A geometry problem by Karan Ahire

Surface area of a sphere is $4 \pi r^2$ , so the radius of the sphere is

$616=4 \pi r^2$ $\implies$ $r=7$

The volume of the sphere is $\dfrac{4}{3} \pi r^3=\dfrac{4}{3} \pi (7)^3=1437.623$

The volume of the sphere must be equal to the volume of the cone, and the volume of the cone is $\dfrac{1}{3} \pi R^2(h)$ , so we have

$1437.623 = \dfrac{1}{3} \pi R^2 (28)$

$R=7$

The desired answer is $2R=14$

We know, the surface area of a solid sphere is 4 ╥ r^2
Given that, the surface area of the sphere is 616 cm^2 i.e. 616 = 4 ╥ r^2 or, r^2 = 616/(4 ╥ ) or, r=(616/4 3.142 )^(1/2) so, r=7 cm We know volume of a sphere is (4/3) ╥ r^3 so, the volume of the sphere = (4/3) 3.142 7^3 =1436.94 cm^3 the volume will remain the same after melting we know, volume of a cone = ╥r^2 h/3 (where h is the height)
given that, h=28 so, the volume of the cone = ╥r^2 (28/3) or, 1436.94 =3.142 * r^2 (28/3) or, r^2 = 1436.94/(3.142
28/3) or, r^2 = 1436.94 3/(3.142 28) or, r= (1436.94 3/(3.142 28))^(1/2) so, r =7 as the radius = 7, the diameter = 7*2 =14 (Ans)