightarrow \text{ 2 }\pi\text{ rh = 88 c}{{\text{m}}^{\text{2}}}\]

\[\Rightarrow \text{  }\pi\text{ dh = 88 c}{{\text{m}}^{\text{2}}}\text{ }\left( \text{d = 2r} \right)\]

\[\Rightarrow \text{ }\left[ \dfrac{\text{22}}{\text{7}}\text{ }\times\text{ d }\times\text{ 14} \right]\text{ cm = 88 c}{{\text{m}}^{\text{2}}}\]

\[\Rightarrow \text{ d = 2 cm}\]

Hence, the diameter of the base of the cylinder is $\text{2 cm}$.

2. It is required to make a closed cylindrical tank of height $\text{1 m}$ and base diameter $\text{140 cm}$ from a metal sheet. How many square meters of the sheet are required for the same? $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$.

Ans:

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We are given the following:

Height $\left( \text{h} \right)$ of the cylindrical tank $\text{= 1 m}$

Base diameter of the cylindrical tank $\text{= 140 cm}$

So, the base radius $\left( \text{r} \right)$ of the cylindrical tank 

$\text{= }\dfrac{\text{140}}{\text{2}}\text{ cm}$

$\text{= 70 cm}$

$\text{= 0}\text{.7 m}$

The area of sheet required will be the same as total surface area of the tank.

So, the area of the sheet required, $\text{A = 2 }\pi\text{ r}\left( \text{h + r} \right)$

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

$\Rightarrow \text{A = }\left( \text{4}\text{.4  }\times\text{  1}\text{.7} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 7}\text{.48 }{{\text{m}}^{\text{2}}}$

Therefore the area of the sheet required is $\text{7}\text{.48 }{{\text{m}}^{\text{2}}}$.

3. A metal pipe is $\text{77 cm}$long. The inner diameter of a cross section is $\text{4 cm}$, the outer diameter being $\text{4}\text{.4 cm}$. Find the following:

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(i) Inner curved surface area 

Ans: Given inner diameter of pipe $\text{= 4 cm}$

So, the inner radius $\left( {{\text{r}}_{\text{1}}} \right)$ of cylindrical pipe $\text{= }\dfrac{\text{4}}{\text{2}}\text{ cm = 2 cm}$

Given outer diameter of pipe $\text{= 4}\text{.4 cm}$

So, the outer radius $\left( {{\text{r}}_{\text{2}}} \right)$ of cylindrical pipe $\text{= }\dfrac{\text{4}\text{.4}}{\text{2}}\text{ cm = 2}\text{.2 cm}$

The height $\left( \text{h} \right)$ of cylindrical pipe $\text{= 77 cm}$

So, curved surface area of inner pipe, ${{\text{A}}_{\text{1}}}\text{ = 2 }\pi\text{ }{{\text{r}}_{\text{1}}}\text{h}$

$\Rightarrow {{\text{A}}_{\text{1}}}\text{ = }\left( \text{2  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  2  }\times\text{  77} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{1}}}\text{ = 968 c}{{\text{m}}^{\text{2}}}$

So, the inner curved surface area is $\text{968 c}{{\text{m}}^{\text{2}}}$.

(ii) Outer curved surface area 

Ans:

Curved surface area of outer pipe, ${{\text{A}}_{\text{2}}}\text{ = 2 }\pi\text{ }{{\text{r}}_{\text{2}}}\text{h}$

$\Rightarrow {{\text{A}}_{\text{2}}}\text{ = }\left( \text{2  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  2}\text{.2  }\times\text{  77} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{2}}}\text{ = 1064}\text{.8 c}{{\text{m}}^{\text{2}}}$

So the outer curved surface area is $\text{1064}\text{.8 c}{{\text{m}}^{\text{2}}}$.

(iii) Total surface area. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

Ans:

The total surface area of the pipe will be the combined area of the curved surface area of both inner and outer surface and also the area of circular ends of the pipe.

Total surface area of the pipe, ${{\text{A}}_{\text{T}}}\text{ = 2 }\pi\text{ }{{\text{r}}_{\text{1}}}\text{h + 2 }\pi\text{ }{{\text{r}}_{\text{2}}}\text{h + 2 }\pi\text{ }\left( {{\text{r}}_{\text{2}}}^{\text{2}}\text{- }{{\text{r}}_{\text{1}}}^{\text{2}} \right)$

$\Rightarrow {{\text{A}}_{\text{T}}}\text{ = 2 }\times\text{ }\dfrac{\text{22}}{\text{7}}\text{ }\times\text{ 2 }\times\text{ 77 + 2 }\times\text{ }\dfrac{\text{22}}{\text{7}}\text{ }\times\text{ 2}\text{.2 }\times\text{ 77 + 2 }\times\text{ }\dfrac{\text{22}}{\text{7}}\left( {{\left( \text{2}\text{.2} \right)}^{\text{2}}}\text{- }{{\left( \text{2} \right)}^{\text{2}}} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{T}}}\text{= }\left[ \text{968 + 1064}\text{.8 + 2 }\times\text{ }\dfrac{\text{22}}{\text{7}}\text{ }\times\text{ 0}\text{.84} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{T}}}\text{= }\left[ \text{2032}\text{.8 + 5}\text{.28} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{T}}}\text{= 2038}\text{.08 c}{{\text{m}}^{\text{2}}}$

Therefore the total surface area is $\text{2038}\text{.08 c}{{\text{m}}^{\text{2}}}$.

4. The diameter of a roller is $\text{84 cm}$ and its length is $\text{120 cm}$. It takes $\text{500}$ complete revolutions to move once over to level a playground. Find the area of the playground in ${{\text{m}}^{\text{2}}}$? $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$.

Ans: The given roller is a cylinder.

The height $\left( \text{h} \right)$ of the roller $\text{= 120 cm}$

Given the diameter of circular end $\text{= 84 cm}$

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So, the radius $\left( \text{r} \right)$ of circular end $\text{= }\dfrac{\text{84}}{\text{2}}\text{ cm = 42 cm}$

The curved surface area of the roller, ${{\text{A}}_{\text{CSA}}}\text{ = 2 }\pi\text{ rh}$

$\Rightarrow {{\text{A}}_{\text{CSA}}}\text{ = }\left( \text{ 2 }\times\text{ }\dfrac{\text{22}}{\text{7}}\text{ }\times\text{ 42 }\times\text{ 120} \right)\text{c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{CSA}}}\text{ = 31680 c}{{\text{m}}^{\text{2}}}$

We can write the area of the field, ${{\text{A}}_{\text{F}}}\text{ = 500 }\times\text{ }{{\text{A}}_{\text{CSA}}}$

$\Rightarrow {{\text{A}}_{\text{F}}}\text{= }\left( \text{500 }\times\text{ 31680} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{F}}}\text{= 15840000 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{F}}}\text{= 1584 }{{\text{m}}^{\text{2}}}$

Therefore the area of the playground is $\text{1584 }{{\text{m}}^{\text{2}}}$.

5. A cylindrical pillar is $\text{50 cm}$ in diameter and $\text{3}\text{.5 m}$ in height. Find the cost of painting the curved surface of the pillar at the rate of $\text{Rs}\text{. 12}\text{.50}$ per ${{\text{m}}^{\text{2}}}$. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$.

Ans: The height $\left( \text{h} \right)$ of the cylindrical pillar $\text{= 3}\text{.5 m}$

Given the diameter of cylindrical pillar $\text{= 50 cm}$

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So, the radius $\left( \text{r} \right)$ of cylindrical pillar $\text{= }\dfrac{\text{50}}{\text{2}}\text{ cm = 25 cm = 0}\text{.25 m}$

The curved surface area of the pillar, ${{\text{A}}_{\text{CSA}}}\text{ = 2 }\pi\text{ rh}$

$\Rightarrow {{\text{A}}_{\text{CSA}}}\text{ = }\left( \text{2 }\times\text{ }\dfrac{\text{22}}{\text{7}}\text{ }\times\text{ 0}\text{.25 }\times\text{ 3}\text{.5} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{CSA}}}\text{ = 5}\text{.5 }{{\text{m}}^{\text{2}}}$

It is given the cost of painting $\text{1 }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 12}\text{.50}$

So, the cost of painting $\text{5}\text{.5 }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \text{5}\text{.5 }\times\text{ 12}\text{.50} \right)\text{ = Rs}\text{. 68}\text{.75}$

Therefore, the cost of painting the curved surface area of the cylindrical pillar is $\text{Rs}\text{. 68}\text{.75}$.

6. Curved surface area of a right circular cylinder is $\text{4}\text{.4 }{{\text{m}}^{\text{2}}}$. If the radius of the base of the cylinder is $\text{0}\text{.7 m}$, find its height. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$.

Ans:

We are given the following:

Let $\left( \text{h} \right)$ be the height of the circular cylinder.

The base radius $\left( \text{r} \right)$ of the cylinder$\text{= 0}\text{.7 m}$

(Image will be Uploaded Soon)

The curved surface area of the cylinder, $\text{A = 4}\text{.4 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{2 }\pi\text{ rh = 4}\text{.4}$

$\Rightarrow \text{2}\times \dfrac{22}{7}\times \text{0}\text{.7}\times \text{h = 4}\text{.4}$

$\Rightarrow 4.4\text{h = 4}\text{.4}$

$\Rightarrow \text{h = 1 m}$

Therefore, the height of the circular cylinder is $\text{1 m}$.

7. The inner diameter of a circular well is $\text{3}\text{.5 m}$. It is $\text{10 m}$ deep. Find 

(i) Its inner curved surface area, 

Ans: The depth $\left( \text{h} \right)$ of the circular well $\text{= 10 m}$

Given the inner diameter of circular well $\text{= 3}\text{.5 m}$

(Image will be Uploaded Soon)

So, the radius $\left( \text{r} \right)$ of circular well $\text{= }\dfrac{\text{3}\text{.5}}{\text{2}}\text{ m = 1}\text{.75 m }$

The inner curved surface area of circular well, ${{\text{A}}_{\text{CSA}}}\text{ = 2 }\pi\text{ rh}$

$\Rightarrow {{\text{A}}_{\text{CSA}}}\text{ = }\left( \text{2 }\times\text{ }\dfrac{\text{22}}{\text{7}}\text{ }\times\text{ 1}\text{.75 }\times\text{ 10} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{CSA}}}\text{ = 110 }{{\text{m}}^{\text{2}}}$

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

(ii) The cost of plastering this curved surface at the rate of $\text{Rs}\text{. 40}$per ${{\text{m}}^{\text{2}}}$.

Ans: It is given the cost of plastering $\text{1 }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 40}$

So, the cost of plastering $\text{110 }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \text{40 }\times\text{ 110} \right)\text{ = Rs}\text{. 4400}$

Therefore the cost of plastering the curved surface area of the circular well is $\text{Rs}\text{. 4400}$.

8. In a hot water heating system, there is a cylindrical pipe of length $\text{28 m}$ and diameter 5 cm. Find the total radiating surface in the system. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$.

Ans: The height $\left( \text{h} \right)$ of the cylindrical pipe $\text{= 28 m}$

Given the diameter of cylindrical pipe $\text{= 5 cm}$

(Image will be Uploaded Soon)

So, the radius $\left( \text{r} \right)$ of cylindrical pipe $\text{= }\dfrac{\text{5}}{\text{2}}\text{ cm = 2}\text{.5 cm = 0}\text{.025 m}$

The inner curved surface area of cylindrical pipe, ${{\text{A}}_{\text{CSA}}}\text{ = 2 }\pi\text{ rh}$

$\Rightarrow {{\text{A}}_{\text{CSA}}}\text{ = }\left( \text{2 }\times\text{ }\dfrac{\text{22}}{\text{7}}\text{ }\times\text{ 0}\text{.025 }\times\text{ 28} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{CSA}}}\text{ = 4}\text{.4 }{{\text{m}}^{\text{2}}}$

Therefore, the total radiating surface in the system is $\text{4}\text{.4 }{{\text{m}}^{\text{2}}}$.

