A spherical iron ball is dropped into a cylindrical vessel of base diameter 14 cm containing water. The water level is increased by $9\frac{1}{3}$ cm. What is radius of the ball?

### Concept & Volume Displacement When a solid object is submerged in a fluid inside a uniform container, the volume of the displaced fluid equals the volume of the object. $$Volume\ of\ Object = Area\ of\ Base \times Rise\ in\ Height$$ ### Step-by-Step Solution 1. Identify the container (cylinder) dimensions: Base diameter = $14 \text{ cm} \implies$ radius $r = 7 \text{ cm}$. 2. Identify the rise in water level: $h = 9\frac{1}{3} \text{ cm} = \frac{28}{3} \text{ cm}$. 3. Calculate the volume of the displaced water: $V_{water} = \pi r^2 h = \pi (7)^2 (\frac{28}{3}) = 49\pi (\frac{28}{3}) = \frac{1372\pi}{3} \text{ cm}^3$. 4. This displaced volume equals the volume of the spherical iron ball. Let the ball's radius be $R$. 5. $V_{sphere} = \frac{4}{3}\pi R^3$. 6. Equate the volumes: $\frac{4}{3}\pi R^3 = \frac{1372\pi}{3}$. 7. Cancel $\frac{\pi}{3}$ from both sides: $4R^3 = 1372$. 8. Divide by 4: $R^3 = \frac{1372}{4} = 343$. 9. Find the cube root: $R = \sqrt[3]{343} = 7 \text{ cm}$. ### Exam Strategy & Shortcut Set up the displacement equation directly: $\frac{4}{3}\pi R^3 = \pi (7^2)(\frac{28}{3})$. The $\pi$ and the denominator $3$ cancel out instantly. $4R^3 = 49 \times 28$. Divide $28$ by $4$ to get $R^3 = 49 \times 7 = 343$. Thus, $R = 7$. ### Common Pitfall Misconverting the mixed fraction $9\frac{1}{3}$ into an improper fraction. Always multiply the whole number by the denominator and add the numerator: $\frac{(9 \times 3) + 1}{3} = \frac{28}{3}$. ### Final Answer Therefore, the correct answer is **7 cm**.