Question

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.

Hint:

Using fractions like \(\frac{21}{2}\) instead of decimals like \(10.5\) makes simplifying with \(\frac{22}{7}\) much easier because factors of 7 and 2 cancel out perfectly.

Answer 1

Step 1: Understanding the Question:
This case study involves a composite shape model.
We have:
- A spherical ball of diameter \(21\text{ cm}\) (radius \(r = 10.5\text{ cm}\)).
- A cylindrical pool with outer diameter \(50\text{ cm}\) (outer radius \(R = 25\text{ cm}\)) and inner diameter \(40\text{ cm}\) (inner radius \(r_{in} = 20\text{ cm}\)).
- The solid base height is \(14\text{ cm}\).
- The pool's filled water height is \(7\text{ cm}\).

Step 2: Key Formula or Approach:
1. Total height: The sum of the height of the base and the diameter of the ball.
2. Volume of the ball:
\[ V = \frac{4}{3}\pi r^3 \]
3. Volume of water: Calculate the total cylindrical water volume of radius \(20\text{ cm}\) and height \(7\text{ cm}\), and subtract the submerged \(\frac{1}{3}\text{rd}\) volume of the ball.
4. Surface Area: Cylindrical outer curved surface area is \(2\pi R h\), and sphere surface area is \(4\pi r^2\).

Step 3: Detailed Explanation:
1. Part (i): Total height of the fountain:
- The solid base has a height of \(14\text{ cm}\).
- The spherical ball has a diameter of \(21\text{ cm}\) and sits on top of this base.
- Total height = \(14 + 21 = 35\text{ cm}\).
2. Part (ii): Volume of the ball:
- Radius \(r = \frac{21}{2} = 10.5\text{ cm}\).
- Volume:
\[ V = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{21}{2}\right)^3 \]
\[ V = \frac{4}{3} \times \frac{22}{7} \times \frac{21 \times 21 \times 21}{8} \]
\[ V = 11 \times 21 \times 21 = 4851\text{ cm}^3 \]
3. Part (iii)(a): Volume of water filled in the pool:
- Inner radius of pool \(r_{in} = \frac{40}{2} = 20\text{ cm}\).
- Height of water \(h_w = 7\text{ cm}\).
- Total volume of the cylindrical pool space up to the water level:
\[ V_{\text{pool}} = \pi r_{in}^2 h_w = \frac{22}{7} \times 20^2 \times 7 = 22 \times 400 = 8800\text{ cm}^3 \]
- Volume of the submerged \(\frac{1}{3}\text{rd}\) part of the ball:
\[ V_{\text{submerged}} = \frac{1}{3} \times 4851 = 1617\text{ cm}^3 \]
- Volume of water:
\[ V_{\text{water}} = 8800 - 1617 = 7183\text{ cm}^3 \]
4. Part (iii)(b) (Alternative): Sum of curved surface areas:
- Outer radius of cylindrical base \(R = 25\text{ cm}\).
- Total height of cylindrical part \(h = 14 + 7 = 21\text{ cm}\).
- Curved Surface Area of Cylinder:
\[ \text{CSA}_{\text{cyl}} = 2\pi R h = 2 \times \frac{22}{7} \times 25 \times 21 = 3300\text{ cm}^2 \]
- Surface Area of the Sphere:
\[ \text{SA}_{\text{sphere}} = 4\pi r^2 = 4 \times \frac{22}{7} \times \left(\frac{21}{2}\right)^2 = 1386\text{ cm}^2 \]
- Sum of surface areas = \(3300 + 1386 = 4686\text{ cm}^2\).

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 filled is \(7183\text{ cm}^3\).
(iii)(b) Sum of surface areas is \(4686\text{ cm}^2\).