9. Find 

a) The lateral or curved surface area of a closed cylindrical petrol storage tank that is $\text{4}\text{.2 m}$ in diameter and $\text{4}\text{.5 m}$ high

Ans: The height $\left( \text{h} \right)$ of the cylindrical tank $\text{= 4}\text{.5 m}$

Given the diameter of cylindrical tank $\text{= 4}\text{.2 m}$

(Image will be Uploaded Soon)

So, the radius $\left( \text{r} \right)$ of cylindrical tank $\text{= }\dfrac{\text{4}\text{.2}}{\text{2}}\text{ m = 2}\text{.1 m}$

The curved surface area of cylindrical tank, ${{\text{A}}_{\text{CSA}}}\text{ = 2 }\pi\text{ rh}$

$\Rightarrow {{\text{A}}_{\text{CSA}}}\text{ = }\left( \text{2 }\times\text{ }\dfrac{\text{22}}{\text{7}}\text{ }\times\text{ 2}\text{.1 }\times\text{ 4}\text{.5} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{CSA}}}\text{ = 59}\text{.4 }{{\text{m}}^{\text{2}}}$

Therefore, the lateral or curved surface area of the tank is $\text{59}\text{.4 }{{\text{m}}^{\text{2}}}$.

b) How much steel was actually used, if $\dfrac{\text{1}}{\text{12}}$ of the steel actually used was wasted in making the tank. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

Ans: The total surface area of the tank ${{\text{A}}_{\text{T}}}\text{= 2 }\pi\text{ r}\left( \text{h + r} \right)$

$\Rightarrow {{\text{A}}_{\text{T}}}\text{= }\left[ \text{2 }\times\text{ }\dfrac{\text{22}}{\text{7}}\text{ }\times\text{ 2}\text{.1}\left( \text{4}\text{.5 + 2}\text{.1} \right) \right]\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{T}}}\text{= }\left( \text{13}\text{.2  }\times\text{  6}\text{.6} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{T}}}\text{= 87}\text{.12 }{{\text{m}}^{\text{2}}}$

Let us assume that $\text{x }{{\text{m}}^{\text{2}}}$ of steel was used in making the cylindrical tank.

So according to the question $\dfrac{\text{1}}{\text{12}}$ steel is used in making the tank.

$\therefore \text{x}\left( \text{1-}\dfrac{\text{1}}{\text{12}} \right)\text{ = 87}\text{.12 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{x = }\left( \dfrac{\text{12}}{\text{11}}\text{ }\times\text{ 87}\text{.12} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{x = 95}\text{.04 }{{\text{m}}^{\text{2}}}$

Hence, the actual amount of steel used in making the tank is $\text{95}\text{.04 }{{\text{m}}^{\text{2}}}$.

10. In the given figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of $\text{20 cm}$and height of $\text{30 cm}$. A margin of $\text{2}\text{.5 cm}$ is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

(Image will be Uploaded Soon)

Ans: We will redraw the diagram with the margin.

(Image will be Uploaded Soon)

From the figure, we get the following:

The height $\left( \text{h} \right)$ of the frame $\text{= }\left( \text{2}\text{.5 + 30 + 2}\text{.5} \right)\text{ cm = 35 cm}$

The diameter of the base $\text{= 20 cm}$

So, the radius $\left( \text{r} \right)$ of the base $\text{= }\dfrac{\text{20}}{\text{2}}\text{ cm = 10 cm}$

The cloth required for covering the lampshade is the same as the curved surface area of the cylindrical frame.

So, the cloth required for covering the lampshade $\text{A = 2 }\pi\text{ rh}$

$\Rightarrow \text{A = }\left( \text{2 }\times\text{ }\dfrac{\text{22}}{\text{7}}\text{ }\times\text{ 10 }\times\text{ 35} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 2200 c}{{\text{m}}^{\text{2}}}$

Therefore, we need $\text{2200 c}{{\text{m}}^{\text{2}}}$ of cloth to cover the frame of the lampshade.

11. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius $\text{3 cm}$ and height $\text{10}\text{.5 cm}$. The Vidyalaya was to supply the competitors with cardboard. If there were $\text{35}$ competitors, how much cardboard was required to be bought for the competition? $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

Ans: We are given the following:

The height $\left( \text{h} \right)$ of penholder $\text{= 10}\text{.5 cm}$

The radius $\left( \text{r} \right)$ of the circular end $\text{= 3 cm}$

(Image will be Uploaded Soon)

The surface area of a penholder is the combination of the curved surface area of the penholder and the area of base of the penholder.

Surface area of a penholder, $\text{A = 2 }\pi\text{ rh +  }\pi\text{ }{{\text{r}}^{\text{2}}}$ 

$\Rightarrow \text{A = }\left[ \text{2 }\times\text{ }\dfrac{\text{22}}{\text{7}}\text{ }\times\text{ 3 }\times\text{ 10}\text{.5 + }\dfrac{\text{22}}{\text{7}}\text{ }\times\text{ }{{\left( \text{3} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{132  }\times\text{  1}\text{.5 + }\dfrac{\text{198}}{\text{7}} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{198 + }\dfrac{\text{198}}{\text{7}} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\dfrac{1584}{7}\text{ c}{{\text{m}}^{\text{2}}}$

The area of cardboard used by a single competitor $\text{= }\dfrac{1584}{7}\text{ c}{{\text{m}}^{\text{2}}}$

So, the area used by $35$ competitors $\text{= }\left( \dfrac{\text{1584}}{\text{7}}\text{ }\times\text{ 35} \right)\text{ c}{{\text{m}}^{\text{2}}}\text{ = 7920 c}{{\text{m}}^{\text{2}}}$

Therefore, $\text{7920 c}{{\text{m}}^{\text{2}}}$ of cardboard is required to be bought for the competition.

Exercise (13.3)

1. Diameter of the base of a cone is $\text{10}\text{.5 cm}$ and its slant height is $\text{10 cm}$. Find its curved surface area. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

Ans: We are given the following:

The slant height $\left( \text{l} \right)$ of the cone $\text{= 10 cm}$

The diameter of the base of cone $\text{= 10}\text{.5 cm}$

(Image will be Uploaded Soon)

So, the radius $\left( \text{r} \right)$ of the base of cone $\text{= }\dfrac{\text{10}\text{.5}}{\text{2}}\text{ cm = 5}\text{.25 cm}$

The curved surface area of cone, $\text{A =  }\pi\text{ rl}$

$\Rightarrow \text{A = }\left( \dfrac{\text{22}}{\text{7}}\text{ }\times\text{ 5}\text{.25 }\times\text{ 10} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{22 }\times\text{ 0}\text{.75 }\times\text{ 10} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 165 c}{{\text{m}}^{\text{2}}}$

Therefore, the curved surface area of the cone is $\text{165 c}{{\text{m}}^{\text{2}}}$.

2. Find the total surface area of a cone, if its slant height is $\text{21 m}$and diameter of its base is $\text{24 m}$. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

Ans: We are given the following:

The slant height $\left( \text{l} \right)$ of the cone $\text{= 21 m}$

The diameter of the base of cone $\text{= 24 m}$

(Image will be Uploaded Soon)

So, the radius $\left( \text{r} \right)$ of the base of cone $\text{= }\dfrac{\text{24}}{\text{2}}\text{ m = 12 m}$

The total surface area of cone, $\text{A =  }\pi\text{ r}\left( \text{l + r} \right)$

$\Rightarrow \text{A = }\left( \dfrac{\text{22}}{\text{7}}\text{  }\times\text{  12  }\times\text{  }\left( \text{21 + 12} \right) \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \dfrac{\text{22}}{\text{7}}\text{  }\times\text{  12  }\times\text{  33} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 1244}\text{.57 }{{\text{m}}^{\text{2}}}$

Therefore, the total surface area of the cone is $\text{1244}\text{.57 }{{\text{m}}^{\text{2}}}$.

3. Curved surface area of a cone is $\text{308 c}{{\text{m}}^{\text{2}}}$ and its slant height is $\text{14 cm}$. Find

(i) Radius of the base and

Ans: It is given that the slant height $\left( \text{l} \right)$ of the cone $\text{= 14 cm}$

The curved surface area of the cone $\text{= 308 c}{{\text{m}}^{\text{2}}}$

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

(Image will be Uploaded Soon)

We know that curved surface area of the cone $\text{=  }\pi\text{ rl}$

$\therefore \text{ }\pi\text{ rl = 308 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \left( \dfrac{\text{22}}{\text{7}}\text{  }\times\text{  r  }\times\text{  14} \right)\text{ cm = 308 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{r = }\dfrac{\text{308}}{\text{44}}\text{ cm}$

$\Rightarrow \text{r = 7 cm}$

Hence, the radius of the base is $\text{7 cm}$.

(ii) Total surface area of the cone. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

Ans: The total surface area of the cone is the sum of its curved surface area and the area of the base.

Total surface area of cone, $\text{A =  }\pi\text{ rl +  }\pi\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{308 + }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{7} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{308 + 154} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 462 c}{{\text{m}}^{\text{2}}}$

Hence, the total surface area of the cone is $\text{462 c}{{\text{m}}^{\text{2}}}$.

4. A conical tent is $\text{10 m}$ high and the radius of its base is $\text{24 m}$. Find

(i) slant height of the tent

Ans:

(Image will be Uploaded Soon)

From the figure we can say that $\text{ABC}$ is a conical tent.

It is given that the height $\left( \text{h} \right)$ of conical tent $\text{= 10 m}$

The radius $\left( \text{r} \right)$ of conical tent $\text{= 24 m}$

Let us assume the slant height as $\text{l}$.

In $\text{ }\Delta\text{ ABD}$, we will use Pythagorean Theorem.

$\therefore \text{A}{{\text{B}}^{\text{2}}}\text{ = AD}{{\text{ }}^{\text{2}}}\text{ + B}{{\text{D}}^{\text{2}}}$

$\Rightarrow {{\text{l}}^{\text{2}}}\text{ = }{{\text{h}}^{\text{2}}}\text{ + }{{\text{r}}^{\text{2}}}$

\[\Rightarrow {{\text{l}}^{\text{2}}}\text{ = }{{\left( \text{10 m} \right)}^{\text{2}}}\text{ + }{{\left( \text{24 m} \right)}^{\text{2}}}\]

$\Rightarrow {{\text{l}}^{\text{2}}}\text{ = 676 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{l = 26 m}$

The slant height of the tent is $\text{26 m}$.

(ii) cost of canvas required to make the tent, if cost of $\text{1 }{{\text{m}}^{\text{2}}}$ canvas is $\text{Rs}\text{. 70}$. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

Ans: The curved surface area of the tent, $\text{A =  }\pi\text{ rl}$

$\Rightarrow \text{A = }\left( \dfrac{\text{22}}{\text{7}}\text{  }\times\text{  24  }\times\text{  26} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \dfrac{13728}{7} \right)\text{ }{{\text{m}}^{\text{2}}}$

It is given that the cost of $\text{1 }{{\text{m}}^{\text{2}}}$ of canvas $\text{= Rs}\text{. 70}$

So, the cost of $\dfrac{13728}{7}\text{ }{{\text{m}}^{\text{2}}}$ canvas $\text{= Rs}\text{. }\left( \dfrac{\text{13728}}{\text{7}}\text{  }\times\text{  70} \right)\text{ = Rs}\text{. 137280}$

Hence, the cost of canvas required to make the tent is $\text{Rs}\text{. 137280}$.

