1. **Problem Statement:** Nakul has a cylindrical container of volume 3080 cm³, half filled with water. When a spherical ball is dropped in, the water level rises, increasing the volume to 6391 cm³. We need to find: (a) The volume by which the water rises. (b) The volume of the spherical ball. (c) The radius of the ball. 2. **Given Data:** - Cylinder radius $r = 7$ cm - Cylinder height $h = 10$ cm - Initial water volume = half of cylinder volume = $\frac{3080}{2} = 1540$ cm³ - Volume after ball is dropped = 6391 cm³ - Water level rise height = 3.4 cm - Use $\pi = \frac{22}{7}$ 3. **Step (a): Find the volume by which the water rises.** - Volume rise $= \text{final volume} - \text{initial volume} = 6391 - 1540 = 4851$ cm³ 4. **Step (b): Find the volume of the spherical ball.** - The volume of the ball equals the volume of water displaced, which is the volume rise calculated above. - So, volume of ball $V = 4851$ cm³ 5. **Step (c): Find the radius of the ball.** - Volume of a sphere formula: $$V = \frac{4}{3} \pi r^3$$ - Substitute $V = 4851$ and $\pi = \frac{22}{7}$: $$4851 = \frac{4}{3} \times \frac{22}{7} \times r^3$$ - Simplify: $$4851 = \frac{88}{21} r^3$$ - Multiply both sides by $\frac{21}{88}$: $$r^3 = 4851 \times \frac{21}{88}$$ - Calculate: $$r^3 = 4851 \times 0.2386 = 1157.5$$ - Find cube root: $$r = \sqrt[3]{1157.5} \approx 10.5 \text{ cm}$$ **Note:** The problem states the radius is 5 cm, but based on the given volumes and calculations, the radius is approximately 10.5 cm. **Final answers:** - (a) Volume rise = 4851 cm³ - (b) Volume of ball = 4851 cm³ - (c) Radius of ball $\approx 10.5$ cm