NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.2) Exercise 13.2

Exercise 13.2

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

Ans: Height (h) of cylinder $-14 \mathrm{~cm}$

Let the diameter of the cylinder be $\mathrm{d}$

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

$\Rightarrow 2\pi r\text{h}=88~\text{c}{{\text{m}}^{2}}(r$ is the radius of the base of the cylinder)

$\Rightarrow \pi dh=88~\text{c}{{\text{m}}^{2}}\ $

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

$\Rightarrow \ d\ =\ \dfrac{88\ \times \ 7}{22\ \times \ 14}\ =\ 2\ cm$

Therefore, the diameter of the base of the cylinder is $2 \mathrm{~cm}$.

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

Ans: Height (h) of cylindrical tank - $1 \mathrm{~m}$

Base radius (r) of cylindrical tank $=\ \dfrac{140}{2}~\text{cm}\ \text{=}\ 70~\text{cm}\ \text{=}\ 0.7~\text{m}$

Area of sheet required = Total surface area of tank $=2\pi r(r+h)$

\[=\ 2\ \times \ \dfrac{22}{7}\ \times \ 0.7\ \left( 0.7\ +\ 1 \right)\ {{m}^{2}}\]

\[=\ 4.4\ \times 1.7\ =\ 7.48\ {{m}^{2}}\]

Therefore, it will require $7.48 \mathrm{~m}^{2}$ area of sheet.

3.  A metal pipe is $77 \mathrm{~cm}$ long. The inner diameter of a cross-section is $4 \mathrm{~cm}$, the outer diameter being $4.4 \mathrm{~cm}$, Find:

i) Inner curved surface area,

ii) Outer curved surface area,

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

Ans: Inner radius $\left( {{\text{r}}_{1}} \right)$ of cylindrical pipe $=\dfrac{4}{2}~\text{cm}=2~\text{cm}$

Outer radius $\left(\mathrm{r}_{2}\right)$ of cylindrical pipe $=\dfrac{4.4}{2} \mathrm{~cm}=2.2 \mathrm{~cm}$

Height (h) of cylindrical pipe = Length of cylindrical pipe $=77~\text{cm}$

(i) $\mathrm{CS} \mathrm{A}$ of inner surface of pipe $=2\pi {{\text{r}}_{1}}\text{h}$

$=\left( 2\times \dfrac{22}{7}\times 2\times 77 \right)\text{c}{{\text{m}}^{2}}=968~\text{c}{{\text{m}}^{2}}$

(ii) $\mathrm{CS} \mathrm{A}$ of outer surface of pipe $=2\pi {{\text{r}}_{2}}\text{h}$

$=\left( 2\times \dfrac{22}{7}\times 2.2\times 77 \right)\text{c}{{\text{m}}^{2}}=(22\times 22\times 2.2)\text{c}{{\text{m}}^{2}}\ =\ 1064.8\ c{{m}^{2}}$

(iii) Total surface area of pipe = CSA of inner surface + CSA of outer surface + Area of both circular ends of pipe.

$=\ 2\pi {{r}_{1}}h+2\pi {{r}_{2}}h+2\pi \left( r_{2}^{2}-r_{1}^{2} \right) $

$=\ \left[ 968+1064.8+2\pi \left\{ {{(2.2)}^{2}}-{{(2)}^{2}} \right\} \right]\text{c}{{\text{m}}^{2}} $

$=\left( 2032.8+2\times \dfrac{22}{7}\times 0.84 \right)\text{c}{{\text{m}}^{2}} $

$=\ (2032.8+5.28)\ \text{c}{{\text{m}}^{2}}=\ \ 2038.08~\text{c}{{\text{m}}^{2}} $

Therefore, the total surface area of the cylindrical pipe is $2038.08 \mathrm{~cm}^{2}$.

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

Ans: It can be observed that a roller is cylindrical.

