A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?
Answers
The radius of the spherical ball is 9 cm.
• Given,
Radius of the cylindrical tub (r) = 12 cm
Depth of the tub = 20 cm
Rise in level of water on dropping a spherical ball into the tub (h) = 6.75 cm
• Volume of water displaced on dropping the ball = Volume of the ball
Also, since the tub is cylindrical in shape, the volume of water displaced will be equal to the volume of a cylinder having a height of 6.75 cm.
• Therefore,
Volume of the spherical ball = Volume of the cylinder with 6.75 cm height
• Now, volume of a sphere is given by (4 / 3).π.(radius)³
Let the radius of the spherical ball be r'.
∴ Volume of the ball = (4 / 3).π.(r')³
• Volume of a cylinder is given by π.(radius)².height
The radius of the cylindrical portion of water displaced = r = 12 cm
The height of the cylindrical portion of water displaced = h = 6.75 cm
∴ Volume of the cylindrical portion of water displaced = πr²h
• According to the question,
(4 / 3).π.(r')³ = πr²h
=> (4 / 3).π.(r')³ = π × (12 cm)² × 6.75 cm
=> (4 / 3).π.(r')³ = π × 144 cm² × 6.75 cm
=> r'³ = (π × 12 cm × 12 cm × 6.75 cm × 3) / 4π
=> r'³ = 3 cm × 12 cm × 6.75 cm × 3
=> r'³ = 729 cm³
=> r' = 729 cm³
=> r' = 9 cm
∴ The radius of the sphere = 9 cm