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Deepak Bora
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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.

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This is the Important question of class 10 Based on Mensuration Chapter of M.L Aggarwal book for ICSE BOARD.
Here 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 and its volume is 2/3 of the hemisphere.
This is the Question Number 15, Exercise 17.4 of M.L Aggarwal.

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  1. ML Aggarwal Sol Class 10 Maths chapter 17-32

    Given radius of the cone, r = 3.5 cm

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

    Volume of hemisphere = (2/3)r3

    = (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 m3

    Volume of cone = 2/3 of volume of hemisphere

    = (2/3)× 539/6

    = 539/9 m3

    Volume of cone = (1/3)r2h

    (1/3)r2h = 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 = √(h2+r2)

    = √((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

    = rl + 2r2

    = 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 m2

    = 141.17 m2

    Hence the Surface area of the buoy is 141.17 m2.

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