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

Given radius of the cylinder, r = 12 cm

It is also given that a spherical iron ball is dropped into the cylinder and the water level raised by 6.75 cm

Hence volume of water displaced = volume of the iron ball

Height of the raised water level, h = 6.75 m

Volume of water displaced = πr^2h

= π × 12 × 12 × 6.75 cm^3

So, Volume of iron ball = π × 12 × 12 × 6.75 cm^3 ... (1)

But, volume of iron ball =4/3πr^3  ...(2)

From (1) and (2) we get

4/3πr^3 = π × 12 × 12 × 6.75

On solving, we get

r^3 = 729

r^3 = 3^6

Therefore, r' = 9 

Radius of the iron ball is 9 cm