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Volume of Sphere

Class 9Class 10Surface Areas and Volumes

A sphere is a perfectly round three-dimensional solid where every point on its surface is at the same distance (radius) from the centre. The volume of a sphere measures the total space enclosed within it.


The formula for the volume of a sphere involves the cube of the radius, which means that even a small change in radius causes a large change in volume. Doubling the radius increases the volume by a factor of 8. This topic is part of the NCERT Class 9 chapter on Surface Areas and Volumes (Chapter 13).


Spheres are found everywhere in nature and daily life — footballs, cricket balls, marbles, globes, planets, stars, bubbles, oranges, and water droplets are all spherical or approximately spherical. Calculating the volume is essential for determining capacity (how much a spherical tank can hold), displacement (how much water rises when a sphere is immersed), and material requirements (how much metal is needed for a hollow sphere).


A hemisphere (half of a sphere, obtained by cutting through the centre) is also covered. Its volume is exactly half the volume of the full sphere. Hemispherical bowls, domes, and igloos are practical examples.


The ancient Greek mathematician Archimedes (287–212 BC) first derived the volume formula by showing that a sphere occupies exactly two-thirds of the volume of the smallest cylinder that can contain it. He considered this his greatest achievement and requested that a sphere inside a cylinder be engraved on his tombstone.

What is Volume of Sphere?

Definition: The volume of a sphere is the amount of space enclosed within the spherical surface.


Volume = (4/3)πr³


Where:

  • r = radius of the sphere
  • π = 22/7 or 3.14159...

For a hemisphere:

Volume of hemisphere = (2/3)πr³


Important:

  • The volume depends on the cube of the radius — doubling the radius increases the volume eight times.
  • All points on the surface of a sphere are at distance r from the centre.
  • A sphere has the largest volume for a given surface area among all shapes.

Volume of Sphere Formula

Key Formulas:


1. Volume of a sphere:

V = (4/3)πr³


2. Volume of a hemisphere:

V = (2/3)πr³


3. Volume in terms of diameter (d = 2r):

  • V = (4/3)π(d/2)³ = (πd³)/6

4. Finding radius from volume:

  • r³ = 3V / (4π)
  • r = ⌈3V / (4π)⌉1/3

5. Relationship between sphere and cylinder:

  • A sphere fits exactly inside a cylinder of radius r and height 2r.
  • Volume of sphere = (2/3) × Volume of this cylinder.
  • Volume of cylinder = πr² × 2r = 2πr³
  • (4/3)πr³ = (2/3) × 2πr³ ✓

Units:

  • If r is in cm, volume is in cm³.
  • If r is in m, volume is in .
  • 1 litre = 1000 cm³

Derivation and Proof

Derivation of the Volume of a Sphere (Cavalieri’s Principle approach):


Step 1: Set up the comparison

  • Consider a sphere of radius r and a cylinder of radius r and height 2r with two cones removed (each cone has radius r and height r, apex at the centre).

Step 2: Cross-sectional areas at height h from centre

  • Sphere: At height h from the centre, the cross-section is a circle with radius √(r² − h²). Its area = π(r² − h²).
  • Cylinder minus two cones: At height h, the cylinder cross-section has area πr², and the cone cross-section has area πh². The net area = πr² − πh² = π(r² − h²).

Step 3: Apply Cavalieri’s Principle

  • Since the cross-sectional areas are equal at every height, the volumes are equal.

Step 4: Calculate the volume

  1. Volume of cylinder = πr² × 2r = 2πr³
  2. Volume of two cones = 2 × (1/3)πr² × r = (2/3)πr³
  3. Volume of sphere = Volume of cylinder − Volume of two cones
  4. Volume of sphere = 2πr³ − (2/3)πr³ = (6πr³ − 2πr³)/3 = (4/3)πr³

Types and Properties

Types of volume calculations involving spheres:


1. Full sphere

  • V = (4/3)πr³
  • Used for balls, marbles, planets, bubbles.

2. Hemisphere (half-sphere)

  • V = (2/3)πr³
  • Used for bowls, domes, half-balls.

3. Hollow sphere (spherical shell)

  • V = (4/3)π(R³ − r³), where R = outer radius, r = inner radius.
  • Used for hollow balls, shells, tank walls.

