Question

a hollow sphere of external diameter 10 cm and internal diameter 6 cm is melted into a cone of base diameter 6cm what is height of cone​

Answer 1

Answer:

height of the cone = 10.89 cm

Step-by-step explanation:

external radius of the sphere, R = 10cm/2 = 5cm

internal radius of the sphere, r = 6cm / 2 = 3cm

=> volume of the hollow sphere = (4/3)$\pi$ (R³-r³) cu. units

                                                     = (4/3)$\pi$ (5³-3³) cm³

                                                     = (4/3)$\pi$ (125-27) cm³

                                                     = (4/3)$\pi$ (98) cm³

                                                     = 392$\pi$ / 3 cm³

given that radius of the cone so formed by melting the hollow sphere

= 6cm/2 = 3cm

=> volume of the cone = volume of the hollow sphere

=> (1/3)$\pi$r²h = 392$\pi$ / 3 cm³

=> (1/3)$\pi$(6²)h = 392$\pi$ / 3 cm³

=> (1/3)$\pi$(6²)h = 392$\pi$ / 3 cm³

=> 36*h = 392 cm

=> h = 392/36 cm

=> h = 98/9 cm

=> h = 10$\frac{8}{9}$ cm

=> h = 10.89 cm

Answer 2

43.5 cm

Step-by-step explanation:

→ External Diameter of Hollow sphere = 10cm

→ External Radius of Hollow sphere = 10/2 = 5cm

Similarly,

→ Internal Diameter of Hollow sphere = 6cm

→ Internal Radius of Hollow sphere = 6/2 = 3cm

→ Diameter of cone = 6cm

→ Radius of cone = 6/2 = 3cm

We are given that, a hollow sphere is melted into a cone. Since the shape is melted, their volumes does not change.

→ V (hollow sphere) = V (cone)

→ 4π(R³-r³)/3 = πr²h/3

Here,

• R = External Radius of sphere
• r (LHS) = Internal Radius of sphere
• r (RHS) = Radius of cone
• h = Height of cone

→ 4π(R³-r³)/3 = πr²h/3

→ 4π(R³-r³) = 3πr²h/3

→ 4π(R³-r³) = πr²h

→ 4(R³-r³) = πr²h/π

→ 4(R³-r³) = r²h

→ 4(5³-3³) = 3²×h

→ 4(125-27) = 9×h

→ 4(98) = 9×h

→ 392 = 9×h

→ 392/9 = h

→ 43.5cm = h

Hence, the height of the cone is 43.5cm.