A solid hemisphere of radius 6 cm is melted and recast into small solid spherical balls, each of radius 1 cm. Assuming no loss of material, how many such spherical balls can be formed?

Introduction / Context:
This is a solid geometry problem involving volumes and the principle of conservation of material. A hemisphere is melted and the molten material is recast into smaller spheres. The key idea is that the total volume of all the small spheres must equal the volume of the original hemisphere, assuming no material is lost. Using standard volume formulas for a sphere and a hemisphere, we can determine how many small spheres fit exactly into the original volume.

Given Data / Assumptions:
- Radius of the hemisphere R = 6 cm. - Radius of each small sphere r = 1 cm. - Volume of a sphere: V_sphere = (4 / 3) * π * r^3. - Volume of a full sphere of radius R is (4 / 3) * π * R^3. - Volume of a hemisphere is half the volume of the corresponding full sphere. - No loss of material during melting and recasting.

Concept / Approach:
Let N be the number of small spheres formed. Then N times the volume of one small sphere must equal the volume of the original hemisphere. That is N * V_small = V_hemi. Substituting the volume formulas, we get an equation in N where many constant factors cancel, leaving a ratio involving only R^3 and r^3. This is a common pattern: when the same shape is scaled by a factor, the number of small shapes we can generate is the ratio of the cubes of their characteristic lengths.

Step-by-Step Solution:
Step 1: Compute the volume of the hemisphere. Volume of full sphere of radius 6 cm is V_full = (4 / 3) * π * 6^3. Step 2: The hemisphere volume is half of that: V_hemi = (1 / 2) * (4 / 3) * π * 6^3. Step 3: Simplify 6^3 = 216, so V_hemi = (1 / 2) * (4 / 3) * π * 216. Step 4: Compute (4 / 3) * 216 = 288, then multiply by 1 / 2 to get V_hemi = 144π. Step 5: Volume of one small sphere with radius 1 cm is V_small = (4 / 3) * π * 1^3 = (4 / 3)π. Step 6: Let N be the number of small spheres. Then total small sphere volume is N * V_small = N * (4 / 3)π. Step 7: Set total small volume equal to hemisphere volume: N * (4 / 3)π = 144π. Step 8: Cancel π from both sides and solve for N: N * (4 / 3) = 144. Step 9: Multiply both sides by 3 / 4 to get N = 144 * (3 / 4) = 108.

Verification / Alternative check:
Rather than computing the full numbers first, we can look at the ratio approach. Volume of hemisphere is (2 / 3) * π * R^3, so V_hemi = (2 / 3) * π * 6^3 = (2 / 3) * π * 216 = 144π as before. The volume of a small sphere of radius 1 is (4 / 3)π, so the ratio V_hemi / V_small is (144π) / ((4 / 3)π) = 144 * (3 / 4) = 108. This ratio method confirms the same result efficiently and validates our calculation.

Why Other Options Are Wrong:
Option 112 exceeds the maximum number of spheres obtainable from the given volume and would require more volume than the hemisphere provides. Option 116 is also larger than the correct value and does not match the exact volume ratio derived from R^3 and r^3. Option 104 is close but does not correspond to the correct cube ratio and would leave unused hemisphere volume if actually used.

Common Pitfalls:
A frequent mistake is to forget that a hemisphere has half the volume of a full sphere and to use the full sphere formula directly. Others may incorrectly use the radius ratio instead of the volume ratio, forgetting that volume scales with the cube of the radius. Additionally, some students cancel π incorrectly or mismanage fractions such as (4 / 3). Carefully writing each step and dealing with the hemisphere factor first helps maintain accuracy.

Final Answer:
The number of small spheres that can be formed is 108.