Question

hollow sphere of internal and external diameter 4 cm and 8 cm respectively is melted into a cone of base diameter 8 centimetre find the height of the cone

Answer 1

V.OF SPHERE=4/3 PIE ( R cube - r cube)
And v.of cone=1/3 pie R square h.

V.of sphere = v.of cone.
4/3*22/7*(4*4*4-2*2*2)=
1/3*22/7*4*4*4h
4(64-8)=16*h
4*56=16*h
h=(4*56)/16
h=14 cm. ANS.

Answer 2

Answer:

The height of the cone  is 14 cm.

Step-by-step explanation:

Given :  

Internal diameter of hollow sphere = 4 cm

Internal radius of hollow sphere , r = 2 cm

External diameter of hollow sphere = 8 cm

External radius of hollow sphere , R = 4 cm

Diameter of the cone = 8 cm

Radius of the cone , r1 = 4 cm

Volume of the hollow spherical shell = 4/3π(R³ − r³)

Let the height of the cone be h cm.

Volume of the cone = 1/3πr1²h

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

Hence, the height of the cone  is 14 cm.

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