A large solid sphere is melted and shaped into identical right circular cones with the same base radius and height as the sphere's radius. One of these cones is melted and shaped into a smaller solid sphere. Then the surface area ratio of the smaller sphere to the surface area of the larger sphere is calculated.

We would be able to see the surface of the newly formed object.

Only the curved surfaces of the two hemispheres and the cylinder are visible.

As a result, the new solid's total surface area is the sum of the curved surface areas.

areas of each individual component

Radius of longer sphere = R units

Its volume $=\frac{4}{3}\pi R^3$

Now cones are formed with base radius

and height same as the radius of larger sphere

 volume of smaller cone $\frac{1}{3}\pi R^3$

and one of the cone is converted into smaller sphere Therefore volume of smaller sphere

$=\frac{1}{3}\pi R^3$

$\therefore \frac{4}{3}\pi r^3=\frac{1}{3}\pi R^3$

$r^3=1$

$\frac{r^3}{R^3}=\frac{1}{4}$

$\therefore \frac{r}{R}=\frac{1}{\sqrt[3]{4}}$

$\therefore \frac{\text{ surface area of smalley sphere }}{\text{ Surface area of larger sphere }}$

$=\frac{4\pi r^2}{4\pi R^2}=\frac{r^2}{R^2}$

$\Rightarrow \frac{(1{)}^2}{{\left((4{)}^{\frac{1}{3}}\right)}^2}=\frac{(1{)}^2}{{\left((2{)}^{\frac{2}{3}}\right)}^2}=\frac{1}{2\frac{4}{3}}$

$\Rightarrow 1:2^{4/3}$