5. What length of tarpaulin $\text{3 m}$ wide will be required to make conical tent of height $\text{8 m}$ and base radius $\text{6 m}$? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately $\text{20 cm}$. $\left[ \text{Use  }\pi\text{  = 3}\text{.14} \right]$

Ans: We are given the following:

The base radius $\left( \text{r} \right)$ of tent $\text{= 6 m}$

The height $\left( \text{h} \right)$ of tent $\text{= 8 m}$

(Image will be Uploaded Soon)

So the slant height of the tent, $\text{l = }\sqrt{{{\text{r}}^{\text{2}}}\text{ + }{{\text{h}}^{\text{2}}}}$ 

$\Rightarrow \text{l = }\left( \sqrt{{{\text{6}}^{\text{2}}}\text{ + }{{\text{8}}^{\text{2}}}} \right)\text{ m}$

$\Rightarrow \text{l = }\left( \sqrt{\text{100}} \right)\text{ m}$

$\Rightarrow \text{l = 10 m}$

The curved surface area of the tent, $\text{A =  }\pi\text{ rl}$

$\Rightarrow \text{A = }\left( \text{3}\text{.14  }\times\text{  6  }\times\text{  10} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 188}\text{.4 }{{\text{m}}^{\text{2}}}$

It is give the width of tarpaulin $\text{= 3 m}$

Let us assume the length of the tarpaulin sheet required be $\text{x}$.

It is given that there will be a wastage of $\text{20 cm}$.

So, the new length of the sheet $\text{=}\left( \text{x - 0}\text{.2} \right)\text{ m}$

We know that the area of the rectangular sheet required will be the same as curved surface area of the tent.

$\therefore \left[ \left( \text{x - 0}\text{.2} \right)\text{  }\times\text{  3} \right]\text{ m = 188}\text{.4 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{x - 0}\text{.2 m = 62}\text{.8 m}$

$\Rightarrow \text{x = 63 m}$

The length of tarpaulin sheet required is $\text{63 m}$.

6. The slant height and base diameter of a conical tomb are $\text{25 m}$ and $\text{14 m}$ respectively. Find the cost of white-washing its curved surface at the rate of $\text{Rs}\text{. 210}$ per $\text{100 }{{\text{m}}^{\text{2}}}$. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

Ans: We are given the following:

The base radius $\left( \text{r} \right)$ of tomb $\text{= 7 m}$

The slant height $\left( \text{l} \right)$ of tomb $\text{= 25 m}$

(Image will be Uploaded Soon)

The curved surface area of the conical tomb, $\text{A =  }\pi\text{ rl}$

$\Rightarrow \text{A = }\left( \dfrac{22}{7}\text{  }\times\text{  7  }\times\text{  25} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 550 }{{\text{m}}^{\text{2}}}$

It is given that the cost of white-washing $\text{1 }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 210}$

So, the cost of white-washing $550\text{ }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \dfrac{\text{210}}{\text{100}}\text{  }\times\text{  550} \right)\text{ = Rs}\text{. 1155}$

Hence, the cost of white-washing the curved surface area of a conical tomb is $\text{Rs}\text{. 1155}$.

7. A joker’s cap is in the form of right circular cone of base radius $\text{7 cm}$ and the height $\text{24 cm}$. Find the area of sheet required to make $\text{10}$ such caps. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

Ans: We are given the following:

The base radius $\left( \text{r} \right)$ of conical cap $\text{= 7 cm}$

The height $\left( \text{h} \right)$ of conical cap $\text{= 24 cm}$

(Image will be Uploaded Soon)

So the slant height of the tent, $\text{l = }\sqrt{{{\text{r}}^{\text{2}}}\text{ + }{{\text{h}}^{\text{2}}}}$ 

$\Rightarrow \text{l = }\left( \sqrt{{{\text{7}}^{\text{2}}}\text{ + 2}{{\text{4}}^{\text{2}}}} \right)\text{ cm}$

$\Rightarrow \text{l = }\left( \sqrt{625} \right)\text{ cm}$

$\Rightarrow \text{l = 25 cm}$

The curved surface area of one conical cap, $\text{A =  }\pi\text{ rl}$

$\Rightarrow \text{A = }\left( \dfrac{22}{7}\text{  }\times\text{  7  }\times\text{  25} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 550 c}{{\text{m}}^{\text{2}}}$

So, the curved surface area of $\text{10}$ conical caps $\text{= }\left( \text{550  }\times\text{  10} \right)\text{ c}{{\text{m}}^{\text{2}}}\text{ = 5500 c}{{\text{m}}^{\text{2}}}$

Therefore, the total area of sheet required is $\text{5500 c}{{\text{m}}^{\text{2}}}$.

8. A bus stop is barricaded from the remaining part of the road, by using $\text{50}$ hollow cones made of recycled cardboard. Each cone has a base diameter of $\text{40 cm}$ and height $\text{1 m}$. If the outer side of each of the cones is to be painted and the cost of painting is $\text{Rs}\text{. 12}$ per ${{\text{m}}^{\text{2}}}$, what will be the cost of painting all these cones? 

$\left[ \text{Use  }\pi\text{  = 3}\text{.14 and take }\sqrt{\text{1}\text{.02}}\text{=1}\text{.02} \right]$

Ans: We are given the following:

The base radius $\left( \text{r} \right)$ of cone $\text{= }\dfrac{\text{40}}{\text{2}}\text{ = 20 cm = 0}\text{.2 m}$

The height $\left( \text{h} \right)$ of cone $\text{= 1 m}$

(Image will be Uploaded Soon)

So the slant height of the cone, $\text{l = }\sqrt{{{\text{r}}^{\text{2}}}\text{ + }{{\text{h}}^{\text{2}}}}$ 

$\Rightarrow \text{l = }\left( \sqrt{{{\left( \text{0}\text{.2} \right)}^{\text{2}}}\text{ + }{{\left( \text{1} \right)}^{\text{2}}}} \right)\text{ m}$

$\Rightarrow \text{l = }\left( \sqrt{1.04} \right)\text{ m}$

$\Rightarrow \text{l = 1}\text{.02 m}$

The curved surface area of one cone, $\text{A =  }\pi\text{ rl}$

$\Rightarrow \text{A = }\left( \text{3}\text{.14  }\times\text{  0}\text{.2  }\times\text{  1}\text{.02} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 0}\text{.64056 c}{{\text{m}}^{\text{2}}}$

So, the curved surface area of $\text{50}$ cones $\text{= }\left( \text{50  }\times\text{  0}\text{.64056} \right)\text{ }{{\text{m}}^{\text{2}}}\text{ = 32}\text{.028 }{{\text{m}}^{\text{2}}}$

It is given that the cost of painting $\text{1 }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 12}$

So, the cost of painting $32.028\text{ }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \text{32}\text{.028  }\times\text{  12} \right)\text{ = Rs}\text{. 384}\text{.336}$

We can also write the cost approximately as $\text{Rs}\text{. 384}\text{.34}$.

Therefore, the cost of painting all the hollow cones is $\text{Rs}\text{. 384}\text{.34}$.

Exercise (13.4)

1. Find the surface area of a sphere of radius:

(i) $\text{10}\text{.5 cm}$

Ans: Given radius of the sphere $\text{r = 10}\text{.5 cm}$

The surface area of the sphere $\text{A = 4 }\pi\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{10}\text{.5} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{88  }\times\text{  1}\text{.5  }\times\text{  1}\text{.5} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 1386 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{1386 c}{{\text{m}}^{\text{2}}}$.

(ii) $\text{5}\text{.6 cm}$

Ans: Given radius of the sphere $\text{r = 5}\text{.6 cm}$

The surface area of the sphere $\text{A = 4 }\pi\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{5}\text{.6} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{88  }\times\text{  0}\text{.8  }\times\text{  5}\text{.6} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 394}\text{.24 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{394}\text{.24 c}{{\text{m}}^{\text{2}}}$.

(iii) $\text{14 cm}$ $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

Ans: Given radius of the sphere $\text{r = 14 cm}$

The surface area of the sphere $\text{A = 4 }\pi\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{14} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{4  }\times\text{  44  }\times\text{  14} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 2464 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{2464 c}{{\text{m}}^{\text{2}}}$.

2. Find the surface area of a sphere of diameter:

(i) $\text{14 cm}$

Ans: Given diameter of the sphere $\text{= 14 cm}$

So, the radius of the sphere $\text{r = }\dfrac{\text{14}}{\text{2}}\text{ = 7 cm}$

The surface area of the sphere $\text{A = 4 }\pi\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{7} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{88  }\times\text{  7} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 616 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{616 c}{{\text{m}}^{\text{2}}}$.

(ii) $\text{21 cm}$

Ans: Given diameter of the sphere $\text{= 21 cm}$

So, the radius of the sphere $\text{r = }\dfrac{\text{21}}{\text{2}}\text{ = 10}\text{.5 cm}$

The surface area of the sphere $\text{A = 4 }\pi\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{10}\text{.5} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 1386 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{1386 c}{{\text{m}}^{\text{2}}}$.

(iii) $\text{3}\text{.5 m}$ $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

Ans: Given diameter of the sphere $\text{= 3}\text{.5 m}$

So, the radius of the sphere $\text{r = }\dfrac{\text{3}\text{.5}}{\text{2}}\text{ = 1}\text{.75 m}$

The surface area of the sphere $\text{A = 4 }\pi\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{1}\text{.75} \right)}^{\text{2}}} \right]\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 38}\text{.5 }{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{38}\text{.5 }{{\text{m}}^{\text{2}}}$.

3. Find the total surface area of a hemisphere of radius $\text{10 cm}$. $\left[ \text{Use  }\pi\text{  = 3}\text{.14} \right]$

Ans:

(Image will be Uploaded Soon)

Given the radius of hemisphere $\text{r = 10 cm}$

The total surface area of the hemisphere is the sum of its curved surface area and the circular base.

Total surface area of hemisphere $\text{A = 2 }\pi\text{ }{{\text{r}}^{\text{2}}}\text{ +  }\pi\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = 3 }\pi\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{3  }\times\text{  3}\text{.14  }\times\text{  }{{\left( \text{10} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 942 c}{{\text{m}}^{\text{2}}}$

Hence, the total surface area of the hemisphere is $\text{942 c}{{\text{m}}^{\text{2}}}$.

4. The radius of a spherical balloon increases from $\text{7 cm}$ to $\text{14 cm}$ as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Ans: Given the initial radius of the balloon ${{\text{r}}_{1}}\text{ = 10 cm}$

The final radius of the balloon ${{\text{r}}_{2}}\text{ = 14 cm}$

We have to find the ratio of surface areas of the balloon in the two cases.

The required ratio $\text{R = }\dfrac{\text{4 }\pi\text{ }{{\text{r}}_{\text{1}}}^{\text{2}}}{\text{4 }\pi\text{ }{{\text{r}}_{\text{2}}}^{\text{2}}}$

$\Rightarrow \text{R = }{{\left( \dfrac{{{\text{r}}_{\text{1}}}}{{{\text{r}}_{\text{2}}}} \right)}^{\text{2}}}$

$\Rightarrow \text{R = }{{\left( \dfrac{\text{7}}{\text{14}} \right)}^{\text{2}}}$

$\Rightarrow \text{R = }\dfrac{\text{1}}{\text{4}}$

Hence, the ratio of the surface areas of the balloon in both case is $\text{1 : 4}$.

5. A hemispherical bowl made of brass has inner diameter $\text{10}\text{.5 cm}$. Find the cost of tinplating it on the inside at the rate of $\text{Rs}\text{. 16}$ per $\text{100 c}{{\text{m}}^{\text{2}}}$. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

Ans: Given the radius of inner hemispherical bowl $\text{r = }\dfrac{\text{10}\text{.5}}{\text{2}}\text{ = 5}\text{.25 cm}$

(Image will be Uploaded Soon)

The surface area of the hemispherical bowl $\text{A = 2 }\pi\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{2  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{5}\text{.25} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 173}\text{.25 c}{{\text{m}}^{\text{2}}}$

It is given that the cost of tin-plating $\text{100 c}{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 16}$

So, the cost of tin-plating $173.25\text{ c}{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \dfrac{\text{16}}{\text{100}}\text{  }\times\text{  173}\text{.25} \right)\text{ = Rs}\text{. 27}\text{.72}$

Hence, the cost of tin-plating the hemispherical bowl is $\text{Rs}\text{. 27}\text{.72}$.

