A cylindrical tub of radius 12 cm contains water 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. What is the radius of the ball?
Given,
Radius of cylinder = 12 cm
Height = 20 cm
Before drop a ball volume of water = v1 = πr2h = πr2 × 20 cm3
After droping rise in water level = 6.75 cm
New height = 20 + 6.75 = 26.75 cm
New volume = πr2 × 26.75 cm3
Volume of spherical ball = πr2 (26.75 – 20)
= πr2 × 6.75 = 
cm3
= 
πR3 = 3054.85
= R3 = 
= R = 
19