How to Solve Cylinder, Cone and Sphere Conversion Problems
Solve cylinder, cone and sphere melt-and-recast conversion problems with volume equality, hollow-pipe formulas, and practice questions.
Expected Interview Answer
Conversion problems between cylinders, cones and spheres — melting one solid and recasting it as another — are solved by equating the volumes of the two solids and solving for the unknown dimension, since melting and recasting conserves total volume, not surface area or any linear dimension.
The three volume formulas — cylinder πr²h, cone ⅓πr²h, sphere (4/3)πr³ — must be set equal to each other (or a multiple, for splitting into several smaller solids) whenever material is recast. A frequent variant hollows out a solid, requiring the volume of material as the difference of an outer and inner solid, such as a pipe’s volume being π(R²−r²)h. Watch for problems that give a diameter for one solid and a radius for another within the same question — mixing them up is the most common arithmetic trap. Always isolate the single unknown after substitution rather than trying to solve two formulas simultaneously.
- Volume equality is the single principle behind every melt-and-recast problem
- The hollow-solid subtraction pattern extends the same idea to pipes and shells
- Isolating one unknown after substitution keeps the algebra manageable
AI Mentor Explanation
Melting down a cylindrical trophy and recasting the same metal into several small cone-shaped medals conserves total volume, so the cylinder’s πr²h must equal the sum of the cones' ⅓πr²h volumes, exactly the volume-equality principle behind conversion problems. A hollow cricket-bat handle, essentially a pipe, has material volume π(R²−r²)h, the outer radius minus inner radius squared, mirroring the hollow-solid subtraction pattern. If the trophy’s diameter is given but a medal’s radius is asked for, mixing the two up is the classic error the problem tests. Isolating the one unknown dimension, say the medal’s radius, only after substituting all known values keeps the algebra from turning into two equations at once.
Worked example — melt cylinder, recast into spheres
Cylinder volume
- π×6²×12
- = 432π cm³
Sphere volume (each)
- (4/3)π×3³
- = 36π cm³
Number of spheres
- 432π ÷ 36π
- = 12
Step-by-Step Explanation
Step 1
Identify both solids
Note which solid is melted (source) and which is formed (target).
Step 2
Write both volume formulas
Use the correct cylinder, cone, or sphere formula for each solid.
Step 3
Equate total volumes
Source volume = target volume × count, since melting conserves volume.
Step 4
Solve for the unknown
Isolate the single missing dimension or count after substituting all known values.
What Interviewer Expects
- Recognizing that melt-and-recast problems conserve volume, not surface area
- Correct volume formulas for cylinder, cone, and sphere
- Correctly handling hollow solids via outer-minus-inner volume subtraction
- Careful tracking of which given number is a radius and which is a diameter
Common Mistakes
- Equating surface areas instead of volumes when solids are recast
- Forgetting to multiply the target solid’s volume by the count when several are formed
- Using R and r inconsistently in the hollow-pipe formula π(R²−r²)h
- Substituting a diameter directly where a radius was required
Best Answer (HR Friendly)
“The key principle is that melting and recasting a solid never changes its volume, only its shape, so I would write the volume formula for the original solid and for the new solid, then equate them — multiplying by a count if the material is split into several pieces. For hollow solids like pipes, I would subtract the inner volume from the outer volume, and I would always double check which given number is a radius versus a diameter before substituting.”
Follow-up Questions
- How would you solve a problem where a cone and a sphere have equal volumes and find their radius ratio?
- How do you compute the volume of material in a hollow cylindrical pipe?
- How would you find how many smaller spheres come from melting a larger sphere?
- How does a frustum’s volume relate to the full cone it was cut from?
MCQ Practice
1. A sphere of radius 6cm is melted and recast into small spheres of radius 2cm each. How many small spheres are formed?
Volume ratio = (6/2)³ = 27, since volume scales with radius cubed.
2. A cone and a cylinder have equal base radius and equal volume. The cylinder’s height is 4cm. The cone’s height is?
⅓πr²h_cone = πr²×4 → h_cone = 12cm, since the cone needs 3× the height to match the cylinder’s volume at equal radius.
3. A hollow cylindrical pipe has outer radius 5cm, inner radius 3cm, and length 10cm. Its material volume is?
Volume = π(5²−3²)×10 = π×16×10 = 160π cm³.
Flash Cards
What is conserved when a solid is melted and recast? — Total volume — never surface area or any single linear dimension.
Volume of a hollow pipe? — π(R² − r²)h, outer radius squared minus inner radius squared, times length.
How to find the count of smaller solids from a larger one? — Divide the larger solid’s volume by one smaller solid’s volume.
Volume scaling with radius? — Volume scales as the cube of the radius ratio for similar solids.