6. Find the radius of a sphere whose surface area is $\text{154 c}{{\text{m}}^{\text{2}}}$. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

Ans: Let us assume the radius of sphere be $\text{r}$.

We are given the surface area of the sphere, $\text{A = 154 c}{{\text{m}}^{\text{2}}}$.

$\therefore \text{4 }\pi\text{ }{{\text{r}}^{\text{2}}}\text{ = 154 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{r}}^{\text{2}}}\text{ = }\left( \dfrac{\text{154  }\times\text{  7}}{\text{2  }\times\text{  22}} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{r = }\left( \dfrac{\text{7}}{\text{2}} \right)\text{ cm}$

$\Rightarrow \text{r = 3}\text{.5 cm}$

Therefore, the radius of the sphere is $\text{3}\text{.5 cm}$.

7. The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface area.

Ans: Let us assume the diameter of earth is $\text{d}$.

So, the diameter of the moon will be $\dfrac{\text{d}}{\text{4}}$.

The radius of the earth ${{\text{r}}_{\text{1}}}\text{ = }\dfrac{\text{d}}{\text{2}}$

The radius of the moon ${{\text{r}}_{\text{2}}}\text{ = }\dfrac{\text{1}}{\text{2}}\text{  }\times\text{  }\dfrac{\text{d}}{\text{2}}\text{ = }\dfrac{\text{d}}{\text{8}}$

The ratio of surface area of moon and earth $\text{R = }\dfrac{\text{4 }\pi\text{ }{{\text{r}}_{\text{2}}}^{\text{2}}}{\text{4 }\pi\text{ }{{\text{r}}_{\text{1}}}^{\text{2}}}$

$\Rightarrow \text{R = }\dfrac{\text{4 }\pi\text{ }{{\left( \dfrac{\text{d}}{\text{8}} \right)}^{\text{2}}}}{\text{4 }\pi\text{ }{{\left( \dfrac{\text{d}}{\text{2}} \right)}^{\text{2}}}}$

$\Rightarrow \text{R = }\dfrac{\text{4}}{\text{64}}$

$\Rightarrow \text{R = }\dfrac{\text{1}}{\text{16}}$

Therefore, the ratio of surface area of moon and earth is $\text{1 : 16}$.

8. A hemispherical bowl is made of steel, $\text{0}\text{.25 cm}$ thick. The inner radius of the bowl is $\text{5 cm}$. Find the outer curved surface area of the bowl. $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

Ans: Given the inner radius $\text{= 5 cm}$

The thickness of the bowl $\text{= 0}\text{.25 cm}$

(Image will be Uploaded Soon)

So, the outer radius of the hemispherical bowl is $\text{r = }\left( \text{5 + 0}\text{.25} \right)\text{ cm = 5}\text{.25 cm}$

The outer curved surface area of the hemispherical bowl $\text{A = 2 }\pi\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A =}\left[ \text{ 2  }\times\text{  }\dfrac{\text{2}}{\text{7}}\text{  }\times\text{  }{{\left( \text{5}\text{.25} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 173}\text{.25 c}{{\text{m}}^{\text{2}}}$

Therefore, the outer curved surface area of the hemispherical bowl is $\text{173}\text{.25 c}{{\text{m}}^{\text{2}}}$.

9. A right circular cylinder just encloses a sphere of radius $\text{r}$ (see figure). Find 

(Image will be Uploaded Soon)

(i) surface area of the sphere, 

Ans: The surface area of the sphere is $\text{4 }\pi\text{ }{{\text{r}}^{\text{2}}}$.

(ii) curved surface area of the cylinder, 

Ans:

(Image will be Uploaded Soon)

Given the radius of cylinder $\text{= r}$

The height of cylinder $\text{= r + r = 2r}$

The curved surface area of cylinder $\text{A = 2 }\pi\text{ rh}$

$\Rightarrow \text{A = 2 }\pi\text{ r }\left( \text{2r} \right)$

$\Rightarrow \text{A = 4 }\pi\text{ }{{\text{r}}^{\text{2}}}$

Therefore the curved surface area of cylinder is  $\text{4 }\pi\text{ }{{\text{r}}^{\text{2}}}$.

(iii) ratio of the areas obtained in (i) and (ii).

Ans: The ratio of surface area of the sphere and curved surface area of cylinder  $\text{R = }\dfrac{\text{4 }\pi\text{ }{{\text{r}}^{\text{2}}}}{\text{4 }\pi\text{ }{{\text{r}}^{\text{2}}}}$

$\text{R = }\dfrac{\text{1}}{\text{1}}$

Therefore, the required ratio is $\text{1 : 1}$.

Exercise (13.5)

1. A matchbox measures \[\text{4 cm  }\times\text{  2}\text{.5 cm  }\times\text{  1}\text{.5 cm}\]. What will be the volume of a packet containing $\text{12}$ such boxes?

Ans: We are given the following:

Length of matchbox $\text{l = 4 cm}$

Breadth of matchbox $\text{b = 2}\text{.5 cm}$

Height of matchbox $\text{h = 1}\text{.5 cm}$

(Image will be Uploaded Soon)

The volume of single matchbox $\text{V = l  }\times\text{  b  }\times\text{  h}$

$\Rightarrow \text{V = }\left( \text{4  }\times\text{  2}\text{.5  }\times\text{  1}\text{.5} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 15 c}{{\text{m}}^{\text{3}}}$

So, the volume of $\text{12}$ matchboxes $\text{= }\left( \text{15  }\times\text{  12} \right)\text{ c}{{\text{m}}^{\text{3}}}\text{ = 180 c}{{\text{m}}^{\text{3}}}$

Hence, the volume of $\text{12}$ matchboxes is $\text{180 c}{{\text{m}}^{\text{3}}}$.

2. A cuboidal water tank is $\text{6 m}$ long, $\text{5 m}$ wide and $\text{4}\text{.5 m}$ deep. How many litres of water can it hold? $\left( \text{1 }{{\text{m}}^{\text{3}}}\text{ = 1000 litre} \right)$

Ans: We are given the following:

Length of tank $\text{l = 6 m}$

Breadth of tank $\text{b = 5 m}$

Height of tank $\text{h = 4}\text{.5 m}$

(Image will be Uploaded Soon)

The volume of tank $\text{V = l  }\times\text{  b  }\times\text{  h}$

$\Rightarrow \text{V = }\left( \text{6  }\times\text{  5  }\times\text{  4}\text{.5} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 135 }{{\text{m}}^{\text{3}}}$

We know the amount of water with volume $\text{1 }{{\text{m}}^{\text{3}}}$ can hold $\text{= 1000 litres}$

So, the amount of water with volume $\text{135 }{{\text{m}}^{\text{3}}}$ can hold $\text{= }\left( \text{1000  }\times\text{  135} \right)\text{ litres = 135000 litres}$

Therefore, the tank can contain $\text{135000 litres}$ of water.

3. A cuboidal vessel is $\text{10 m}$ long and $\text{8 m}$ wide. How high must it be made to hold $\text{380}$ cubic metres of a liquid?

Ans: We are given the following:

Length of vessel $\text{l = 10 m}$

Width of vessel $\text{b = 8 m}$

Volume of vessel $\text{V = 380 }{{\text{m}}^{\text{3}}}$

Let $\text{h}$ be the height of the vessel.

(Image will be Uploaded Soon)

We know the volume of cuboidal vessel $\text{= l  }\times\text{  b  }\times\text{  h}$

$\therefore \text{l  }\times\text{  b  }\times\text{  h = 380 }{{\text{m}}^{\text{3}}}$

\[\Rightarrow \left[ \text{10  }\times\text{  8  }\times\text{  h} \right]\text{ }{{\text{m}}^{2}}\text{ = 380 }{{\text{m}}^{\text{3}}}\]

$\Rightarrow \text{h = 4}\text{.75 m}$

Therefore, the height of the vessel is $\text{4}\text{.75 m}$.

4. Find the cost of digging a cuboidal pit $\text{8 m}$ long, $\text{6 m}$ broad and $\text{3 m}$deep at the rate of $\text{Rs}\text{. 30}$ per ${{\text{m}}^{\text{3}}}$.

Ans: We are given the following:

Length of tank $\text{l = 8 m}$

Breadth of tank $\text{b = 6 m}$

Depth of tank $\text{h = 3 m}$

(Image will be Uploaded Soon)

The volume of tank $\text{V = l  }\times\text{  b  }\times\text{  h}$

$\Rightarrow \text{V = }\left( \text{8  }\times\text{  6  }\times\text{  3} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 144 }{{\text{m}}^{\text{3}}}$

It is given that the cost of digging per ${{\text{m}}^{\text{3}}}$ volume $\text{= Rs}\text{. 30}$

So, the cost of digging per $\text{144 }{{\text{m}}^{\text{3}}}$ volume $\text{= Rs}\text{. }\left( \text{30  }\times\text{  144} \right)\text{ = Rs}\text{. 4320}$

Therefore, the required cost of digging is $\text{Rs}\text{. 4320}$.

5. The capacity of a cuboidal tank is $\text{5000 litres}$ of water. Find the breadth of the tank, if its length and depth are respectively $\text{2}\text{.5 m}$ and $\text{10 m}$.

Ans: We are given the following:

Length of tank $\text{l = 2}\text{.5 m}$

Depth of tank $\text{h = 10 m}$

Let us assume the breadth of the tank to be $\text{b m}$.

(Image will be Uploaded Soon)

The volume of tank $\text{V = l  }\times\text{  b  }\times\text{  h}$

$\Rightarrow \text{V = }\left( \text{2}\text{.5  }\times\text{  b  }\times\text{  10} \right)\text{ }{{\text{m}}^{\text{3}}}$

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

We can write the capacity of the tank $\text{= 25b }{{\text{m}}^{\text{3}}}\text{ = 25000b litres}$

But it is given that the capacity of the tank is $\text{50000 litres}$

$\therefore \text{25000b = 50000}$

$\Rightarrow \text{b = 2}$

Therefore, the breadth of the cuboidal tank is $\text{2 m}$.

6. A village, having a population of $\text{4000}$, requires $\text{150 litres}$ of water per head per day. It has a tank measuring $\text{20 m  }\times\text{  15 m  }\times\text{  6 m}$. For how many days will the water of this tank last?

Ans: We are given the following:

Length of tank $\text{l = 20 m}$

Breadth of tank $\text{b = 15 m}$

Height of tank $\text{h = 6 m}$

(Image will be Uploaded Soon)

The capacity of tank $\text{V = l  }\times\text{  b  }\times\text{  h}$

$\Rightarrow \text{V = }\left( \text{20  }\times\text{  15  }\times\text{  6} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 1800 }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 1800000 litres}$

The total water consumed by the people in a day $\text{= }\left( \text{4000  }\times\text{  150} \right)\text{ litres = 600000 litres}$

Let us assume that the water in the tank will last for $\text{n}$ days.

$\therefore \text{n  }\times\text{  600000 = 1800000}$

$\Rightarrow \text{n = 3}$

Therefore, the water in the tank will last $\text{3}$ days.

7. A godown measures $\text{40 m  }\times\text{  25 m  }\times\text{  15 m}$. Find the maximum number of wooden crates each measuring $\text{1}\text{.5 m  }\times\text{  1}\text{.25 m  }\times\text{  0}\text{.5 m}$ that can be stored in the godown.

