We aim to bring you a detailed collection of questions and solutions from the exercises with relevant answers, as per the NCERT syllabus. NCERT Solutions for Class 9 Maths are meticulously created by our subject matter experts and experienced teaching faculty. NCERT solutions help students to score high in class, first and second term and competitive examinations. Students can practise and enhance their Math skills by solving the NCERT solutions chapter wise for Class 9 Maths that is provided here. These solutions include questions from the exercises given in the NCERT Textbooks as per the syllabus guidelines. The main aim in creating these questions is to enable the students to score well in Class 9 first and second term exams. NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.7 helps students to face the second term exams with confidence and also score marks as they gain hold on the chapter solving these exercises.

**Access Other Exercise Solutions of Class 9 Maths Chapter 13 Surface Areas and Volumes**

Exercise 13.1 solution (9 questions)

Exercise 13.2 solution (8 questions)

Exercise 13.3 solution (9 questions)

Exercise 13.4 solution (5 questions)

Exercise 13.5 solution (5 questions)

Exercise 13.6 solution (8 questions)

Exercise 13.8 solution (10 questions)

Exercise 13.9 solution (3 questions)

**Access Answers to NCERT Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.7**

**1. 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)**

**Solution:**

Volume of cone = (1/3) πr^{2}h cube units

Where r be radius and h be the height of the cone

(i) Radius of cone, r = 6 cm

Height of cone, h = 7cm

Say, V be the volume of the cone, we have

V = (1/3)×(22/7)×36×7

= (12×22)

= 264

The volume of the cone is 264 cm^{3}.

(ii) Radius of cone, r = 3.5cm

Height of cone, h = 12cm

Volume of cone = (1/3)×(22/7)×3.5^{2}×7 = 154

Hence,

The volume of the cone is 154 cm^{3}.

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

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

**(Assume π = 22/7)**

**Solution:**

(i) Radius of cone, r =7 cm

Slant height of cone, l = 25 cm

or h = 24

Height of the cone is 24 cm

Now,

Volume of cone, V = (1/3) πr^{2}h (formula)

V = (1/3)×(22/7) ×7^{2}×24

= (154×8)

= 1232

So, the volume of the vessel is 1232 cm^{3}

Therefore, capacity of the conical vessel = (1232/1000) liters (because 1L = 1000 cm^{3})

= 1.232 Liters.

(ii) Height of cone, h = 12 cm

Slant height of cone, l = 13 cm

r = 5

Hence, the radius of cone is 5 cm.

Now, Volume of cone, V = (1/3)πr^{2}h

V = (1/3)×(22/7)×52×12 cm^{3}

= 2200/7

Volume of cone is 2200/7 cm^{3}

Now, Capacity of the conical vessel= 2200/7000 litres (1L = 1000 cm^{3})

= 11/35 litres

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

**Solution:**

Height of the cone, h = 15 cm

Volume of cone =1570 cm^{3}

Let r be the radius of the cone

As we know: Volume of cone, V = (1/3) πr^{2}h

So, (1/3) πr^{2}h = 1570

(1/3)×3.14×r^{2 }×15 = 1570

r^{2} = 100

r = 10

Radius of the base of cone 10 cm.

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

**Solution:**

Height of cone, h = 9cm

Volume of cone =48π cm^{3}

Let r be the radius of the cone.

As we know: Volume of cone, V = (1/3) πr^{2}h

So, 1/3 π r^{2}(9) = 48 π

r^{2} = 16

r = 4

Radius of cone is 4 cm.

So, diameter = 2×Radius = 8

Thus, diameter of base is 8cm.

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

**(Assume π = 22/7)**

**Solution:**

Diameter of conical pit = 3.5 m

Radius of conical pit, r = diameter/ 2 = (3.5/2)m = 1.75m

Height of pit, h = Depth of pit = 12m

Volume of cone, V = (1/3) πr^{2}h

V = (1/3)×(22/7) ×(1.75)^{2}×12 = 38.5

Volume of cone is 38.5 m^{3}

Hence, capacity of the pit = (38.5×1) kiloliters = 38.5 kiloliters.

**6. The volume of a right circular cone is 9856cm ^{3}. 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)**

**Solution:**

Volume of a right circular cone = 9856 cm^{3}

Diameter of the base = 28 cm

(i) Radius of cone, r = (28/2) cm = 14 cm

Let the height of the cone be h

Volume of cone, V = (1/3) πr^{2}h

(1/3) πr^{2}h = 9856

(1/3)×(22/7) ×14×14×h = 9856

h = 48

The height of the cone is 48 cm.

Slant height of the cone is 50 cm.

(iii) curved surface area of cone = πrl

= (22/7)×14×50

= 2200

curved surface area of the cone is 2200 cm^{2}.

**7. 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.**

**Solution:**

Height (h)= 12 cm

Radius (r) = 5 cm, and

Slant height (l) = 13 cm

Volume of cone, V = (1/3) πr^{2}h

V = (1/3)×π×5^{2}×12

= 100π

Volume of the cone so formed is 100π cm^{3}.

**8. 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.**

**Solution:**

A right-angled ΔABC is revolved about its side 5cm, a cone will be formed of radius as 12 cm, height as 5 cm, and slant height as 13 cm.

Volume of cone = (1/3) πr^{2}h; where r is the radius and h be the height of cone

= (1/3)×π×12×12×5

= 240 π

The volume of the cones of formed is 240π cm^{3}.

So, required ratio = (result of question 7) / (result of question 8) = (100π)/(240π) = 5/12 = 5:12.

**9. 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.**

**(Assume π = 22/7)**

**Solution:**

Radius (r) of heap = (10.5/2) m = 5.25

Height (h) of heap = 3m

Volume of heap = (1/3)πr^{2}h

= (1/3)×(22/7)×5.25×5.25×3

= 86.625

The volume of the heap of wheat is 86.625 m^{3}.

Again,

= (22/7)×5.25×6.05

= 99.825

Therefore, the area of the canvas is 99.825 m^{2}.

This exercise of Chapter 13 of NCERT Solutions for Class 9 Maths helps students to learn how to find the surface area and volume of various geometrical objects in an easy and smart way. It helps students to tackle the questions in the best possible way. For detailed questions and explanations, students can refer to the exercises in the NCERT chapter-wise solutions.

**Key Features of NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volume Exercise 13.7**

Solving these solutions help students to:

- Improve the speed in answering the problems.
- Helps students to find the volume of the right circular cone with given specifications.
- Aids students to find the capacity of a conical vessel in litres.
- Pupils can self-assess their knowledge about the chapter easily.