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A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is 2/3 of the hemisphere. Calculate the height of the cone and the surface area of the buoy correct to 2 places of decimal.

Answer:

**Solution:**

As buoy comprises of cone and hemisphere so its surface area will be equal to the sum of surface area of cone and hemisphere.

Given radius of the cone, r = 3.5 cm

Radius of hemisphere, r = 3.5 cm = 7/2 cm

Volume of hemisphere = (2/3)π r^{3}

= (2/3)×(22/7)×(7/2)^{3}

= (2/3)×(22/7)×(7/2)×(7/2)×(7/2)

= (22/3)×(7/2)×(7/2)

= 11×49/6

= 539/6 m^{3}

Volume of cone = 2/3 of volume of hemisphere

= (2/3)× 539/6

= 539/9 m^{3}

Volume of cone = (1/3)π r^{2}h

(1/3)π r^{2}h = 539/9

(1/3)×(22/7)×(7/2)^{2}×h = 539/9

h = 539×3×2/9×11×7

h = 14/3

h = 4.667

h = 4.67 m

Hence the height of the cone is 4.67 m.

Slant height of cone, l= √(h^{2}+r^{2})

= √((14/3)^{2}+(7/2)^{2})

= √((196/9)+(49/4))

= √((784/36)+(441/36))

= √(1225/36)

= 35/6 m

Surface area of the buoy = Surface area of cone + surface area of the hemisphere

= π r*l* + 2π r^{2}

=π r(*l*+ 2r)

= (22/7)×(7/2)×((35/6)+2×(7/2))

= 11×((35/6)+7)

= 11×(5.8333+7)

= 11×(12.8333)

= 141.166 m^{2}

= 141.17 m^{2}

**Hence the Surface area of the buoy is 141.17 m ^{2}.**

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