A solid cylindrical iron rod has height equal to its radius. It is melted and recast into solid spherical balls, each of radius equal to half the radius of the original rod. How many such balls are formed?

Introduction / Context:
This question is about conservation of volume when a solid is melted and recast into new shapes. The mass and hence the volume of metal remain the same before and after recasting, assuming no loss. Here a cylindrical rod of iron is transformed into several spherical balls. We must relate the volume of the cylinder to the total volume of the spheres to determine how many spheres are formed.

Given Data / Assumptions:

• Original solid is a cylinder.
• Height of cylinder h equals its radius R.
• New solids are spheres.
• Radius of each sphere r = R / 2.
• No loss of material during melting and recasting.

Concept / Approach:
We use the standard volume formulas:
Volume of cylinder Vc = π * R^2 * h. Volume of sphere Vs = (4 / 3) * π * r^3. Since h = R, the cylinder volume simplifies. Because total volume is conserved, we set:
Vc = n * Vs, where n is the number of spheres. Solving for n gives the required count.

Step-by-Step Solution:
Step 1: Use h = R for the cylinder. Vc = π * R^2 * R = π * R^3. Step 2: Radius of each sphere r = R / 2. Vs = (4 / 3) * π * (R / 2)^3. Step 3: Compute (R / 2)^3 = R^3 / 8. Vs = (4 / 3) * π * R^3 / 8 = (4 * π * R^3) / 24 = (1 / 6) * π * R^3. Step 4: Set total volume of spheres equal to cylinder volume. π * R^3 = n * (1 / 6) * π * R^3. Step 5: Cancel π * R^3 from both sides. 1 = n * (1 / 6) ⇒ n = 6.

Verification / Alternative check:
We can pick a convenient value, such as R = 2 units. Then cylinder volume Vc = π * 2^3 = 8π. Each sphere has radius 1 unit, so Vs = (4 / 3) * π * 1^3 = (4 / 3)π. The number of spheres is Vc / Vs = 8π / ((4 / 3)π) = 8 / (4 / 3) = 8 * 3 / 4 = 6. This numeric example confirms the symbolic computation.

Why Other Options Are Wrong:
3 balls, 4 balls and 5 balls would together have less volume than the original cylinder, which would require some of the metal to vanish, contradicting the conservation of volume assumption.
8 balls would require more volume of iron than available from the original cylinder, which is not possible without adding extra material.

Common Pitfalls:
The main error is forgetting that the radius of each sphere is half the radius of the cylinder, which dramatically reduces the volume of each sphere because volume scales with the cube of the radius. Another pitfall is miscomputing the exponent when evaluating (R / 2)^3. Keeping track of the cube and simplifying carefully is essential in such problems.

Final Answer:
The number of spherical balls formed is 6 balls.