Ans: We are given the following:

Length of godown $\text{L= 40 m}$

Breadth of godown $\text{B = 25 m}$

Height of godown $\text{H = 15 m}$

(Image will be Uploaded Soon)

Length of wooden crate $\text{l = 1}\text{.5 m}$

Breadth of wooden crate $\text{b = 1}\text{.25 m}$

Height of wooden crate $\text{h = 0}\text{.5 m}$

The volume of godown ${{\text{V}}_{\text{g}}}\text{ = L  }\times\text{  B  }\times\text{  H}$

$\Rightarrow {{\text{V}}_{\text{g}}}\text{ = }\left( \text{40  }\times\text{  25  }\times\text{  15} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow {{\text{V}}_{\text{g}}}\text{ = 15000 }{{\text{m}}^{\text{3}}}$

The volume of a wooden crate ${{\text{V}}_{\text{c}}}\text{ = l  }\times\text{  b  }\times\text{  h}$

$\Rightarrow {{\text{V}}_{\text{c}}}\text{ = }\left( \text{1}\text{.5  }\times\text{  1}\text{.25  }\times\text{  0}\text{.5} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow {{\text{V}}_{\text{c}}}\text{ = 0}\text{.9375 }{{\text{m}}^{\text{3}}}$

Let us assume that $\text{n}$ wooden crates can be stored in the godown.

So, we can say that the volume of $\text{n}$ wooden crates will be the same as the volume of godown.

$\therefore \text{0}\text{.9375  }\times\text{  n = 15000}$

$\Rightarrow \text{n = 16000}$

Therefore, we can easily store $\text{16000}$ wooden crates in the godown.

8. A solid cube of side $\text{12 cm}$ is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

Ans: Given side of cube $\text{a = 12 cm}$

(Image will be Uploaded Soon)

Volume of this cube $\text{V = }{{\left( \text{a} \right)}^{\text{3}}}$

$\Rightarrow \text{V = }{{\left( \text{12 cm} \right)}^{\text{3}}}$

$\Rightarrow \text{V = 1728 c}{{\text{m}}^{\text{3}}}$

Let us assume the side of the smaller cube to be $\text{x}$.

So, the volume of smaller cube $\text{= }\dfrac{\text{1728}}{\text{8}}\text{ = 216 c}{{\text{m}}^{\text{3}}}$

$\therefore {{\text{x}}^{\text{3}}}\text{ = 216 c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{x = 6 cm}$

Hence, the side of the smaller cubes will be $\text{6 cm}$.

The ratio of the surface areas $\text{R = }\dfrac{\text{6}{{\text{a}}^{\text{2}}}}{\text{6}{{\text{x}}^{\text{2}}}}$

$\Rightarrow \text{R = }\dfrac{{{\left( \text{12} \right)}^{\text{2}}}}{{{\left( \text{6} \right)}^{\text{2}}}}$

$\Rightarrow \text{R = }\dfrac{\text{4}}{\text{1}}$

Therefore the ratio between the surface areas of the cubes is $\text{4 : 1}$.

9. A river $\text{3 m}$ deep and $\text{40 m}$ wide is flowing at the rate of $\text{3}$ km per hour. How much water will fall into the sea in a minute?

Ans: Given the rate of flow of water $\text{R = 2 km/hr}$

$\Rightarrow \text{R = }\left( \dfrac{\text{2  }\times\text{  1000}}{\text{60}} \right)\text{ m/min}$

$\Rightarrow \text{R = }\left( \dfrac{\text{100}}{3} \right)\text{ m/min}$

Given the depth of the river $\text{h = 3 m}$

Width of the river $\text{b = 40 m}$

The volume of water that will flow in \[\text{1 min}\] $\text{V = }\left( \dfrac{\text{100}}{\text{3}}\text{  }\times\text{  40  }\times\text{  3} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 4000 }{{\text{m}}^{\text{3}}}$

Therefore, $\text{4000 }{{\text{m}}^{\text{3}}}$ of water will fall into the sea in a minute.

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? $\left( \text{1000 c}{{\text{m}}^{\text{3}}}\text{ = 1 litre} \right)$. $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

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

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

(Image will be Uploaded Soon)

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

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

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

$\text{r = 21 cm}$

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

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

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

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

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

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 $\text{0}\text{.6 g}$. $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

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

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

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

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

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

(Image will be Uploaded Soon)

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

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

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

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

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 $\text{= }\left( \dfrac{\text{5720  }\times\text{  0}\text{.6}}{\text{1000}} \right)\text{ kg = 3}\text{.432 kg}$

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? $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: We will calculate for the cuboidal tin can.

(Image will be Uploaded Soon)

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 a circular base.

(Image will be Uploaded Soon)

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

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

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

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

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

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

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

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 the cylinder $\text{h = 5 cm}$.

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

(Image will be Uploaded Soon)

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 $\text{= 2 }\pi\text{ rh}$

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

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

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

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

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 $\left( \dfrac{\text{1}}{\text{20}}\text{  }\times\text{  2200} \right)\text{ }{{\text{m}}^{\text{2}}}\text{ = 110 }{{\text{m}}^{\text{2}}}$

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 $\text{h = 10 m}$.

(Image will be Uploaded Soon)

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

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

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

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

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

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

(iii) Capacity of the vessel

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

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

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

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

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

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? $\left[ \text{Assume }\pi=\dfrac{\text{22}}{\text{7}} \right]$

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

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

(Image will be Uploaded Soon)

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

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

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

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

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

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

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

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

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. $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans:

(Image will be Uploaded Soon)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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? $\left[\text{Assume  }\pi\text{  = }\dfrac{\text{22}}{\text{7}} \right]$

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

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

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

(Image will be Uploaded Soon)

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

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

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

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

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

Exercise (13.7)

1. Find the volume of the right circular cone with

(i) Radius $\text{6 cm}$, height $\text{7 cm}$

Ans: It is given the radius of cone $\text{r = 6 cm}$

The height of the cone $\text{h = 7 cm}$

The volume of the cone $\text{V = }\dfrac{\text{1}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\left[ \dfrac{\text{1}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{6} \right)}^{\text{2}}}\text{  }\times\text{  7} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left( \text{12  }\times\text{  22} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 264 c}{{\text{m}}^{\text{3}}}$

The volume of the right circular cone is $\text{264 c}{{\text{m}}^{\text{3}}}$.

(ii) Radius $\text{3}\text{.5 cm}$, height $\text{12 cm}$ $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: It is given the radius of cone $\text{r = 3}\text{.5 cm}$

The height of the cone $\text{h = 12 cm}$

The volume of the cone $\text{V = }\dfrac{\text{1}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\left[ \dfrac{\text{1}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{3}\text{.5} \right)}^{\text{2}}}\text{  }\times\text{  12} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left( \text{1}\text{.75  }\times\text{  88} \right)\text{ c}{{\text{m}}^{\text{3}}}$

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

The volume of the right circular cone is $\text{154 c}{{\text{m}}^{\text{3}}}$.

2. Find the capacity in litres of a conical vessel with

(i) Radius $\text{7 cm}$, slant height $\text{25 cm}$

Ans: It is given the radius of cone $\text{r = 7 cm}$

The slant height of the cone $\text{l = 25 cm}$

(Image will be Uploaded Soon)

So, the height of the cone $\text{h = }\sqrt{{{\text{l}}^{\text{2}}}\text{ - }{{\text{r}}^{\text{2}}}}$

$\Rightarrow \text{h = }\sqrt{\text{2}{{\text{5}}^{\text{2}}}\text{ - }{{\text{7}}^{\text{2}}}}\text{ cm}$

$\Rightarrow \text{h = 24 cm}$

The volume of the cone $\text{V = }\dfrac{\text{1}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\left[ \dfrac{\text{1}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{7} \right)}^{\text{2}}}\text{  }\times\text{  24} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left( \text{154  }\times\text{  8} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 1232 c}{{\text{m}}^{\text{3}}}$

We know that $\text{1000 c}{{\text{m}}^{\text{3}}}\text{ = 1 litre}$

So, the capacity of the conical vessel $\text{= }\dfrac{\text{1232}}{\text{1000}}\text{ = 1}\text{.232 litres}$

Therefore, the capacity of the conical vessel is $\text{1}\text{.232 litres}$.

(ii) height $\text{12 cm}$, slant height $\text{13 cm}$ $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: It is given the height of cone $\text{h = 12 cm}$

The slant height of the cone $\text{l = 13 cm}$

(Image will be Uploaded Soon)

So, the radius of the cone $\text{r = }\sqrt{{{\text{l}}^{\text{2}}}\text{ - }{{\text{h}}^{\text{2}}}}$

$\Rightarrow \text{r = }\sqrt{\text{1}{{\text{3}}^{\text{2}}}\text{ - 1}{{\text{2}}^{\text{2}}}}\text{ cm}$

$\Rightarrow \text{r = 5 cm}$

The volume of the cone $\text{V = }\dfrac{\text{1}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\left[ \dfrac{\text{1}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{5} \right)}^{\text{2}}}\text{  }\times\text{  12} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left( \text{4  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  25} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\dfrac{2200}{7}\text{ c}{{\text{m}}^{\text{3}}}$

We know that $\text{1000 c}{{\text{m}}^{\text{3}}}\text{ = 1 litre}$

So, the capacity of the conical vessel $\text{= }\dfrac{\text{2200}}{\text{7}}\text{  }\times\text{  }\dfrac{\text{1}}{\text{1000}}\text{ = 0}\text{.314 litres}$

Therefore, the capacity of the conical vessel is $\text{0}\text{.314 litres}$.

3. The height of a cone is $\text{15 cm}$. It its volume is $\text{1570 c}{{\text{m}}^{\text{3}}}$, find the diameter of its base. $\left[ \text{Use  }\pi\text{  = 3}\text{.14} \right]$

Ans: It is given the height of cone $\text{h = 12 cm}$

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

(Image will be Uploaded Soon)

The volume of the cone is $\text{V = 1570 c}{{\text{m}}^{\text{3}}}$

We know the formula for the volume of the cone $\text{= }\dfrac{\text{1}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\therefore \dfrac{\text{1}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{2}}}\text{h = 1570 c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \left[ \dfrac{\text{1}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{r} \right)}^{\text{2}}}\text{  }\times\text{  12} \right]\text{ cm = 1570 c}{{\text{m}}^{\text{3}}}$

$\Rightarrow {{\text{r}}^{\text{2}}}\text{ = 100 c}{{\text{m}}^{\text{2}}}$

\[\Rightarrow \text{r = 10 cm}\]

Diameter of base \[\text{= 2r = 20 cm}\]

Therefore, the diameter of the cone is  \[\text{20 cm}\].

4. If the volume of right circular cone of height $\text{9 cm}$ is $\text{48 }\pi\text{  c}{{\text{m}}^{\text{3}}}$, find the diameter of its base.

Ans: It is given the height of cone $\text{h = 9 cm}$

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

(Image will be Uploaded Soon)

The volume of the cone is $\text{V = 48 }\pi\text{  c}{{\text{m}}^{\text{3}}}$

We know the formula for the volume of the cone $\text{= }\dfrac{\text{1}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\therefore \dfrac{\text{1}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{2}}}\text{h = 48 }\pi\text{  c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \left[ \dfrac{\text{1}}{\text{3}}\text{  }\times\text{   }\pi\text{   }\times\text{  }{{\left( \text{r} \right)}^{\text{2}}}\text{  }\times\text{  9} \right]\text{ cm = 48 }\pi\text{  c}{{\text{m}}^{\text{3}}}$

$\Rightarrow {{\text{r}}^{\text{2}}}\text{ = 16 c}{{\text{m}}^{\text{2}}}$

\[\Rightarrow \text{r = 4 cm}\]

Diameter of base \[\text{= 2r = 8 cm}\]

Therefore, the diameter of the base of the cone is \[\text{8 cm}\].

5. A conical pit of top diameter $\text{3}\text{.5 m}$ is $\text{12 m}$ deep. What is the capacity in kilolitres? $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: It is given the height of conical pit $\text{h = 12 m}$

The radius of conical pit $\text{r = }\dfrac{\text{3}\text{.5}}{\text{2}}\text{ m = 1}\text{.75 m}$

(Image will be Uploaded Soon)

We know the volume of the conical pit $\text{V = }\dfrac{\text{1}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\left[ \dfrac{\text{1}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{1}\text{.75} \right)}^{\text{2}}}\text{  }\times\text{  12} \right]\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 38}\text{.5 }{{\text{m}}^{\text{3}}}$

We know that $\text{1 kilolitre = 1 }{{\text{m}}^{\text{3}}}$

So, the capacity of the pit $\text{= }\left( \text{38}\text{.5  }\times\text{  1} \right)\text{ kilolitres = 38}\text{.5 kilolitres}$

Therefore, the capacity of the conical pit is $\text{38}\text{.5 kilolitres}$.

6. The volume of a right circular cone is $\text{9856 c}{{\text{m}}^{\text{3}}}$. If the diameter of the base is $\text{28 cm}$, find

(i) Height of the cone

Ans: It is given the diameter of base of cone $\text{= 28 cm}$

So, the radius $\text{r = }\dfrac{\text{28}}{\text{2}}\text{ = 14 cm}$

Let us assume the height of the cone be $\text{h}$.

(Image will be Uploaded Soon)

The volume of the cone is $\text{V = 9856 c}{{\text{m}}^{\text{3}}}$

We know the formula for the volume of the cone $\text{= }\dfrac{\text{1}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\therefore \dfrac{\text{1}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{2}}}\text{h = 9856 c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \left[ \dfrac{\text{1}}{\text{3}}\text{  }\times\text{  }\dfrac{22}{7}\text{  }\times\text{  }{{\left( \text{14} \right)}^{\text{2}}}\text{  }\times\text{  h} \right]\text{ c}{{\text{m}}^{\text{2}}}\text{ = 9856 c}{{\text{m}}^{\text{3}}}$

\[\Rightarrow \text{h = }\left( \dfrac{\text{9856  }\times\text{  21}}{\text{22  }\times\text{  196}} \right)\text{ cm}\]

\[\Rightarrow \text{h = 48 cm}\]

Therefore, the height of the cone is \[\text{48 cm}\].

(ii) Slant height of the cone

Ans: The slant height of the cone $\text{l = }\sqrt{{{\text{h}}^{\text{2}}}\text{ + }{{\text{r}}^{\text{2}}}}$

$\Rightarrow \text{l = }\sqrt{\text{4}{{\text{8}}^{\text{2}}}\text{ + 1}{{\text{4}}^{\text{2}}}}\text{ cm}$

$\Rightarrow \text{l = }\sqrt{\text{2304 + 196}}\text{ cm}$

$\Rightarrow \text{l = 50 cm}$

Therefore, the slant height of the cone is $\text{50 cm}$.

(iii) Curved surface area of the cone. $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: The curved surface area of the cone $\text{A =  }\pi\text{ rl}$

$\Rightarrow \text{A = }\left( \dfrac{\text{22}}{\text{7}}\text{  }\times\text{  14  }\times\text{  50} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 2200 c}{{\text{m}}^{\text{2}}}$

Therefore, the curved surface area of the cone is $\text{2200 c}{{\text{m}}^{\text{2}}}$.

7. A right triangle $\text{ }\Delta\text{ ABC}$ with sides $\text{5 cm}$,$\text{12 cm}$ and \[\text{13 cm}\] is revolved about the side $\text{12 cm}$. Find the volume of the solid so obtained.

Ans: We will draw the given figure.

(Image will be Uploaded Soon)

If the triangle is revolved about the side $\text{12 cm}$, we will get a cone with:

Radius $\text{r = 5 cm}$

Slant height $\text{l = 13 cm}$

Height $\text{h = 12 cm}$

We know the volume of the cone $\text{V = }\dfrac{\text{1}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\left[ \dfrac{\text{1}}{\text{3}}\text{  }\times\text{   }\pi\text{   }\times\text{  }{{\left( \text{5} \right)}^{\text{2}}}\text{  }\times\text{  12} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 100 }\pi\text{  c}{{\text{m}}^{\text{3}}}$

Therefore, the volume of the cone will be $\text{100 }\pi\text{  c}{{\text{m}}^{\text{3}}}$.

8. If the triangle $\text{ }\Delta\text{ ABC}$ in the Question $\text{7}$ above is revolved about the side $\text{5 cm}$, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions $\text{7}$ and $\text{8}$.

Ans:

(Image will be Uploaded Soon)

If the triangle is revolved about the side $\text{5 cm}$, we will get a cone with:

Radius $\text{r = 12 cm}$

Slant height $\text{l = 13 cm}$

Height $\text{h = 5 cm}$

We know the volume of the cone $\text{V = }\dfrac{\text{1}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\left[ \dfrac{\text{1}}{\text{3}}\text{  }\times\text{   }\pi\text{   }\times\text{  }{{\left( \text{12} \right)}^{\text{2}}}\text{  }\times\text{  5} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 240 }\pi\text{  c}{{\text{m}}^{\text{3}}}$

Therefore, the volume of the cone will be $\text{240 }\pi\text{  c}{{\text{m}}^{\text{3}}}$.

The ratio of volume of cone from previous question an the one we obtained above $\text{= }\dfrac{\text{100 }\pi\text{ }}{\text{240 }\pi\text{ }}\text{ = }\dfrac{\text{5}}{\text{12}}\text{ = 5 : 12}$

Therefore, the required ratio is $\text{5 : 12}$.

9. A heap of wheat is in the form of a cone whose diameter is $\text{10}\text{.5 m}$and height is $\text{3 m}$. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

$\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: It is given that diameter of the heap $\text{= 10}\text{.5 m}$

So, the radius of heap $\text{r = }\dfrac{\text{10}\text{.5}}{\text{2}}\text{ = 5}\text{.25 m}$

Height of heap $\text{h = 3 m}$

(Image will be Uploaded Soon)

We know the volume of the cone $\text{V = }\dfrac{\text{1}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\left[ \dfrac{\text{1}}{\text{3}}\text{  }\times\text{  }\dfrac{22}{7}\text{  }\times\text{  }{{\left( \text{5}\text{.25} \right)}^{\text{2}}}\text{  }\times\text{  3} \right]\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 86}\text{.625 }{{\text{m}}^{\text{3}}}$

Hence, the volume of heap is $\text{86}\text{.625 }{{\text{m}}^{\text{3}}}$.

The area of canvas required is same as curved surface area of the cone.

$\therefore \text{A =  }\pi\text{ rl}$

$\Rightarrow \text{A =  }\pi\text{ r}\sqrt{{{\text{h}}^{\text{2}}}\text{ + }{{\text{r}}^{\text{2}}}}$

$\Rightarrow \text{A = }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  5}\text{.25  }\times\text{  }\sqrt{{{\left( \text{3} \right)}^{\text{2}}}\text{ + }{{\left( \text{5}\text{.25} \right)}^{\text{2}}}}\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \dfrac{\text{22}}{\text{7}}\text{  }\times\text{  5}\text{.25  }\times\text{  6}\text{.05} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 99}\text{.825 }{{\text{m}}^{\text{2}}}$

Therefore, to protect the heap from the rain, the amount of canvas required is $\text{99}\text{.825 }{{\text{m}}^{\text{2}}}$.

Exercise (13.8)

1. Find the volume of the sphere whose radius is

(i) $\text{7 cm}$

Ans: It is given the radius of sphere $\text{r = 7 cm}$

The volume of the sphere $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V =}\left[ \text{ }\dfrac{\text{4}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{7} \right)}^{\text{3}}} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\dfrac{\text{4312}}{\text{3}}\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 1437}\text{.33 c}{{\text{m}}^{\text{3}}}$

Therefore, the volume of the sphere is $\text{1437}\text{.33 c}{{\text{m}}^{\text{3}}}$.

(ii) $\text{0}\text{.63 m}$ $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: It is given the radius of sphere $\text{r = 0}\text{.63 m}$

The volume of the sphere $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V =}\left[ \text{ }\dfrac{\text{4}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{0}\text{.63} \right)}^{\text{3}}} \right]\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 1}\text{.0478 }{{\text{m}}^{\text{3}}}$

Therefore, the volume of the sphere is $\text{1}\text{.0478 }{{\text{m}}^{\text{3}}}$.

2. Find the amount of water displaced by a solid spherical ball of diameter

(i) $\text{28 cm}$

Ans: It is given the diameter of ball $\text{= 28 cm}$

So, the radius of ball $\text{r = }\dfrac{\text{28}}{\text{2}}\text{ = 14 cm}$

The volume of the ball $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V =}\left[ \text{ }\dfrac{\text{4}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{14} \right)}^{\text{3}}} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 11498 c}{{\text{m}}^{\text{3}}}$

Therefore, volume of the sphere is $\text{11498 c}{{\text{m}}^{\text{3}}}$.

(ii) $\text{0}\text{.21 m}$ $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: It is given the diameter of ball $\text{= 0}\text{.21 m}$

So, the radius of ball $\text{r = }\dfrac{\text{0}\text{.21}}{\text{2}}\text{ = 0}\text{.105 m}$

The volume of the sphere $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V =}\left[ \text{ }\dfrac{\text{4}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{0}\text{.105} \right)}^{\text{3}}} \right]\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 0}\text{.004851 }{{\text{m}}^{\text{3}}}$

Therefore, the volume of the sphere is $\text{0}\text{.004851 }{{\text{m}}^{\text{3}}}$.

3. The diameter of a metallic ball is $\text{4}\text{.2 cm}$. What is the mass of the ball, if the density of the metal is $\text{8}\text{.9 g per c}{{\text{m}}^{\text{3}}}$? $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: It is given the diameter of metallic ball $\text{= 4}\text{.2 cm}$

(Image will be Uploaded Soon)

So, the radius of ball $\text{r = }\dfrac{\text{4}\text{.2}}{\text{2}}\text{ = 2}\text{.1 cm}$

The volume of the sphere $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

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

$\Rightarrow \text{V = 38}\text{.808 c}{{\text{m}}^{\text{3}}}$

We know that $\text{Density = }\dfrac{\text{Mass}}{\text{Volume}}$

$\Rightarrow \text{Mass = Density  }\times\text{  Volume}$

$\Rightarrow \text{Mass = }\left( \text{8}\text{.9  }\times\text{  38}\text{.808} \right)\text{ g}$

$\Rightarrow \text{Mass = 345}\text{.39 g}$

Therefore, the mass of the metallic ball is $\text{345}\text{.39 g}$.

4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Ans: Let us assume the diameter of earth be $\text{d}$.

So, the radius of earth will be $\text{R = }\dfrac{\text{d}}{\text{2}}$.

From the question, we can write the diameter of the moon as $\dfrac{\text{d}}{\text{4}}$.

So, the radius of moon will be $\text{r = }\dfrac{\text{d}}{\text{8}}$.

The volume of earth $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{R}}^{\text{3}}}$

$\Rightarrow \text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\left( \dfrac{\text{d}}{\text{2}} \right)}^{\text{3}}}$

$\Rightarrow \text{V = }\dfrac{\text{1}}{8}\text{  }\times\text{  }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{d}}^{\text{3}}}$

The volume of moon $\text{{V}' = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{{V}' = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\left( \dfrac{\text{d}}{\text{8}} \right)}^{\text{3}}}$

$\Rightarrow \text{{V}' = }\dfrac{\text{1}}{\text{512}}\text{  }\times\text{  }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{d}}^{\text{3}}}$

The ratio of volume of moon and that of earth $\text{= }\dfrac{\dfrac{\text{1}}{\text{512}}\text{  }\times\text{  }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{d}}^{\text{3}}}}{\dfrac{\text{1}}{\text{8}}\text{  }\times\text{  }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{d}}^{\text{3}}}}\text{ = }\dfrac{\text{1}}{\text{64}}$

So, $\dfrac{\text{Volume of moon}}{\text{Volume of earth}}\text{=}\dfrac{\text{1}}{\text{64}}$

$\Rightarrow \text{Volume of moon = }\left( \dfrac{\text{1}}{\text{64}} \right)\text{ Volume of earth}$

Therefore, the volume of moon is $\dfrac{\text{1}}{\text{64}}$ times the volume of earth.

