One thousand solid metallic spheres, each of diameter 6 centimetres, are melted and recast into a single new solid sphere. Find the diameter of the new sphere in centimetres.

Introduction / Context:
This mensuration problem is about conservation of volume when smaller solid spheres are melted and recast into a larger sphere. It requires using the volume formula for a sphere and handling scaling when many identical small solids are combined into one larger solid of the same material. Such problems are common in aptitude exams and help reinforce volume relationships.

Given Data / Assumptions:

• Number of small spheres = 1000.
• Diameter of each small sphere = 6 cm, so radius r_small = 3 cm.
• All small spheres are metallic and are completely melted.
• The molten metal is recast into one large solid sphere.
• No loss of material occurs, so total volume of metal is conserved.
• We must find the diameter of the new sphere.

Concept / Approach:
The volume of a sphere of radius r is:
V = (4 / 3) * pi * r^3.
The total volume of 1000 identical small spheres is 1000 times the volume of one small sphere. This equals the volume of the single large sphere. By equating these volumes, we can solve for the radius of the large sphere and then compute its diameter, which is twice the radius.

Step-by-Step Solution:
Step 1: Compute the volume of one small sphere with radius r_small = 3 cm. Step 2: V_small = (4 / 3) * pi * 3^3 = (4 / 3) * pi * 27 = 36 * pi cubic centimetres. Step 3: Total volume of 1000 small spheres is V_total = 1000 * 36 * pi = 36000 * pi cubic centimetres. Step 4: Let the radius of the new large sphere be R. Step 5: Volume of the large sphere is V_large = (4 / 3) * pi * R^3. Step 6: Since material volume is conserved, set V_large = V_total. Step 7: (4 / 3) * pi * R^3 = 36000 * pi. Step 8: Cancel pi from both sides to get (4 / 3) * R^3 = 36000. Step 9: Multiply both sides by 3 / 4: R^3 = 36000 * (3 / 4) = 27000. Step 10: Find R by taking the cube root: R^3 = 27000 implies R = 30 cm, since 30^3 = 27000. Step 11: Diameter of the new sphere = 2 * R = 2 * 30 = 60 cm.

Verification / Alternative check:
We can verify the calculation by checking the total volume. Volume of the large sphere with R = 30 cm is V_large = (4 / 3) * pi * 30^3 = (4 / 3) * pi * 27000 = 4 * 9000 * pi = 36000 * pi, which matches the total volume of the 1000 small spheres. This confirms that the radius and hence the diameter are correct.

Why Other Options Are Wrong:
30 cm and 45 cm: These values are too small for the diameter; they would correspond to radii with volumes much less than the combined volume of 1000 small spheres.
90 cm and 15 cm: These are either excessively large or too small, giving volumes that do not match the total initial volume when substituted into the sphere volume formula.

Common Pitfalls:
Errors often occur when cubing the radius or when multiplying by the factor 1000. Another common mistake is forgetting that diameter is twice the radius and reporting the radius instead. Some students may also incorrectly cancel factors or mishandle the fraction 4 / 3. Keeping track of each algebraic step and clearly marking radius versus diameter helps avoid these errors.

Final Answer:
The diameter of the new sphere is 60 cm.