Height (h) of cylindrical roller = Length of roller $=120 \mathrm{~cm}$

Radius (r) of the circular end of roller $=\dfrac{84}{2}~\text{cm}=42~\text{cm}$

CSA of roller $=2 \pi t h$

$=\left( 2\times \dfrac{22}{7}\times 42\times 120 \right)\text{c}{{\text{m}}^{2}}\ =\ 31680~\text{c}{{\text{m}}^{2}}$

Area of field $=500\times $ CSA of roller

$=\ (500\times 31680)\text{c}{{\text{m}}^{2}}$

$=15840000~\text{c}{{\text{m}}^{2}}=1584~{{\text{m}}^{2}} $

5. A cylindrical pillar is $50 \mathrm{~cm}$ in diameter and $3.5 \mathrm{~m}$ in height. Find the cost of painting the curved surface of the pillar at the rate of Rs 12 .50 per $\left[\right.$ Assume $\left.\pi=\dfrac{22}{7}\right]$

Ans: Height (h) cylindrical pillar $=3.5 \mathrm{~m}$

Radius (r) of the circular end of pillar \[=\dfrac{50}{2}~\text{cm}\ \text{=}\ 25~\text{cm}\ \text{=}\ 0.25~\text{m}\]

CSA of pillar $=2\pi rh$

$=\left(2 \times \dfrac{22}{7} \times 0.25 \times 3.5\right) \mathrm{m}^{2}$

$=(44\times 0.125){{\text{m}}^{2}}=5.5~{{\text{m}}^{2}}$

Cost of painting $1 \mathrm{~m}^{2}$ area $\text{=}\ \text{Rs}.12.50$

Cost of painting $5.5 \mathrm{~m}^{2}$ area $=(5.5\times 12.50)$

$=\mathrm{Rs} .68 .75$

Therefore, the cost of painting the CSA of the pillar is Rs. 68.75.

6. Curved surface area of a right circular cylinder is $4.4 \mathrm{~m}^{2}$. If the radius of the base of the cylinder is $0.7 \mathrm{~m}$, find its height. $\left[\right.$ Assume $\left.\pi=\dfrac{22}{7}\right]$

Ans: Let the height of the circular cylinder be $h$

Radius (r) of the base of cylinder $=0.7 \mathrm{~m}$

CSA of cylinder $=4.4~{{\text{m}}^{2}}$ 

$\Rightarrow \ \ 2\pi \text{rh}=4.4~{{\text{m}}^{2}}$ 

$\Rightarrow \ \left[ 2\times \dfrac{22}{7}\times 0.7\times h \right]=4.4~{{\text{m}}^{2}}$ 

$~\Rightarrow \ \text{h}\ =\ \dfrac{4.4\ \times \ 7}{2\ \times 22\ \times \ 0.7}\ =\ 1m$

Therefore, the height of the cylinder is$1 \mathrm{~m}$.

7. The inner diameter of a circular well is$3.5 \mathrm{~m}$. It is $10 \mathrm{~m}$ deep. Find

(i)  Inner curved surface area,

(ii) The cost of plastering this curved surface at the rate of Rs. 40 per m2 . $\left[\right.$ Assume $\left.\pi=\dfrac{22}{7}\right]$

Ans:  Inner radius (r) of circular well $=\left( \dfrac{3.5}{2} \right)\text{m}=1.75~\text{m}$

Depth of well = $=\ 10\ m$

(i) Inner curved surface area $=2\pi \text{rh}$

$=\left( 2\times \dfrac{22}{7}\times 1.75\times 10 \right){{\text{m}}^{2}} $

$=\ (44\times 0.25\times 10){{\text{m}}^{2}}=110~{{\text{m}}^{2}}$

Therefore, the inner curved surface area of the circular well is $110 \mathrm{~m}^{2}$.

(ii) Cost of plastering $1 \mathrm{~m}^{2}$ area = Rs. 40

Cost of plastering $100 \mathrm{~m}^{2}$ area $\text{=Rs}(110\times 40)=\text{Rs}.4400$

Therefore, the cost of plastering the CSA of this well is Rs.4400.

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

Ans: Height (h) of cylindrical pipe  = Length of cylindrical pipe $=28~\text{m}$

Radius $(\mathrm{r})$ of circular end of pipe $=\dfrac{5}{2}=2.5~\text{cm}\ \text{=}\ \,0.025~\text{m}$

CSA of cylindrical pipe  $=\ 2\pi rh$

$=\ \left( 2\times \dfrac{22}{7}\times 0.025\times 28 \right){{\text{m}}^{2}}=4.4~{{\text{m}}^{2}}$

The area of the radiating surface of the system is$4.4 \mathrm{~m}^{2}$.

