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Question 45
If the diameter of the base of a right circular cylinder is reduced by $$33\frac{1}{3}\%$$ and its height is doubled, then the volume of the cylinder will:
Solution
Let the radius be r.
Intently volume of right circular cylinder = $$\pi r^2 h$$
Final volume of right circular cylinder = $$\pi (r - \frac{r}{3})^2 \times 2h$$ = $$ \frac{8}{9} \pi r^2 h$$
Decrement =Â $$\pi r^2 h -Â \frac{8}{9} \pi r^2 h =Â \frac{1}{9} \pi r^2 h$$
Percentage decrement = $$\frac{\frac{1}{9} \pi r^2 h}{\pi r^2 h} \times 100$$ = $$11\frac{1}{9}\%$$
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