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A model of Leafy Ball Fountain is made to be kept on the tabletop. Water
ngently cascades down the ball into a decorative cylindrical pool where it is
nrecycled.
nThe diameter of spherical ball is $21$ cm.
nCylindrical pool - Outer diameter is $50$ cm and inner diameter is $40$ cm.
nHeight of solid base is $14$ cm.
nHeight of water filled is $7$ cm.
n(i) Determine the total height of the fountain.
n(ii) Find the volume of the ball.
n(iii) (a) If one-third of the ball is submerged in the water, find the
nvolume of the water filled in the pool.
nOR
n(iii) (b) Find the sum of the outer curved surface area of the
ncylindrical part and surface area of the ball.
ngently cascades down the ball into a decorative cylindrical pool where it is
nrecycled.
nThe diameter of spherical ball is $21$ cm.
nCylindrical pool - Outer diameter is $50$ cm and inner diameter is $40$ cm.
nHeight of solid base is $14$ cm.
nHeight of water filled is $7$ cm.
n(i) Determine the total height of the fountain.
n(ii) Find the volume of the ball.
n(iii) (a) If one-third of the ball is submerged in the water, find the
nvolume of the water filled in the pool.
nOR
n(iii) (b) Find the sum of the outer curved surface area of the
ncylindrical part and surface area of the ball.
Show SolutionHide Solution↓
(i) Total height of the fountain $= 14 + 21 = 35$ cm (1 Mark)
(ii) Volume of the ball $= \frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}$ (1/2 Mark)
$= 4851$ cm$^3$ (1/2 Mark)
(iii) (a) Volume of water = Volume of inner upper part - $\frac{1}{3} \times$ Volume of the ball
$= \frac{22}{7} \times 20 \times 20 \times 7 - \frac{1}{3} \times 4851$ (1 Mark)
$= 7183$ cm$^3$ (1 Mark)
OR
(iii) (b) Required area = Outer CSA of cylinderical part + Surface area of the ball
$= 2 \times \frac{22}{7} \times 25 \times (14 + 7) + 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}$ (1 Mark)
$= 4686$ cm$^2$ (1 Mark)
(ii) Volume of the ball $= \frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}$ (1/2 Mark)
$= 4851$ cm$^3$ (1/2 Mark)
(iii) (a) Volume of water = Volume of inner upper part - $\frac{1}{3} \times$ Volume of the ball
$= \frac{22}{7} \times 20 \times 20 \times 7 - \frac{1}{3} \times 4851$ (1 Mark)
$= 7183$ cm$^3$ (1 Mark)
OR
(iii) (b) Required area = Outer CSA of cylinderical part + Surface area of the ball
$= 2 \times \frac{22}{7} \times 25 \times (14 + 7) + 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}$ (1 Mark)
$= 4686$ cm$^2$ (1 Mark)