**NCERT Solutions for Class 9 Maths Exercise 13.7 of Chapter 13 Surface areas and Volumes(Term 2)**

**NCERT Solutions for Class 9 Maths Exercise 13.7 Chapter 13 Surface Areas and Volumes (Term 2)** CBSE Board exams are created here for aiding the students in their ground preparation for the** term 2** CBSE Board exam.All questions of the **class 9 NCERT maths** course book **exercise 13.7 chapter 13 Surface areas and Volumes** are solved here by an accomplished CBSE **Maths** instructor. **Exercise 13.5** depends on the **Surface areas and volumes** of the three-dimensional conical figures faced by us in daily life. These **NCERT Solutions** for **class 9 maths chapter 13** are valuable for competitive entrance exams moreover.

**NCERT Solutions of class 9 maths** is the main review material in the preparation of the CBSE Board exam since **chapter 13 – Surface areas and Volumes** accomplishes every one of the kinds of questions in light of our genuine daily life problems.

**Click for online shopping**

**Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc**

**NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Term 2 )CBSE Board exam**

**NCERT Solutions Class 9 Maths-All Chapters**

**NCERT Solutions for Class 9 Maths Exercise 13.7 of Chapter 13 Surface areas and Volumes(Term 2)**

**Q1. Find the volume of the right circular cone with**

**(i)radius 6cm, height 7 cm (ii) radius 3.5 cm, height 12 cm (Assume π = 22/7)**

Ans.(i) Radius(r) of the right circular cone is 6 cm and height(h) is 7 cm

Volume of the right circular cone

=(1/3) πr²h

=(1/3)×(22/7)×6×6×7

=22×2×6

=264

**Hence the volume of the cone is 264 cm³**

(ii) Radius(r) of the right circular cone is 3.5 cm and height(h) is 12 cm

Volume of the right circular cone

=(1/3) πr²h

=(1/3)×(22/7)×3.5×3.5×12

=22×0.5×3.5×4

=154

**Hence the volume of the cone is 154 cm³**

**Q2.Find the capacity in liters of a conical vessel with **

**(i) radius 7cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm**

**(Assume π = 22/7)**

Ans.(i) Radius (r) of the conical vessel is 7cm, slant height(l) is 25 cm

height(h) =√(l²-r²) = √(25²-7²) =√(625-49) =√576=24 cm

The capacity of the conical vessel is

=(1/3)πr²h

=(1/3)×(22/7)×7×7×24

=22×7×8

=1232

The volume of the vessel is 1232 cm³

**Hence the capacity of the conical vessel is 1232 cm³=[1232 ÷1000]litres=1.232 l**

(ii) Height (h) of the conical vessel is 12cm, slant height(l) is 13 cm

Radius(r) =√(l²-h²) = √(13²-12²) =√(169-144) =√25=5 cm

The volume of the conical vessel is

=(1/3)πr²h

=(1/3)×(22/7)×5×5×12

=(22/7)×5×5×4

=2200/7

Volume of the vessel is 2200/7 cm³

**Hence the capacity of the conical vessel is 2200/7 cm³=[2200/7 ÷1000]litres=11/35 l**

** Q3.The height of a cone is 15cm. If its volume is 1570cm ^{3}, find the diameter of its base. (Use π = 3.14)**

Ans.**The height(h) of a cone is 15cm**

Volume of the cone is

=πr²h

=(22/7)×r²×15

**Volume of the cone is given 1570cm³1570×21)/(22×15)**

∴(1/3)(22/7)×r²×15 = 1570

(1/3)(3.14)×r²×15 = 1570

3.14 ×r²×5= 1570

r² = (1570)/(3.14×5)

r² = 1570/15.7 =100

r = √100 =10

**Hence the diameter of base of the cone is 2r =2×10 =20 cm**

**Q4.** **If the volume of a right circular cone of height 9cm is 48πcm ^{3}, find the diameter of its base.**

Ans.Height (h)of the right circular cone is 9 cm

Volume of the right circular cone is given 48πcm³

Volume of the right circular cone = πr²h

(1/3)πr²h = 48π

(1/3)πr²×9 = 48π

r² = (48×3)/9=16

r = √16 =4

**Hence the diameter of the base of right circular cone is 2r =2×4 =8 cm**

**Q5.A conical pit of top diameter 3.5m is 12m deep. What is its capacity in kiloliters?**

**(Assume π = 22/7)**

Ans. The diameter of the top of the conical pit is 3.5 m,then its radius is 3.5/2 =1.75 m

