A cylindrical vessel is filled with water up to some height. When a sphere of diameter 8 cm is dropped into the cylinder the water level rises by half of the initial level. When a sphere of diameter 16 cm is dropped the water level rises by some other value. What percentage of this new height is the initial level of water?

Answer: 20%

Explanation:

Let the radius of the cylinder be ‘r’.

Let the initial height up to which water is filled be ‘h’.

Let the height after the sphere of diameter 8 cm is dropped by ‘h1’.

Also h1 = 1.5h. Therefore, rise in height is ‘h1-h = 0.5h’.

Hence volume displaced by the sphere = pi*r*r*0.5h

This should be equal to the volume of the sphere. Hence,

(4/3)*pi*4*4*4 = pi*r*r*0.5h – – – – – – – – – – (1)

Again, let the height after the sphere of diameter 16 cm is dropped by ‘h2’. Hence,

(4/3)*pi*8*8*8 = pi*r*r*(h2-h) – – – – – – – – – – (2)

Dividing (1) by (2), we get

1/22 = h/(h2-h)

h2 – h = 4h

h = (1/5)h2

Hence, h is 20% of h2.

## Questions, answers, comments welcome