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Deepak Bora
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The internal and external diameters of a hollow hemispherical shell are 6 cm and 10 cm, respectively. It is melted and recast into a solid cone of base diameter 14 cm. Find the height of the cone so formed.

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This is one of the important question from the cbse board exam point of view because in 2005 cbse board exam this question was asked. Question number 6 from RS Aggarwal book page number 810, exercise 17B, chapter volume and surface area of solid.

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  1. Given data,

    Internal diameter of the hemispherical shell = 6 cm

    Internal radius of the hemispherical shell = r= 3 cm

    External diameter of the hemispherical shell = 10 cm

    External radius of the hemispherical shell = R= 5 cm

    Volume of hemispherical shell = [2/3] π [ R³ – r³ ]

    =  [2/3] π [ 5³ – 3³ ]

    = 616/3

    Volume of hemispherical shell  = 205.33 cm³

    Radius of cone = 7 cm

    Let the height of the cone be h cm.

    Volume of cone = [1/3] πr²h

    = [1/3] * [22/7] * 7² *h

    = [154/3] h cm³

    Volume of the cone = [154/3] h cm³
    The volume of the hemispherical shell must be equal to the volume of the cone.

    ∴ 616/3= [154/3] h

    ∴ h = 4 m

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