A solid sphere of radius 3 cm is melted and recast into a hollow right circular cylindrical tube of length 4 cm and external radius 5 cm. Assuming no loss of material and uniform thickness throughout, what is the thickness (in cm) of the cylindrical tube?

Introduction / Context:
This question combines volume conservation with the geometry of a hollow cylinder. A solid sphere is melted and reshaped into a cylindrical tube with a given external radius and length, but unknown internal radius. The difference between the outer and inner radii represents the thickness of the tube. Since no material is lost in the process, the volume of the sphere must equal the volume of the hollow cylinder. We use volume formulas to find the internal radius and then compute the thickness.

Given Data / Assumptions:

• Radius of the sphere, R = 3 cm.
• The sphere is melted and formed into a hollow cylinder (tube).
• Length (height) of the cylindrical tube, h = 4 cm.
• External radius of the tube, R_outer = 5 cm.
• Internal radius of the tube, R_inner = r (unknown).
• No loss of volume during melting and recasting.

Concept / Approach:
The volume of the material in the hollow cylinder equals the volume of the sphere. The volume of the sphere is (4 / 3) * π * R^3. The volume of a hollow cylinder is π * (R_outer^2 − R_inner^2) * h. By equating these two volumes, we can solve for R_inner^2, then find R_inner. Finally, the thickness of the tube is the difference between the outer and inner radii: thickness = R_outer − R_inner.

Step-by-Step Solution:
Volume of the sphere = (4 / 3) * π * R^3, with R = 3 cm.Compute R^3: 3^3 = 27.So volume of sphere = (4 / 3) * π * 27 = 4 * 9 * π = 36π cubic centimetres.Volume of hollow cylinder = π * (R_outer^2 − R_inner^2) * h.Here R_outer = 5 cm, h = 4 cm, R_inner = r.So volume of hollow cylinder = π * (25 − r^2) * 4 = 4π(25 − r^2).Equate the volumes: 36π = 4π(25 − r^2).Cancel π from both sides to get 36 = 4(25 − r^2).Divide both sides by 4: 9 = 25 − r^2.So r^2 = 25 − 9 = 16.Thus r = 4 cm (internal radius).Thickness of the tube = external radius − internal radius = 5 − 4 = 1 cm.

Verification / Alternative check:
Substitute r = 4 cm back into hollow cylinder volume to check. Volume = 4π(25 − 16) = 4π * 9 = 36π cubic centimetres, which matches the volume of the sphere. This confirms that the inner radius is correct and that the derived thickness of 1 cm is consistent with the conservation of volume.

Why Other Options Are Wrong:
A thickness of 9 cm or 1.5 cm would imply an internal radius that is negative or unrealistically small, and the resulting hollow cylinder would not have the same volume as the sphere. A thickness of 0.6 cm or 2 cm similarly does not satisfy the volume equality and would give either too little or too much volume in the cylindrical shell. Only a thickness of 1 cm correctly balances the volumes and matches all the geometric constraints.

Common Pitfalls:
Common errors include forgetting to subtract the inner cylinder volume from the outer cylinder volume, miscomputing 25 − r^2, or confusing the sphere radius with the cylinder radius. Some students also forget to cancel π, which can complicate calculations unnecessarily. Being systematic about equating volumes and carefully handling squares and subtractions prevents these mistakes.

Final Answer:
The thickness of the hollow cylindrical tube is 1 cm.