Answer: A
Let the original height be \(h\) units \(\therefore\) New height \(= h + h \times \cfrac{10}{100} = \cfrac{11h}{10}\) units \(\therefore\) Percentage increase in volume \(= \cfrac{\cfrac{1}{3}\pi r^2 \times \cfrac{11h}{10} - \cfrac{1}{3}\pi r^2 h}{\cfrac{1}{3}\pi r^2 h} \times 100\%\) \(= (\cfrac{11}{10} - 1) \times 100\%\) \(= 10\%\)
Let the original height be \(h\) units \(\therefore\) New height \(= h + h \times \cfrac{10}{100} = \cfrac{11h}{10}\) units \(\therefore\) Percentage increase in volume \(= \cfrac{\cfrac{1}{3}\pi r^2 \times \cfrac{11h}{10} - \cfrac{1}{3}\pi r^2 h}{\cfrac{1}{3}\pi r^2 h} \times 100\%\) \(= (\cfrac{11}{10} - 1) \times 100\%\) \(= 10\%\)