**NCERT Solution For Class 9, Maths, Chapter 13, Surface Areas And Volumes, Exercise 13.2** involves questions related to curved surface area of cylinder with the formula. Class 9 , maths chapter 13 , Ex13.2 consists of total 11 questions to study. all questions are formula based which are given below.

Table of Contents

Toggle## Class 9, Maths, Chapter 13, Exercise 13.2 Solutions (Page no. 216)

**Q.1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm ^{2}. Find the **

**diameter of the base of the cylinder.**

**Ans: **Let r be the radius of the base and h = 14 cm be the height of the cylinder. Then, Curved surface area =

⟹ $88=2\times \frac{22}{7}\times r\times 14$

⟹ $r=\frac{88\times 7}{2\times 22\times 14}=1$ cm

Diameter of the base = 2r

⇒ 2 × 1= 2 cm

Hence, the diameter of the base of the cylinder is 2 cm.

**Q.2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm ****from a metal sheet. How many square metres of the sheet are required for the same?**

**Ans:** Let r be the radius of the base and h be the height of the cylinder.

Metal sheet required to make a closed cylindrical tank = Its total surface area

= 2πr (h + r)

= 2 × 22/7 × 0.7 (1+ 0.70) m^{2}

[∵ h =1m, r =140/2 cm = 70 cm =0.70 m]

= 2 × 22 × 0.1 × 1.70 m^{2}

= 7.48 cm^{2}

Hence, the sheet required is 7.48 m^{2}

**Q.3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm in the given figure. Find its**

**(i) inner curved surface area,**

**(ii) outer curved surface area,**

**(iii) total surface area.**

**Ans: **We have, R = external radius =$\frac{4.4}{2}$ cm = 2.2 cm

r = internal radius = $\frac{4}{2}$ = 2 cm

h = length of the pipe = 77 cm

(i) Inner curved surface = 2πrh

= $2\times \frac{22}{7}\times 2\times 77\,\,c{{m}^{2}}$

= 968 cm^{2}

(ii) Outer curved surface =

= $2\times \frac{22}{7}\times 2.2\times 77\,\,c{{m}^{2}}$

= 1064.8 cm^{2}

(iii) Total surface area of a pipe

= Inner curved surface area + outer curved surface area + areas of two bases

= 2πrh + 2πRh + 2π (R^{2} – r^{2})

= 968 + 1064.8 + $2\times \frac{22}{7}\times \left[ {{(2.2)}^{2}}-{{(2)}^{2}} \right]$

= 2032.8 + $\frac{44}{7}\times \left[ 4.84-4 \right]$

= 2032.8 +(44 x 0.12)

= 2032.8 + 5.28

= 2038.08 cm^{2}

**Q.4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground ****in m ^{2}.**

**Ans:** The length of the roller is 120 cm i.e., h = 1.2 m

and radius of the cylinder (i.e., roller) = $\frac{84}{2}$cm = 42 cm = 0.42 m.

**Distance covered by roller in one revolution** = Its curved surface area

= 2πrh

= (2 × $\frac{22}{7}$× 0.42 ×1.2) m^{2}

= 3.168 m^{2}

**Area of the playground **= Distance covered by roller in 500 revolutions

= (500 × 3.168) m^{2} = 1584 m^{2}

**Q.5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m ^{2}.**

**Ans:** Let r be the radius of the base and h be the height of the pillar

∴ r = $\frac{50}{2}$ cm = 25 cm = 0.25 m and h = 3.5 m

**Curved surface = 2πrh **

= (2 × × 0.20 × 3.5) m^{2}

= 5.5 m^{2}

Cost of painting the curved surface @ ₹ 12.50 per m^{2}

= ₹ (5.5 ×12.5) = ₹ 68.75

**Q.6. Curved surface area of a right circular cylinder is 4.4 m ^{2}. If the radius of the base of the **

**cylinder is 0.7 m, find its height.**

**Ans:** Let r be the radius of the base and h be the height of the cylinder. Then,

Curved surface area = 4.4 m^{2}

⇒ 2πrh = 4.4

⇒ (2 × × 0.7 × h) = 4.4 [∵ r = 0.7 m]

⇒ h = = 1m

Thus, the height of the cylinder = 1 meter.

