A solid sphere of radius 10 cm is melted and recast into 8 solid spherical balls of equal radius. What is the surface area of each small spherical ball in square centimetres?

Introduction / Context:
This question involves conservation of volume when a solid object is melted and recast into several smaller objects. A large solid sphere is melted and formed into 8 smaller solid spheres of equal radius. Because no material is lost or gained, the total volume remains the same. We use this idea to find the radius of each smaller sphere, and then compute its surface area. This problem tests understanding of volume–surface area relationships for spheres.

Given Data / Assumptions:

• Radius of original sphere, R = 10 cm.
• Number of smaller spheres = 8.
• Each smaller sphere is solid and has equal radius r.
• All material is conserved; no loss in melting or recasting.

Concept / Approach:
The volume of a sphere of radius x is:
V = (4/3) * pi * x^3 Since the large sphere is recast into 8 equal small spheres, total volume is conserved:
(4/3) * pi * R^3 = 8 * (4/3) * pi * r^3 This leads to R^3 = 8 * r^3, so r^3 = R^3 / 8. After finding r, the surface area of one small sphere is:
Surface area = 4 * pi * r^2

Step-by-Step Solution:
Original radius R = 10 cm. Volume of original sphere = (4/3) * pi * R^3 = (4/3) * pi * 10^3. Total volume of 8 small spheres = 8 * (4/3) * pi * r^3. Equate volumes: (4/3) * pi * 10^3 = 8 * (4/3) * pi * r^3. Cancel common factors (4/3 * pi) to get 10^3 = 8 * r^3. So r^3 = 1000 / 8 = 125. Therefore r = cube root of 125 = 5 cm. Surface area of one small sphere = 4 * pi * r^2 = 4 * pi * 5^2. Surface area = 4 * pi * 25 = 100 pi square centimetres.

Verification / Alternative check:
We can check total surface area increase. Original surface area = 4 * pi * 10^2 = 400 pi. Each small sphere has surface area 100 pi, and there are 8 of them, so combined surface area is 8 * 100 pi = 800 pi. The total surface area has doubled, which is expected when splitting a solid into multiple pieces of smaller size, confirming the internal logic.

Why Other Options Are Wrong:
Values like 80 pi, 50 pi, 64 pi and 75 pi sq.cm correspond to guessing a smaller radius than 5 cm or miscomputing r^3. For example, 80 pi would occur if r were wrongly taken as 4 cm, giving 4 * pi * 16 = 64 pi, still not listed. The only value consistent with r = 5 cm is 100 pi sq.cm.

Common Pitfalls:
A common mistake is to assume that the radius scales linearly with the number of spheres, which is incorrect because volume depends on the cube of the radius. Another error is to forget that volume is conserved and instead try to directly relate surface areas. Correct reasoning requires working through the volume equation first, then computing surface area from the derived radius.

Final Answer:
The surface area of each small spherical ball is 100 pi sq.cm.