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.