A solid metallic sphere of radius $r$ is converted into a solid right circular cylinder of radius $R$. If the height of the cylinder is twice the radius of the sphere, then

### Concept & Algebraic Volume Conservation When reforming solids, equate their volume formulas. $$V_{sphere} = \frac{4}{3}\pi r^3$$ $$V_{cylinder} = \pi R^2h$$ ### Step-by-Step Solution 1. Given the sphere has radius $r$, its volume is $\frac{4}{3}\pi r^3$. 2. Given the cylinder has radius $R$ and its height $h = 2r$. 3. The volume of the cylinder is $\pi R^2(2r)$. 4. Equating the volumes: $\frac{4}{3}\pi r^3 = \pi R^2 (2r)$. 5. Cancel $\pi$ and $r$ from both sides: $\frac{4}{3}r^2 = 2R^2$. 6. Divide by 2: $R^2 = \frac{2}{3}r^2$. 7. Take the square root of both sides: $R = \sqrt{\frac{2}{3}r^2} = r\sqrt{\frac{2}{3}}$. ### Exam Strategy & Shortcut Substitute the height into the cylinder volume formula immediately and equate: $2\pi R^2 r = \frac{4}{3}\pi r^3$. Canceling terms on sight yields $R^2 = \frac{2}{3}r^2$, skipping intermediate algebraic steps. ### Common Pitfall Misinterpreting the wording "height of the cylinder is twice the radius of the sphere" and assigning $h = 2R$ instead of $h = 2r$. ### Final Answer Therefore, the correct answer is **$R = r\sqrt{\frac{2}{3}}$**.