Keep cone volume fixed when base radius doubles — find new height: For a right circular cone, if the base radius is doubled and the volume is to remain unchanged, what must be the new height as a fraction of the previous height?

Introduction / Context:
The cone volume depends on the square of the radius and linearly on the height. If volume is to remain unchanged while radius changes, the height must adjust to compensate for the squared effect of radius.

Given Data / Assumptions:

• Original volume V = (1/3)π r^2 h.
• New radius r′ = 2r.
• Volume constant: V′ = V.

Concept / Approach:
Since V ∝ r^2 h, keeping V fixed implies r^2 h = constant. If r becomes 2r, then r^2 grows by a factor of 4. Therefore, height must shrink by the same factor (1/4) so that the product r^2 h remains the same.

Step-by-Step Solution:
V′ = (1/3)π (2r)^2 h′ = (1/3)π * 4r^2 * h′Set V′ = V = (1/3)π r^2 h ⇒ 4r^2 h′ = r^2 hh′ = h / 4

Verification / Alternative check:
Insert any numbers (e.g., r = 3, h = 12): V ∝ 9 * 12 = 108. With r′ = 6, h′ should be 3: 36 * 3 = 108, same product.

Why Other Options Are Wrong:
Half compensates for a linear change, not quadratic; one-third and 1/√2 do not neutralize the 4× increase from radius doubling.

Common Pitfalls:
Forgetting that radius is squared in the formula; trying to keep height inversely proportional to r rather than r^2.

Final Answer:
one-fourth of the previous height