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.

Answer 1

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

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

= `2/3 xx 22/7 xx 7/2 xx 7/2 xx 7/2` 

= `539/6 m^3` 

Volume of conical part = `2/3 xx 539/6 m^3`  ...`(2/3 "of hemisphere")` 

Let height of the cone = h

Then, 

`1/3pir^2h = (2 xx 539)/(3 xx 6)` 

`=> 1/3 xx 22/7 xx 7/2 xx 7/2 xx h = (2 xx 539)/(3 xx 6)` 

`=> h =(2 xx 539 xx 2 xx 2 xx 7 xx 3)/(3 xx 6 xx 22 xx 7 xx 7)` 

`=> h = 14/3m = 4 2/3m  = 4.67 m` 

Height of the cone = 4.67 m

Surface area of buoy = 2πr² + πrl 

But `l = sqrt(r^2 + h^2)` 

`l = sqrt((7/2)^2 + (14/3)^2` 

= `sqrt(49/4 + 196/9)`

= `sqrt(1225/36)`

= `35/6m`  

Therefore, Surface area

= `(2 xx 22/7 xx 7/2 xx 7/2) + (22/7 xx 7/2 xx 35/6)m^2` 

= `77/1 + 385/6`

= `847/6` 

= 141.17 m² 

Surface Area = 141.17 m²