1. **Problem statement:** Kobe's toy consists of a cone on top of a hemisphere. The cone's height is 4 cm, and the hemisphere's diameter is 6 cm. (i) Find the volume of the toy in terms of $\pi$. (ii) Find the volume of one spherical foam bead of radius 2 mm (0.2 cm) in terms of $\pi$. (iii) Given 1500 beads fill the toy, find the percentage of empty space inside. --- 2. **Formulas and important rules:** - Volume of a cone: $$V_{cone} = \frac{1}{3} \pi r^2 h$$ - Volume of a hemisphere: $$V_{hemisphere} = \frac{2}{3} \pi r^3$$ - Volume of a sphere: $$V_{sphere} = \frac{4}{3} \pi r^3$$ - Percentage empty space formula: $$\text{Empty space \%} = \left(1 - \frac{\text{Volume occupied by beads}}{\text{Volume of toy}}\right) \times 100$$ --- 3. **Step (i): Volume of the toy** - Radius of hemisphere and cone base: $$r = \frac{6}{2} = 3\text{ cm}$$ - Volume of hemisphere: $$V_{hemisphere} = \frac{2}{3} \pi (3)^3 = \frac{2}{3} \pi 27 = 18\pi$$ - Volume of cone: $$V_{cone} = \frac{1}{3} \pi (3)^2 (4) = \frac{1}{3} \pi 9 \times 4 = 12\pi$$ - Total volume of toy: $$V_{toy} = V_{hemisphere} + V_{cone} = 18\pi + 12\pi = 30\pi$$ --- 4. **Step (ii): Volume of one bead** - Radius of bead: $$r = 2\text{ mm} = 0.2\text{ cm}$$ - Volume of one bead (sphere): $$V_{bead} = \frac{4}{3} \pi (0.2)^3 = \frac{4}{3} \pi 0.008 = \frac{4}{3} \times 0.008 \pi = 0.010666\pi$$ --- 5. **Step (iii): Percentage of empty space** - Total volume of beads: $$V_{beads} = 1500 \times 0.010666\pi = 16\pi$$ - Volume of toy: $30\pi$ - Empty space volume: $$V_{empty} = 30\pi - 16\pi = 14\pi$$ - Percentage empty space: $$\frac{14\pi}{30\pi} \times 100 = \frac{14}{30} \times 100 = 46.67\%$$ --- **Final answers:** (i) Volume of toy = $30\pi$ cm³ (ii) Volume of one bead = $\frac{4}{3} \pi (0.2)^3 = \frac{4}{3} \pi 0.008 = 0.010666\pi$ cm³ (iii) Percentage of empty space inside the toy = $46.67\%$