Here is the original question:-
Here is my attempt:-
As you can see, the answer that I found is not even given in the option. I did find the correct solution online. However, I do not understand what I did wrong. I'd appreciate if someone can point out my mistake.
In my opinion, perhaps the radii of the new spheres are calculated wrong (like I did in the figure). Is there a reason?
Your handwriting is illegible and you do not explain your reasoning. I don't understand where you got $$2\pi (r - h)^2 + 2\pi (r+h)^2 = 4\pi r^2 + \frac{1}{4}(4\pi r^2).$$ The right-hand side I assume to represent the part of the question that says "$25\%$ more than the surface area of the sphere" but the left-hand side makes no sense at all. It doesn't represent any recognizable geometric property.
You seem to be under the impression that there is more than one sphere involved; however, this is not correct. What the question is saying is, the sum of the total surface areas of the cut pieces equals $25\%$ more than the surface area of the original sphere. When the original sphere is cut, the resulting pieces are not spherical.
The conclusion we are intended to draw from the statement of the problem is that twice the area of the circle that is formed by the cross-section of the cut must equal one quarter of the surface area of the sphere. To see why, we first observe that the radius $r'$ of the circular cross-section is obtained via the Pythagorean theorem: $$(r')^2 + h^2 = r^2,$$ hence $$r' = \sqrt{r^2 - h^2}$$ and the area of this cross-section is simply $$\pi (r')^2 = \pi (r^2 - h^2).$$ Consequently, the surface area of the smaller cut piece is equal to $A + \pi (r^2 - h^2)$, where $A$ is the surface area of the spherical cap that excludes the area of the cross-section; and the surface area of the larger cut piece is equal to $B + \pi(r^2 - h^2)$, where $B$ is the surface area of the spherical cap on the other side of the cut. And we know that $A + B = 4 \pi r^2$ is the total surface area of the sphere.
The given condition in the problem states that the sum of these two surface areas is $25\%$ more than the surface area of the sphere; i.e., $$\left(A + \pi (r^2 - h^2)\right) + \left(B + \pi (r^2 - h^2)\right) = \frac{5}{4}(A + B),$$ from which we conclude $$2\pi(r^2 - h^2) = \frac{A+B}{4} = \pi r^2.$$ This is precisely the claim we made above: twice the area of the circular cross-section equals a quarter of the surface area of the sphere. From here, it is trivial to solve for $h$ in terms of $r$.