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