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.
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding stimulus
Model Answer
(i) Total height of the fountain:
Total height = Height of solid base + Height of water filled + Radius of ball
= 14 + 7 + 10.5 = 31.5 cm
(ii) Volume of the ball:
Radius of ball = 21/2 = 10.5 cm
$$V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (10.5)^3 = \frac{4}{3} \times \frac{22}{7} \times 1157.625 = \textbf{4851 cm}^3$$
(iii) Volume of water filled in the pool:
Volume of water = Volume of cylindrical annular region – Volume of submerged part of ball
Inner radius = 20 cm, Outer radius = 25 cm → Inner pool radius = 20 cm
$$\text{Volume of water region (cylinder)} = \pi r^2 h = \frac{22}{7} \times (20)^2 \times 7 = 8800 \text{ cm}^3$$
Volume of ball submerged = $\dfrac{1}{3} \times 4851 = 1617 \text{ cm}^3$
$$\text{Volume of water} = 8800 - 1617 = \textbf{7183 cm}^3$$
Source: Mensuration – Spheres and Cylinders
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Explanation
- Part (i): The ball sits on top of the water level, so total height adds the radius of the ball (not the diameter) to the base and water heights.
- Part (ii): Standard formula $\frac{4}{3}\pi r^3$; use $r = 10.5$ cm and $\pi = \frac{22}{7}$.
- Part (iii): The water fills the inner cylindrical region (radius 20 cm, height 7 cm) minus the portion of the ball dipped in. One-third of ball volume = $\frac{4851}{3}$ = 1617 cm³. Subtract from the cylindrical water volume. Examiners expect both steps shown clearly for full 2 marks.