Class 9 Maths NCERT Solutions Chapter 11 Surface Area and Volume Exercise 11.4 focuses on solving problems related to the volume and surface area of spheres and hemispheres. This exercise introduces students to new formulas and helps them apply these concepts to practical questions involving solid shapes. It is designed to strengthen their understanding of how to calculate curved and total surface areas, as well as volumes, making the chapter both interesting and essential for geometry skills.
These solutions offer detailed, step-by-step explanations to help students grasp the methods easily and avoid common mistakes in calculations. Practicing Exercise 11.4 not only prepares students for their exams but also builds a strong foundation for higher-level mathematics. Following the latest NCERT guidelines, these solutions serve as a valuable resource for effective learning and revision.
Get the complete and trusted NCERT Solutions Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.4 in a downloadable PDF format. The solutions are prepared by subject experts in alignment with the latest CBSE guidelines so that students can easily revise and practice.
Here is a quick overview and access to solutions for all exercises from Chapter 11, Surface Area and Volume.
(Assume π=22/7 unless stated otherwise)
1. Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m
Sol.
The volume of a sphere is given by V = (4/3)πr³.
(i) Radius (r) = 7 cm.
Volume = (4/3) × (22/7) × (7)³ cm³
= (4/3) × (22/7) × 343 cm³
= (4/3) × 22 × 49 cm³
= (88 × 49) / 3 cm³
= 4312 / 3 cm³
= 1437 1/3 cm³ (or approximately 1437.33 cm³).
(ii) Radius (r) = 0.63 m.
Volume = (4/3) × (22/7) × (0.63)³ m³
= (4/3) × (22/7) × 0.250047 m³
= (88/21) × 0.250047 m³
= 4.19047... × 0.250047 m³
= 1.047816 m³ (approximately 1.05 m³).
2. Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m
Sol.
The amount of water displaced by a spherical ball is equal to its volume. The volume of a sphere is V = (4/3)πr³.
(i) Diameter = 28 cm.
Radius (r) = 28/2 cm = 14 cm.
Amount of water displaced = (4/3) × (22/7) × (14)³ cm³
= (4/3) × (22/7) × 2744 cm³
= (88/21) × 2744 cm³
= 241472 / 21 cm³
= 11498 2/3 cm³ (or approximately 11498.67 cm³).
(ii) Diameter = 0.21 m.
Radius (r) = 0.21/2 m = 0.105 m.
Amount of water displaced = (4/3) × (22/7) × (0.105)³ m³
= (4/3) × (22/7) × 0.001157625 m³
= (88/21) × 0.001157625 m³
= 0.004851 m³ (approximately).
3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?
Sol.
Diameter of metallic ball = 4.2 cm.
Radius of metallic ball (r) = 4.2/2 cm = 2.1 cm.
Volume of sphere = (4/3)πr³
= (4/3) × (22/7) × (2.1)³ cm³
= (4/3) × (22/7) × 9.261 cm³
= (88/21) × 9.261 cm³
= 4.19047... × 9.261 cm³
= 38.808 cm³.
Density = Mass / Volume.
So, Mass = Density × Volume.
Mass = 8.9 g/cm³ × 38.808 cm³
= 345.3912 g.
Hence, the mass of the ball is 345.39 g (approximately).
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?
Sol.
Let d_moon and d_earth be the diameters of the moon and the earth respectively.
Given, d_moon = (1/4) d_earth.
This implies r_moon = (1/4) r_earth (since radius is half the diameter).
Volume of a sphere = (4/3)πr³.
Volume of moon (V_moon) = (4/3)π(r_moon)³
Volume of earth (V_earth) = (4/3)π(r_earth)³
Fraction of the volume of the earth that is the volume of the moon = V_moon / V_earth
= [(4/3)π(r_moon)³] / [(4/3)π(r_earth)³]
= (r_moon / r_earth)³
Substitute r_moon = (1/4) r_earth:
= ((1/4) r_earth / r_earth)³ = (1/4)³ = 1/64.
The volume of the moon is 1/64 fraction of the volume of the earth.
5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Sol.
Diameter = 10.5 cm.
Radius (r) = 10.5/2 cm = 5.25 cm = 21/4 cm.
