Question

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

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

Given :

Two solid spheres off radii 2cm and 4cm are melted and recast into a cone of height 8cm.

To Find :

Find the radius of the cone so formed.

Solution:

Radius of sphere 1 = 2 cm

Volume of sphere = $\frac{4}{3} \pi r^3$

Volume of sphere 1 =$\frac{4}{3} \times \frac{22}{7} \times 2^3$

Radius of sphere 2= 4 cm

Volume of sphere = $\frac{4}{3} \pi r^3$

Volume of sphere 2$= \frac{4}{3} \time \frac{22}{7} \times 4^3$

We are given that two solid spheres off radii 2cm and 4cm are melted and recast into a cone of height 8cm.

Height of cone = 8 cm

Volume of cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 8$

ATQ

$\frac{4}{3} \times \frac{22}{7} \times 2^3 + \frac{4}{3} \times \frac{22}{7} \times 4^3 = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 8$

$4 \times 2^3 + 4 \times 4^3 = r^2 \times 8$

$\frac{4 \times 2^3 + 4 \times 4^3}{8}=r^2$

$36=r^2$

6 = r

Hence the radius of the cone so formed is 6 cm