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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# 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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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