Question

a hollow sphere of internal and external radii 2cm and 4 cm respectively is melted and recast into a cone of diameter 8 cm find the height of the cone with full explanation​

Answer 1

Answer:

nternal radius of hollow sphere ,r = 2 cm

External radius of hollow sphere , R = 4 cm

Radius of the cone , r1 = 4 cm

Since, the hollow spherical shell is melted into a cone , so volume of both are equal

Volume of the hollow spherical shell = Volume of the cone

4/3π(R³ − r³) = 1/3πr1²h

4(R³ − r³) = r1²h

4(4³ - 2³) = 4²h

4(64 - 8) = 16h

4(56) = 16h

h = (4 × 56) /16

h= 56/4 = 14

h= 14 cm

h =  14 cm.

Slant height of the cone, l = √ r² + h²

l = √8² + 14²  

l = √64 + 196

l = √260

l = 16.124 cm 

Slant height of the cone = 16.124cm.

Hence, the height of the cone is 14 cm & Slant height of the cone is 16.124cm.

Answer 2

Answer:

14 cm

Step-by-step explanation:

Given a hollow sphere internal and external radii 2 cm and 4 cm respectively.

If this sphere is melted and react into another shape its volume remains constant.

We know that volume of a sphere is 4πr³/3 and that of cone is πr²h/3.

Now to find volume of hollow sphere, we need to imagine that a small sphere of radius 2 cm is covered by a big sphere of radius 4 cm.

When we subtract the volume of small sphere from big sphere we get volume of hollow cylinder.

i.e volume of hollow cylinder is 4/3 π(4)³-4/3 π(2)³

⇒4/3 π 56cm³.

Let the height of cone be h.

Now its volume is π(4)²h/3=π 16 h/3   (since diameter of cone is 8 cm its radius is 8/2=4 cm),

Volume of cone=volume of sphere

4/3 π 56 = π 16 h/3

h=56/4

h=14 cm