5. How many litres of milk can a hemispherical bowl of diameter $\text{10}\text{.5 cm}$ can hold? $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: It is given the diameter of the hemispherical bowl $\text{= 10}\text{.5 cm}$.

(Image will be Uploaded Soon)

So, the radius of the bowl $\text{r = }\dfrac{\text{10}\text{.5}}{\text{2}}\text{ = 5}\text{.25 cm}$.

The volume of the hemispherical bowl $\text{V = }\dfrac{\text{2}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V =}\left[ \text{ }\dfrac{\text{2}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{5}\text{.25} \right)}^{\text{3}}} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 303}\text{.1875 c}{{\text{m}}^{\text{3}}}$

We know that $\text{1000 c}{{\text{m}}^{\text{3}}}\text{ = 1 litre}$

So, the capacity of the bowl $\text{= }\dfrac{\text{303}\text{.1875}}{\text{1000}}\text{ = 0}\text{.303 litre}$

Therefore, the volume of the hemispherical bowl is $\text{0}\text{.303 litre}$.

6. A hemispherical tank is made up of an iron sheet $\text{1 cm}$thick. If the inner radius is $\text{1 m}$, then find the volume of the iron used to make the tank. $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: The inner radius of hemispherical tank $\text{r = 1 m}$

The thickness of iron sheet $\text{= 1 cm = 0}\text{.01 m}$.

(Image will be Uploaded Soon)

So, the outer radius of the hemispherical tank $\text{R = }\left( \text{1 + 0}\text{.01} \right)\text{ = 1}\text{.01 m}$

The volume of iron sheet required to make the tank $\text{V = }\dfrac{\text{2}}{\text{3}}\text{ }\pi\text{ }\left( {{\text{R}}^{\text{3}}}\text{ - }{{\text{r}}^{\text{3}}} \right)$

$\Rightarrow \text{V = }\dfrac{\text{2}}{\text{3}}\times \dfrac{22}{7}\times \left( {{\left( 1.01 \right)}^{\text{3}}}\text{ - }{{\left( 1 \right)}^{\text{3}}} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\dfrac{\text{44}}{\text{21}}\text{  }\times\text{  }\left( \text{1}\text{.030301 - 1} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 0}\text{.06348 }{{\text{m}}^{\text{3}}}$

Therefore, the volume of iron sheet required to make the hemispherical tank is $\text{0}\text{.06348 }{{\text{m}}^{\text{3}}}$.

7. Find the volume of a sphere whose surface area is $\text{154 c}{{\text{m}}^{\text{2}}}$. $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: Let us assume the radius of the sphere be $\text{r}$.

It is given the surface area of the sphere $\text{= 154 c}{{\text{m}}^{\text{2}}}$.

$\therefore \text{4 }\pi\text{ }{{\text{r}}^{\text{2}}}\text{ = 154 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{r}}^{\text{2}}}\text{ = }\left( \dfrac{\text{154  }\times\text{  7}}{\text{4  }\times\text{  22}} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{r}}^{\text{2}}}\text{ = }\left( \dfrac{\text{49}}{\text{4}} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{r = }\left( \dfrac{\text{7}}{\text{2}} \right)\text{ cm}$

The volume of the sphere $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

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

$\Rightarrow \text{V = }\left[ \dfrac{\text{49  }\text{ }\times\text{ }\text{  11}}{\text{3}} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 179}\text{.67 c}{{\text{m}}^{\text{3}}}$

Therefore, the volume of the sphere is $\text{179}\text{.67 c}{{\text{m}}^{\text{3}}}$.

8. A dome of a building is in the form of a hemisphere. From inside, it was whitewashed at the cost of $\text{Rs}\text{. 498}\text{.96}$. If the cost of white-washing is $\text{Rs}\text{. 2}\text{.00}$per square meter, find the

(i) Inside surface area of the dome,

Ans: It is given that it costs $\text{Rs}\text{. 2}\text{.00}$ to whitewash an area $\text{= 1 }{{\text{m}}^{\text{2}}}$

So, it costs $\text{Rs}\text{. 498}\text{.96}$ to whitewash an area $\text{= }\dfrac{\text{498}\text{.96}}{\text{2}}\text{ }{{\text{m}}^{\text{2}}}\text{ = 249}\text{.48 }{{\text{m}}^{\text{2}}}$.

Therefore, the inner surface area of the dome is $\text{249}\text{.48 }{{\text{m}}^{\text{2}}}$.

(ii) Volume of the air inside the dome. $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: Let us assume the radius of the hemispherical dome be $\text{r}$.

We obtained the curved surface area of the inner dome $\text{= 249}\text{.48 }{{\text{m}}^{\text{2}}}$

$\therefore \text{2}{{\text{ }\pi\text{ }}^{\text{2}}}\text{ = 249}\text{.48 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{2  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\text{r}}^{\text{2}}}\text{ = 249}\text{.48 }{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{r}}^{\text{2}}}\text{ = }\left( \dfrac{\text{249}\text{.48  }\times\text{  7}}{\text{2  }\times\text{  22}} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{r}}^{\text{2}}}\text{ = 39}\text{.69 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{r = 6}\text{.3 m}$

Volume of hemispherical dome $\text{V = }\dfrac{\text{2}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V = }\left[ \dfrac{\text{2}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{6}\text{.3} \right)}^{\text{3}}} \right]\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 523}\text{.908 }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 523}\text{.9  }{{\text{m}}^{\text{3}}}\left( \text{approximately} \right)$

Therefore, the volume of air inside the hemispherical dome is $\text{523}\text{.9 }{{\text{m}}^{\text{3}}}$.

9. Twenty-seven solid iron spheres, each of radius $\text{r}$ and surface area $\text{S}$ are melted to form a sphere with surface area $\text{{S}'}$. Find the 

(i) radius $\text{{r}'}$ of the new sphere, 

Ans: It is given the radius of one iron sphere $\text{= r}$.

The volume of one iron sphere $\text{= }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

So, the volume of $\text{27}$ iron spheres $\text{= 27  }\times\text{  }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

These spheres are melted to form one big sphere.

Let us assume the radius of this new sphere be $\text{{r}'}$.

The volume of new iron sphere $\text{= }\dfrac{\text{4}}{\text{3}}\text{  }\pi\text{ }{{\text{{r}'}}^{\text{3}}}$

We can now equate the volumes.

$\Rightarrow \dfrac{\text{4}}{\text{3}}\text{  }\pi\text{ }{{\text{{r}'}}^{\text{3}}}\text{ = 27  }\times\text{  }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow {{\text{{r}'}}^{\text{3}}}\text{ = 27}{{\text{r}}^{\text{3}}}$

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

Therefore, the radius of the new sphere is $\text{3r}$.

(ii) ratio of $\text{S}$ and $\text{{S}'}$. 

Ans: The surface area of an iron sphere of $\text{r}$is $\text{S = 4 }\pi\text{ }{{\text{r}}^{\text{2}}}$.

The surface area of an iron sphere of $\text{{r}'}$is $\text{{S}' = 4 }\pi\text{ }{{\text{{r}'}}^{\text{2}}}$.

$\Rightarrow \text{{S}' = 4 }\pi\text{ }{{\left( \text{3r} \right)}^{\text{2}}}$

$\Rightarrow \text{{S}' = 36 }\pi\text{ }{{\text{r}}^{\text{2}}}$

The ratio of $\dfrac{\text{S}}{{\text{{S}'}}}\text{ = }\dfrac{\text{4 }\pi\text{ }{{\text{r}}^{\text{2}}}}{\text{36 }\pi\text{ }{{\text{r}}^{\text{2}}}}\text{ = }\dfrac{\text{1}}{\text{9}}\text{ = 1 : 9}$

Therefore, the required ratio is $\text{1 : 9}$.

10. A capsule of medicine is in the shape of a sphere of diameter $\text{3}\text{.5 mm}$. How much medicine $\left( \text{in m}{{\text{m}}^{\text{3}}} \right)$ is needed to fill this capsule? $\left[ \text{Assume  }\pi\text{  =}\dfrac{\text{22}}{\text{7}} \right]$

Ans: It is given that the diameter of the capsule $\text{= 3}\text{.5 mm}$.

So, the radius will be $\text{r = }\left( \dfrac{\text{3}\text{.5}}{\text{2}} \right)\text{ = 1}\text{.75 mm}$.

(Image will be Uploaded Soon)

Volume of spherical capsule $\text{V = }\dfrac{\text{4}}{\text{3}}\text{ }\pi\text{ }{{\text{r}}^{\text{3}}}$

$\Rightarrow \text{V = }\left[ \dfrac{\text{4}}{\text{3}}\text{  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{1}\text{.75} \right)}^{\text{3}}} \right]\text{ m}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 22}\text{.458 m}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 22}\text{.46 m}{{\text{m}}^{\text{3}}}\left( \text{approx} \right)$

Hence, the amount of medicine required to fill the capsule is $\text{22}\text{.46 m}{{\text{m}}^{\text{3}}}$.

Exercise (13.9)

1. A wooden bookshelf has external dimensions as follows: Height $\text{= 110 cm}$, Depth $\text{= 25 cm}$, Breadth $\text{= 85 cm}$ (see the given figure). The thickness of the plank is $\text{5 cm}$ everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is $\text{20 paise per c}{{\text{m}}^{\text{2}}}$ and the rate of painting is $\text{10 paise per c}{{\text{m}}^{\text{2}}}$, find the total expenses required for polishing and painting the surface of the bookshelf.

(Image will be Uploaded Soon)

Ans: We will redraw the given diagram.

(Image will be Uploaded Soon)

We are given the following:

The external height of bookshelf $\text{h = 110 cm}$

The external length of bookshelf $\text{l = 85 cm}$

The external breadth of bookshelf $\text{b = 25 cm}$

The external surface area of the shelf $\text{V = lh + 2 }\left( \text{bh + lb} \right)$

$\Rightarrow \text{V = }\left( \text{85  }\times\text{  110} \right)\text{ + 2 }\left( \text{25  }\times\text{  110 + 85  }\times\text{  25} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{V = }\left( \text{9350 + 9750} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{V = 19100 c}{{\text{m}}^{\text{2}}}$

The area of the front face $\text{A=}\left[ \left( \text{85  }\times\text{  110} \right)\text{ - }\left( \text{75  }\times\text{  100} \right)\text{ + 2}\left( \text{75  }\times\text{  5} \right) \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{1850 + 750} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 2600 c}{{\text{m}}^{\text{2}}}$

The external faces will be polished.

The area to be polished $\text{= }\left( \text{19100 + 2600} \right)\text{ c}{{\text{m}}^{\text{2}}}\text{ = 21700 c}{{\text{m}}^{\text{2}}}$

It is given that the cost of polishing $\text{1 c}{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 0}\text{.20}$

So, the cost of polishing $\text{21700 c}{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \text{0}\text{.20  }\times\text{  21700} \right)\text{ = Rs}\text{. 4340}$

Therefore, the total expenses required for polishing the surface of bookshelf is $\text{Rs}\text{. 4340}$.

