A sphere of maximum volume is cut out from a solid hemisphere of radius $r$. The ratio of the volume of the hemisphere to that of the cut out sphere is :

### Concept & Inscribed Sphere in Hemisphere To fit a sphere of maximum volume inside a hemisphere of radius $r$, the sphere must sit perfectly within the height and width constraints. The maximum possible diameter of this sphere equals the radius of the hemisphere $r$. Therefore, the radius of the new sphere is $r/2$. Volume of hemisphere: $$V = \frac{2}{3}\pi r^3$$ Volume of sphere: $$V = \frac{4}{3}\pi R^3$$ ### Step-by-Step Solution 1. The original solid is a hemisphere with radius $r$. Its volume is $V_{hemi} = \frac{2}{3}\pi r^3$. 2. The largest sphere that can be carved out will have its diameter equal to the hemisphere's height (which is $r$). So, the radius of this inscribed sphere is $R = \frac{r}{2}$. 3. Calculate the volume of this new sphere: $V_{sphere} = \frac{4}{3}\pi (\frac{r}{2})^3 = \frac{4}{3}\pi (\frac{r^3}{8}) = \frac{1}{6}\pi r^3$. 4. Find the required ratio of the hemisphere's volume to the cut-out sphere's volume: $\frac{V_{hemi}}{V_{sphere}} = \frac{\frac{2}{3}\pi r^3}{\frac{1}{6}\pi r^3}$. 5. Cancel out $\pi r^3$ and simplify the fraction: $\frac{2/3}{1/6} = \frac{2}{3} \times \frac{6}{1} = \frac{12}{3} = 4$. 6. The ratio is $4 : 1$. ### Exam Strategy & Shortcut Visualize the dimensions. A sphere inscribed in a hemisphere has half the radius. Because volume scales with the cube of the radius, halving the radius means the sphere's volume is $\frac{1}{8}$ of a *full* sphere of radius $r$. Since a hemisphere is half a full sphere, the ratio is $\frac{1/2}{1/8} = \frac{1}{2} \times 8 = 4$. ### Common Pitfall Students often guess that the inscribed sphere has a radius of $r$ divided by $\sqrt{2}$ or make an error assuming it shares the same base area. Recognizing that the height $r$ strictly limits the diameter is key. ### Final Answer Therefore, the correct answer is **4 : 1**.