How many smaller solid balls of radius 2 cm can be made by completely melting and recasting one larger solid ball of diameter 16 cm? (Take π = 22/7 and assume no loss of material.)

Introduction / Context:
This is a standard volume and mensuration problem involving spheres. A larger solid sphere is melted and reshaped into many smaller solid spheres. Since there is no loss of material, the total volume of all the small spheres must equal the volume of the original large sphere. Using the volume formula for a sphere and the concept of conservation of volume, we can determine how many smaller balls are formed.

Given Data / Assumptions:
- Radius of each small ball is 2 cm. - Diameter of the large ball is 16 cm, so its radius is 8 cm. - Volume of a sphere is V = (4 / 3) * π * r^3. - π is taken as 22 / 7. - No material is lost during melting and recasting.

Concept / Approach:
The key idea is conservation of volume. If N small spheres are formed, then N times the volume of one small sphere equals the volume of the large sphere. Because π and the constant 4 / 3 appear in both volumes, they cancel when we take the ratio. Therefore, the number of small balls N is simply the ratio of the cube of the large radius to the cube of the small radius. This avoids heavy arithmetic with π and fractions.

Step-by-Step Solution:
Step 1: Compute the radius of the large sphere. Its diameter is 16 cm, so radius R = 16 / 2 = 8 cm. Step 2: Radius of each small sphere is r = 2 cm. Step 3: Volume of the large sphere is V_large = (4 / 3) * π * R^3. Step 4: Volume of one small sphere is V_small = (4 / 3) * π * r^3. Step 5: Let N be the number of small spheres. Then N * V_small = V_large. Step 6: Substitute the volumes: N * (4 / 3) * π * r^3 = (4 / 3) * π * R^3. Step 7: Cancel (4 / 3) * π from both sides, giving N * r^3 = R^3. Step 8: So N = R^3 / r^3 = 8^3 / 2^3. Step 9: Compute 8^3 = 512 and 2^3 = 8. Therefore N = 512 / 8 = 64.

Verification / Alternative check:
If you wish, you can compute actual numerical volumes using π = 22 / 7. Large sphere volume: V_large = (4 / 3) * (22 / 7) * 8^3. Small sphere volume: V_small = (4 / 3) * (22 / 7) * 2^3. The ratio V_large / V_small simplifies to 8^3 / 2^3 as before, which is 64. This confirms that all constant factors cancel and our calculation is correct.

Why Other Options Are Wrong:
Option 128 would correspond to doubling the correct number of spheres and would require twice the volume that the large sphere provides. Option 32 is exactly half of the correct number and would leave unused volume if only 32 small spheres were formed. Option 96 would imply a ratio of volumes that does not match the cube ratio of the radii 8 and 2.

Common Pitfalls:
A common mistake is to think the number of small spheres is simply the ratio of radii or the ratio of surface areas, not the ratio of volumes. Because volume of a sphere is proportional to the cube of the radius, the correct ratio uses R^3 and r^3. It is also easy to miscalculate 8^3 or 2^3, so always check these small powers carefully. Remember to focus on volume when dealing with melted and recast solid shapes.

Final Answer:
The number of smaller balls that can be made is 64.