A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 9 cm. Find the radius of the ball. (Use π =22/7).
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The radius of the ball = 12 cm.
Given that the radius(r) of the cylinder = 16 cm and water is present upto a depth of 30 cm.
When the spherical ball of radius r' is dropped , the water level in the cylinder is raised by 9 cm.
So the volume of the ball is equal to the volume of the water level raised.
Volume of 9 cm increase of water level in cylinder = volume of spherical ball
=> π × r² × 9 = (4/3) π × r'³
=> (16)² × 9 = (4/3)r'³
=> r'³ = 1728
=> r' = 12 cm
Radius of the ball = 12 cm
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