Class 9 NCERT Solutions- Chapter 13 Surface Areas And Volumes - Exercise 13.8

Chapter 13 of the Class 9 NCERT Mathematics textbook focuses on the Surface Areas and Volumes of essential topics in geometry. Exercise 13.8 provides practice problems that help students apply their understanding of the surface areas and volumes of the various 3D shapes such as cuboids, cylinders, cones, and spheres. This exercise aims to solidify students' skills in calculating and solving real-world problems related to these geometric concepts.

What are Surface Areas and Volumes?

The Surface Areas refer to the total area of the outer surfaces of the three-dimensional object. It is the sum of the areas of all faces or curved surfaces of the object. Volumes measure the amount of space enclosed within the 3D shape. It represents the capacity of the object. The Surface areas and volumes are crucial in understanding the properties of the solids and in practical applications like packaging, construction, and material estimation.

Question 1: Find the volume of a sphere whose radius is

(i) 7 cm (ii) 0.63 m    (Assume π =22/7)

Solution:

(i) Given: Radius of sphere, r = 7 cm

By using the formula, volume of sphere = (4/3) πr³

= (4/3)×(22/7)×(7)³

= 4312/3 cm³

Therefore, the volume of the sphere is 4312/3 cm³

(ii) Given: Radius of sphere, r = 0.63 m

By using the formula, volume of sphere = (4/3) πr³

= (4/3)×(22/7)×(0.63)³

= 1.0478 m³

Therefore, the volume of the sphere is 1.05 m³ (approx).

Question 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) Given: Diameter = 28 cm

Therefore, the radius, r = 28/2 cm = 14cm

By using the formula, volume of the solid spherical ball = (4/3) πr³

= (4/3)×(22/7)×(14)³

= 34496/3 cm³

Therefore, the volume of the ball is 34496/3 cm³

(ii) Given: Diameter = 0.21 m

Therefore, the radius of the ball =0.21/2 m= 0.105 m

By using the formula, volume of the ball = (4/3 )πr³

= (4/3)× (22/7)×(0.105)³ 

= 0.004851 m³

Therefore, the volume of the ball = 0.004851 m³

Question 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 cm³? (Assume π=22/7)

Solution:

Given: Diameter of a metallic ball = 4.2 cm

Therefore, the radius of the metallic ball, r = 4.2/2 cm = 2.1 cm

By using the formula, volume of the metallic ball = 4/3 πr³

= (4/3)×(22/7)×(2.1)³ 

= 38.808 cm³

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

Density = Mass/Volume

Mass = Density × volume

= (8.9×38.808) g

= 345.3912 g

Therefore, the mass of the ball is 345.39 g (approx.)

Question 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 us assume the diameter of the earth as “d”. 

Therefore, the radius of the earth will be d/2

The diameter of the moon will be d/4 

The radius of the moon will be d/8

Finding the volume of the moon.

By using the formula, volume of the moon = (4/3) πr³ 

= (4/3) π (d/8)³ 

= 4/3π(d³/512)

Finding the volume of the earth :

By using the formula, volume of the earth = (4/3) πr³

= (4/3) π (d/2)³ 

= 4/3π(d³/8)

The fraction of the volume of the earth to the volume of the moon

Volume of moon/volume of earth = 4/3π(d³/512)/4/3π(d³/8)

= 1/64

Therefore, the volume of moon is 1/64 ^th of the volume of earth.

Question 5: How many liters of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume π = 22/7)

Solution:

Given: Diameter of hemispherical bowl = 10.5 cm

Therefore, the radius of hemispherical bowl, r = 10.5/2 cm = 5.25 cm

By using the formula, volume of the hemispherical bowl = (2/3) πr³

= (2/3)×(22/7)×(5.25)³ 

= 303.1875 cm³

Therefore, the volume of the hemispherical bowl is 303.1875 cm³

Capacity of the bowl = (303.1875)/1000 L 

= 0.303 liters(approx.)

Therefore, the hemispherical bowl can hold 0.303 liters of milk.

Question 6: A hemispherical 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:

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

Also, the thickness of hemispherical tank = 0.01m

Therefore, the outer Radius (R) = 1 + 0.01 = 1.01m

By using the formula, volume of the iron used in the tank = (2/3) π(R³– r³)

= (2/3)×(22/7) × (1.01³– 1³) 

= 0.06348 m³

Therefore, the volume of the iron used in the hemispherical tank is 0.06348 m³.

Question 7: Find the volume of a sphere whose surface area is 154 cm². (Assume π = 22/7)

Solution:

Given: Surface area of sphere =154 cm²

Let us assume r to be the radius of a sphere.

By using the formula, Surface area of sphere = 4πr²

4πr² = 154 cm²

r² = (154×7)/(4 ×22)

r = 7/2

Radius r = 7/2 cm

Now,

By using the formula, volume of the sphere = (4/3) πr³

= (4/3)×(22/7)×(7/2)³

= 539/3 cm³

Therefore, the volume of a sphere is 539/3 cm³

Question 8: A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs. 4989.60. If the cost of white-washing is Rs20 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) Given: Cost of white-washing the dome from inside = Rs 4989.60

Also, Cost of white-washing 1m² area = Rs 20

Therefore, CSA of the inner side of dome = 498.96/2 m²  

= 249.48 m²

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

CSA of the inner side of dome = 249.48 m²

By using the formula, CSA of a hemisphere = 2πr²

2πr² = 249.48

2×(22/7)×r² = 249.4

r² = (249.48×7)/(2×22)

r² = 39.6

r = 6.3

Hence, the radius r = 6.3 m

Volume of air inside the dome = Volume of hemispherical dome

By using the formula, volume of the hemisphere = 2/3 πr³

= (2/3)×(22/7)×(6.3)³

= 523.908 m³

= 523.9(approx.)

Therefore, the volume of air inside the dome is 523.9 m³.

Question 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 S and S’.

Solution:

By using the formula, volume of the solid sphere = (4/3)πr³

Therefore, the volume of twenty-seven solid sphere = 27×(4/3)πr³ = 36 πr³

(i) Given: new solid iron sphere radius = r’

By using the formula, volume of this new sphere = (4/3)π(r’)³

(4/3)π(r’)³ = 36 πr³

(r’)³ = 27r³

r’= 3r

Radius of new sphere = 3r 

(ii) We know that,

Surface area of iron sphere of radius r, S =4πr²

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

S/S’ = (4πr²)/( 4π(r’)²)

S/S’ = r²/(3r’)² 

= r²/9r²

= 1/9

Therefore, the ratio of S and S’ is 1: 9.

Question 10: A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm³) is needed to fill this capsule? (Assume π = 22/7)

Solution:

Given: Diameter of capsule = 3.5 mm

Therefore, the radius of capsule, r = (3.5/2) mm = 1.75mm

By using the formula, volume of spherical capsule = 4/3 πr³

= (4/3)×(22/7)×(1.75)³ 

= 22.458 mm³

Therefore, the volume of the spherical capsule is 22.46 mm³.

Conclusion

Exercise 13.8 in Chapter 13 of the Class 9 NCERT textbook provides the essential practice for the mastering surface areas and volumes of the various solids. By solving these problems, students enhance their ability to the calculate and apply these concepts to the real-world situations reinforcing their understanding of the geometry.