The depth(h) of the pit is 12 m

The volume of the pit

=(1/3)πr²h

=(1/3)(22/7)×1.75×1.75×12

=22×0.25**×**1.75×4

=38.5

The volume of pit is 38.5 m³

1 Kl = 1m³

Hence the capacity of the pit is 38.5 ×1 =38.5 Kilo liters

**Q6. The volume of a right circular cone is 9856cm³. If the diameter of the base is 28cm, find**

**(i) height of the cone**

**(ii) slant height of the cone**

**(iii) curved surface area of the cone**

**(Assume π = 22/7)**

Ans.The diameter of the base of a right circular cone is 28 cm then its radius(r) is 14 cm

The volume of a right circular cone is

**=**(1/3)πr²h

The given volume of the right circular cone is 9856cm³

(1/3)×(**22/7**)×14×14h = 9856

(1/3) ×22×2×14h = 9856

h = ( 9856×3)/(44×14)

h=48

**Hence the height of the cone is 48 cm**

**Q7.A right triangle ABC with sides 5cm, 12cm and 13cm is revolved about the side 12 cm. Find the volume of the solid so obtained.**

Ans. Since right Δ ABC is is revolved about the side 12 cm,therefore the height(h) of so formed cone is 12 cm and radius(r) is 5 cm and slant height(l) is 13 cm

Volume of the so formed cone =(1/3)πr²h

Where r =5 cm,h =12 cm

Volume

= (1/3)π×5×5×12

=100π

**Hence the volume of the so formed cone is 100π cm³**

**Q8.If the triangle ABC in the Question 7 is revolved about the side 5cm, then find the volume of the solids so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.**

Ans. Since right Δ ABC is is revolved about the side 5 cm,therefore the height(h) of so formed cone is 5 cm and radius(r) is 12 cm and slant height(l) is 13 cm

Volume of the so formed cone =(1/3)πr²h

Where r =12 cm,h =5 cm

Volume

= (1/3)π×12×12×5

=240π

**Hence the volume of the so formed cone is 240π cm³**

Ratio of the volumes of the two solids obtained in Questions 7 and 8

100π : 240π

=5 : 12

**Hence the** r**atio of the volumes of the two solids is 5 : 12**

**Q9.A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas.**

Ans.The diameter of conical heap is 10.5 m therefore its radius(r) is 10.5/2 =5.25 m and height(h) is 3 m

The volume of the heap

=(1/3)πr²h

=(1/3)×(22/7)×5.25 ×5.25 ×3

=22×0.75 ×5.25

=86.625

**The volume of the heap is 86.625 m³**

**The heap is to be covered by canvas to protect it from rain**

= Curved surface area of the heap

= πrl,where r =5.25 m

Slant height(l) =√(h²+r²) =√(3²+5.25²)=√(9+27.5625)=√(36.5625)=6.05 m

Curved surface area of the heap

=(22/7)×5.25×6.05

=22×0.75×6.05

=99.825

**Hence the canvas required to covered the heap is 99.825 m²**

**NCERT Solutions** for Class 9 Maths Exercise 13.7 of Chapter 13 Surface areas and Volumes(Term 2) help the student in clearing the concepts on Surface areas and Volumes which are useful not only for the students of class 9 maths but also for the candidates who are preparing for the competitive entrance exams.

**You can compensate us by donating any amount of money for our survival**

**Our Paytm No 9891436286**

**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions of class 9 maths**

**NCERT Solutions of class 9 science **

**CBSE Class 9-Question paper of science 2020 with solutions**

**CBSE Class 9-Sample paper of science**

**CBSE Class 9-Unsolved question paper of science 2019**

**NCERT Solutions of class 10 maths**

**CBSE Class 10-Question paper of maths 2021 with solutions**

**CBSE Class 10-Half yearly question paper of maths 2020 with solutions**

**CBSE Class 10 -Question paper of maths 2020 with solutions**

**CBSE Class 10-Question paper of maths 2019 with solutions**

**NCERT solutions of class 10 science**

**Solutions of class 10 last years Science question papers**

**CBSE Class 10 – Question paper of science 2020 with solutions**

**CBSE class 10 -Latest sample paper of science**

**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT solutions of class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |