Answer: C
Let the original height and radius of the cone be \(h\) units and \(r\) units respectively. \(\therefore\) The new height and radius of the cone are \(2h\) units and \(2r\) units respectively. Original volume = \(\cfrac{1}{3}\pi r^2 h\) cubic units New volume = \(\cfrac{1}{3}\pi (2r)^2 \times (2h)\) cubic units \(= \cfrac{1}{3} \times 8\pi r^2 h\) cubic units \(\therefore\) Increase in volume = \(\cfrac{1}{3} \times 8\pi r^2 h - \cfrac{1}{3}\pi r^2 h\) \(= 7 \times \cfrac{1}{3}\pi r^2 h\) cubic units \(= 7 \times\) the original volume.
Let the original height and radius of the cone be \(h\) units and \(r\) units respectively. \(\therefore\) The new height and radius of the cone are \(2h\) units and \(2r\) units respectively. Original volume = \(\cfrac{1}{3}\pi r^2 h\) cubic units New volume = \(\cfrac{1}{3}\pi (2r)^2 \times (2h)\) cubic units \(= \cfrac{1}{3} \times 8\pi r^2 h\) cubic units \(\therefore\) Increase in volume = \(\cfrac{1}{3} \times 8\pi r^2 h - \cfrac{1}{3}\pi r^2 h\) \(= 7 \times \cfrac{1}{3}\pi r^2 h\) cubic units \(= 7 \times\) the original volume.