Question

A conical vessel of radius 12 cm and height 16 cm is comletely filled with water .A sphere is lowered into the water and its size is such that ,when it touches the sides ,it is just immersed .What fraction of water ovrflows?

Answer 1

The figure is as shown below. Lets assume the radius as r cm and the length below it as x cm. We can calculate the length of the side as 10 cm using pythagoras theorem. We know that r is incident at 90 degrees in both the sides thus both the triangles are congruent. Therefore. the part above where the sphere touches the side is 6 cm.


Applying pythagoras theorem in the lower part.

(r+x)2=r2+42(r+x)2=r2+42
r2+2rx+x2=r2+42r2+2rx+x2=r2+42
x2+2rx−16=0x2+2rx−16=0

Thus solving for x we get
x=−2r±4r2+4×16√2x=−2r±4r2+4×162
x=−2r±2×r2+16√2x=−2r±2×r2+162
x=−r±r2+16−−−−−−√x=−r±r2+16

Since x can't be negative we avoid the negative case

x=−r+r2+16−−−−−−√x=−r+r2+16

Substituting for x in

8=2r+x8=2r+x
8=2r−r+r2+16−−−−−−√8=2r−r+r2+16
8−r=r2+16−−−−−−√8−r=r2+16
64−16r+r2=r2+1664−16r+r2=r2+16
64−16r=1664−16r=16
48=16r48=16r
r=3cmr=3cm

Thus volume of sphere is

Vs=4×πr33Vs=4×πr33
Vs=4×π333Vs=4×π333

Volume of cone is

Vc=πr2×h3Vc=πr2×h3
Vc=π62×83Vc=π62×83

Dividing Vs by Vc we get the ratio of water flowing out which is

4×π333/π62×834×π333/π62×83

which evaluates to

38