Question 50

If the radius and height of a solid cone are increased by 10% and 20% respectively. By what percentage the volume of the cone is increased?

Solution

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Let r and h be the radius and height of the cone.
Initial volume = $$\frac{1}{3}*\pi*r^2*h$$
Final volume = $$\frac{1}{3}*\pi*(1.1r)^2*(1.2h)$$
Therefore, final volume = $$1.452*\frac{1}{3}*\pi*r^2*h$$
Therefore, percentage increase = $$\frac{1.452*\frac{1}{3}*\pi*r^2*h - \frac{1}{3}*\pi*r^2*h}{\frac{1}{3}*\pi*r^2*h}*100 = 45.20$$%
Hence option D is the correct answer.

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