Case Study - 1 : A model of Leafy Ball Fountain is made to be kept on the tabletop. Water gently cascades down the ball into a decorative cylindrical pool where it is recycled. The diameter of spherical ball is 21 cm. Cylindrical pool - Outer diameter is 50 cm and inner diameter is 40 cm. Height of solid base is 14 cm. Height of water filled is 7 cm. Observe the figure and answer the following questions : (i) Determine the total height of the fountain. (ii) Find the volume of the ball. (iii) (a) If one-third of the ball is submerged in the water, find the volume of the water filled in the pool. OR (iii) (b) Find the sum of the outer curved surface area of the cylindrical part and surface area of the ball.
Show Hint
Write fractional values like \( \frac{21}{2} \) instead of decimal values like \( 10.5 \) during calculation.
This allows easy cancellations with factors of 7 and 3 in volume and surface area formulas, saving time and improving mathematical accuracy.
Step 1: Understanding the Question:
The topic of this question is Surface Areas and Volumes.
We are given a composite structure: a tabletop fountain.
The components include a spherical ball and a cylindrical base/pool with given outer/inner diameters and heights.
We need to answer various sub-questions regarding height, volume, and surface areas of the components.
Step 2: Key Formula or Approach:
- Diameter of the sphere is \( 21\text{ cm} \), so its radius \( R = \frac{21}{2} = 10.5\text{ cm} \).
- Volume of a sphere:
\[ V_{\text{sphere}} = \frac{4}{3}\pi R^3 \]
- Surface area of a sphere:
\[ \text{SA}_{\text{sphere}} = 4\pi R^2 \]
- Volume of a cylinder:
\[ V_{\text{cyl}} = \pi r^2 h \]
- Curved Surface Area of a cylinder:
\[ \text{CSA}_{\text{cyl}} = 2\pi rh \]
Step 3: Detailed Explanation:
1. Part (i): Determine total height of the fountain:
The total height is the sum of the height of the solid base and the diameter of the ball:
\[ \text{Total Height} = \text{Height of solid base} + \text{Diameter of ball} \]
\[ \text{Total Height} = 14 + 21 = 35\text{ cm} \]
2. Part (ii): Find the volume of the ball:
Using the sphere volume formula with \( R = 10.5\text{ cm} = \frac{21}{2}\text{ cm} \):
\[ V = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{21}{2}\right)^3 \]
\[ V = \frac{4}{3} \times \frac{22}{7} \times \frac{21 \times 21 \times 21}{8} \]
\[ V = \frac{4 \times 22 \times 3 \times 21 \times 21}{3 \times 7 \times 8} \]
Simplify terms:
\[ V = 11 \times 21 \times 21 = 11 \times 441 = 4851\text{ cm}^3 \]
3. Part (iii)(a): Volume of water filled in the pool:
The water is filled inside the inner cylinder.
Inner diameter is \( 40\text{ cm} \), so inner radius \( r_i = 20\text{ cm} \).
Height of water filled is \( h = 7\text{ cm} \).
The total space occupied by the water and the submerged ball is the cylinder volume up to height 7 cm:
\[ V_{\text{total space}} = \pi r_i^2 h = \frac{22}{7} \times (20)^2 \times 7 = 22 \times 400 = 8800\text{ cm}^3 \]
One-third of the ball is submerged, which displaces an equal volume:
\[ V_{\text{submerged}} = \frac{1}{3} \times V_{\text{ball}} = \frac{1}{3} \times 4851 = 1617\text{ cm}^3 \]
Therefore, the actual volume of water filled is:
\[ V_{\text{water}} = V_{\text{total space}} - V_{\text{submerged}} = 8800 - 1617 = 7183\text{ cm}^3 \]
4. Part (iii)(b): Sum of outer CSA of cylinder and surface area of the ball:
- Outer radius of pool \( r_o = 25\text{ cm} \) (since outer diameter is 50 cm).
- Total height of the outer cylindrical part is \( 14 + 7 = 21\text{ cm} \).
- Outer CSA of cylinder:
\[ \text{CSA}_{\text{cyl}} = 2\pi r_o H = 2 \times \frac{22}{7} \times 25 \times 21 = 2 \times 22 \times 25 \times 3 = 3300\text{ cm}^2 \]
- Surface Area of the ball:
\[ \text{SA}_{\text{ball}} = 4\pi R^2 = 4 \times \frac{22}{7} \times \left(\frac{21}{2}\right)^2 = 4 \times \frac{22}{7} \times \frac{441}{4} = 22 \times 63 = 1386\text{ cm}^2 \]
- Sum of areas:
\[ \text{Sum} = 3300 + 1386 = 4686\text{ cm}^2 \]
Step 4: Final Answer:
(i) Total height is \(35\text{ cm}\).
(ii) Volume of the ball is \(4851\text{ cm}^3\).
(iii)(a) Volume of water is \(7183\text{ cm}^3\).
(iii)(b) Sum of areas is \(4686\text{ cm}^2\).