4. Volume of material in a hollow sphere

  • Same as hollow sphere: V = (4/3)π(R³ − r³)
  • This gives the volume of metal or material used.

5. Melting and recasting problems

  • When a solid sphere is melted and recast into smaller spheres, the total volume is conserved.
  • Volume of large sphere = n × Volume of small sphere.

6. Water displacement

  • When a sphere is immersed in water, the water displaced equals the volume of the sphere.
  • Used to find the rise in water level in a container.

Solved Examples

Example 1: Example 1: Volume of a sphere given radius

Problem: Find the volume of a sphere of radius 7 cm. (Use π = 22/7)


Solution:

Given:

  • r = 7 cm

Using V = (4/3)πr³:

  • V = (4/3) × (22/7) × 7³
  • V = (4/3) × (22/7) × 343
  • V = (4/3) × 22 × 49
  • V = (4/3) × 1078
  • V = 4312/3 = 1437.33 cm³

Answer: Volume = 1437.33 cm³ (approx.).

Example 2: Example 2: Volume of a hemisphere

Problem: Find the volume of a hemisphere of radius 10.5 cm. (Use π = 22/7)


Solution:

Given:

  • r = 10.5 cm = 21/2 cm

Using V = (2/3)πr³:

  • V = (2/3) × (22/7) × (21/2)³
  • V = (2/3) × (22/7) × 9261/8
  • V = (2/3) × (22 × 1323)/8
  • V = (2/3) × (29106/8)
  • V = (2/3) × 3638.25
  • V = 2425.5 cm³

Answer: Volume = 2425.5 cm³.

Example 3: Example 3: Finding radius from volume

Problem: The volume of a sphere is 4851 cm³. Find its radius. (Use π = 22/7)


Solution:

Given:

  • V = 4851 cm³

Using V = (4/3)πr³:

  1. 4851 = (4/3) × (22/7) × r³
  2. 4851 = (88/21) × r³
  3. r³ = 4851 × 21/88 = 101871/88 = 1157.625
  4. r³ = 1157.625
  5. r = (1157.625)1/3 = 10.5 cm

Answer: Radius = 10.5 cm.

Example 4: Example 4: Volume of a sphere given diameter

Problem: A spherical ball has a diameter of 28 cm. Find its volume. (Use π = 22/7)


Solution:

Given:

  • Diameter = 28 cm ⇒ r = 14 cm

Using V = (4/3)πr³:

  • V = (4/3) × (22/7) × 14³
  • V = (4/3) × (22/7) × 2744
  • V = (4/3) × 22 × 392
  • V = (4/3) × 8624
  • V = 34496/3 = 11498.67 cm³

Answer: Volume ≈ 11498.67 cm³.

Example 5: Example 5: Melting and recasting

Problem: A metallic sphere of radius 21 cm is melted and recast into smaller spheres of radius 7 cm each. How many smaller spheres are formed?


Solution:

Given:

  • Radius of large sphere (R) = 21 cm
  • Radius of small sphere (r) = 7 cm

Volume of large sphere:

  • V_large = (4/3)πR³ = (4/3)π(21)³ = (4/3)π × 9261

Volume of one small sphere:

  • V_small = (4/3)πr³ = (4/3)π(7)³ = (4/3)π × 343

Number of small spheres:

  • n = V_large / V_small = 9261 / 343 = 27

Answer: 27 smaller spheres are formed.

Example 6: Example 6: Water displacement

Problem: A sphere of radius 3.5 cm is dropped into a cylindrical vessel of radius 7 cm partly filled with water. Find the rise in water level.


Solution:

Given:

  • Radius of sphere (r) = 3.5 cm
  • Radius of cylinder (R) = 7 cm

Volume of sphere:

  • V = (4/3)π(3.5)³ = (4/3) × (22/7) × 42.875 = (4/3) × 134.75 = 179.67 cm³

Rise in water level (h):

  • Volume of water displaced = πR²h
  • 179.67 = (22/7) × 49 × h
  • 179.67 = 154h
  • h = 179.67/154 = 1.167 cm (approx.)

Answer: The water level rises by approximately 1.17 cm.

