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# Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.  Answer Verified
Hint: Volume of the 3 spheres when melted to form a single sphere will be equal. Equate the sum of volume of the 3 spheres with the volume of the new sphere to find the radius of the new sphere.

Three metallic spheres of radii 6 cm, 8 cm and 10 cm are to be melted and formed into a single sphere. Let the radius of the new sphere be ‘r’.

Let the radius of the three metallic spheres be $r1,{\text{ }}r2$ and $r3$.
So we have, $r1 = 6,r2 = 8,r3 = 10$

When the three spheres are melted and formed into a single sphere, their volumes will remain the same. So, we can write it as,
(Volume of sphere 1 + Volume of sphere 2 + Volume of sphere 3)=Volume of the new sphere ..(1)

Volume of sphere 1= $\dfrac{4}{3}\pi r{1^3}$
$= \dfrac{4}{3} \times \pi \times {(6)^3} \\ = \dfrac{{864}}{3}\pi c{m^3} \\$

Volume of sphere 2= $\dfrac{4}{3}\pi r{2^3}$
$= \dfrac{4}{3} \times \pi \times {(8)^3} \\ = \dfrac{{2048}}{3}\pi c{m^3} \\$

Volume of sphere 3= $\dfrac{4}{3}\pi r{3^3}$
$= \dfrac{4}{3} \times \pi \times {(10)^3} \\ = \dfrac{{4000}}{3}\pi c{m^3} \\$

So as per equation (1), we need to add all the three volumes and equate it to the volume of the new sphere with radius ‘r’

Volume of new sphere= $\dfrac{{864}}{3}\pi + \dfrac{{2048}}{3}\pi + \dfrac{{4000}}{3}\pi$
$= \dfrac{\pi }{3}\left( {864 + 2048 + 4000} \right) \\ = \dfrac{{6912}}{3}\pi \\ = 2304\pi c{m^3} \\$

Volume of new sphere= $\dfrac{4}{3}\pi {r^3}$
So, by equating both we get,
$\dfrac{4}{3}\pi {r^3} = 2304\pi \\ {r^3} = \dfrac{{6912}}{4} \\ {r^3} = 1728 \\$
Expressing in terms of exponents where the exponent needs to be 3, we get
${r^3} = {12^3}$
When exponents are the same, the bases are equal.
$r = 12$ cm
Hence the radius of the new sphere formed is 12 cm.

Note: In these types of problems, it is better to keep $\pi$ till the end because it will be easier to calculate and in the end it might get cancelled. This also avoids the possibility of calculation errors.
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