Question

A sphere has a radius of 6 meters. A second sphere has a radius of 3 meters. What is the difference of the volumes of the spheres?

Answer 1

Given

Radius of 1st sphere = 6 m.

Radius of 2nd sphere = 3 m.

To find

Difference b/w volumes of two spheres.

Solution

Here, radius of two spheres given as 6 m & 3 m.

➝ r₁ = 6 m.

➝ r₂ = 3 m.

We know that,

➥ Volume of sphere = 4/3 × πr³

Putting values we get :

➻ Volume₁ = 4/3 × (22/7 × 6³)

➻ Volume₁ = 4/3 × (22/7 × 216)

➻ Volume₁ = 4 × 22/7 × 72

➻ Volume₁ = 6336/7

➻ Volume₁ = 905.14 m³

Now finding volume of 2nd sphere :

➾ Volume₂ = 4/3 × (22/7 × 3³)

➾ Volume₂ = 4/3 × (22/7 × 27)

➾ Volume₂ = 4 × 22/7 × 9

➾ Volume₂ = 792/7

➾ Volume₂ = 113.14 m³

Now finding difference b/w two volumes :

⟿ Difference = Volume₁ - Volume₂

⟿ Difference = 905.14 - 113.14

⟿ Difference = 792 m³

Therefore,

Difference b/w volumes of two spheres is 792 m³ .

Answer 2

Given :-

• Radius of First Sphere = 6m.
• Radius of second sphere = 3m.

To Find :-

• Difference b/w their volumes ?

Formula used :-

• Volume of sphere = (4/3) * π * (radius)³ .

Solution :-

Let us Assume that, radius of first sphere with greater radius is r₁ & radius of second sphere with smaller radius is r₂ .

So,

→ Required V = [(4/3) * π * (r₁)³] - [(4/3) * π * (r₂)³]

→ Required V = (4/3) * π * [ (r₁)³ - (r₂)³]

→ Required V = (4/3) * π * [ (6)³ - (3)³]

→ Required V = (4/3) * π [ 216 - 27 ]

→ Required V = (4/3) * (22/7) * 189

→ Required V = (4 * 22 * 189) / 21

→ Required V = 4 * 22 * 9

→ Required V = 88 * 9

→ Required V = 792cm³ (Ans.)

Hence, Required Difference b/w their volume will be 792cm³.

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Extra :-

→ CSA = TSA of sphere = 4 * π * (radius)² .

→ CSA of Hemi-Sphere = 2 * π * (radius)².

→ TSA of Hemi-Sphere = 3 * π * (radius)².

→ Volume of Hemi-Sphere = (2/3) * π * (radius)³.