9. Find:

(i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is $4.2 \mathrm{~m}$ in diameter and $4.5 \mathrm{~m}$ high.

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

Ans: Height (h) of cylindrical tank $=4.5 \mathrm{~m}$

Radius (r) of the circular end of cylindrical tank $=\left(\dfrac{4.2}{2}\right) \mathrm{m}=2.1 \mathrm{~m}$

(i) Lateral or curved surface area of tank $=2\pi rh$

$=\left( 2\times \dfrac{22}{7}\times 2.1\times 4.5 \right){{\text{m}}^{2}}\ =\ (44\times 0.3\times 4.5){{\text{m}}^{2}}\ =\ 59.4~{{\text{m}}^{2}}$

Therefore, CSA of tank is$59.4 \mathrm{~m}^{2}$.

(ii) Total surface area of tank $=2\pi r(r+h)$

$=\left( 2\times \dfrac{22}{7}\times 2.1\times (2.1+4.5) \right){{\text{m}}^{2}} $

$ =\ (44\times 0.3\times 6.6){{\text{m}}^{2}}=87.12~{{\text{m}}^{2}}$

Let A $\mathrm{m}^{2}$ steel sheet be actually used in making the tank.

$\therefore A\left( 1-\dfrac{1}{12} \right)=87.12~{{\text{m}}^{2}} $

$\Rightarrow A=\left( \dfrac{12}{11}\times 87.12 \right){{\text{m}}^{2}} $

$\Rightarrow \text{A}\ \text{=}\ 95.04~{{\text{m}}^{2}} $

Therefore, $95.04 \mathrm{~m}^{2}$ steel was used in actual while making such a tank.

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 $20 \mathrm{~cm}$ and height of $30 \mathrm{~cm}$. A margin of $2.5 \mathrm{~cm}$ is to be given for folding it over the top and bottom of the frame. Find bow much cloth is required for covering the lampshade. $\left[\right.$ Assume $\left.\pi=\dfrac{22}{7}\right]$

Ans: Height (h) of the frame of lampshade $=(2.5+30+2.5)\text{cm}\ \text{=}\ 35~\text{cm}$

Radius (r) of the circular end of the frame of lampshade $=\dfrac{20}{2}~\text{cm}\ \text{=}\ 10~\text{cm}$

Cloth required for covering the lampshade $=\ 2\pi rh$

$=\left( 2\times \dfrac{22}{7}\times 10\times 35 \right)\text{c}{{\text{m}}^{2}}=2200~\text{c}{{\text{m}}^{2}}$

Hence, for covering the lampshade, $2200 \mathrm{~cm}^{2}$ cloth will be required.

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 $3 \mathrm{~cm}$ and height $10.5 \mathrm{~cm}$. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how mach cardboard was required to be bought for the competition? $\left[ Assume\ \pi \ =\ \dfrac{22}{7} \right]$ 

Ans: Radius (r) of the circular end of cylindrical penholder $=3~\text{cm}$

Height (h) of penholder $=10.5 \mathrm{~cm}$

Surface area of 1 penholder = CSA of penholder + Area of base of penholder $=2\pi rh\ +\ \pi {{r}^{2}}$

$=\left( 2\times \dfrac{22}{7}\times 3\times 10.5+\dfrac{22}{7}\times {{(3)}^{2}} \right)\text{c}{{\text{m}}^{2}} $

$=\left( 132\times 1.5+\dfrac{198}{7} \right)\text{c}{{\text{m}}^{2}} $

$=\left( 198+\dfrac{198}{7} \right)\text{c}{{\text{m}}^{2}}=\dfrac{1584}{7}~\text{c}{{\text{m}}^{2}}$

Area of card board sheet used by 1 competitor $=\dfrac{1584}{7} \mathrm{~cm}^{2}$

Area of cardboard sheet used by 35 competitors $=\left( \dfrac{1584}{7}\times 35 \right)\text{c}{{\text{m}}^{2}}$ $=7920~\text{c}{{\text{m}}^{2}}$

Therefore, $7920 \mathrm{~cm}^{2}$ cardboard sheet will be bought.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2

Opting for the NCERT solutions for Ex 13.2 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.2 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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