Question 101
The radius of the base and the height of a right circular cone are doubled. The volume of the cone will be
Solution
Let original radius be $$r$$ and height be $$h$$
=> Original volume of cone = $$\frac{1}{3} \pi r^2h$$
New radius = $$2r$$ and new height = $$2h$$
=> New volume = $$\frac{1}{3} \pi (4r^2) (2h)$$
= $$8 * \frac{1}{3} \pi r^2h$$
=> New volume of cone = 8 times the previous one.
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