Question

A hollow sphere of the internal and external radius is 6cm and 8cm respectively, is melted and recast into small cones of base 2cm and height 8cm. Find the no. of cones

Answer 1

Given:-

• A hollow sphere of the Internal and external radius is 6cm and 8cm.

• It is melted and recast into small cones of base 2cm and Height 8cm.

To Find:-

The No. of cones.

FORMULAE USED:-

Volume of Sphere = $\rm{\dfrac{4}{3}\pi r^3}$

Volume of cone = $\rm{\dfrac{1}{3}\pi r^2h}$

Now,

Inner radius = r¹ , Outer radius = r²

→ Volume of Sphere = 4/3π(r²)³ - 4/3π(r¹)³

→ Volume of Sphere = 4/3π{(r²)³ - (r¹)³}

→ $\rm{\dfrac{4}{3}\times{\pi}\times{(8^3)} - (6)^3}}$

→ $\rm{\dfrac{4}{3}\times{\pi}\times{512 - 216}}$

→ $\rm{\dfrac{4}{3}\times{\pi}\times{ 296}}$

→ $\rm{\dfrac{1184\pi}{3}}$.

Therefore,

→ Volume of cone = 1/3πr²h

→ 1/3π× (2)² × 8

→ 1/3π × 4 × 8

→ 1/3π × 32

→ 32π/3

Now,

→ No. of cones = $\rm{\dfrac{Volume\:of\:Sphere}{Volume\:of\:cone}}$

→ No. of cone = $\rm{\dfrac{\dfrac{1184\pi}{3}}{\dfrac{32\pi}{3}}}$

→ No. of cone = $\rm{\dfrac{1184\pi}{32\pi}}$

→ No. of cone = $\rm{ 37}$.

Hence, The no. of cone is 37.

Answer 2

Answer:

Inner radius = r¹ , Outer radius = r²

→ Volume of Sphere = 4/3π(r²)³ - 4/3π(r¹)³

→ Volume of Sphere = 4/3π{(r²)³ - (r¹)³}

→  

→  

→  

→ .

Therefore,

→ Volume of cone = 1/3πr²h

→ 1/3π× (2)² × 8

→ 1/3π × 4 × 8

→ 1/3π × 32

→ 32π/3

Now,

→ No. of cones =  

→ No. of cone =  

→ No. of cone =  

→ No. of cone = .

Hence, The no. of cone is 37.

Step-by-step explanation: