NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volume Exercise 13.8 is made available here for the students in PDF format. These problems in NCERT Solutions have been solved by experienced Maths experts considering the understanding and grasping capabilities of the students. All the solutions for 9th Maths subjects are in accordance with NCERT syllabus and guidelines and are helpful for the students to score good marks.

### Download PDF of NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.8

### Access Other Exercise Solutions of 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 13.7 solution (9 questions)

Exercise 13.9 solution (3 questions)

### Access Answers to NCERT Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.8

**1. Find the volume of a sphere whose radius is **

**(i) 7 cm (ii) 0.63 m**

**(Assume π =22/7)**

**Solution:**

(i) Radius of sphere, r = 7 cm

Using, Volume of sphere = (4/3) πr^{3}

= (4/3)×(22/7)×7^{3}

= 4312/3

Hence, volume of the sphere is 4312/3 cm^{3}

(ii) Radius of sphere, r = 0.63 m

Using, volume of sphere = (4/3) πr^{3}

= (4/3)×(22/7)×0.63^{3}

= 1.0478

Hence, volume of the sphere is 1.05 m^{3 }(approx).

**2. Find the amount of water displaced by a solid spherical ball of diameter**

**(i) 28 cm (ii) 0.21 m **

**(Assume π =22/7)**

**Solution:**

(i) Diameter = 28 cm

Radius, r = 28/2 cm = 14cm

Volume of the solid spherical ball = (4/3) πr^{3}

Volume of the ball = (4/3)×(22/7)×14^{3 }= 34496/3

Hence, volume of the ball is 34496/3 cm^{3}

(ii) Diameter = 0.21 m

Radius of the ball =0.21/2 m= 0.105 m

Volume of the ball = (4/3 )πr^{3}

Volume of the ball = (4/3)× (22/7)×0.105^{3} m^{3}

Hence, volume of the ball = 0.004851 m^{3}

**3.The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm ^{3}? (Assume π=22/7)**

**Solution:**

Given,

Diameter of a metallic ball = 4.2 cm

Radius(r) of the metallic ball, r = 4.2/2 cm = 2.1 cm

Volume formula = 4/3 πr^{3}

Volume of the metallic ball = (4/3)×(22/7)×2.1 cm^{3}

Volume of the metallic ball = 38.808 cm^{3}

Now, using relationship between, density, mass and volume,

Density = Mass/Volume

Mass = Density × volume

= (8.9×38.808) g

= 345.3912 g

Mass of the ball is 345.39 g (approx).

**4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?**

**Solution:**

Let the diameter of earth be “d”. Therefore, the radius of earth will be will be d/2

Diameter of moon will be d/4 and the radius of moon will be d/8

Find the volume of the moon :

Volume of the moon = (4/3) πr^{3 }= (4/3) π (d/8)^{3} = 4/3π(d^{3}/512)

Find the volume of the earth :

Volume of the earth = (4/3) πr^{3}= (4/3) π (d/2)^{3} = 4/3π(d^{3}/8)

Fraction of the volume of the earth is the volume of the moon

Answer: Volume of moon is of the 1/64 volume of earth.

**5. How many litres of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume π = 22/7)**

**Solution:**

Diameter of hemispherical bowl = 10.5 cm

Radius of hemispherical bowl, r = 10.5/2 cm = 5.25 cm

Formula for volume of the hemispherical bowl = (2/3) πr^{3}

Volume of the hemispherical bowl = (2/3)×(22/7)×5.25^{3} = 303.1875

Volume of the hemispherical bowl is 303.1875 cm^{3}

Capacity of the bowl = (303.1875)/1000 L = 0.303 litres(approx.)

Therefore, hemispherical bowl can hold 0.303 litres of milk.

**6. A hemi spherical tank is made up of an iron sheet 1cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. (Assume π = 22/7)**

**Solution:**

Inner Radius of the tank, (r ) = 1m

Outer Radius (R ) = 1.01m

Volume of the iron used in the tank = (2/3) π(R^{3}– r^{3})

Put values,

Volume of the iron used in the hemispherical tank = (2/3)×(22/7)×(1.01^{3}– 1^{3}) = 0.06348

So, volume of the iron used in the hemispherical tank is 0.06348 m^{3}.

