A hollow right circular cone of height h and of semivertical angle `theta` is placed with its base on a horizontal table. If the cone is filled with a liquid of density `rho` and the weight of the empty cone is equal to the weight of the liquid that it contains, find the thrust of the liquid on the base of the cone and the pressure exerted on the table.

Explanation

According to Fig. 2.7, `r=htantheta` = radius of the circular base of the cone.

Volume of the cone, `V=1/3pir^(2)h`
Mass of the liquid inside the cone, `M=1/3pirhor^(2)h`
Liquid pressure on the base of the cone = `hrhog`
`therefore` Thrust exerted by the liquid on the base of the cone
= `"pressure"xx"area"=hrhogxxpir^(2)=pirhogh^(3)tan^(2)theta`
Again, mass of the empty cone = M = `1/3pirhor^(2)h`
`therefore` Total force exerted on the table
= weight of the cone + weight of the liquid
`=1/3pirhor^(2)h*g+1/3pirhor^(2)h*g=2/3pirhogr^(2)h`
`therefore` Pressure on the table = `("force")/("area")=(2/3pirhogr^(2)h)/(pir^(2))=2/3hrhog`.