If the radius of the base of a right circular cylinder is halved, keeping the height same, what is the ratio of the volume of the reduced cylinder to that of the original one?

### Concept & Formula The volume of a cylinder is proportional to the square of its radius when the height is constant. $$V = \pi r^2 h$$ ### Step-by-Step Solution 1. **Define the original state:** - Original Volume ($V_1$) = $\pi r^2 h$ 2. **Define the new state:** - New radius ($r_2$) = $\frac{r}{2}$ - New Volume ($V_2$) = $\pi \left(\frac{r}{2}\right)^2 h$ - $V_2 = \pi \left(\frac{r^2}{4}\right) h = \frac{1}{4} \pi r^2 h$ 3. **Calculate the ratio:** - $V_2 : V_1 = \left(\frac{1}{4} \pi r^2 h\right) : (\pi r^2 h)$ - $V_2 : V_1 = \frac{1}{4} : 1 = 1 : 4$ ### Exam Strategy & Shortcut Since volume varies as the square of the radius ($V \propto r^2$), halving the radius scales the volume by $(\frac{1}{2})^2 = \frac{1}{4}$. The ratio is instantly $1 : 4$. ### Common Pitfall A common mistake is assuming the relationship is linear, leading to the incorrect answer $1 : 2$. Always remember the radius term in the volume formula is squared. ### Final Answer Therefore, the correct answer is **$1 : 4$**.