**7. Find the volume of a sphere whose surface area is 154 cm ^{2}. (Assume π = 22/7)**

**Solution:**

Let r be the radius of a sphere.

Surface area of sphere = 4πr^{2}

4πr^{2 }= 154 cm^{2} (given)

r^{2} = (154×7)/(4 ×22)

r = 7/2

Radius is 7/2 cm

Now,

Volume of the sphere = (4/3) πr^{3}

^{}

**8. A dome of a building is in the form of a hemi sphere. From inside, it was white-washed at the cost of Rs. 4989.60. If the cost of white-washing isRs20 per square meter, find the **

**(i) inside surface area of the dome (ii) volume of the air inside the dome**

** (Assume π = 22/7)**

**Solution:**

(i) Cost of white-washing the dome from inside = Rs 4989.60

Cost of white-washing 1m^{2} area = Rs 20

CSA of the inner side of dome = 498.96/2 m^{2 }= 249.48 m^{2}

(ii) Let the inner radius of the hemispherical dome be r.

CSA of inner side of dome = 249.48 m^{2} (from (i))

Formula to find CSA of a hemi sphere = 2πr^{2}

2πr^{2} = 249.48

2×(22/7)×r^{2 }= 249.48

r^{2 } = (249.48×7)/(2×22)

r^{2 }= 39.69

r = 6.3

So, radius is 6.3 m

Volume of air inside the dome = Volume of hemispherical dome

Using formula, volume of the hemisphere = 2/3 πr^{3}

= (2/3)×(22/7)×6.3×6.3×6.3

= 523.908

= 523.9(approx.)

Answer: Volume of air inside the dome is 523.9 m^{3}.

**9. Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the **

**(i) radius r’ of the new sphere, **

**(ii) ratio of Sand S’.**

**Solution:**

Volume of the solid sphere = (4/3)πr^{3}

Volume of twenty seven solid sphere = 27×(4/3)πr^{3} = 36 π r^{3}

(i) New solid iron sphere radius = r’

Volume of this new sphere = (4/3)π(r’)^{3}

(4/3)π(r’)^{3 }= 36 π r^{3}

(r’)^{3 }= 27r^{3}

r’= 3r

Radius of new sphere will be 3r (thrice the radius of original sphere)

(ii) Surface area of iron sphere of radius r, S =4πr^{2}

Surface area of iron sphere of radius r’= 4π (r’)^{2}

Now

S/S’ = (4πr^{2})/( 4π (r’)^{2})

S/S’ = r^{2}/(3r’)^{2} = 1/9

The ratio of S and S’ is 1: 9.

**10. A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm ^{3}) is needed to fill this capsule? (Assume π = 22/7)**

**Solution:**

Diameter of capsule = 3.5 mm

Radius of capsule, say r = diameter/ 2 = (3.5/2) mm = 1.75mm

Volume of spherical capsule = 4/3 πr^{3}

Volume of spherical capsule = (4/3)×(22/7)×(1.75)^{3} = 22.458

Answer: The volume of the spherical capsule is 22.46 mm^{3}.

This exercise features ten questions based on the volume of the sphere. For all exercise questions for NCERT Solutions for Class 9 Maths Chapter 13, reach us at BYJU’S. Solve previous year questions papers along with solutions, to get an idea of question pattern. You can also get here, notes, and books along with tips and tricks of learning for the 9th standard syllabus of Maths and Science. The questions in NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 includes short and long answers question which should be practised for a better score. NCERT solutions provide answers to all the problems of subjects like Maths and Science for all the classes.

### Key Features of NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volume Exercise 13.8

Solving these solutions help students to:

- Improve the speed in answering the problems.
- Pupils can assess their problem-solving ability and prepare accordingly.
- Helps students to find the volume of the right circular cone with given specifications.
- The method of finding the radius of the sphere can be known.