A cylindrical tub of radius 12 cm contains water up to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. The radius of the ball is .......
Solution
When the ball is dropped in a cylindrical tub, it will displace the water equal to its volume based on Archimedes' Principle, which states that the volume of the displaced fluid is the same as the volume of the object submerged.
We know the radius of the cylinder = 12 cm.
The height of water in the cylinder is increased by 6.75 cm.
We need to find the radius of the spherical ball. (Assume r)
Volume of cylinder = $$\pi R^2H$$
The volume of the cylinders increased because of the iron ball by a height of 6.75 cm
Volume increased = $$\pi R^2H$$ = $$\pi\left(12\right)^26.75$$
The volume of spherical iron ball = $$\dfrac{4}{3}\pi r^3$$
Comparing both,
$$\dfrac{4}{3}\pi r^3$$ = $$\pi\left(12\right)^26.75$$
$$r^3=9\times9\times9$$
r = 9 cm.