Example 7: Example 7: Hollow sphere

Problem: A hollow sphere has an outer radius of 10 cm and inner radius of 8 cm. Find the volume of the material used. (Use π = 3.14)


Solution:

Given:

  • R = 10 cm (outer), r = 8 cm (inner)

Volume of material:

  • V = (4/3)π(R³ − r³)
  • V = (4/3) × 3.14 × (1000 − 512)
  • V = (4/3) × 3.14 × 488
  • V = 4.1867 × 488
  • V = 2043.12 cm³ (approx.)

Answer: Volume of material ≈ 2043.12 cm³.

Example 8: Example 8: Comparing volumes of sphere and hemisphere

Problem: A sphere and a hemisphere have the same radius of 6 cm. Find the ratio of their volumes.


Solution:

Volume of sphere:

  • V_sphere = (4/3)πr³

Volume of hemisphere:

  • V_hemi = (2/3)πr³

Ratio:

  • V_sphere / V_hemi = [(4/3)πr³] / [(2/3)πr³] = 4/2 = 2 : 1

Answer: The ratio of volumes is 2 : 1.

Example 9: Example 9: Volume in litres

Problem: A hemispherical bowl has an internal radius of 10.5 cm. Find the volume of milk it can hold in litres. (Use π = 22/7)


Solution:

Given:

  • r = 10.5 cm

Volume of hemisphere:

  • V = (2/3)πr³ = (2/3) × (22/7) × (10.5)³
  • V = (2/3) × (22/7) × 1157.625
  • V = (2/3) × 3637.5
  • V = 2425 cm³

Converting to litres:

  • 1 litre = 1000 cm³
  • V = 2425/1000 = 2.425 litres

Answer: The bowl can hold 2.425 litres of milk.

Example 10: Example 10: Sphere inscribed in a cube

Problem: A sphere is inscribed in a cube of side 14 cm. Find the volume of the sphere and the volume of the cube not occupied by the sphere. (Use π = 22/7)


Solution:

Given:

  • Side of cube = 14 cm
  • Diameter of sphere = side of cube = 14 cm ⇒ r = 7 cm

Volume of sphere:

  • V_sphere = (4/3) × (22/7) × 7³ = (4/3) × 22 × 49 = 4312/3 = 1437.33 cm³

Volume of cube:

  • V_cube = 14³ = 2744 cm³

Unoccupied volume:

  • V = 2744 − 1437.33 = 1306.67 cm³

Answer: Volume of sphere = 1437.33 cm³; unoccupied volume = 1306.67 cm³.

Real-World Applications

Applications of Volume of Sphere:


  • Sports Equipment Manufacturing: The volume formula determines how much rubber, leather, or synthetic material is needed to manufacture balls for cricket, football, basketball, tennis, and golf. A cricket ball has a radius of approximately 3.6 cm, giving a volume of about 195 cm³ of cork, twine, and leather. Football manufacturers use the formula to ensure consistent size and weight across all match balls.
  • Medicine and Pharmacology: Spherical pills, capsules, and drug-delivery microspheres are designed using volume calculations. Doctors estimate tumour sizes by measuring their radius on CT scans and calculating volume using V = (4/3)πr³. This helps in treatment planning and monitoring disease progression.
  • Astronomy and Space Science: The volume of planets, moons, and stars is calculated using the sphere formula. Earth’s volume is approximately (4/3)π(6,371)³ ≈ 1.083 × 10¹² km³. Comparing the volumes of planets helps scientists understand their densities and compositions.
  • Storage Tanks and Water Supply: Spherical and hemispherical water tanks are used in elevated water supply systems because a sphere provides maximum volume for a given surface area (minimum material cost). Engineers calculate tank capacity using the volume formula to ensure adequate water supply for a community.
  • Metallurgy and Foundries: When metal objects are melted and recast into different shapes, the total volume is conserved. Foundries use volume calculations to determine how many smaller balls, rods, or sheets can be produced from a given sphere of molten metal. This is a standard application of the "melting and recasting" principle.
  • Food Industry: Ice cream scoops are hemispherical, and their volume determines the serving size. Spherical candies, laddoos, and fruits (oranges, watermelons) all involve volume estimation for packaging, pricing, and nutritional labelling.
  • Environmental Science: Raindrops, hailstones, and bubbles are approximately spherical. Scientists calculate the volume of water in a raindrop (radius ~1 mm, volume ~4.19 mm³) to estimate total rainfall from the number of drops per square metre.
  • Nanotechnology: At the nanoscale, spherical nanoparticles are used in drug delivery, catalysis, and electronics. Volume calculations at the nanometre scale determine the number of atoms or molecules that can be encapsulated inside a nanoparticle.