(Image will be Uploaded Soon)

From the figure, we can write the following:

The height of each row of bookshelf $\text{h = 30 cm}$

The length of each row of bookshelf $\text{l =  75 cm}$

The breadth of each row of bookshelf $\text{b = 20 cm}$

The area to be painted $\text{{V}' = lh + 2 }\left( \text{l + h} \right)\text{b}$

$\Rightarrow \text{{V}' = }\left[ \text{2 }\left( \text{75  +  30} \right)\times \text{20 + }\left( 75\times 30 \right) \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{{V}' = }\left( \text{4200 + 2250} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{{V}' = 6450 c}{{\text{m}}^{\text{2}}}$

So, the area to be painted in three rows $\text{= }\left( \text{3  }\times\text{  6450} \right)\text{ c}{{\text{m}}^{\text{2}}}\text{ = 19350 c}{{\text{m}}^{\text{2}}}$

It is given that the cost of painting $\text{1 c}{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 0}\text{.10}$

So, the cost of painting $\text{19350 c}{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \text{0}\text{.1  }\times\text{  19350} \right)\text{ = Rs}\text{. 1935}$

Therefore, the total expenses required for painting the surface of bookshelf is $\text{Rs}\text{. 1935}$.

The total expenses required for both polishing and painting the bookshelf$\text{= Rs}\text{. }\left( \text{4340 + 1935} \right)\text{ = Rs}\text{. 6275}$

Therefore, the total cost for polishing and painting the surface of the bookshelf is $\text{Rs}\text{. 6275}$.

2. The front compound wall of a house is decorated by wooden spheres of diameter $\text{21 cm}$, placed on small supports as shown in the given figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius $\text{1}\text{.5 cm}$and height $\text{7 cm}$and is to be painted black. Find the cost of paint required if silver paint costs $\text{25 paise per c}{{\text{m}}^{\text{2}}}$ and black paint costs $\text{5 paise per c}{{\text{m}}^{\text{2}}}$.

Ans:

(Image will be Uploaded Soon)

The diameter of wooden sphere $\text{= 21 cm}$

So, the radius of wooden sphere $\text{r = }\dfrac{\text{21}}{\text{2}}\text{ = 10}\text{.5 cm}$

The surface area of the wooden sphere $\text{A = 4 }\pi\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{10}\text{.5} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 1386 c}{{\text{m}}^{\text{2}}}$

The radius of the cylindrical support \[{{\text{r}}_{\text{1}}}\text{ = 1}\text{.5 cm}\]

The height of cylindrical support $\text{h = 7 cm}$

The curved surface area of the cylindrical support ${{\text{A}}_{\text{1}}}\text{ = 2 }\pi\text{ rh}$

$\Rightarrow {{\text{A}}_{\text{1}}}\text{ = }\left[ \text{2  }\times\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\times\text{  1}\text{.5 }\times \text{ 7} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{1}}}\text{ = 66 c}{{\text{m}}^{\text{2}}}$

The area of circular end of cylindrical support ${{\text{A}}_{\text{2}}}\text{ =  }\pi\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{2}}}\text{ = }\left[ \dfrac{\text{22}}{\text{7}}\text{  }\times\text{  }{{\left( \text{1}\text{.5} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{2}}}\text{ = 7}\text{.07 c}{{\text{m}}^{\text{2}}}$

The area to be painted in silver ${{\text{A}}_{\text{s}}}\text{ = }\left[ \text{8  }\times\text{  }\left( \text{1386 - 7}\text{.07} \right) \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{s}}}\text{ = }\left[ \text{8  }\times\text{  1378}\text{.93} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{s}}}\text{ = 11031}\text{.44 c}{{\text{m}}^{\text{2}}}$

It is given that to paint $\text{1 c}{{\text{m}}^{\text{2}}}$ of area in silver, it costs $\text{= Rs}\text{. 0}\text{.25}$

So, to paint $\text{11031}\text{.44 c}{{\text{m}}^{\text{2}}}$ area in silver, it will cost $\text{= Rs}\text{. }\left( \text{11031}\text{.44  }\times\text{  0}\text{.25} \right)\text{ = Rs}\text{. 2757}\text{.86}$

The area to be painted black ${{\text{A}}_{\text{b}}}\text{ = }\left( \text{8  }\times\text{  66} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{b}}}\text{ = 528 c}{{\text{m}}^{\text{2}}}$

It is given that to paint $\text{1 c}{{\text{m}}^{\text{2}}}$ of area in black, it costs $\text{= Rs}\text{. 0}\text{.05}$

So, to paint $\text{528 c}{{\text{m}}^{\text{2}}}$ area in black, it will cost $\text{= Rs}\text{. }\left( \text{528  }\times\text{  0}\text{.05} \right)\text{ = Rs}\text{. 26}\text{.40}$

Hence, the total cost of painting will be $\text{= Rs}\text{. }\left( \text{2757}\text{.86 + 26}\text{.40} \right)\text{ = Rs}\text{. 2784}\text{.26}$

Therefore, the total cost for the painting will be $\text{Rs}\text{. 2784}\text{.26}$.

3. The diameter of a sphere is decreased by $\text{25  }\%\text{ }$. By how much percent does its curved surface area decrease?

Ans: Let us assume the diameter of the sphere to be $\text{d}$.

So, the radius of the sphere will be ${{\text{r}}_{\text{1}}}\text{ = }\dfrac{\text{d}}{\text{2}}$

Now the diameter is decreased by $\text{25  }\%\text{ }$.

So, the new radius will be ${{\text{r}}_{\text{2}}}\text{ = }\dfrac{\text{d}}{\text{2}}\left( \text{1 - }\dfrac{\text{25}}{\text{100}} \right)\text{ = }\dfrac{\text{3}}{\text{8}}\text{d}$

The curved surface area of initial sphere ${{\text{A}}_{\text{1}}}\text{ = 4 }\pi\text{ }{{\text{r}}_{\text{1}}}^{\text{2}}$

$\Rightarrow {{\text{A}}_{\text{1}}}\text{ = 4 }\pi\text{ }{{\left( \dfrac{\text{d}}{\text{2}} \right)}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{1}}}\text{ =  }\pi\text{ }{{\text{d}}^{\text{2}}}$

The curved surface area of new decreased sphere ${{\text{A}}_{\text{2}}}\text{ = 4 }\pi\text{ }{{\text{r}}_{\text{2}}}^{\text{2}}$

$\Rightarrow {{\text{A}}_{\text{2}}}\text{ = 4 }\pi\text{ }{{\left( \dfrac{\text{3d}}{\text{8}} \right)}^{\text{2}}}$

$\Rightarrow {{\text{A}}_{\text{2}}}\text{ = }\dfrac{9}{16}\text{ }\pi\text{ }{{\text{d}}^{\text{2}}}$

So, the percentage decrease in the surface area $\text{P = }\dfrac{{{\text{A}}_{\text{1}}}\text{ - }{{\text{A}}_{\text{2}}}}{{{\text{A}}_{\text{1}}}}\text{  }\times\text{  100  }\%\text{ }$

$\Rightarrow \text{P = }\dfrac{\text{ }\pi\text{ }{{\text{d}}^{\text{2}}}\text{ - }\dfrac{\text{9}}{\text{16}}\text{ }\pi\text{ }{{\text{d}}^{\text{2}}}}{\text{ }\pi\text{ }{{\text{d}}^{\text{2}}}}\text{  }\times\text{  100  }\%\text{ }$

$\Rightarrow \text{P = }\dfrac{\text{7 }\pi\text{ }{{\text{d}}^{\text{2}}}}{\text{16 }\pi\text{ }{{\text{d}}^{\text{2}}}}\text{  }\times\text{  100  }\%\text{ }$

$\Rightarrow \text{P = }\dfrac{\text{700}}{\text{16}}\text{  }\%\text{ }$

$\Rightarrow \text{P = 43}\text{.75  }\%\text{ }$

Therefore, the curved surface area of the sphere will be decreased by $\text{43}\text{.75  }\%\text{ }$.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes - PDF Download

Surface Area Volumes Class 9 free PDF 

NCERT Solutions for Class 9 Chapter 13 are available in the PDF format on the official website of CoolGyan for free. As the PDF contains several previous year papers, students can understand how the questions will be asked along with an in-depth explanation of the concepts.

You can opt for Chapter 13 - Surface Areas and Volumes NCERT Solutions for Class 9 Maths PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 9 Maths

• Chapter 1 - Number System

• Chapter 2 - Polynomials 

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclids Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Herons formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

Surface Area and Volumes Class 9 NCERT Solutions

13.1 Introduction

The PDF of Class 9 Maths Surface Areas and Volumes recalls the knowledge of students at this introduction part. Also explained here are the plane shapes and solid shapes with a brief description. According to the PDF, it is essential to understand the shapes to explore their surface areas and volumes.

13.2 Surface Area of a Cuboid and a Cube

In this section, the Surface Area and Volumes Class 9 chapter asks the students to observe the shape of the cuboid keenly. The students need to perform an activity with brown paper to understand more clearly. After the observation, we can conclude that it is made up of six rectangles. So to find out the surface area of cuboids, we need to find 6 areas of different rectangles. After performing all these steps by step procedures, the formula derived for the area of a cuboid is,

Surface Area of Cuboid= 2(lb+bh+hl).

In the case of cuboid, the edges like l, b, H mary. But in the case of a cube, all sides are equal so it is denoted by a. The surface area of the cube = 6a2.

Exercise 13.1 has 8 questions. 4 are short answer type and 4 are long answer type.

13.3 Surface Area of a Right Circular Cylinder

The NCERT Solutions for Class 9 Maths Chapter 13 PDF, in this section, helps the students to understand how to find the surface area of a right circular cylinder. We need to split the cylinder into two parts: the middle part and the bottom part.

The curved surface area of cylinder= 2πrh

Total surface area of cylinder = πr2(r+h)

Exercise 13.2 has 11 questions. 5 are short answer type questions and 6 are long answer type questions.

13.4 Surface Area of a Right Circular Cone

This section explains about the cone which has a triangle shape and circle shape. By combining these two areas, Class 9 Surfaces Area and Volumes, the formula of the cone is derived as below,

Curved Surface Area of a Cone = ½  × l ×2πr = πrl

Total Surface Area of a Cone = πrl + πr2 = πr (l + r)

Exercise 13.3 has 8 questions.

13.5 Surface Area of Sphere

As the entire chapter deals with the surface areas of different shapes, students can understand the formula and process for finding the surface area of the sphere in this section. Maths NCERT Class 9 Chapter 13 explains the concept well and provides an activity for the students to understand.

Surface Area of a Sphere = 4 πr2. 

Exercise 13.4 has 9 questions. 4 are short answer type and 5 are long answer type questions

13.6 Volume of the Cuboid

The NCERT Solutions Class 9 Chapter 13 explains volumes of different solid shapes from this section of the chapter. Here the students can understand what volume is and how to find out the volume of a cuboid and cube.

Volume of cuboid = l*b*h

The volume of cube = a3

Exercise 13.5 has 9 questions. 5 are long answer type questions.

13.7 Volume of a Cylinder

The next part of the chapter covers the volume of a cylinder. Students have already understood the meaning of volume, the NCERT Solutions of Class 9 Chapter 13 directly explains the formula of volume of a cylinder with several examples.

Volume of a Cylinder = πr2h

Exercise 13.6 contains 8 problems. All the questions are short answer type.

13.8 Volume of a Right Circular Cone

The NCERT Chapter 13 Maths Class 9 gave an activity to the students to understand the difference between cylinder and cone. From that, they derived a formula for the volume of a cone is ⅓ πr2h.

Exercise 13.7 has nine long answer type questions.

13.9 Volume of a Sphere

The last section of Maths Chapter 13 Class 9 Ncert Solutions is about finding the volume of a sphere using the formula 4/3πr3. It is also explained with multiple examples. 

Exercise 13.8 has 10 questions. 5 are short answer type and 5 are long answer type.

13.10 Summary

Exercise 13.9 has 3 questions with solutions in PDF.

Key Takeaways of NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes free PDF

The NCERT Chapter 13 Class 9 Solutions has several takeaways for the students. Some of them are given below-

• The PDFs are available for free on the main website.

• Students can learn and understand slowly as per their convenience with the help of physical copy.

• Students can improve their knowledge as well as score. It gives command on the subject.