Courses for Kids
Free study material
Offline Centres
Store Icon

A sphere is melted and half of the molten liquid is used to form 11 identical cubes, whereas the remaining half is used to form 7 identical small spheres. The ratio of the side of the cube to the radius of the new small sphere is –
A.${\left( {\dfrac{4}{3}} \right)^{\dfrac{1}{3}}}$
B.${\left( {\dfrac{8}{3}} \right)^{\dfrac{1}{3}}}$
C.${\left( 3 \right)^{\dfrac{1}{3}}}$

Last updated date: 13th Jun 2024
Total views: 412.5k
Views today: 9.12k
412.5k+ views
Hint:In order to solve this problem we need to calculate the volume of 11 identical cubes and 7 identical small spheres. Then we have to equate the volume of 11 cubes with 7 smaller cubes to get the ratio of side of the cube to radius of the new sphere.

Complete step-by-step answer:
Radius of the 7 spherical ball = r
The volume of the sphere = \[\dfrac{4}{3}{\text{ }}\pi {r^3}\]
As there are 7 spherical balls so total volume =\[7\left( {\dfrac{4}{3}{\text{ }}\pi {r^3}} \right) = 7\left( {\dfrac{4}{3}{\text{ }}\left( {\dfrac{{22}}{7}} \right){r^3}} \right) = \dfrac{{88}}{3}{r^3}\]
Side of the identical cubes = a
So, the volume of the cube = ${a^3}$
As there are 11 identical cubes so total volume = $11{a^3}$
Here it is given 11 identical cubes are made from half sphere melted and similarly 7 identical small spheres are made from half sphere melted. So, both the volumes are equal.
Hence, we get the below equation: -
  \dfrac{{88}}{3}{r^3} = 11{a^3} \\
   \Rightarrow \dfrac{{{a^3}}}{{{r^3}}} = \dfrac{8}{3} \\
   \Rightarrow \dfrac{a}{r} = {\left( {\dfrac{8}{3}} \right)^{\dfrac{1}{3}}} \\
Hence, the ratio of the side of the cube to the radius of the new small sphere is \[{\left( {\dfrac{8}{3}} \right)^{\dfrac{1}{3}}}\]

So, the correct answer is “Option B”.

Note:To solve this type of question, remember the formula of volume of all shapes. In this question the sphere is casted into smaller balls and cubes whereas in other questions they may convert into any other three-dimensional figures. The approach remains the same to solve such kind of questions.