Question

A sphere is melted and half of the melted liquid is used
to form 11 identical cubes, where as the remaining half
is used to form 7 identical smaller spheres. The ratio
of the side of the cube to the radius of the new small
sphere is??? with step calculations .​

Answer 1

Answer:

4:3

hope this is helpful....

Answer 2

Therefore the ratio of the side of cube to the radius of the new sphere is

$=2:\sqrt{3}$

Step-by-step explanation:

Given that a sphere is melted and half of the melted liquid is  used to form 11 identical cubes and remaining half is used to form of 7 identical smaller sphere.

Then

The volume of 11 cubes = The volume of 7 smaller sphere.

Consider the length of side of the square be x.

Consider the length of radius of the spheres be y.

The volume of a square is = side³

the volume one sphere is =$\frac{4}{3} \pi r^3$

The volume of 11 cubes =( 11×x³) cubic unit

The volume of 7 spheres = $(7\times \frac{4}{3} \pi y^3)$ cubic unit

According to the problem,

$11x^3=7\times \frac{4}{3} \pi y^3$

$\Rightarrow 11x^3=7\times \frac{4}{3} \times \frac{22}{7} \times y^3$

$\Rightarrow \frac{x^3}{y^3} =\frac{7\times 4\times 22 }{11\times3\times 7}$

$\Rightarrow \frac{x^3}{y^3}=\frac{8}{3}$

$\Rightarrow \frac{x}{y} =\sqrt[3]{\frac{8}{3} }$

$\Rightarrow \frac{x}{y} =\frac{2}{\sqrt{3} }$

Therefore the ratio of the side of cube to the radius of the new sphere is

$=2:\sqrt{3}$