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Question
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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.
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Observe the figure and answer the following questions:
- Determine the total height of the fountain. [1]
- Find the volume of the ball. [1]
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- If one-third of the ball is submerged in the water, find the volume of the water filled in the pool. [2]
OR - Find the sum of the outer curved surface area of the cylindrical part and surface area of the ball. [2]
- If one-third of the ball is submerged in the water, find the volume of the water filled in the pool. [2]
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Solution
Given: Diameter of spherical ball = 21 cm
Outer diameter of cylindrical pool = 50 cm
Inner diameter of cylindrical pool = 40 cm
Height of solid base = 14 cm
Height of water filled = 7 cm
(i) Radius of ball, r = `d/2`
= `21/2`
= 10.5 cm
Total height = Height of solid base + Height of water + Radius of ball (since ball is on top)
= 14 + 7 + 10.5
= 31.5 cm
(ii) The volume of the ball.
Volume of sphere = `4/3 pi r^3`
V = `4/3 pi (10.5)^3`
= `4/3 xx 22/7 xx 1157.625`
= `(4 xx 22 xx 1157.625)/21`
= `101871/21`
= 4851 cm3
(iii) (a) Inner radius of cylindrical pool = `40/2`
= 20
Volume of cylinder = πr2h
= `22/7 xx (20)^2 xx 7`
= `22/7 xx 400 xx 7`
= 22 × 400
= 8800 cm3
Volume of submerged part of ball = `1/3 xx "Volume of ball"`
= `1/3 xx 4851`
= 1617
Volume of water filled = 8800 − 1617
= 7183 cm3
(iii) (b) Outer radius cylinder = `50/2`
= 25
Curved surface area of cylinder = 2πrh
= `2 xx 22/7 xx 25 xx 14`
= 2200 cm2
Surface area of sphere = 4πr2
= `4 xx 22/7 xx (10.5)^2`
= `4 xx 22/7 xx 110.25`
= `(4 xx 22 xx 110.25)/7`
= `9702/7`
= 1386 cm2
Sum = 2200 + 1386
= 3586 cm2


