Question

Metallic spheres of radii \[6cm,8cm\] and \[10cm\] respectively, are melted to form a single solid sphere. Find the radius (in cm) of the resulting sphere.

Answer 1

Hint: Here we are given 3 small metallic spheres, which would be melted to form a single solid sphere. So, the sum of the volume of all three small spheres will be equal to the volume of the single big solid sphere, hence we equate the volumes and on simplification we get, the radius of the big sphere.

Complete step by step solution: As we know that the radius of the smaller metallic sphere is 6 cm,8 cm and 10 cm and hence mould it into one larger sphere.
And so the volume of one larger sphere is equal to the sum of the volume of all the smaller spheres and volume of any sphere is given as \[v = \dfrac{4}{3}\pi {r^3}\].

Let the radius of smaller metallic sphere is \[r' = 6cm,\,\,r'' = 8cm\] and \[r''' = 10cm\].

And the volume of the larger sphere be r

Hence, equating the volume of the spheres as,

\[\dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi r{'^3} + \dfrac{4}{3}\pi r'{'^3} + \dfrac{4}{3}\pi r''{'^3}\]

On substituting the value of radius we get,

\[\dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {6^3} + \dfrac{4}{3}\pi {8^3} + \dfrac{4}{3}\pi {100^3}\]

As,

\[
  {6^3} = 216 \\
  {8^3} = 512 \\
  {10^3} = 1000 \\
 \]

We Cancel out common terms from both the sides and on substituting the above values, we get,

\[{r^3} = 216 + 512 + 1000\]

Hence, \[{r^3} = 1728\]

Taking cube root on both sides we can say that \[r = 12cm\]

Hence, the radius of a single solid sphere is 12cm.

Note: A sphere is a geometrical object in three-dimensional space that is the surface of a ball. Like a circle in a two-dimensional space, a sphere is defined mathematically as the set of points that are all at the same distance r from a given point in a three-dimensional space.

Volume of a sphere is given by \[v = \dfrac{4}{3}\pi {r^3}\], and surface area of sphere is \[S = 4\pi {r^2}\].