55
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.
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 SolutionHide Solution↓
Sol. (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
$= \pi \times 20 \times 20 \times 7 - \frac{1}{3} \times 4851$ (1 Mark)
$= 7183$ cm$^3$ (1 Mark)
OR
(iii) (b) Required area = Outer CSA of cylindrical 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
$= \pi \times 20 \times 20 \times 7 - \frac{1}{3} \times 4851$ (1 Mark)
$= 7183$ cm$^3$ (1 Mark)
OR
(iii) (b) Required area = Outer CSA of cylindrical 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)