Question

Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.

Please give step by step solution.
Its is urgent so plz send the solution fast.

Answer 1

Let the radius of first sphere be r1 , second sphere be r2 and radius of the cone be r3 , then -

4/3πr1^3 + 4/3πr2^3 = 1/3πr3^3h
4/3π(r1^3 + r2^3) = 1/3πr3^3h
4(8+64) = 8r3^3
72 = 2r3^3
r3^3 = 36
r3 = 6

Thus , radius of the cone = 6 cm.

Answer 2

Answer: 2 cm

Step-by-step explanation:

Let the radius of first sphere be r1 , second sphere be r2 and radius of the cone be r3 , t

4/3πr1^3 + 4/3πr2^3 = 1/3πr3^3h

4/3π(r1^3 + r2^3) = 1/3πr3^3h

4(8+64) = 8r3^3

72 = 2r3^3

r3^3 = 36

r3 = 6^1/3 = 1.9=2cm approx

radius of the cone =2 cm