#### Question

A buoy is made in the form of hemisphere surmounted by a right cone whose circular base coincides with the plane surface of hemisphere. The radius of the base of the cone is 3.5 metres and its volume is two third of the hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two places of decimal

#### Solution

Radius of hemispherical part (r)=`3.5 m=7/2 m`

Therefore, Volume of hemisphere=`2/3pir^3`

`=2/3xx22/7xx7/2xx7/2xx7/2`

`=539/6 m^3`

Volume of conical part`2/3xx539/6m^3(2/3 "of hemisphere")`

Let height of the cone = h

Then,

`1/3pir^2h=(2xx539)/(3xx6)`

⇒ `1/3xx22/7xx7/2xx7/2xxh=(2xx539)/(3xx6)`

⇒ `=(2xx539xx2xx2xx7xx3)/(3xx6xx22xx7xx7)`

⇒`h=14/3m=4 2/3m =4.67 m`

Height of the cone = 4.67 m

Surface area of buoy=`2pir^2+pirl`

But l`=sqrt(r^2+h^2)`

`l=sqrt((7/2)^2+(14/3)^2`

`=sqrt(49/4+196/9)=sqrt1225/36=35/6m`

Therefore, Surface area =

=`(2xx22/7xx7/2xx7/2)+(22/7xx7/2xx35/6)m^2`

`=77/1+385/6=847/6`

`=141.17 m^2`

Surface Area`=141.17 m^2`