Question

The surface of a solid metallic sphere is $616{\rm{ }}c{m^2}$. It is melted and recast into a cone of height $28cm$. Find the diameter of the base of the cone so formed. (Use $\pi = \dfrac{{22}}{7}$).

Answer 1

Hint: In this question we have to find the diameter of the cone. It is given that, surface of a solid metallic sphere is melted and recast into a cone, that means their volume is the same. After equating their volumes, the radius of the cone can be found and the diameter of the cone can be found using $d=2R$ formula.

Complete step-by-step answer:
Here in this question, a solid metallic sphere is melted and recast into a cone. Then, their volume will be equal.
Volume of sphere = Volume of cone
We know that,
Surface area of sphere = $4\pi {r^2}$
Given,
Surface area of sphere=$616{\rm{ }}c{m^2}$ and $\pi = \dfrac{{22}}{7}$.
Then equation (2) becomes,
$ \Rightarrow 616 = 4 \times \dfrac{{22}}{7} \times {r^2}$
On simplification,
$ \Rightarrow {r^2} = \dfrac{{7 \times 616}}{{4 \times 22}}$
$ \Rightarrow {r^2} = 49$
$ \Rightarrow r = \sqrt {49} $
$ \Rightarrow r = 7cm$
Then, the radius of the sphere is $7cm$.
Let the radius of the cone be ’R’.
Volume of sphere = $\dfrac{4}{3}\pi {r^3}$
Volume of cone = $\dfrac{1}{3}\pi {R^2}h$
Substituting these formulas in equation (1), we get,
$ \Rightarrow \dfrac{4}{3}\pi {r^3} = \dfrac{1}{3}\pi {R^2}h$
Dividing both the sides by $\dfrac{\pi }{3}$
$ \Rightarrow 4{r^3} = {R^2}h$
Substituting $r = 7cm$ and $h = 28cm$ (given)
$ \Rightarrow 4{(7)^3} = {R^2}(28)$
$ \Rightarrow {R^2} = \dfrac{{4{{(7)}^3}}}{{28}}$
$ \Rightarrow {R^2} = 49$
$ \Rightarrow R = 7cm$
We know that, ${\rm{diameter}} = 2R$
$\Rightarrow {\rm{diameter}} = 2(7)$
$\Rightarrow {\rm{diameter}} = 14cm$
Hence, the diameter of the cone is $14cm$.

Note: Whenever you face such a type of problem the key concept is the volume of the object given. One should know the volume of sphere, cone, cylinder and use them as per the question. Here, we have used the surface area of sphere, volume of cone and sphere.