Capacity of the hemispherical bowl = (2/3)πr³
= (2/3) × (22/7) × (21/4)³ cm³
= (2/3) × (22/7) × (21/4) × (21/4) × (21/4) cm³
= (2 × 22 × 21 × 21 × 21) / (3 × 7 × 4 × 4 × 4) cm³
= (4851 × 2) / 32 cm³ (Incorrect multiplication in original)
Let's recalculate:
= (2 × 22 × 3 × 21 × 21) / (3 × 4 × 4 × 4) cm³
= (22 × 21 × 21) / (4 × 4) cm³
= (22 × 441) / 16 cm³
= 9702 / 16 cm³ = 606.375 cm³.
Capacity in litres: 1 L = 1000 cm³.
Capacity = 606.375 / 1000 L = 0.606375 L (approximately 0.606 L).
6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Sol.
Inner radius (r) = 1 m.
Thickness of iron sheet = 1 cm = 0.01 m.
Outer radius (R) = Inner radius (r) + Thickness of iron sheet = 1 m + 0.01 m = 1.01 m.
The volume of the iron used is the difference between the outer volume and the inner volume of the hemisphere.
Volume of iron used = (2/3)π(R³ − r³)
= (2/3) × (22/7) × ((1.01)³ − (1)³) m³
= (44/21) × (1.030301 − 1) m³
= (44/21) × 0.030301 m³
= 0.06348 m³ (approximately).
7. Find the volume of a sphere whose surface area is 154 cm².
Sol.
Surface area of the sphere = 154 cm².
First, find the radius (r) from the surface area:
4πr² = 154
4 × (22/7) × r² = 154
(88/7) × r² = 154
r² = (154 × 7) / 88
r² = (7 × 7) / 4 = 49/4
r = √(49/4) = 7/2 cm.
Now, find the volume of the sphere:
Volume = (4/3)πr³
= (4/3) × (22/7) × (7/2)³ cm³
= (4/3) × (22/7) × (343/8) cm³
= (4 × 22 × 49) / (3 × 8) cm³
= (11 × 49) / 3 cm³
= 539 / 3 cm³
= 179 2/3 cm³ (or approximately 179.67 cm³).
7. A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of Rs. 4989.6. If the cost of white washing is Rs. 20 per square metre, find the
(i) Inside surface area of the dome,
(ii) Volume of the air inside the dome.
Sol.
(i) Total cost of white washing = Rs. 4989.6.
Cost of white washing per m² = Rs. 20.
Inside surface area of the dome = Total Cost / Cost per m²
= 4989.6 / 20 m²
= 249.48 m².
The inside surface area of a hemispherical dome is 2πr².
2πr² = 249.48
2 × (22/7) × r² = 249.48
(44/7) × r² = 249.48
r² = (249.48 × 7) / 44
r² = 1746.36 / 44
r² = 39.69
r = √39.69 = 6.3 m.
(ii) The volume of air inside the dome is the volume of the hemisphere.
Volume = (2/3)πr³
= (2/3) × (22/7) × (6.3)³ m³
= (44/21) × 250.047 m³
= 523.908 m³ (approximately 523.9 m³).
8. Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find:
(i) radius r' of the new sphere,
(ii) ratio of S and S'.
Sol.
(i) When spheres are melted and recast, their total volume remains the same.
Volume of 27 small spheres = Volume of the new large sphere.
27 × (4/3)πr³ = (4/3)π(r')³
Divide both sides by (4/3)π:
27r³ = (r')³
Take the cube root of both sides:
r' = ³√(27r³)
r' = 3r.
The radius of the new sphere is 3r.
(ii) Surface area of one small sphere (S) = 4πr².
Surface area of the new large sphere (S') = 4π(r')².
Substitute r' = 3r:
S' = 4π(3r)² = 4π(9r²) = 36πr².
Ratio of S and S' = S / S' = (4πr²) / (36πr²)
= 1/9.
The ratio S:S' is 1:9.
9. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?
Sol.
Diameter = 3.5 mm.
Radius (r) = 3.5/2 mm = 1.75 mm.
Capacity of the capsule (volume of the sphere) = (4/3)πr³
= (4/3) × (22/7) × (1.75)³ mm³
= (4/3) × (22/7) × 5.359375 mm³
= (88/21) × 5.359375 mm³
= 22.458 mm³ (approximately).
(Session 2025 - 26)