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Question
A buoy is made in the form of a hemisphere surmounted by a right circular cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 m and its volume is two-third the volume of hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two decimal places.
Solution 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`m3
Volume of conical part = `2/3 xx 539/6` m3 (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 = (539 xx 2 xx 2 xx 7 xx 3)/(3 xx 6 xx 22 xx 7 xx 7)`
⇒ h = `14/3`m = 4`2/3`m = 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) = sqrt(1225/36) = 35/6`m
Therefore , surface area =
= `(2 xx 22/7 xx 7/2 xx 7/2) + (22/7 xx 7/2 xx 35/6)`m2
= `77/1 + 385/6 = 847/6`
= 141.17 m2
Solution 2
According to question.
`2/3` (Volume of hemisphere) = Volume of Cone
`2/3 ( 2/3 πr^3 ) = 1/3 πr^2h`
`4/9 (3.5)^3 = 1/3 (3.5)^2.h`
h = `(4 xx 3.5 xx 3.5 xx 3.5 xx 3)/(3.5 xx 3.5 xx 9)`
= `42.0/9 = 14/3 "m" = 4.67 "m"`
l = `sqrt (r^2 + h^2)`
= `sqrt((3.5)^2 + (4.67)^2)`
= `35/6` m
Now surface area of buoy = Surface area of right cone + Surface area of hemisphere
= πrl + 2πr2
= πr (l + 2r )
= `22/7 xx 3.5 (35/6 + 2 xx 3.5 )`
= 11 x (5.83 + 7)
= 11 x 12.83
= 141.13 sq.m.
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