Key Points to Remember

  • Volume of sphere = (4/3)πr³, where r is the radius of the sphere.
  • Volume of hemisphere = (2/3)πr³ = exactly half the volume of the full sphere.
  • Volume depends on the cube of the radius — doubling the radius makes the volume 8 times larger (2³ = 8). Tripling the radius makes it 27 times larger.
  • Hollow sphere (spherical shell): Volume of material = (4/3)π(R³ − r³), where R is the outer radius and r is the inner radius.
  • A sphere fits exactly inside a cylinder of height 2r and radius r. The sphere’s volume is 2/3 of the cylinder’s volume. Archimedes discovered this remarkable relationship.
  • In melting and recasting problems, the total volume is conserved. Volume of original = sum of volumes of new shapes.
  • Volume of water displaced by an immersed sphere = volume of the sphere. This is Archimedes’ principle applied to spheres.
  • To convert volume from cm³ to litres: divide by 1000. Since 1 litre = 1000 cm³, a volume of 4,500 cm³ = 4.5 litres.
  • Among all solids with the same surface area, a sphere has the maximum volume. This is why bubbles are spherical — nature minimises surface area for a given volume of air.
  • To find the radius when volume is given: r = (3V/(4π))^(1/3). Take the cube root of 3V/(4π).
  • The volume formula can be derived using Cavalieri’s Principle by comparing cross-sections of a sphere with a cylinder minus two cones.
  • The volume of a sphere is covered in Chapter 13 (Surface Areas and Volumes) of the CBSE Class 9 NCERT textbook.

Practice Problems

  1. Find the volume of a sphere of radius 3.5 cm. (Use π = 22/7)
  2. Find the volume of a hemisphere of diameter 21 cm.
  3. A sphere has a volume of 36π cm³. Find its radius.
  4. A metallic sphere of radius 6 cm is melted and recast into 8 equal smaller spheres. Find the radius of each smaller sphere.
  5. A hemispherical tank of internal radius 1.4 m is full of water. Find the volume of water in litres.
  6. A hollow sphere has outer diameter 14 cm and inner diameter 10 cm. Find the volume of the metal used.
  7. The radius of a sphere is increased by 50%. By what percentage does its volume increase?
  8. A sphere of radius 4 cm is dropped into a cylindrical vessel of radius 8 cm. Find the rise in water level.

Frequently Asked Questions

Q1. What is the formula for the volume of a sphere?

The volume of a sphere is V = (4/3)πr³, where r is the radius of the sphere.

Q2. What is the volume of a hemisphere?

The volume of a hemisphere is V = (2/3)πr³, which is exactly half the volume of a full sphere with the same radius.

Q3. How does doubling the radius affect the volume?

Doubling the radius increases the volume by a factor of 2³ = 8. Since volume depends on r³, any change in radius has a cubic effect on volume.

Q4. What is a hollow sphere?

A hollow sphere has two concentric spherical surfaces with an outer radius R and inner radius r. The volume of material = (4/3)π(R³ − r³).

Q5. How do you convert volume from cm³ to litres?

Divide by 1000. Since 1 litre = 1000 cm³, a volume of 5000 cm³ = 5 litres.

Q6. What is the relationship between the volumes of a sphere and a cylinder?

A sphere of radius r fits inside a cylinder of radius r and height 2r. The volume of the sphere is (2/3) of the volume of this cylinder.

Q7. How do you find the radius if the volume is given?

Rearrange the formula: r³ = 3V/(4π), then r = (3V/(4π))^(1/3). Take the cube root of the result.

Q8. Is the volume of a sphere topic in CBSE Class 9?

Yes. Volume of sphere and hemisphere is part of Chapter 13 (Surface Areas and Volumes) in the CBSE Class 9 syllabus.

Q9. What is the volume of a sphere in terms of diameter?

If the diameter is d, then r = d/2. Substituting: V = (4/3)π(d/2)³ = πd³/6.

Q10. Why is a sphere the most efficient shape for volume?

Among all closed surfaces with a given surface area, the sphere encloses the maximum volume. This is why bubbles form spheres — nature minimises surface area for a given volume of air.

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