**NCERT Solution For Class 9, Maths, Chapter 13, Surface ****Areas And Volumes, Exercise 13.6** has questions related to cylindrical shapes in which diameter and length of any cylinder is given with which we need to find volume of the given figure. Exercise 13.6 class 9 maths chapter 13 contains total eight questions to discuss.

Table of Contents

Toggle## Class 9, Maths, Chapter 13, Exercise 13.6 Solutions (Page no. 230)

**Q.1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. ****How many liters of water can it hold? (1000 cm ^{3} = 1l) **

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

Circumference of the base = 132 cm

⇒ 2πr = 132

⇒ $2\times \frac{22}{7}\times r=132$

⇒ $r=\frac{132\times 7}{2\times 22}$= 21 cm

Volume of the cylinder = πr^{2}h cm^{3}

= $\frac{22}{7}\times 21\times 21\times 25$ = 34650 cm^{3}

Vessel can hold = $\frac{34650}{1000}Liters$

i.e., 3465 liters of water.

**Q.2. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm ^{3} of wood has a mass of 0.6 g.**

**Ans:**

**Q.3. A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?**

**Ans:** (i) Capacity of tin can = lbh cm^{3}

= (5 × 4 × 15) cm^{3} = 300 cm^{3}

(ii) Capacity of plastic cylinder = πr^{2}h cm^{3}

= $\left( \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times 10 \right)cm$ = 385 cm^{3}

Thus, the plastic cylinder has greater capacity by 385 cm^{3}

**Q.4. If the lateral surface of a cylinder is 94.2 cm ^{2} and its height is 5 cm, then find**

**(i) radius of its base **

**(ii) its volume. (Use π = 3.14)**

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

Lateral surface = 94.2 cm^{3}

⇒ 2 × 3.14 × r × 5 = 94.2

⇒ $r=\frac{94.2}{2\times 3.14\times 5}$= 3

Thus, the radius of its base = 3 cm

(ii) Volume of the cylinder = πr^{2}h

= (3.14 × 3^{2 }× 5) cm^{3}

= (3.14 x 3 x 3 x 5) cm^{3}

= 141. 3 cm^{3}

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

**(i) inner curved surface area of the vessel,**

**(ii) radius of the base,**

**(iii) capacity of the vessel.**

**Ans:**

**Q.6. The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?**

**Ans:**

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

**Ans:**

**Q.8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the ****bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare ****daily to serve 250 patients?**

**Ans: **** **Diameter of the cylindrical bowl = 7 cm

Radius = $\frac{7}{2}cm$

Height of serving bowl = 4 cm

Soup saved in on serving = Volume of the bowl

= πr^{2}h

= cm^{3 }=1.54 cm^{3}

Soup served to 250 patient = (250×1.54) cm^{3}

= 38500 cm^{3} i.e., 38.5 liters.

## 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**