4. If the radius and height of a cone are both increased by 100%, then its volume is
increased by
a. 400%
b. 300%
c. 200%
d. 100%
If Radius and Height of Cone Increased by 100%, Volume Increase Percentage
The volume of a cone increases by 700% when both its radius and height are doubled (increased by 100%). The correct answer is none of the given options (a. 400%, b. 300%, c. 200%, d. 100%), as the actual increase is 700%.
Cone Volume Formula
The volume \(V\) of a cone is \(V = \frac{1}{3}\pi r^2 h\), where \(r\) is the base radius and \(h\) is the height. Assume initial values \(r = 1\) and \(h = 1\), so the original volume is \(V_1 = \frac{1}{3}\pi (1)^2 (1) = \frac{\pi}{3}\). When radius and height are each increased by 100%, the new radius is \(r’ = 2r\) and the new height is \(h’ = 2h\). The new volume is
\(V_2 = \frac{1}{3}\pi (2r)^2 (2h) = \frac{1}{3}\pi (4r^2)(2h) = \frac{8}{3}\pi r^2 h = 8V_1\).
Percentage Increase Calculation
The percentage increase in volume is
\[
\text{Percentage increase} =
\frac{V_2 – V_1}{V_1} \times 100\% =
\frac{8V_1 – V_1}{V_1} \times 100\% =
700\%.
\]
Doubling the radius multiplies the volume by \(2^2 = 4\) because of the \(r^2\) term, and doubling the height multiplies the volume by 2, so the total factor is \(4 \times 2 = 8\) times the original, which is 700% more.
Option Analysis
- a. 400%: Corresponds to 5 times the original (500% of original), which is incorrect because the actual factor is 8.
- b. 300%: Corresponds to 4 times the original (400% of original), which could occur if only the radius were doubled, but here both radius and height change.
- c. 200%: Corresponds to 3 times the original (300% of original), which could occur in other mixed-change scenarios but not when both radius and height are doubled.
- d. 100%: Corresponds to 2 times the original (200% of original), which would be the case if only the height were doubled or only one linear dimension changed.
Why Volume Scales by 800%
Cone volume depends on the product \(r^2 h\), so doubling both radius and height yields \((2)^2 \times 2 = 8\) times the original volume. A numerical check with \(V_1 \approx 1.047\) and \(V_2 \approx 8.378\) confirms that the increase is 700%, since \(8.378 \approx 8 \times 1.047\).
Options like 400% or 300% often result from neglecting that the radius is squared, while the correct approach recognizes that all linear dimensions scale together, causing volume to scale with the cube of the linear factor for similar solids.
Exam Tips for Similar Questions
- Identify the type of scaling: linear measures scale directly (e.g., doubling gives a 100% increase), areas scale with the square of the factor, and volumes of similar solids typically scale with the cube of the factor.
- Practice variants: if only the height of a cone is doubled while radius is fixed, volume doubles, giving a 100% increase.
- Avoid common traps: a 100% increase in both radius and height does not mean a simple 200% total increase; the effects multiply through the formula.
- For geometry questions in exams like CSIR NET, remember that when all relevant linear dimensions of a 3D figure are scaled by \(k\), the volume scales by \(k^3\), leading to large percentage increases.