**Q.7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find**

**(i) its inner curved surface area,**

**(ii) the cost of plastering this curved surface at the rate of ` 40 per m ^{2}.**

**Ans:** **(i) Let r be the radius of the face and h be depth of the well.** Then,

Curved surface area = 2πrh

⇒ $\left( 2\times \frac{22}{7}\times \frac{3.5}{2}\times 10 \right)$ m^{2} = 110 m^{2}

**(ii) Cost of plastering is ₹ 40 per m ^{2}**

∴ Cost of plastering the curved surface = ₹ (110 × 40) = ₹ 4400.

**Q.8. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter ****5 cm. Find the total radiating surface in the system.**

**Ans:** Total radiating surface in the system

= Curved surface area of the pipe

Where $r=\frac{5}{2}cm=2.5cm=\frac{2.5}{100}m=0.025\,m$ and h = 28 m

⇒ $\left( 2\times \frac{22}{7}\times 0.025\times 28 \right)$ m^{2 }= 4.4 m^{2}

**Q.9. Find : (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.**

**(ii) how much steel was actually used, if **** of the steel actually used was wasted in making the tank.**

**Ans: **(i) Here, $r=\frac{4.2}{2}m=2.1\,\,m$ and h = 4.5 m

Lateral surface area = 2πrh

^{⟹ }$\left( 2\times \frac{22}{7}\times 2.1\times 4.5 \right){{m}^{2}}$

= 59.4 cm^{2}

(ii) Since $\frac{1}{12}$of the actual steel used was wasted, the area of the steel which has gone into the tank = $\left( 1-\frac{1}{12} \right)$of x = $\frac{1}{12}$ of x.

Steel used = (2πrh+ 2πr^2) cm^3

= (59.4 + 2 × × 2.1 × 2.1) cm^{2}

= (59.4 + 27.72) cm = 87.12 cm^{2}

∴ $\frac{11}{12}\times x=87.12$

⟹ $x=\frac{87.12\times 12}{11}=95.04\,\,{{m}^{2}}$

Hence, the actual area of the steel used = 95.04 m^{2}.

**Q.10. In given the 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 cm and height of 30 cm. A margin of 2.5 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.**

**Ans:** Here, $r=\frac{20}{2}cm=10$ cm and h = (30 cm + 2× 2.5) cm (i.e, margin)= 35 cm.

Cloth required for covering the lampshade

= Its curved surface area = 2πrh

= $\left( 2\times \frac{22}{7}\times 10\times 35 \right)\,c{{m}^{2}}$

= 2200 cm^{2}

**Q.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 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?**

**Ans:** Cardboard required by each competitor

= Curved surface area of one penholder + base area

=2πrh + πr^{2} where r = 3 cm, h = 10.5 cm

= $\left[ \left( 2\times \frac{22}{7}\times 3\times 10.5 \right)+\frac{22}{7}\times 9 \right]\,c{{m}^{2}}$

= (198+28.28) cm^{2}

= 226.28 cm2 (approx.)

Cardboard required for 35 competitors

= (35×226.28) cm^{2}

= 7920 cm^{2} (approx.)

## NCERT Solutions For Class 9, Maths, Chapter 13, Surface Areas And Volumes (All Exercises)

**Class 9, Maths, Surface Areas And Volumes, Exercise 13.1****Class 9, Maths, Surface Areas And Volumes, Exercise 13.2****Class 9, Maths, Surface Areas And Volumes, Exercise 13.3****Class 9, Maths, Surface Areas And Volumes, Exercise 13.4****Class 9, Maths, Surface Areas And Volumes, Exercise 13.5****Class 9, Maths, Surface Areas And Volumes, Exercise 13.6****Class 9, Maths, Surface Areas And Volumes, Exercise 13.7****Class 9, Maths, Surface Areas And Volumes, Exercise 13.8****Class 9, Maths, Surface Areas And Volumes, Exercise 13.9**